Hi, I'm Beau with Free Code Camp. This course is taught by Dr. Linda green. She teaches at the University of North Carolina at Chapel Hill, and she has been teaching calculus two for many years to undergraduate students, we recommend using a paper and pencil so you can follow along at home. Let's start. Hi, I'm Beau with Free Code Camp. This course is taught by Dr. Linda green. She teaches at the University of North Carolina at Chapel Hill, and she has been teaching calculus two for many years to undergraduate students, we recommend using a paper and pencil so you can follow along at home. Let's start. This video introduces the idea of finding the area between two curves. Let's start with some review. in calculus one, you approximate the area between a curve and the x axis by dividing it up into tall skinny rectangles. You represented the width of one of these rectangles using the symbols delta x, here delta x means a small distance along the x axis or a small change in x values. You picked out an x value called a sample point from each of these little sub intervals along the x axis, one sample point for each rectangle. The sample point on the x axis for rectangle number i is denoted by x i star. You use the sample points and the function f to figure out the height of each rectangle. The height of rectangle number i is given by the functions value on the sample point f of x star. From this, you calculate the area of each rectangle, the area of a rectangle is base times height. So the area of the first rectangle is delta x times f of x one star, the area of the second rectangle is delta x times f of x two star. And the area of rectangle number i is going to be its base delta x times its height, f of x i star, and so on, if there are n rectangles, than the last rectangle will have base delta x and height f of x and star. So the approximate area under the curve is given by adding up all these areas of all these rectangles. In sigma notation, this can be written as sigma, the psalm from i equals one to n, the number of rectangles of the area of the rectangle, delta x times f of x i star. The exact area is then given by the limit of these approximating areas, as the number of rectangles goes to infinity. That's the limit as n goes to infinity of this Riemann sum This video introduces the idea of finding the area between two curves. Let's start with some review. in calculus one, you approximate the area between a curve and the x axis by dividing it up into tall skinny rectangles. You represented the width of one of these rectangles using the symbols delta x, here delta x means a small distance along the x axis or a small change in x values. You picked out an x value called a sample point from each of these little sub intervals along the x axis, one sample point for each rectangle. The sample point on the x axis for rectangle number i is denoted by x i star. You use the sample points and the function f to figure out the height of each rectangle. The height of rectangle number i is given by the functions value on the sample point f of x star. From this, you calculate the area of each rectangle, the area of a rectangle is base times height. So the area of the first rectangle is delta x times f of x one star, the area of the second rectangle is delta x times f of x two star. And the area of rectangle number i is going to be its base delta x times its height, f of x i star, and so on, if there are n rectangles, than the last rectangle will have base delta x and height f of x and star. So the approximate area under the curve is given by adding up all these areas of all these rectangles. In sigma notation, this can be written as sigma, the psalm from i equals one to n, the number of rectangles of the area of the rectangle, delta x times f of x i star. The exact area is then given by the limit of these approximating areas, as the number of rectangles goes to infinity. That's the limit as n goes to infinity of this Riemann sum of areas. of areas. A limit of a Riemann sum like this is by definition, an integral. So we can rewrite this using the integral sign as the integral of f of x dx. And the bounds of integration here, based on our picture are from x equals a 2x equals b. Notice that when we convert the limit of a Riemann sum to integral notation, the sample points x vystar just become our variable x, and the delta x becomes our dx. Now that we've reviewed these ideas of calculating the area under a curve, we're going to use these same ideas to calculate the area between two curves. To compute the area between two curves, y equals f of x and y equals g of x, in between the x values of a and b, we can again divide up the region and a tall skinny rectangles as shown in this picture. Once again, let's let delta x be the width of each rectangle. And let's let x i star represent a sample point in the ife interval. So x sub i star is a point on the x axis that lives in the rectangle number i. Now if we want to compute the area of one of these tall skinny rectangles, as always, the area of a rectangle is the base times the height. The base of any of these rectangles is given by delta x. But the height is different for each rectangle. If I focus on in on rectangle number, I assume this is rectangle number i right here. It stretches all the way up to F of the sample point f of x by star and it stretches all the way down. To add g of that x, y star. So the height of that rectangle is the gap between f of x and g of x at that sample point. In other words, it's the difference, f of x sub i star minus g of x, y star. Now that we have an expression for the area of one of these rectangles, we can add up all those areas as before, to get an expression for the approximate area between the curves, it's just the sum of these areas. And as before, we can get the exact area. By making these rectangles skinnier and skinnier. By taking more of them, we take the limit as the number of rectangles goes to infinity of this Riemann sum. And as always, the limit of a Riemann sum is given by the integral where the exabyte stars a sample points just become our variable x, and the delta x becomes our dx, let's put in our bounds of integration, we're told that x goes between A and B, and we have an expression for the area between our two curves. Actually, this formula only works if f of x is greater than or equal to g of x on the interval from a to b. That inequality guarantees that this expression, f of x i star minus g of x i star will be a positive number, we want a positive height for our rectangle, so that will get a positive area. If instead, f of x is less than or equal to g of x, then you'll need to switch around your subtraction and take the integral of g of x minus f of x dx instead, in order to get a positive area. One way to remember what to do is just to write the integral as the integral from a to b of the top y value minus the bottom y value, dx. Remembering that you'll need to replace the top y value and the bottom y value with functions of x before you can integrate. A limit of a Riemann sum like this is by definition, an integral. So we can rewrite this using the integral sign as the integral of f of x dx. And the bounds of integration here, based on our picture are from x equals a 2x equals b. Notice that when we convert the limit of a Riemann sum to integral notation, the sample points x vystar just become our variable x, and the delta x becomes our dx. Now that we've reviewed these ideas of calculating the area under a curve, we're going to use these same ideas to calculate the area between two curves. To compute the area between two curves, y equals f of x and y equals g of x, in between the x values of a and b, we can again divide up the region and a tall skinny rectangles as shown in this picture. Once again, let's let delta x be the width of each rectangle. And let's let x i star represent a sample point in the ife interval. So x sub i star is a point on the x axis that lives in the rectangle number i. Now if we want to compute the area of one of these tall skinny rectangles, as always, the area of a rectangle is the base times the height. The base of any of these rectangles is given by delta x. But the height is different for each rectangle. If I focus on in on rectangle number, I assume this is rectangle number i right here. It stretches all the way up to F of the sample point f of x by star and it stretches all the way down. To add g of that x, y star. So the height of that rectangle is the gap between f of x and g of x at that sample point. In other words, it's the difference, f of x sub i star minus g of x, y star. Now that we have an expression for the area of one of these rectangles, we can add up all those areas as before, to get an expression for the approximate area between the curves, it's just the sum of these areas. And as before, we can get the exact area. By making these rectangles skinnier and skinnier. By taking more of them, we take the limit as the number of rectangles goes to infinity of this Riemann sum. And as always, the limit of a Riemann sum is given by the integral where the exabyte stars a sample points just become our variable x, and the delta x becomes our dx, let's put in our bounds of integration, we're told that x goes between A and B, and we have an expression for the area between our two curves. Actually, this formula only works if f of x is greater than or equal to g of x on the interval from a to b. That inequality guarantees that this expression, f of x i star minus g of x i star will be a positive number, we want a positive height for our rectangle, so that will get a positive area. If instead, f of x is less than or equal to g of x, then you'll need to switch around your subtraction and take the integral of g of x minus f of x dx instead, in order to get a positive area. One way to remember what to do is just to write the integral as the integral from a to b of the top y value minus the bottom y value, dx. Remembering that you'll need to replace the top y value and the bottom y value with functions of x before you can integrate. Let's look at an example. We want to find the area between the curve y equals x squared plus x and y equals three minus x squared. x squared plus x is a parabola pointing upwards. So that must be the red curve here, while three minus x squared is a parabola pointing downwards. So that must be this blue curve. We know that area is given by the integral from our starting x value to our ending x value of our top y values minus our bottom y values, all integrated with respect to x, our top y values are given by the equation three minus x squared, and our bottom y values are given by the equation x squared plus x. Now we still need to figure out the values of a and b, the bounds of integration. And from the graph, it looks like a should be maybe about negative one and a half, and B should be about one, since that's where the green area starts and ends in the horizontal direction. But to find exact values of a and b, the easiest thing to do is to set these two equations equal to each other and solve for x. So I'll set three minus x squared equal to x squared plus x, I can add x squared to both sides to get to x squared plus x minus three equals zero. This factors into 2x plus three, and x minus one, and therefore x has to equal negative three halves and or x equals one, just like we thought from the graph. So we can finish doing our setup, our bounds of integration are from negative three halves to one, and I can simplify my integrand here. This is three minus x squared minus x squared minus x. Or in other words, the integral from negative three halves to one of minus 2x squared minus x plus three dx. This integrates to minus 2x cubed over three my minus x squared over two plus 3x, evaluated between one and negative three halves. Now I just need to plug in my bounds of integration and then simplify to get 125 20 fourths as the area. Let's look at an example. We want to find the area between the curve y equals x squared plus x and y equals three minus x squared. x squared plus x is a parabola pointing upwards. So that must be the red curve here, while three minus x squared is a parabola pointing downwards. So that must be this blue curve. We know that area is given by the integral from our starting x value to our ending x value of our top y values minus our bottom y values, all integrated with respect to x, our top y values are given by the equation three minus x squared, and our bottom y values are given by the equation x squared plus x. Now we still need to figure out the values of a and b, the bounds of integration. And from the graph, it looks like a should be maybe about negative one and a half, and B should be about one, since that's where the green area starts and ends in the horizontal direction. But to find exact values of a and b, the easiest thing to do is to set these two equations equal to each other and solve for x. So I'll set three minus x squared equal to x squared plus x, I can add x squared to both sides to get to x squared plus x minus three equals zero. This factors into 2x plus three, and x minus one, and therefore x has to equal negative three halves and or x equals one, just like we thought from the graph. So we can finish doing our setup, our bounds of integration are from negative three halves to one, and I can simplify my integrand here. This is three minus x squared minus x squared minus x. Or in other words, the integral from negative three halves to one of minus 2x squared minus x plus three dx. This integrates to minus 2x cubed over three my minus x squared over two plus 3x, evaluated between one and negative three halves. Now I just need to plug in my bounds of integration and then simplify to get 125 20 fourths as the area. In this video, we saw that the area between two curves in between the x values of a and b can be given by the integral from a to b have the top y values minus the bottom y values. dx, were the top y values and bottom y values are functions of x. More specifically, if the top curve is given by y equals f of x, and the bottom curve is y equals g of x, then this integral is the integral from a to b of f of x minus g of x, dx. In this video, we'll calculate the volumes of solids of revolution. A solid of revolution is a three dimensional object that can be formed by rotating a region of the plane around an axis. volumes formed by rotating a region of the plane around a line are called solids of revolution. In this figure on the left, this three dimensional object is formed by rotating a region of the plane shaped like this, around the x axis, the solid on the right can be formed by rotating a crescent shaped region of the plane like this, again, around the x axis. If we slice these two solids of revolution, using planes perpendicular to the x axis, on the left side, our cross sections look like disks. In this video, we saw that the area between two curves in between the x values of a and b can be given by the integral from a to b have the top y values minus the bottom y values. dx, were the top y values and bottom y values are functions of x. More specifically, if the top curve is given by y equals f of x, and the bottom curve is y equals g of x, then this integral is the integral from a to b of f of x minus g of x, dx. In this video, we'll calculate the volumes of solids of revolution. A solid of revolution is a three dimensional object that can be formed by rotating a region of the plane around an axis. volumes formed by rotating a region of the plane around a line are called solids of revolution. In this figure on the left, this three dimensional object is formed by rotating a region of the plane shaped like this, around the x axis, the solid on the right can be formed by rotating a crescent shaped region of the plane like this, again, around the x axis. If we slice these two solids of revolution, using planes perpendicular to the x axis, on the left side, our cross sections look like disks. A disc here means the inside of a circle. The figure on the right is different because it's hollow inside. And when we slice it by planes perpendicular to the x axis, we get cross sections that are shaped like washers. A washer here means the region in between two concentric circles. So for solids of revolution, the cross sections can have the shape of a disk, or the shape of a washer. The area of a disk is given by the familiar formula, pi r squared, where r is the radius. And the area of a washer can be written as pi times r outer squared minus pi times our inner squared, where our outer is the radius of the big circle, and our inner is the radius of the little circle. This formula works because the area of the washer is just the area of the larger circle minus the area of the inside smaller circle. Now we know that the volume of any solid that can be sliced into cross sections using planes perpendicular to the x axis is given by v is the integral from x equals a to x equals B of the area of the cross section at point x, integrated dx. If our cross sections are disks, this formula becomes the integral of pi r squared dx, where the radius R is a function of x. If instead, the cross sections look like washers, then the volume formula becomes the integral of pi r outer squared minus pi r inner squared dx. Again, our outer and our inner are functions of x here. These formulas work when they solid a revolution is formed by rotating a region around the x axis or any horizontal line. When we rotate around a horizontal line, then our cross sectional discs or washers are perpendicular to the x axis and are thin in the x direction. So it makes sense to integrate dx. If instead, we want to rotate around the y axis or a vertical line, then our disk or wash our cross sections are going to be perpendicular to the y axis and are going to be thin in the y direction. So when rotating around the y axis or a vertical line, we'll need to do our integral with respect to y. Our cross sectional area will be a function of y will integrate d y and our bounds of integration will have to be Y values, our formulas will look pretty much the same. We'll just have to calculate our Radia and bounds of integration in terms of y instead of x. As our first example, let's consider the region bounded by the curve y equals a cube root of x, the x axis and the line x equals eight. We want to find the volume of the solid of revolution found by rotating this region around the x axis. or cross sections here are going to be discs. And these discs are thin in the extraction. So we're going to be integrating dx, our smallest x value is zero, and our largest is eight. So those are our bounds of integration. And we want to integrate pi times the radius squared dx. Now the radius of our disks is given by the y coordinate on this curve. So we can write r is equal to y, which is equal to the cube root of x according to our equation. So we can rewrite the volume as the integral from zero to eight of pi times the cube root of x squared dx, I can pull out the PI and rewrite this integral using exponents and integrate and then evaluate using bounds of integration to get three fifths pi times eight to the five thirds minus zero. Now, eight to the five thirds means eight to the 1/3 raised to the fifth power, eight to the 1/3 is two and two to the fifth is 32. So this expression simplifies to three fifths pi times 32, or 96/5 A disc here means the inside of a circle. The figure on the right is different because it's hollow inside. And when we slice it by planes perpendicular to the x axis, we get cross sections that are shaped like washers. A washer here means the region in between two concentric circles. So for solids of revolution, the cross sections can have the shape of a disk, or the shape of a washer. The area of a disk is given by the familiar formula, pi r squared, where r is the radius. And the area of a washer can be written as pi times r outer squared minus pi times our inner squared, where our outer is the radius of the big circle, and our inner is the radius of the little circle. This formula works because the area of the washer is just the area of the larger circle minus the area of the inside smaller circle. Now we know that the volume of any solid that can be sliced into cross sections using planes perpendicular to the x axis is given by v is the integral from x equals a to x equals B of the area of the cross section at point x, integrated dx. If our cross sections are disks, this formula becomes the integral of pi r squared dx, where the radius R is a function of x. If instead, the cross sections look like washers, then the volume formula becomes the integral of pi r outer squared minus pi r inner squared dx. Again, our outer and our inner are functions of x here. These formulas work when they solid a revolution is formed by rotating a region around the x axis or any horizontal line. When we rotate around a horizontal line, then our cross sectional discs or washers are perpendicular to the x axis and are thin in the x direction. So it makes sense to integrate dx. If instead, we want to rotate around the y axis or a vertical line, then our disk or wash our cross sections are going to be perpendicular to the y axis and are going to be thin in the y direction. So when rotating around the y axis or a vertical line, we'll need to do our integral with respect to y. Our cross sectional area will be a function of y will integrate d y and our bounds of integration will have to be Y values, our formulas will look pretty much the same. We'll just have to calculate our Radia and bounds of integration in terms of y instead of x. As our first example, let's consider the region bounded by the curve y equals a cube root of x, the x axis and the line x equals eight. We want to find the volume of the solid of revolution found by rotating this region around the x axis. or cross sections here are going to be discs. And these discs are thin in the extraction. So we're going to be integrating dx, our smallest x value is zero, and our largest is eight. So those are our bounds of integration. And we want to integrate pi times the radius squared dx. Now the radius of our disks is given by the y coordinate on this curve. So we can write r is equal to y, which is equal to the cube root of x according to our equation. So we can rewrite the volume as the integral from zero to eight of pi times the cube root of x squared dx, I can pull out the PI and rewrite this integral using exponents and integrate and then evaluate using bounds of integration to get three fifths pi times eight to the five thirds minus zero. Now, eight to the five thirds means eight to the 1/3 raised to the fifth power, eight to the 1/3 is two and two to the fifth is 32. So this expression simplifies to three fifths pi times 32, or 96/5 pie. pie. As our next example, let's consider the region in the first quadrant bounded by two curves, the curve y equals the cube root of x and y equals 1/4 times x. We'll start by rotating this region around the x axis to get this sort of hollow base shape. Notice that our cross sections this time are shaped like washers, where the outer circle of the washer is swept out by the curve y equals cube root of x, and the inner circle is swept out by the curve y equals 1/4 X. We know our volume is given by the integral of pi times our outer squared minus pi times our inner squared. And since our washers are thin in the x direction, we know we'll need to integrate dx, our bounds of integration are just our lowest x value of zero, and our largest x value, which is where these two curves intersect, which is an x value of eight. We can confirm that the two curves intersect when x equals eight by setting them equal to each other. And solving for x, dividing both sides by x to the 1/3. And multiplying both sides by four gives us four equals x to the two thirds. And raising both sides to the three halves power gives us x equals four to three halves, or x equals eight. As our next example, let's consider the region in the first quadrant bounded by two curves, the curve y equals the cube root of x and y equals 1/4 times x. We'll start by rotating this region around the x axis to get this sort of hollow base shape. Notice that our cross sections this time are shaped like washers, where the outer circle of the washer is swept out by the curve y equals cube root of x, and the inner circle is swept out by the curve y equals 1/4 X. We know our volume is given by the integral of pi times our outer squared minus pi times our inner squared. And since our washers are thin in the x direction, we know we'll need to integrate dx, our bounds of integration are just our lowest x value of zero, and our largest x value, which is where these two curves intersect, which is an x value of eight. We can confirm that the two curves intersect when x equals eight by setting them equal to each other. And solving for x, dividing both sides by x to the 1/3. And multiplying both sides by four gives us four equals x to the two thirds. And raising both sides to the three halves power gives us x equals four to three halves, or x equals eight. This confirms we have the correct bound of integration here. Now we need to figure out a formula for the outer radius as a function of x and a formula for the inner radius. Also, This confirms we have the correct bound of integration here. Now we need to figure out a formula for the outer radius as a function of x and a formula for the inner radius. Also, since the outer circle is swept out by the curve y equals cube root of x, the outer radius is just given by the y coordinate of this curve as a function of x, that's the cube root of x. Now the inner radius is given by the y coordinate of this line, and the y coordinate of this line as a function of x is 1/4 x. So we set up an equation for the volume of our solid and I routine computation gives us a volume of 128 fifteenth's times pi. Now let's switch gears and rotate this region around the y axis instead. Our cross sections are still washers, but this time the washers are thin in the y direction. So we're going to be integrating with respect to y. Our bounds of integration are also y values and start at the minimum y value of zero and the maximum Y value of two, that corresponds to this intersection, where x equals eight, and y, which is the cube root of x, or 1/4 times x is then equal to two. For this problem, we need our our outer and our inner to the functions of y. From the picture, we see that our outer is actually the x coordinate on this line. The line has the equation y equals 1/4 x. And so x is equal to four y. So that gives us our outer as a function of y. defined our inner, we can do the same sort of thing using the x coordinate of the curve, y equals cube root of x. But writing that x coordinate in terms of y, we have y equals the cube root of x means that x is equal to y cubed. And so our inner is equal to y cubed as a function of y. putting this all together, we have an equation for volume. We can simplify this and integrate to get an answer of 512 pi over 21. In this video, we calculated the volumes of solids of revolution, using the disk and the washer methods, and the following formulas. One way to calculate the volume of a loaf of bread is to slice it into slices, and calculate the volume of each slice separately. That's the idea to keep in mind as we cover this topic, on volumes using cross sections. A lot of three dimensional objects are kind of like loaves of bread, and can be naturally sliced into slabs or slices like this. Let's suppose we've sliced our object into n slabs, we'll call them s one s two through sn in this picture, and is seven, here's s one, here's s two, here's s3, and so on, we'll assume we've sliced in such a way that each slice is the same thickness, we'll call that thickness delta x. Well, the volume of the solid is just the sum of the volumes of the slices. And sigma notation, we can write this as the sum from i equals one to n, the number of slabs volume of s by the I slab. Now the volume of the slab is approximately its cross sectional area times its width. By cross sectional area of the slab, I mean the area of the front or the back part of the slice, where you might spread peanut butter. Well, now, as you've noticed from making peanut butter sandwiches, the area of the front of the slice might be slightly different from the area of the back of the slice. So what we'll do is for each interval, I, each slab will just pick a sample point x vystar, that's in that I've interval, it could be on the left endpoint, or it could be on the right endpoint, or it could be in the middle somewhere. And we'll just look at the cross sectional area, I'll call it a at x, y star that represents the area if I were to go karate chop right at that x y star and figure out the cross sectional area at that point. since the outer circle is swept out by the curve y equals cube root of x, the outer radius is just given by the y coordinate of this curve as a function of x, that's the cube root of x. Now the inner radius is given by the y coordinate of this line, and the y coordinate of this line as a function of x is 1/4 x. So we set up an equation for the volume of our solid and I routine computation gives us a volume of 128 fifteenth's times pi. Now let's switch gears and rotate this region around the y axis instead. Our cross sections are still washers, but this time the washers are thin in the y direction. So we're going to be integrating with respect to y. Our bounds of integration are also y values and start at the minimum y value of zero and the maximum Y value of two, that corresponds to this intersection, where x equals eight, and y, which is the cube root of x, or 1/4 times x is then equal to two. For this problem, we need our our outer and our inner to the functions of y. From the picture, we see that our outer is actually the x coordinate on this line. The line has the equation y equals 1/4 x. And so x is equal to four y. So that gives us our outer as a function of y. defined our inner, we can do the same sort of thing using the x coordinate of the curve, y equals cube root of x. But writing that x coordinate in terms of y, we have y equals the cube root of x means that x is equal to y cubed. And so our inner is equal to y cubed as a function of y. putting this all together, we have an equation for volume. We can simplify this and integrate to get an answer of 512 pi over 21. In this video, we calculated the volumes of solids of revolution, using the disk and the washer methods, and the following formulas. One way to calculate the volume of a loaf of bread is to slice it into slices, and calculate the volume of each slice separately. That's the idea to keep in mind as we cover this topic, on volumes using cross sections. A lot of three dimensional objects are kind of like loaves of bread, and can be naturally sliced into slabs or slices like this. Let's suppose we've sliced our object into n slabs, we'll call them s one s two through sn in this picture, and is seven, here's s one, here's s two, here's s3, and so on, we'll assume we've sliced in such a way that each slice is the same thickness, we'll call that thickness delta x. Well, the volume of the solid is just the sum of the volumes of the slices. And sigma notation, we can write this as the sum from i equals one to n, the number of slabs volume of s by the I slab. Now the volume of the slab is approximately its cross sectional area times its width. By cross sectional area of the slab, I mean the area of the front or the back part of the slice, where you might spread peanut butter. Well, now, as you've noticed from making peanut butter sandwiches, the area of the front of the slice might be slightly different from the area of the back of the slice. So what we'll do is for each interval, I, each slab will just pick a sample point x vystar, that's in that I've interval, it could be on the left endpoint, or it could be on the right endpoint, or it could be in the middle somewhere. And we'll just look at the cross sectional area, I'll call it a at x, y star that represents the area if I were to go karate chop right at that x y star and figure out the cross sectional area at that point. Now we can go back and write our volume as the cross sectional area at x star times delta x the thickness of the slice. Now this is just an approximation of volume, because this expression here gives you the volume as if the slices are very regular, and have the same area from one side to the other. But it's a good approximation if the slices are thin. And in fact, we can calculate the exact volume by taking the limit as the slices get thinner and thinner. Or in other words, as the number of slices goes to infinity. What we have here is the limit of a Riemann sum. And therefore, we can rewrite it as an integral where the x i star becomes our variable x and the delta x becomes r dx. For the bounds of integration, we'll just use the abstract x values of a and b for a real problem. We would fill these in based on In the context of the problem, this gives us an abstract expression for the volume of a three dimensional object. But in practice, in order to compute volumes like this, we'll first need a formula for a of x, the cross sectional area as a function of x. As an example, let's try to find the volume of a solid whose base is an ellipse given by this equation, and whose cross sections perpendicular to the x axis are squares. First, let me graph the base, it looks like an ellipse that's thinner in the extraction than in the y direction. So something like this. Now sitting above this base, are a bunch of squares. And the squares are oriented in such a way that they're perpendicular to the x axis. So they're oriented sort of like this, I'll try to draw a square here, that's supposed to be coming out of the picture here. And here's another square, again, coming out of the picture. Here's a slightly better picture that I drew using Mathematica. It's tilted. So we're looking at it from below, where you can see the ellipse here. And you can see the square cross sections, the x axis is going in this direction, that's x and the y axis is in that direction. This picture is actually an approximation of the solid, where they're only about eight or 10 slices, each one's kind of sick and has the same area on the front and the back. better picture of the solid is this one. Here, the slices are infinitely thin, but they're still square shaped. And they're still oriented in such a way that they're perpendicular to the x axis. Now we know that volume is given by the integral from a to b of area as a function of x dx. On our lips, the minimum x value is negative two, and the maximum x value is two. So I can write those in for my bounds of integration. Also, I know that the area of a square is just the side length squared. So I can write my cross sectional area as s of x squared, where s of x is the side length of the square as a function of x. Notice that for different x values, my side lengths are different. But my side length is always twice the y value twice the distance from the x axis to the y value on the ellipse. So I'll rewrite my formula as two times y as a function of x squared. I can simplify this a little bit as the integral of four times y of x squared, dx. Now all I need to do is find a formula for y as a function of x. And since I've got this equation up here, relating y and x, all I have to do is solve for y in terms of x. In fact, I can get by solving for y squared, since I've really got y squared in my formula. solving for y squared, I have y squared over nine is equal to one minus x squared over four, which means that y squared is equal to nine times one minus x squared Now we can go back and write our volume as the cross sectional area at x star times delta x the thickness of the slice. Now this is just an approximation of volume, because this expression here gives you the volume as if the slices are very regular, and have the same area from one side to the other. But it's a good approximation if the slices are thin. And in fact, we can calculate the exact volume by taking the limit as the slices get thinner and thinner. Or in other words, as the number of slices goes to infinity. What we have here is the limit of a Riemann sum. And therefore, we can rewrite it as an integral where the x i star becomes our variable x and the delta x becomes r dx. For the bounds of integration, we'll just use the abstract x values of a and b for a real problem. We would fill these in based on In the context of the problem, this gives us an abstract expression for the volume of a three dimensional object. But in practice, in order to compute volumes like this, we'll first need a formula for a of x, the cross sectional area as a function of x. As an example, let's try to find the volume of a solid whose base is an ellipse given by this equation, and whose cross sections perpendicular to the x axis are squares. First, let me graph the base, it looks like an ellipse that's thinner in the extraction than in the y direction. So something like this. Now sitting above this base, are a bunch of squares. And the squares are oriented in such a way that they're perpendicular to the x axis. So they're oriented sort of like this, I'll try to draw a square here, that's supposed to be coming out of the picture here. And here's another square, again, coming out of the picture. Here's a slightly better picture that I drew using Mathematica. It's tilted. So we're looking at it from below, where you can see the ellipse here. And you can see the square cross sections, the x axis is going in this direction, that's x and the y axis is in that direction. This picture is actually an approximation of the solid, where they're only about eight or 10 slices, each one's kind of sick and has the same area on the front and the back. better picture of the solid is this one. Here, the slices are infinitely thin, but they're still square shaped. And they're still oriented in such a way that they're perpendicular to the x axis. Now we know that volume is given by the integral from a to b of area as a function of x dx. On our lips, the minimum x value is negative two, and the maximum x value is two. So I can write those in for my bounds of integration. Also, I know that the area of a square is just the side length squared. So I can write my cross sectional area as s of x squared, where s of x is the side length of the square as a function of x. Notice that for different x values, my side lengths are different. But my side length is always twice the y value twice the distance from the x axis to the y value on the ellipse. So I'll rewrite my formula as two times y as a function of x squared. I can simplify this a little bit as the integral of four times y of x squared, dx. Now all I need to do is find a formula for y as a function of x. And since I've got this equation up here, relating y and x, all I have to do is solve for y in terms of x. In fact, I can get by solving for y squared, since I've really got y squared in my formula. solving for y squared, I have y squared over nine is equal to one minus x squared over four, which means that y squared is equal to nine times one minus x squared over four. over four. Now I plug this into my volume equation. And I get the integral from negative two to four times nine, one minus x squared over four dx. I'll pull out the four times nine that's 36. and integrate. Plugging in values and simplifying, we get a final answer of 96. Now, how would this problem be different if we used cross sections that were perpendicular to the y axis instead of the x axis? Well, for one thing, our picture would look a little bit different. Since our squares would now be running in the other direction. Since our squares are now fin in the y direction, instead of the extraction, it makes sense to have the width of a slab B delta y, and to compute our volume as an integral with respect to y. Our bounds of integration now and they also need to be Y values. So they run from the minimum y value of negative three to the maximum y value of three. And our cross sectional area should also be written in terms of y. area is still side length squared. But now our side length is actually twice our x value, instead of twice our y value. And we can write our x value squared in terms of our y value as four times one minus y squared over nine. Therefore, our area, which is our side length squared, which is 2x, quantity squared, or 4x squared, is going to be equal to 16 times one minus y squared over nine, we'll need to calculate our volume by taking the integral from negative three to three of 16 times one minus y squared over nine d y. If we integrate this, we get an answer of 64. a different answer from the answer to our first problem, and it makes sense that we get a different answer, because we now have a different three dimensional object with a different shape and a different volume. In this video, we saw that if we divided three dimensional objects into slices, then the volume of the three dimensional object is the integral of the cross sectional area, dx. In this video, I'll derive a formula to calculate the length of a curve given as a function y equals f of x. For a curve like this one that's made up of a bunch of straight line segments, it's easy to calculate the length just by using the distance formula to find the length of each line segment. The distance formula says the distance between two points x one, y one, and x two, y two is given by the square root of x two minus x one squared plus y two minus y one squared. Applying this formula to the first line segment, connecting the points one, two, and two, three, we get a length of the square root of two minus one squared plus three minus two squared, which is the square root of two, the length of the next line segment can be calculated similarly, and the next piece has length two, we don't even need the distance formula for that one. And the last line segment has a length of the square to five. If we add up all the lengths of these four line segments, we get a total length of the curve of two times the square root of five, plus the square root of two plus two, Now I plug this into my volume equation. And I get the integral from negative two to four times nine, one minus x squared over four dx. I'll pull out the four times nine that's 36. and integrate. Plugging in values and simplifying, we get a final answer of 96. Now, how would this problem be different if we used cross sections that were perpendicular to the y axis instead of the x axis? Well, for one thing, our picture would look a little bit different. Since our squares would now be running in the other direction. Since our squares are now fin in the y direction, instead of the extraction, it makes sense to have the width of a slab B delta y, and to compute our volume as an integral with respect to y. Our bounds of integration now and they also need to be Y values. So they run from the minimum y value of negative three to the maximum y value of three. And our cross sectional area should also be written in terms of y. area is still side length squared. But now our side length is actually twice our x value, instead of twice our y value. And we can write our x value squared in terms of our y value as four times one minus y squared over nine. Therefore, our area, which is our side length squared, which is 2x, quantity squared, or 4x squared, is going to be equal to 16 times one minus y squared over nine, we'll need to calculate our volume by taking the integral from negative three to three of 16 times one minus y squared over nine d y. If we integrate this, we get an answer of 64. a different answer from the answer to our first problem, and it makes sense that we get a different answer, because we now have a different three dimensional object with a different shape and a different volume. In this video, we saw that if we divided three dimensional objects into slices, then the volume of the three dimensional object is the integral of the cross sectional area, dx. In this video, I'll derive a formula to calculate the length of a curve given as a function y equals f of x. For a curve like this one that's made up of a bunch of straight line segments, it's easy to calculate the length just by using the distance formula to find the length of each line segment. The distance formula says the distance between two points x one, y one, and x two, y two is given by the square root of x two minus x one squared plus y two minus y one squared. Applying this formula to the first line segment, connecting the points one, two, and two, three, we get a length of the square root of two minus one squared plus three minus two squared, which is the square root of two, the length of the next line segment can be calculated similarly, and the next piece has length two, we don't even need the distance formula for that one. And the last line segment has a length of the square to five. If we add up all the lengths of these four line segments, we get a total length of the curve of two times the square root of five, plus the square root of two plus two, we can use the same process to approximate the length of any curve by dividing it up into n small pieces, and approximate the length of one small curved piece with a straight line segment. And using the distance formula to calculate the length of these line segments, in this picture, the curve is divided up into nine sub intervals. I'll label the corresponding points on the x axis, x zero for a x 1x, two, all the way through x eight and x nine is B. And I can label the points on my curve. So here's the sixth point P six has x coordinates xx, and then y coordinate will be f of x x, since my curve is given by the equation y equals f of x. More generally, I have N sub intervals. And I'll label an arbitrary point piece of I. The point before it is then a piece of i minus one. And the length of the ice segment is given by the distance between P sub i and P sub i minus one. So that's the square root of x sub i minus x sub i minus one squared plus f of x sub i minus f of x sub i minus one squared by the distance formula. The total length of the curve can be approximated by adding the lengths of all these line segments up. So that's the sum from i equals one to N for the N line segments of these lengths. This is starting to look a little bit like a Riemann sum because of the Sigma sign, but it's missing the delta x That I usually have out here. So I'm going to use a trick, I'm going to multiply each term of this expression by x sub i minus x sub i minus one divided by x sub i minus x sub i minus one. This doesn't change the value of my expression, but it does introduce a delta x over delta x into my equation, because delta x represents the width of a sub interval. And that's equal to x by minus x, y minus one, I'm going to suck the factor in the denominator inside the square root sign, notice I have to square it, when I pull it inside the square root sign. Now I'm going to rewrite my fraction inside the square root sign. Now this first fraction is just one, and the second fraction can be rewritten with a single squared sign. And the second expression should look familiar to you. It's the expression for the slope of a secant line. At effects, I n x i minus one are very close to each other, the slope of that secant line is very close to the slope of a tangent line for at a point in that interval. In fact, you might recall that the mean value theorem says that the slope of the secant line is actually exactly equal to the slope of a tangent line. For some point, I'll call it x i star in that interval. So I'll rewrite my expression for the approximate length of the curve. And since x i minus x i minus one can be written as delta x, I have a Riemann sum here. So if I want to find the exact arc length, I just need to take the limit of this Riemann sum. This is the limit as the number of intervals goes to infinity. We know that the limit of a Riemann sum is given by an integral. So the arc length is given by the integral of the square root of one plus f prime of x squared dx, starting from the first x value of A to the last x value of b. And so we've derived a formula for arc length. Sometimes, this formula is also written with the alternative notation of dydx, instead of f prime of x. Let's use the arc length formula to find the length of the curve y equals x to the three halves between x equals one and x equals four. That's this section of the curve drawn. we can use the same process to approximate the length of any curve by dividing it up into n small pieces, and approximate the length of one small curved piece with a straight line segment. And using the distance formula to calculate the length of these line segments, in this picture, the curve is divided up into nine sub intervals. I'll label the corresponding points on the x axis, x zero for a x 1x, two, all the way through x eight and x nine is B. And I can label the points on my curve. So here's the sixth point P six has x coordinates xx, and then y coordinate will be f of x x, since my curve is given by the equation y equals f of x. More generally, I have N sub intervals. And I'll label an arbitrary point piece of I. The point before it is then a piece of i minus one. And the length of the ice segment is given by the distance between P sub i and P sub i minus one. So that's the square root of x sub i minus x sub i minus one squared plus f of x sub i minus f of x sub i minus one squared by the distance formula. The total length of the curve can be approximated by adding the lengths of all these line segments up. So that's the sum from i equals one to N for the N line segments of these lengths. This is starting to look a little bit like a Riemann sum because of the Sigma sign, but it's missing the delta x That I usually have out here. So I'm going to use a trick, I'm going to multiply each term of this expression by x sub i minus x sub i minus one divided by x sub i minus x sub i minus one. This doesn't change the value of my expression, but it does introduce a delta x over delta x into my equation, because delta x represents the width of a sub interval. And that's equal to x by minus x, y minus one, I'm going to suck the factor in the denominator inside the square root sign, notice I have to square it, when I pull it inside the square root sign. Now I'm going to rewrite my fraction inside the square root sign. Now this first fraction is just one, and the second fraction can be rewritten with a single squared sign. And the second expression should look familiar to you. It's the expression for the slope of a secant line. At effects, I n x i minus one are very close to each other, the slope of that secant line is very close to the slope of a tangent line for at a point in that interval. In fact, you might recall that the mean value theorem says that the slope of the secant line is actually exactly equal to the slope of a tangent line. For some point, I'll call it x i star in that interval. So I'll rewrite my expression for the approximate length of the curve. And since x i minus x i minus one can be written as delta x, I have a Riemann sum here. So if I want to find the exact arc length, I just need to take the limit of this Riemann sum. This is the limit as the number of intervals goes to infinity. We know that the limit of a Riemann sum is given by an integral. So the arc length is given by the integral of the square root of one plus f prime of x squared dx, starting from the first x value of A to the last x value of b. And so we've derived a formula for arc length. Sometimes, this formula is also written with the alternative notation of dydx, instead of f prime of x. Let's use the arc length formula to find the length of the curve y equals x to the three halves between x equals one and x equals four. That's this section of the curve drawn. Here's the general formula for arc length. And since dydx, for our curve is three halves x to the one half, we get that arc length is the integral from one to four of the square root of one plus three halves x to the one half squared dx. After simplifying a little bit, we can use use substitution to rewrite this. So we get the integral from 13 fourths, to 10 of the square root of u times four ninths do which integrates to u to the three halves divided by three halves, times four ninths, evaluated between 10 and 13 fourths, which after some computation works out to 1/27 times at the square root of 10 minus 13, the squared of 13 or approximately 7.6 units, which seems about right, based on the graph, taking into account the fact that this scale here is by twos on the y axis. In this video, we found the formula for the arc length of a curve. If the curve stretches from x equals a to x equals b, then the arc length is given by the integral from a to b of the square root of one plus f prime of x squared dx. This video introduces the idea of work from physics and the key role of integration in doing work calculations. If a constant force f is applied to move an object a distance d, then the work done to move the object is defined to be work equals force times distance, or in symbols, w equals F times d. The units of force can be given in metric units, or in bold fashion, English units. Since force is mass times acceleration, the units of force are going to be units of mass, which are kilograms, times units of acceleration, meters per second squared. This collection of units is also called a Newton. In English units, force is given typically, in terms of pounds. Now the units of work since work is force times distance, and distance in metric units is in meters. That gives units for work of kilograms meters squared per second squared, or we can write it work as Newton meters. And these collection of units is also given its own name, which is the joule. If we're using English units for work, work again is force times distance. So the units become pounds times feet are usually this is written instead, as a foot pound. Now, if I give you my weight in pounds, I weigh about 140 pounds. Those pounds are already and I get it a force. So that 140 pounds, my weight is also telling the force of gravity on my body. But if I tell you instead, that my mass is 63.5 kilograms, well, that's a unit of mass not a unit of force. So if I want to know the force, due to gravity on my body, I'm gonna have to multiply that 63.5 by the acceleration due to gravity, which is 9.8 meters per second squared. That product works out to be 622.3 kilogram meters per second squared. Notice that we're now have the right units for force, we could also say, that's the force of gravity on my body is 622.3 Newtons. Here's the general formula for arc length. And since dydx, for our curve is three halves x to the one half, we get that arc length is the integral from one to four of the square root of one plus three halves x to the one half squared dx. After simplifying a little bit, we can use use substitution to rewrite this. So we get the integral from 13 fourths, to 10 of the square root of u times four ninths do which integrates to u to the three halves divided by three halves, times four ninths, evaluated between 10 and 13 fourths, which after some computation works out to 1/27 times at the square root of 10 minus 13, the squared of 13 or approximately 7.6 units, which seems about right, based on the graph, taking into account the fact that this scale here is by twos on the y axis. In this video, we found the formula for the arc length of a curve. If the curve stretches from x equals a to x equals b, then the arc length is given by the integral from a to b of the square root of one plus f prime of x squared dx. This video introduces the idea of work from physics and the key role of integration in doing work calculations. If a constant force f is applied to move an object a distance d, then the work done to move the object is defined to be work equals force times distance, or in symbols, w equals F times d. The units of force can be given in metric units, or in bold fashion, English units. Since force is mass times acceleration, the units of force are going to be units of mass, which are kilograms, times units of acceleration, meters per second squared. This collection of units is also called a Newton. In English units, force is given typically, in terms of pounds. Now the units of work since work is force times distance, and distance in metric units is in meters. That gives units for work of kilograms meters squared per second squared, or we can write it work as Newton meters. And these collection of units is also given its own name, which is the joule. If we're using English units for work, work again is force times distance. So the units become pounds times feet are usually this is written instead, as a foot pound. Now, if I give you my weight in pounds, I weigh about 140 pounds. Those pounds are already and I get it a force. So that 140 pounds, my weight is also telling the force of gravity on my body. But if I tell you instead, that my mass is 63.5 kilograms, well, that's a unit of mass not a unit of force. So if I want to know the force, due to gravity on my body, I'm gonna have to multiply that 63.5 by the acceleration due to gravity, which is 9.8 meters per second squared. That product works out to be 622.3 kilogram meters per second squared. Notice that we're now have the right units for force, we could also say, that's the force of gravity on my body is 622.3 Newtons. Now that we familiarize ourselves with units a bit, let's do some examples. As our first example, how much work is done to lift a two pound book off the floor onto a shelf that's five feet high? Well, we know that work is force times distance, and two pounds is already a unit of force. And distance is five feet. So the work done is 10 foot pounds. Now let's do the same problem in metric units. The two pound book is actually a naught point nine kilogram book, and we're lifting it off the floor onto a shelf that's about 1.5 meters high. Well, work is still force times distance. But now the force is point nine kilograms times the acceleration due to gravity 9.8 meters per second squared, times our distance of 1.5 meters. That gives us a product of 13.23 kilograms meters squared per second squared, or in other words, 13.23 jewels. In the previous two examples, force was constant, so we could just multiply force by distance to get work. Now let's consider the case where force is not constant. Let's say a particle moves along the x axis from a point x equals a to a point x equals b. According to a force, F of x, that's a function of x and varies with x. How much work is done in moving a particle although the force is not constant, on the whole interval from a to b, if we divide up that interval, into a bunch of little sub intervals, each of width delta x, then on any particular sub interval The force is going to be approximately constant, it's not going to change a whole lot on a tiny little sub interval. As usual, let's pick a sample point exabyte star in the eye sub interval for each little sub interval. exabytes star could be the left endpoint of the sub interval, the right endpoint or any point in the middle. Now, on the AI sub interval, the force is approximately constant is approximately equal to f at EXA biostar. Therefore, the work on that I sub interval is approximately equal to this constant f x sub i star times the distance that the particle is going on that sub interval, that that distance is just the length of the sub interval delta x. Instead of thinking of the big picture of the particle going all the way from A to B, I'm thinking of it going just along the first sub interval with the approximately constant force at a distance of delta x. And then it's going to go the second sub interval, again, the forces approximately constant times the distance of delta x. And then we'll do some more work getting along the third sub interval, another proximately, constant force times delta x. And so on each little tiny bit of the way, I'll get another little chunk of work. And then I can get the entire amount of work by adding all those little chunks of work up. So the total work done is going to be the sum from i equals one to n, where n is the number of sub intervals of the work done on each sub interval, which is f of x sub i star times distance delta x. I should say this is approximately the total work, in order to get the actual total work, we'll need to take a limit as we use more and more skinnier and skinnier subintervals. So the limit as n goes to infinity of this Riemann sum, the limit of a Riemann sum is an integral. So we've got the integral of the force f of x, dx, and we'll integrate between the minimum x value of a and the maximum x value of b. And that's our formula for work. Now that we familiarize ourselves with units a bit, let's do some examples. As our first example, how much work is done to lift a two pound book off the floor onto a shelf that's five feet high? Well, we know that work is force times distance, and two pounds is already a unit of force. And distance is five feet. So the work done is 10 foot pounds. Now let's do the same problem in metric units. The two pound book is actually a naught point nine kilogram book, and we're lifting it off the floor onto a shelf that's about 1.5 meters high. Well, work is still force times distance. But now the force is point nine kilograms times the acceleration due to gravity 9.8 meters per second squared, times our distance of 1.5 meters. That gives us a product of 13.23 kilograms meters squared per second squared, or in other words, 13.23 jewels. In the previous two examples, force was constant, so we could just multiply force by distance to get work. Now let's consider the case where force is not constant. Let's say a particle moves along the x axis from a point x equals a to a point x equals b. According to a force, F of x, that's a function of x and varies with x. How much work is done in moving a particle although the force is not constant, on the whole interval from a to b, if we divide up that interval, into a bunch of little sub intervals, each of width delta x, then on any particular sub interval The force is going to be approximately constant, it's not going to change a whole lot on a tiny little sub interval. As usual, let's pick a sample point exabyte star in the eye sub interval for each little sub interval. exabytes star could be the left endpoint of the sub interval, the right endpoint or any point in the middle. Now, on the AI sub interval, the force is approximately constant is approximately equal to f at EXA biostar. Therefore, the work on that I sub interval is approximately equal to this constant f x sub i star times the distance that the particle is going on that sub interval, that that distance is just the length of the sub interval delta x. Instead of thinking of the big picture of the particle going all the way from A to B, I'm thinking of it going just along the first sub interval with the approximately constant force at a distance of delta x. And then it's going to go the second sub interval, again, the forces approximately constant times the distance of delta x. And then we'll do some more work getting along the third sub interval, another proximately, constant force times delta x. And so on each little tiny bit of the way, I'll get another little chunk of work. And then I can get the entire amount of work by adding all those little chunks of work up. So the total work done is going to be the sum from i equals one to n, where n is the number of sub intervals of the work done on each sub interval, which is f of x sub i star times distance delta x. I should say this is approximately the total work, in order to get the actual total work, we'll need to take a limit as we use more and more skinnier and skinnier subintervals. So the limit as n goes to infinity of this Riemann sum, the limit of a Riemann sum is an integral. So we've got the integral of the force f of x, dx, and we'll integrate between the minimum x value of a and the maximum x value of b. And that's our formula for work. Let's look at a physical example. How much work is required to lift 1000 kilograms satellite from the Earth's surface to an altitude of two times 10 to the six meters above the Earth's surface. We're given that the gravitational force is F equals g times capital M times lowercase m divided by r squared, where m is the mass of the Earth, lowercase m is the mass of the satellite, r is the distance between the satellite and the center of the earth, and g is the gravitational constant. We're also given numbers for the radius of the Earth, the mass of the Earth, and the gravitational constant. This problem is different from the problem of lifting the calculus book. When we lifted a book over just a few meters, the force of gravity was essentially constant over such a small distance. So we could use the equation work equals force times distance. But in this problem, since we're moving the satellite a larger distance, the force of gravity changes with distance. And so we need to use work as the integral of this force with respect to distance. The distance variable in this problem is R. So I'll rewrite this using the equation for force and integrate with respect to Dr. Let's look at a physical example. How much work is required to lift 1000 kilograms satellite from the Earth's surface to an altitude of two times 10 to the six meters above the Earth's surface. We're given that the gravitational force is F equals g times capital M times lowercase m divided by r squared, where m is the mass of the Earth, lowercase m is the mass of the satellite, r is the distance between the satellite and the center of the earth, and g is the gravitational constant. We're also given numbers for the radius of the Earth, the mass of the Earth, and the gravitational constant. This problem is different from the problem of lifting the calculus book. When we lifted a book over just a few meters, the force of gravity was essentially constant over such a small distance. So we could use the equation work equals force times distance. But in this problem, since we're moving the satellite a larger distance, the force of gravity changes with distance. And so we need to use work as the integral of this force with respect to distance. The distance variable in this problem is R. So I'll rewrite this using the equation for force and integrate with respect to Dr. I'm starting at the Earth's surface. So that's a distance of 6.4 times 10 to the six meters from the center of the earth, since that's the earth radius. And I'm ending at a height of two times 10 to the six meters above the Earth's surface. So that's a distance r of 6.4 plus two, or 8.4 times 10 to the six meters from the earth center. My only variable for this integral is R. So let me pull out the constants. And I'll rewrite the one over r squared as r to the minus two. Now I can integrate an R to the minus two becomes r to the minus one over minus one Rewrite one more time and substitute in for R to get a preliminary answer of negative GE, capital M lowercase n times negative 3.72024 times 10 to the minus eight. Now I still need to plug in for capital G, capital M and lowercase m, my negatives cancel here. And I have G is 6.67 times 10 to the minus 11. capital M, the mass of the Earth is six times 10 to the 24, and lowercase and the mass the satellite was 1000 kilograms. Multiplying all these numbers together, gives us a final answer of approximately 1.5 times 10 to the 10th. jewels. To put this number in perspective, this is about the same amount of work done by a car in a year, or by the human heart beating for about 400 years. In this video, we saw that for a constant force work is just equal to the force times distance. But for a variable force work is equal to the integral of force with respect to distance. This video introduces the idea of an average value of a function. To take the average of a finite list of numbers, we just add the numbers up and divide by n, the number of numbers. In summation notation, we write the sum from i equals one to n of Q i all divided by n. But defining the average value of a continuous function is a little different. Because a function can take on infinitely many values on an interval from a to be, we could estimate the average value of the function by sampling it at a finite Li many evenly spaced x values. I'll call them x one through x n. And let's assume that they're spaced a distance of delta x apart, then the average value of f at these sample points is just the sum of the values of f divided by n, the number of values are in summation notation, the sum from i equals one to n of f of x i all divided by n. This is an approximate average value of f, since we're just using n sample points. But the approximation gets better as the number of sample points n gets bigger and bigger. So we could define the average as the limit as n goes to infinity of the sample average. I'd like to make this look more like a Riemann sum. So I need to get delta x in there. So I'm just going to multiply the top and the bottom by delta x. And notice that n times delta x is just the length of the interval b minus a. Now as the number of sample points goes to infinity, delta x, the distance between them goes to zero. So I can rewrite my limit as the limit as delta x goes to zero of the sum of FX II times delta x divided by b minus a. Now the limit of this Riemann sum in the numerator is just the integral from a to b of f of x dx. And so the average value of the function is given by the integral on the interval from a to b divided by the length of the interval. I'm starting at the Earth's surface. So that's a distance of 6.4 times 10 to the six meters from the center of the earth, since that's the earth radius. And I'm ending at a height of two times 10 to the six meters above the Earth's surface. So that's a distance r of 6.4 plus two, or 8.4 times 10 to the six meters from the earth center. My only variable for this integral is R. So let me pull out the constants. And I'll rewrite the one over r squared as r to the minus two. Now I can integrate an R to the minus two becomes r to the minus one over minus one Rewrite one more time and substitute in for R to get a preliminary answer of negative GE, capital M lowercase n times negative 3.72024 times 10 to the minus eight. Now I still need to plug in for capital G, capital M and lowercase m, my negatives cancel here. And I have G is 6.67 times 10 to the minus 11. capital M, the mass of the Earth is six times 10 to the 24, and lowercase and the mass the satellite was 1000 kilograms. Multiplying all these numbers together, gives us a final answer of approximately 1.5 times 10 to the 10th. jewels. To put this number in perspective, this is about the same amount of work done by a car in a year, or by the human heart beating for about 400 years. In this video, we saw that for a constant force work is just equal to the force times distance. But for a variable force work is equal to the integral of force with respect to distance. This video introduces the idea of an average value of a function. To take the average of a finite list of numbers, we just add the numbers up and divide by n, the number of numbers. In summation notation, we write the sum from i equals one to n of Q i all divided by n. But defining the average value of a continuous function is a little different. Because a function can take on infinitely many values on an interval from a to be, we could estimate the average value of the function by sampling it at a finite Li many evenly spaced x values. I'll call them x one through x n. And let's assume that they're spaced a distance of delta x apart, then the average value of f at these sample points is just the sum of the values of f divided by n, the number of values are in summation notation, the sum from i equals one to n of f of x i all divided by n. This is an approximate average value of f, since we're just using n sample points. But the approximation gets better as the number of sample points n gets bigger and bigger. So we could define the average as the limit as n goes to infinity of the sample average. I'd like to make this look more like a Riemann sum. So I need to get delta x in there. So I'm just going to multiply the top and the bottom by delta x. And notice that n times delta x is just the length of the interval b minus a. Now as the number of sample points goes to infinity, delta x, the distance between them goes to zero. So I can rewrite my limit as the limit as delta x goes to zero of the sum of FX II times delta x divided by b minus a. Now the limit of this Riemann sum in the numerator is just the integral from a to b of f of x dx. And so the average value of the function is given by the integral on the interval from a to b divided by the length of the interval. Notice the similarity between the formula for the average value of a function and the formula for the average value of a list of numbers, the integral for the function corresponds to the summation sign for the list of numbers. And the length of the interval B minus A for the function corresponds to n, the number of numbers in the list of numbers. Now let's work an example. For the function g of x equals one over one minus 5x. On the interval from two to five. We know that the average value of G is given by the integral from two to five of one over one minus 5x dx divided by the length of that interval. I'm going to use use of the tuition to integrate. So I'm going to set u equal to one minus 5x. So d u is negative five dx. In other words, dx is negative 1/5 times do. Looking at my bounds of integration, when x is equal to two, u is equal to one minus five times two, which is negative nine. And when x is equal to five, u is equal to negative 24. substituting into my integral, I get the integral from negative nine to negative 24 of one over u times negative 1/5. Do and that's divided by three. Now dividing by three is same as multiplying by 1/3. And as I integrate, I'm going to pull the negative 1/5 out, and then take the integral of one over u, that's ln of the absolute value of u, evaluated in between negative 24 and negative nine. The absolute value signs are important here, because they prevent me from trying to take the natural log of negative numbers. To evaluate, I get negative 1/15 times ln of 24 minus ln of nine, I can use my log rules to simplify and get negative 1/15 ln of 24 over nine, that's negative 1/15 ln of eight thirds, and as a decimal, that's approximately negative 0.0654. So I found the average value of G. Now my next question is, does g ever achieve that average value, in other words, is there a number c in the interval from two to five for which GFC equals its average value? Well, one way to find out is just to set GSC equal to GS average value. In other words, set one over one minus five c equal to negative 1/15 ln of eight thirds, and try to solve for C. There are lots of ways to solve this equation. But I'm going to take the reciprocal of both sides, subtract one from both sides and divide by negative five. This simplifies to three over ln of eight thirds, plus 1/5. What which is approximately 3.25. And that x value does lie inside the interval from two to five. So we've demonstrated that g does achieve its average value over the interval. But in fact, we could have predicted this to be true. Gs average value has to lie somewhere between GS minimum value and maximum value on this interval. And since G is continuous on the interval from two to five, it has to achieve every value that lies in between as minimum and maximum, including its average value. The same argument shows that for any continuous function, the function must achieve its average value on an interval. And this is known as the mean value theorem for integrals. Namely, for any continuous function f of x on an interval from a to b, there has to be at least one number c, between A and B, such that f of c equals its average value, or, in symbols, f of c equals the integral from a to b of f of x dx divided by b minus a. This video gave the definition of an average value of a function, and stated the mean value theorem for integrals. If we rewrite the formula for average value a little, then we can see a geometric interpretation for average value. The area of the box with height the average value Notice the similarity between the formula for the average value of a function and the formula for the average value of a list of numbers, the integral for the function corresponds to the summation sign for the list of numbers. And the length of the interval B minus A for the function corresponds to n, the number of numbers in the list of numbers. Now let's work an example. For the function g of x equals one over one minus 5x. On the interval from two to five. We know that the average value of G is given by the integral from two to five of one over one minus 5x dx divided by the length of that interval. I'm going to use use of the tuition to integrate. So I'm going to set u equal to one minus 5x. So d u is negative five dx. In other words, dx is negative 1/5 times do. Looking at my bounds of integration, when x is equal to two, u is equal to one minus five times two, which is negative nine. And when x is equal to five, u is equal to negative 24. substituting into my integral, I get the integral from negative nine to negative 24 of one over u times negative 1/5. Do and that's divided by three. Now dividing by three is same as multiplying by 1/3. And as I integrate, I'm going to pull the negative 1/5 out, and then take the integral of one over u, that's ln of the absolute value of u, evaluated in between negative 24 and negative nine. The absolute value signs are important here, because they prevent me from trying to take the natural log of negative numbers. To evaluate, I get negative 1/15 times ln of 24 minus ln of nine, I can use my log rules to simplify and get negative 1/15 ln of 24 over nine, that's negative 1/15 ln of eight thirds, and as a decimal, that's approximately negative 0.0654. So I found the average value of G. Now my next question is, does g ever achieve that average value, in other words, is there a number c in the interval from two to five for which GFC equals its average value? Well, one way to find out is just to set GSC equal to GS average value. In other words, set one over one minus five c equal to negative 1/15 ln of eight thirds, and try to solve for C. There are lots of ways to solve this equation. But I'm going to take the reciprocal of both sides, subtract one from both sides and divide by negative five. This simplifies to three over ln of eight thirds, plus 1/5. What which is approximately 3.25. And that x value does lie inside the interval from two to five. So we've demonstrated that g does achieve its average value over the interval. But in fact, we could have predicted this to be true. Gs average value has to lie somewhere between GS minimum value and maximum value on this interval. And since G is continuous on the interval from two to five, it has to achieve every value that lies in between as minimum and maximum, including its average value. The same argument shows that for any continuous function, the function must achieve its average value on an interval. And this is known as the mean value theorem for integrals. Namely, for any continuous function f of x on an interval from a to b, there has to be at least one number c, between A and B, such that f of c equals its average value, or, in symbols, f of c equals the integral from a to b of f of x dx divided by b minus a. This video gave the definition of an average value of a function, and stated the mean value theorem for integrals. If we rewrite the formula for average value a little, then we can see a geometric interpretation for average value. The area of the box with height the average value is the same as the area under the curve. This video gives two proofs of the mean value theorem for integrals. the mean value theorem for integrals says the for continuous function f of x, defined on an interval from a to b, there's some number c between A and B, such that f of c is equal to the average value of f. The first proof that I'm going to give us is the intermediate value theorem. Recall that the intermediate value theorem says that if we have a continuous function f defined on an interval, which I'll call x 1x, two. If we have some number l in between f of x one and f of x two, then f has to achieve the value l somewhere between x one and x two Keeping in mind the intermediate value theorem, let's turn our attention back to the mean value theorem for integrals. Now, it's possible that our function f of x might be constant on the interval from a to b. But if that's true, then our mean value theorem for integrals holds easily. Because f AF is just equal to that constant, which is equal to f of c for any c between A and B. So let's assume that f is not constant. Well, like continuous function on a closed interval has to have a minimum value and a maximum value, which I'll call little m, and big M. Now, we know that F's average value on the interval has to be between its maximum value and its minimum value. If you don't believe this, consider the fact that all of us values on the interval have to lie between big M and little m. And if we integrate this inequality, we get little m times b minus a is less than or equal to the integral of f is less than or equal to big M times b minus a. Notice that the first and the last integrals, were just integrating a constant. Now if I divide all three sides by b minus a, I can see that little m is less than or equal to the average value of f is less than or equal to big M as I wanted. Now, I just need to apply the intermediate value theorem with F average as my number L and little m and big M as my values of f of x one and f of x two. The intermediate value theorem says that F average is achieved by f of c for some C in between my x one and x two. And therefore, for some C in my interval a b. And that proves the mean value theorem for integrals. Now I'm going to give a second proof for the mean value theorem for integrals. And this time, it's going to be as a corollary to the regular mean value theorem for functions. Recall that the mean value theorem for functions, says that if g of x is continuous on a closed interval, and differentiable on the interior of that interval, then there's some number c in the interval, such that the derivative of g at C is equal to the average rate of change of G, across the whole interval from a to b. Let's keep the mean value theorem for functions in mind, and turn our attention back to the mean value theorem for integrals. I'm going to define a function g of x to be the integral from a to x of f of t dt, where F is the function given to us in the statement of the mean value theorem for integrals. Notice that g of A is just the integral from a to a, which is zero. Well, g of B is the integral from a to b of our function. Now, by the fundamental theorem of calculus, our function g of x is continuous and differentiable on the interval a, b, and g prime of x is equal to f of x. is the same as the area under the curve. This video gives two proofs of the mean value theorem for integrals. the mean value theorem for integrals says the for continuous function f of x, defined on an interval from a to b, there's some number c between A and B, such that f of c is equal to the average value of f. The first proof that I'm going to give us is the intermediate value theorem. Recall that the intermediate value theorem says that if we have a continuous function f defined on an interval, which I'll call x 1x, two. If we have some number l in between f of x one and f of x two, then f has to achieve the value l somewhere between x one and x two Keeping in mind the intermediate value theorem, let's turn our attention back to the mean value theorem for integrals. Now, it's possible that our function f of x might be constant on the interval from a to b. But if that's true, then our mean value theorem for integrals holds easily. Because f AF is just equal to that constant, which is equal to f of c for any c between A and B. So let's assume that f is not constant. Well, like continuous function on a closed interval has to have a minimum value and a maximum value, which I'll call little m, and big M. Now, we know that F's average value on the interval has to be between its maximum value and its minimum value. If you don't believe this, consider the fact that all of us values on the interval have to lie between big M and little m. And if we integrate this inequality, we get little m times b minus a is less than or equal to the integral of f is less than or equal to big M times b minus a. Notice that the first and the last integrals, were just integrating a constant. Now if I divide all three sides by b minus a, I can see that little m is less than or equal to the average value of f is less than or equal to big M as I wanted. Now, I just need to apply the intermediate value theorem with F average as my number L and little m and big M as my values of f of x one and f of x two. The intermediate value theorem says that F average is achieved by f of c for some C in between my x one and x two. And therefore, for some C in my interval a b. And that proves the mean value theorem for integrals. Now I'm going to give a second proof for the mean value theorem for integrals. And this time, it's going to be as a corollary to the regular mean value theorem for functions. Recall that the mean value theorem for functions, says that if g of x is continuous on a closed interval, and differentiable on the interior of that interval, then there's some number c in the interval, such that the derivative of g at C is equal to the average rate of change of G, across the whole interval from a to b. Let's keep the mean value theorem for functions in mind, and turn our attention back to the mean value theorem for integrals. I'm going to define a function g of x to be the integral from a to x of f of t dt, where F is the function given to us in the statement of the mean value theorem for integrals. Notice that g of A is just the integral from a to a, which is zero. Well, g of B is the integral from a to b of our function. Now, by the fundamental theorem of calculus, our function g of x is continuous and differentiable on the interval a, b, and g prime of x is equal to f of x. And by the mean value theorem for functions, we know that g prime of c has to equal g of b minus g of a over b minus a, for some number c and the interval a b, if we substitute in the three facts above, into our equation below, we get f of c is equal to the integral from a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we wanted to reach. This shows that the mean value theorem for integrals really is the mean value theorem for functions where our function is an integral. And this completes the second proof of the mean value theorem for integrals. So now I've proved the mean value theorem for integrals in two different ways. And I've used a lot of the great theorems of calculus along the way. In this video, we'll learn a technique of integration called inner By parts. integration by parts is based on the product rule for taking derivatives. Recall that the product rule says that when you take the derivative of a product of two functions, that's equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. If we rearrange this formula, by solving for this last term, we get f of x times g prime of x is equal to f of x g of x prime minus f prime of x g of x. Now let's integrate both sides of this equation with respect to x. The integral of a difference is the difference of the integrals. So I can break up this right hand side into two integrals. Now the integral of the derivative of f times g is just equal to f times g, by the fundamental theorem of calculus. And I carry the rest of the formula down. And now I have a formula relating the integral of f times g prime to the integral of f prime times G. This formula allows us to rewrite something that might be tricky to integrate, in terms of something that's hopefully easier to integrate. Although I've written this formula, using indefinite integrals with no bounds of integration, it would be just as easy to put on bounds of integration. Now, the fundamental theorem tells us that the integral of the derivative is the original product of functions evaluated from A to B. There's another version of this formula that might be a little easier to remember. If we let u equal f of x, and v equal g of x, then do is equal to f prime of x dx using differential notation, and dv is equal to g prime of x dx. In this notation, we can rewrite the formula as the integral of u times dv is equal to u times v minus the integral of v, d U. since V is our G of X, and D, u is our f prime of x dx. Again, we can include bounds of integration, if we're working with definite integrals. This will be our key formula for this section. and integrating using this formula is called integration by parts. Let's use integration by parts to find the integral of x e to the x dx. Or for formula for integration by parts, says the integral of u dv is equal to u times v minus the integral of the do. So we need to split up our inner grand x e to the x dx into the part that we're going to call you, and a part that we're going to call dv. One natural way to split it up is to let u equal x and dv equal e to the x dx. But another option might be to make u equal to e to the x and dv equal to x dx bar, we might decide to make you be the whole product, x e to the x and leave dv as just dx. And by the mean value theorem for functions, we know that g prime of c has to equal g of b minus g of a over b minus a, for some number c and the interval a b, if we substitute in the three facts above, into our equation below, we get f of c is equal to the integral from a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we wanted to reach. This shows that the mean value theorem for integrals really is the mean value theorem for functions where our function is an integral. And this completes the second proof of the mean value theorem for integrals. So now I've proved the mean value theorem for integrals in two different ways. And I've used a lot of the great theorems of calculus along the way. In this video, we'll learn a technique of integration called inner By parts. integration by parts is based on the product rule for taking derivatives. Recall that the product rule says that when you take the derivative of a product of two functions, that's equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. If we rearrange this formula, by solving for this last term, we get f of x times g prime of x is equal to f of x g of x prime minus f prime of x g of x. Now let's integrate both sides of this equation with respect to x. The integral of a difference is the difference of the integrals. So I can break up this right hand side into two integrals. Now the integral of the derivative of f times g is just equal to f times g, by the fundamental theorem of calculus. And I carry the rest of the formula down. And now I have a formula relating the integral of f times g prime to the integral of f prime times G. This formula allows us to rewrite something that might be tricky to integrate, in terms of something that's hopefully easier to integrate. Although I've written this formula, using indefinite integrals with no bounds of integration, it would be just as easy to put on bounds of integration. Now, the fundamental theorem tells us that the integral of the derivative is the original product of functions evaluated from A to B. There's another version of this formula that might be a little easier to remember. If we let u equal f of x, and v equal g of x, then do is equal to f prime of x dx using differential notation, and dv is equal to g prime of x dx. In this notation, we can rewrite the formula as the integral of u times dv is equal to u times v minus the integral of v, d U. since V is our G of X, and D, u is our f prime of x dx. Again, we can include bounds of integration, if we're working with definite integrals. This will be our key formula for this section. and integrating using this formula is called integration by parts. Let's use integration by parts to find the integral of x e to the x dx. Or for formula for integration by parts, says the integral of u dv is equal to u times v minus the integral of the do. So we need to split up our inner grand x e to the x dx into the part that we're going to call you, and a part that we're going to call dv. One natural way to split it up is to let u equal x and dv equal e to the x dx. But another option might be to make u equal to e to the x and dv equal to x dx bar, we might decide to make you be the whole product, x e to the x and leave dv as just dx. Whatever choice we make, we need a u times dv to be our entire integrand here, and we need dx to be part of dv in order to use proper differential notation. Let's try using the first choice first. If u is equal to x, then d u is equal to dx. And if dv is e to the x dx, then we can find V by integrating this and the integral of e to the x dx is e to the x. plugging into our integration by parts formula, we have the integral of u dv that's x, e to the x dx is equal to u times v x into dx minus the integral of vdu. That's e to the x dx. Well, this is looking very promising, because I know how to integrate e to the x dx. It's just e to the x plus a constant of integration. And so integration by part has allowed us to compute our integral, let's check our answer by taking the derivative of what we got the derivative of x, e to the x minus e to the x plus C is going to be the derivative of x, that's one, times e to the x plus x times the derivative of e to the x, which is dx, minus the derivative of iliacs, which is again, the x plus the derivative of C, which is just zero. And since this term and this term cancel out, we're left with x e to the x, which is exactly what we started out with as our integrand. So we've checked that our work is correct. Notice that in order to take the derivative of our answer, we ended up having to use the product rule. And that's no coincidence, because the formula for integration by parts is really just the product rule used in reverse. So we were successful in computing this integral using our first choice of u and dv. But before we leave this problem, let's see what would have happened if we used a different choice. Instead, if we used you've equal to either the X and dv equal to x dx, if we done that, then we would have gotten do to be either the x dx, and we would have computed V by integrating x dx two that get x squared over two. plugging this into our formula for integration by parts, we would get the integral of udv, that's e to the x times x dx is equal to u times v, that's e to the x times x squared over two, minus the integral of vdu. That's x squared over two times e to the x, dx. Now in order to go any further, we need to be able to compute the integral of x squared over two times e to the x dx. dividing by two poses no problem, that's just a constant, but the integral of x squared e to the x that's more complicated than the integral of e to the x times x that we started with. So we're going the wrong direction here. And this turns out to be a poor choice of u and dv. I'll let you check that the third choice of u and dv that I suggested also ends up making things more complicated instead of simpler. So the first choice turned out to be the best choice for you, and dv. integration by parts works by replacing an integral of the form u dv by the equivalent expression, u times v minus the integral of vdu. That's hopefully easier to compute. There are a lot of trig identities that come in handy when you're doing calculus. But the good news is, if you're willing to do a little bit of algebra, there are only a few that you really have to memorize. In this video, I'll tell you my three favorite trig identities, and show you how to derive a bunch more with very little effort from these three. So I'm going to tell you the three trig identities that I think everybody should know. Fortunately, all three of them are very easy to remember. The first one is the Pythagorean identity. That's the identity that says sine squared of theta plus cosine squared of theta is equal to one. That one's easy to remember, because it's really just the Pythagorean theorem for triangles in the context of the unit circle. Whatever choice we make, we need a u times dv to be our entire integrand here, and we need dx to be part of dv in order to use proper differential notation. Let's try using the first choice first. If u is equal to x, then d u is equal to dx. And if dv is e to the x dx, then we can find V by integrating this and the integral of e to the x dx is e to the x. plugging into our integration by parts formula, we have the integral of u dv that's x, e to the x dx is equal to u times v x into dx minus the integral of vdu. That's e to the x dx. Well, this is looking very promising, because I know how to integrate e to the x dx. It's just e to the x plus a constant of integration. And so integration by part has allowed us to compute our integral, let's check our answer by taking the derivative of what we got the derivative of x, e to the x minus e to the x plus C is going to be the derivative of x, that's one, times e to the x plus x times the derivative of e to the x, which is dx, minus the derivative of iliacs, which is again, the x plus the derivative of C, which is just zero. And since this term and this term cancel out, we're left with x e to the x, which is exactly what we started out with as our integrand. So we've checked that our work is correct. Notice that in order to take the derivative of our answer, we ended up having to use the product rule. And that's no coincidence, because the formula for integration by parts is really just the product rule used in reverse. So we were successful in computing this integral using our first choice of u and dv. But before we leave this problem, let's see what would have happened if we used a different choice. Instead, if we used you've equal to either the X and dv equal to x dx, if we done that, then we would have gotten do to be either the x dx, and we would have computed V by integrating x dx two that get x squared over two. plugging this into our formula for integration by parts, we would get the integral of udv, that's e to the x times x dx is equal to u times v, that's e to the x times x squared over two, minus the integral of vdu. That's x squared over two times e to the x, dx. Now in order to go any further, we need to be able to compute the integral of x squared over two times e to the x dx. dividing by two poses no problem, that's just a constant, but the integral of x squared e to the x that's more complicated than the integral of e to the x times x that we started with. So we're going the wrong direction here. And this turns out to be a poor choice of u and dv. I'll let you check that the third choice of u and dv that I suggested also ends up making things more complicated instead of simpler. So the first choice turned out to be the best choice for you, and dv. integration by parts works by replacing an integral of the form u dv by the equivalent expression, u times v minus the integral of vdu. That's hopefully easier to compute. There are a lot of trig identities that come in handy when you're doing calculus. But the good news is, if you're willing to do a little bit of algebra, there are only a few that you really have to memorize. In this video, I'll tell you my three favorite trig identities, and show you how to derive a bunch more with very little effort from these three. So I'm going to tell you the three trig identities that I think everybody should know. Fortunately, all three of them are very easy to remember. The first one is the Pythagorean identity. That's the identity that says sine squared of theta plus cosine squared of theta is equal to one. That one's easy to remember, because it's really just the Pythagorean theorem for triangles in the context of the unit circle. If I draw a right triangle with angle theta on a unit circle, it's hi partners has length one, because a unit circle means a circle of radius one. The base of this triangle has length cosine of theta, because by definition, cosine of theta is the x coordinate of the point on the unit circle at angle theta. Similarly, the height of the triangle is sine of theta, because sine of theta means the y coordinate on the unit circle. So by the Pythagorean theorem for right triangles, we know that sine theta squared plus cosine theta squared equals one, which is exactly the bit beggary and identity. The second identities that everybody should know are the even and odd identities. I guess technically, these are two different identities but they go together, so I'm counting them as one. The even identity says that cosine of negative theta is equal to cosine of Beta. In other words, cosine is an even function. And the identity says that sine of negative theta is equal to negative sine of theta. In other words, sine is an odd function. One way to remember these identities is by looking at the graphs of sine and cosine. Y equals cosine x has this bilateral symmetry, so it's an even function. And if I look at the value of cosine at theta, and cosine at negative theta, my graph has the same height. So cosine has the same value for both of these angles. The graph of y equals sine x does not have bilateral symmetry, instead, it's got 180 degree rotational symmetry, which is characteristic of an odd function. And if I compare the y values, that theta and negative theta, I see that the y value at negative theta is exactly the opposite as the y value at theta, and therefore, sine of negative theta is equal to negative sine of theta. Another way to understand that, even in odd identities, is by going back to the unit circle. If I look at an angle of theta, compared to an angle of negative theta, the x coordinate, which is cosine has the same value for both of these angles. So that means cosine of negative theta is equal to cosine of theta. the y coordinate instead is the opposite. For negative theta, versus theta, one of them is positive and one of them is negative, but they have the same magnitude. Therefore, sine of negative theta is the negative of sine of theta. My third favorite trig identity is the angle sum formula. Again, there are really two, one for sine and one for cosine, but they go together, so I'll consider them a single identity. These identities are easy to remember because there's a song that goes with them. And the song goes, sine, cosine, cosine, sine, cosine, cosine, minus sign sign, you may recognize the tune, please sing along with me this time. So remember it 123 go sine cosine cosine sine, cosine cosine minus sine sine. The only thing you have to remember when you sing the song is that it gives the sine of A plus B first, and then the cosine of A plus B. So these are the three trig identities that I think everybody should memorize. Next, I'll show you how to derive a bunch more trig identities. Pretty simply, from these three. I've written my three favorite trig identities across the top for easy reference. First of all, we can derive a couple of good identities just from the beggary and identity. Let's start by dividing both sides of this identity by cosine squared theta. We can break up the fraction on the left into sine squared theta over cosine squared theta plus cosine squared theta over cosine squared theta equals one over cosine squared theta. But sine squared theta over cosine squared theta is just tan squared theta. and cosine squared theta over cosine squared theta is one. And one over cosine squared theta is seacon squared theta, since secant theta by definition is one over cosine. If I draw a right triangle with angle theta on a unit circle, it's hi partners has length one, because a unit circle means a circle of radius one. The base of this triangle has length cosine of theta, because by definition, cosine of theta is the x coordinate of the point on the unit circle at angle theta. Similarly, the height of the triangle is sine of theta, because sine of theta means the y coordinate on the unit circle. So by the Pythagorean theorem for right triangles, we know that sine theta squared plus cosine theta squared equals one, which is exactly the bit beggary and identity. The second identities that everybody should know are the even and odd identities. I guess technically, these are two different identities but they go together, so I'm counting them as one. The even identity says that cosine of negative theta is equal to cosine of Beta. In other words, cosine is an even function. And the identity says that sine of negative theta is equal to negative sine of theta. In other words, sine is an odd function. One way to remember these identities is by looking at the graphs of sine and cosine. Y equals cosine x has this bilateral symmetry, so it's an even function. And if I look at the value of cosine at theta, and cosine at negative theta, my graph has the same height. So cosine has the same value for both of these angles. The graph of y equals sine x does not have bilateral symmetry, instead, it's got 180 degree rotational symmetry, which is characteristic of an odd function. And if I compare the y values, that theta and negative theta, I see that the y value at negative theta is exactly the opposite as the y value at theta, and therefore, sine of negative theta is equal to negative sine of theta. Another way to understand that, even in odd identities, is by going back to the unit circle. If I look at an angle of theta, compared to an angle of negative theta, the x coordinate, which is cosine has the same value for both of these angles. So that means cosine of negative theta is equal to cosine of theta. the y coordinate instead is the opposite. For negative theta, versus theta, one of them is positive and one of them is negative, but they have the same magnitude. Therefore, sine of negative theta is the negative of sine of theta. My third favorite trig identity is the angle sum formula. Again, there are really two, one for sine and one for cosine, but they go together, so I'll consider them a single identity. These identities are easy to remember because there's a song that goes with them. And the song goes, sine, cosine, cosine, sine, cosine, cosine, minus sign sign, you may recognize the tune, please sing along with me this time. So remember it 123 go sine cosine cosine sine, cosine cosine minus sine sine. The only thing you have to remember when you sing the song is that it gives the sine of A plus B first, and then the cosine of A plus B. So these are the three trig identities that I think everybody should memorize. Next, I'll show you how to derive a bunch more trig identities. Pretty simply, from these three. I've written my three favorite trig identities across the top for easy reference. First of all, we can derive a couple of good identities just from the beggary and identity. Let's start by dividing both sides of this identity by cosine squared theta. We can break up the fraction on the left into sine squared theta over cosine squared theta plus cosine squared theta over cosine squared theta equals one over cosine squared theta. But sine squared theta over cosine squared theta is just tan squared theta. and cosine squared theta over cosine squared theta is one. And one over cosine squared theta is seacon squared theta, since secant theta by definition is one over cosine. So we found a new identity, a variation on the Pythagorean identity that involves tangent and secant. Now, what if we wanted an identity that involved cotangent and cosecant instead? Well, as you might be thinking, we could just start with the theory and identity. And this time, divide both sides by sine squared of theta. If we break up the fraction like before, we get one plus cotangent squared theta equals cosecant squared theta. Since cotangent is cosine over sine, and cosecant is one of our sign. Next, it's possible to get a bunch of great identities from the angle some formulas. Let's start by deriving an angle difference formula. If we want a formula for sine of A minus B, all we have to do is plug in negative B for B In our angle some formula. So we get sine of A cosine of negative b plus cosine of A sine of negative B. But we know that cosine of negative B is cosine of B, since cosine is 11. And sine of negative b is negative sine of B, so that turns this positive sign to a negative sign. And we have an angle difference formula for sine. Similarly, we can use the angle sum formula for cosine, plugging in negative B for b. and use the even odd properties to get an angle difference formula for cosine. Now let's derive a super useful formula, the double angle formula. If we want a formula for sine of two theta, we can think of that as being the sine of theta plus theta. So let's plug theta in for a and theta in for B in our angle sum formula. That gives us sine cosine, cosine sine, which simplifies to two sine theta cosine theta. That's the double angle formula for sine. That worked out so well, let's try the same thing for cosine. cosine of two theta is cosine of theta plus theta. And using the angle sum formula, that's cosine, cosine minus sine sine, we can rewrite this as cosine squared theta minus sine squared theta, and get our first version of the double angle formula for cosine. But we could also use the Pythagorean identity to replace cosine squared theta by one minus sine squared theta. Or if we prefer, we can replace sine squared theta by one minus cosine squared theta. If we replace cosine squared, we get cosine of two theta is one minus sine squared theta minus sine squared theta. In other words, cosine of two theta is one minus twice sine squared theta. And if we go back to the original, and replace sine squared theta by one minus cosine squared theta, we get cosine of two theta is cosine squared theta minus quantity one minus cosine squared theta, which simplifies to cosine of two theta is twice cosine squared theta minus one. So we found one version of a double angle formula for sine, and three versions for cosine, just using the angle sum formula and the Pythagorean identity. These last two formulas are particularly useful for integration when they're rewritten slightly. For this formula, let me solve for sine squared of theta, I get two sine squared of theta is equal to one minus cosine of two theta. So sine squared of theta is one half minus one half cosine of two theta. I'll do the analogous thing with this formula and solve for cosine squared theta So we found a new identity, a variation on the Pythagorean identity that involves tangent and secant. Now, what if we wanted an identity that involved cotangent and cosecant instead? Well, as you might be thinking, we could just start with the theory and identity. And this time, divide both sides by sine squared of theta. If we break up the fraction like before, we get one plus cotangent squared theta equals cosecant squared theta. Since cotangent is cosine over sine, and cosecant is one of our sign. Next, it's possible to get a bunch of great identities from the angle some formulas. Let's start by deriving an angle difference formula. If we want a formula for sine of A minus B, all we have to do is plug in negative B for B In our angle some formula. So we get sine of A cosine of negative b plus cosine of A sine of negative B. But we know that cosine of negative B is cosine of B, since cosine is 11. And sine of negative b is negative sine of B, so that turns this positive sign to a negative sign. And we have an angle difference formula for sine. Similarly, we can use the angle sum formula for cosine, plugging in negative B for b. and use the even odd properties to get an angle difference formula for cosine. Now let's derive a super useful formula, the double angle formula. If we want a formula for sine of two theta, we can think of that as being the sine of theta plus theta. So let's plug theta in for a and theta in for B in our angle sum formula. That gives us sine cosine, cosine sine, which simplifies to two sine theta cosine theta. That's the double angle formula for sine. That worked out so well, let's try the same thing for cosine. cosine of two theta is cosine of theta plus theta. And using the angle sum formula, that's cosine, cosine minus sine sine, we can rewrite this as cosine squared theta minus sine squared theta, and get our first version of the double angle formula for cosine. But we could also use the Pythagorean identity to replace cosine squared theta by one minus sine squared theta. Or if we prefer, we can replace sine squared theta by one minus cosine squared theta. If we replace cosine squared, we get cosine of two theta is one minus sine squared theta minus sine squared theta. In other words, cosine of two theta is one minus twice sine squared theta. And if we go back to the original, and replace sine squared theta by one minus cosine squared theta, we get cosine of two theta is cosine squared theta minus quantity one minus cosine squared theta, which simplifies to cosine of two theta is twice cosine squared theta minus one. So we found one version of a double angle formula for sine, and three versions for cosine, just using the angle sum formula and the Pythagorean identity. These last two formulas are particularly useful for integration when they're rewritten slightly. For this formula, let me solve for sine squared of theta, I get two sine squared of theta is equal to one minus cosine of two theta. So sine squared of theta is one half minus one half cosine of two theta. I'll do the analogous thing with this formula and solve for cosine squared theta to cosine squared theta is equal to one plus cosine of two theta. So cosine squared theta is equal to one half, plus one half cosine of two theta. That's all the triggers in these we're gonna need. It may seem like a lot, but remember, they all follow very naturally from my three favorites at the top. Well, all these formulas are very useful. The ones that are of particular importance for techniques of integration are the first one, that factory and identity and its various forms. And the last two. In this video, I told you my three favorite trig identities, the Pythagorean identity, the even an odd identities and the angle some formulas. From these I derived a bunch of other identities, including a handful that will be particularly useful as we do techniques of integration. Remember the angle some formulas, those are the formulas for computing sine of A plus B, and cosine of A plus B. I like to sing them, sine, cosine, cosine, sine, cosine, cosine minus sine sine. This video gives a geometric proof of those formulas. There are many great proofs of the angle some formulas, but I'd like to share with you one of my favorites for those who are interested. Alright, the angle some formulas up here, so we'll know what we're trying to prove. To prove these formulas, let me start by drawing an angle A and angle B on top of that. Next, I'm going to draw a line perpendicular to this middle line. And I'm going to extend the top line until it meets that perpendicular, making a right triangle. Finally, I'll draw a rectangle around that right triangle that just touches its vertices. My rectangle is now divided up into four right triangles. And I'm going to choose units of measurement, so that the high partners of my middle triangle has length one. Now let's stop for a minute to think about the angles of these other triangles. Since the top and the bottom edge of the rectangle are parallel lines, and this high partners is a transversal. This angle up here must have the same measure as a plus b down here. Also, this skinny angle here must have the same measure as a down here. Because this angle is 180 degrees, minus 90 degrees minus this angle here. And this angle A is also 180 degrees, the measure of the angles in a triangle minus 90 degrees minus that same angle. to cosine squared theta is equal to one plus cosine of two theta. So cosine squared theta is equal to one half, plus one half cosine of two theta. That's all the triggers in these we're gonna need. It may seem like a lot, but remember, they all follow very naturally from my three favorites at the top. Well, all these formulas are very useful. The ones that are of particular importance for techniques of integration are the first one, that factory and identity and its various forms. And the last two. In this video, I told you my three favorite trig identities, the Pythagorean identity, the even an odd identities and the angle some formulas. From these I derived a bunch of other identities, including a handful that will be particularly useful as we do techniques of integration. Remember the angle some formulas, those are the formulas for computing sine of A plus B, and cosine of A plus B. I like to sing them, sine, cosine, cosine, sine, cosine, cosine minus sine sine. This video gives a geometric proof of those formulas. There are many great proofs of the angle some formulas, but I'd like to share with you one of my favorites for those who are interested. Alright, the angle some formulas up here, so we'll know what we're trying to prove. To prove these formulas, let me start by drawing an angle A and angle B on top of that. Next, I'm going to draw a line perpendicular to this middle line. And I'm going to extend the top line until it meets that perpendicular, making a right triangle. Finally, I'll draw a rectangle around that right triangle that just touches its vertices. My rectangle is now divided up into four right triangles. And I'm going to choose units of measurement, so that the high partners of my middle triangle has length one. Now let's stop for a minute to think about the angles of these other triangles. Since the top and the bottom edge of the rectangle are parallel lines, and this high partners is a transversal. This angle up here must have the same measure as a plus b down here. Also, this skinny angle here must have the same measure as a down here. Because this angle is 180 degrees, minus 90 degrees minus this angle here. And this angle A is also 180 degrees, the measure of the angles in a triangle minus 90 degrees minus that same angle. So I'll label this skinny angle with a. Next, let's figure out as many side lengths as we can. So I'll label this skinny angle with a. Next, let's figure out as many side lengths as we can. Based on the middle right triangle with high partners one, we know that this side length down here must be cosine of B, since adjacent over hypotenuse is cosine B. Similarly, this side length here must be sine of B. Since opposite of our hypothesis is sine of B, now we see that sine of B is the hypotenuse of this right triangle, which means that this little side here has measure sign a time sign B. That's because the opposite over the hypothesis of this angle has to equal sign a. A similar argument shows that this side has to have measure cosine A time sign B. Please pause the video and take a moment to fill in the side length of this right triangle. And this right triangle, you should be getting sine A cosine B cosine A cosine B sine of A plus B and cosine of A plus B. But remember, we have a rectangle here. So the opposite sides have equal length. This tells us that sine of A plus B has to equal sine of A cosine of B plus cosine of A sine of B, which is exactly the first angle sum formula. Also, cosine of A plus B, which is this side length, is exactly the difference of this side length cosine A cosine B minus this side length sine of A sine B. And that's the second angle sum formula. So I think that's a pretty great geometric proof of the angle some formulas. This video is about integrating trig functions that involve at least one odd power of sine or cosine. Let's start by evaluating the integral of sine to the fourth x times cosine x dx. This integral is a good candidate for u substitution. If we let u equal sine of x Then d u is equal to cosine of x dx. And we can rewrite the integral as the integral of u to the fourth do this integrates easily to 1/5 times u to the fifth plus C. And I can convert things back to x by plugging in sine of x for you, to get 1/5 sine to the fifth of x plus C. The next integral, the integral of sine to the fourth x times cosine cubed x dx can also be handled with a similar use of the tuition with a bit more work. First, I'm going to rewrite the integral as the integral of sine to the fourth x times cosine squared x times cosine of x dx. I'm rewriting the cosine cubed of x as cosine squared of x times cosine x in the hopes that I can use cosine of x dx as my d u, like I did in the previous problem. Now if we have d u, as cosine of x, dx, we're going to need you to be sine of x. But unlike the previous problem, I can't just replace sine of the fourth x with you to the fourth and cosine x dx with D, U and be done. Because I've got this cosine squared x hanging around that I need to deal with. I wish I didn't have that cosine squared x, I wish everything else over here was just written in terms of sine. Now, as you might realize, it's not hard for me to get my wish. We know from the Pythagorean identity, that cosine squared of x plus sine squared of x is equal to one. So cosine squared of x can be written as one minus sine squared x. Making that substitution, I can now rewrite my integral entirely in terms of you. And do now if I multiply things out, this becomes easy to integrate, I get 1/5 u to the fifth minus 1/7 u to the seventh plus C. And I can rewrite this in terms of x. Based on the middle right triangle with high partners one, we know that this side length down here must be cosine of B, since adjacent over hypotenuse is cosine B. Similarly, this side length here must be sine of B. Since opposite of our hypothesis is sine of B, now we see that sine of B is the hypotenuse of this right triangle, which means that this little side here has measure sign a time sign B. That's because the opposite over the hypothesis of this angle has to equal sign a. A similar argument shows that this side has to have measure cosine A time sign B. Please pause the video and take a moment to fill in the side length of this right triangle. And this right triangle, you should be getting sine A cosine B cosine A cosine B sine of A plus B and cosine of A plus B. But remember, we have a rectangle here. So the opposite sides have equal length. This tells us that sine of A plus B has to equal sine of A cosine of B plus cosine of A sine of B, which is exactly the first angle sum formula. Also, cosine of A plus B, which is this side length, is exactly the difference of this side length cosine A cosine B minus this side length sine of A sine B. And that's the second angle sum formula. So I think that's a pretty great geometric proof of the angle some formulas. This video is about integrating trig functions that involve at least one odd power of sine or cosine. Let's start by evaluating the integral of sine to the fourth x times cosine x dx. This integral is a good candidate for u substitution. If we let u equal sine of x Then d u is equal to cosine of x dx. And we can rewrite the integral as the integral of u to the fourth do this integrates easily to 1/5 times u to the fifth plus C. And I can convert things back to x by plugging in sine of x for you, to get 1/5 sine to the fifth of x plus C. The next integral, the integral of sine to the fourth x times cosine cubed x dx can also be handled with a similar use of the tuition with a bit more work. First, I'm going to rewrite the integral as the integral of sine to the fourth x times cosine squared x times cosine of x dx. I'm rewriting the cosine cubed of x as cosine squared of x times cosine x in the hopes that I can use cosine of x dx as my d u, like I did in the previous problem. Now if we have d u, as cosine of x, dx, we're going to need you to be sine of x. But unlike the previous problem, I can't just replace sine of the fourth x with you to the fourth and cosine x dx with D, U and be done. Because I've got this cosine squared x hanging around that I need to deal with. I wish I didn't have that cosine squared x, I wish everything else over here was just written in terms of sine. Now, as you might realize, it's not hard for me to get my wish. We know from the Pythagorean identity, that cosine squared of x plus sine squared of x is equal to one. So cosine squared of x can be written as one minus sine squared x. Making that substitution, I can now rewrite my integral entirely in terms of you. And do now if I multiply things out, this becomes easy to integrate, I get 1/5 u to the fifth minus 1/7 u to the seventh plus C. And I can rewrite this in terms of x. To recap, we separated out one copy of cosine x to be part of our D U. And then we converted the rest of the cosines in the signs isn't a Pythagorean identity. This allowed us to do use of the tuition with u equal to sine x and evaluate the integral. This same technique with some modifications works on a lot of other problems. In this next problem, if we tried separating out one copy of cosine of x to be part of our do, we'd run into problems, because we just have one copy of cosine x left. And we can't use the Pythagorean identity can to convert a single cosine into signs. Well, I guess technically, we could do something like this. But then we'd have to introduce a plus or minus sign and a square root sign into our integrand, which would make things difficult to integrate. It's a lot easier if we can just use the cosine squared x is equal to one minus sine squared x identity to replace our cosines with signs. But this identity only applies if we have an even power of cosines leftover that we want to convert. So instead of trying to save out a copy of cosine x, let's say about a copy of sine x instead. So we'll rewrite our integral as the integral of sine to the fourth of x times a sine of x. And we'll keep the cosine squared x and the dx. Now we want the sine x to be part of our D U. So let's set u equal to cosine of x. That way d u is equal to negative sine of x dx. And so sine of x dx is equal to negative d U. since u is equal to cosine x, this time, we want to replace all of our stray signs with cosines. We know that sine squared of x is equal to one minus cosine squared of x by the Pythagorean identity. So let's rewrite this sine to the fourth as sine squared of x squared. That allows us to substitute in one minus cosine squared of x for sine squared of x And now we can replace everything with us and d is I'll bring out the negative sign and multiply things out. One minus u squared squared is one minus two u squared plus u to the fourth. And this multiplies out to us squared minus two u to the fourth plus u to the sixth. Now I can integrate, distribute my negative sign. And finally, plug in cosine of x for you. That completes this problem. In this video, we'll use use substitution and the Pythagorean identity. To evaluate integrals with at least one odd power of sine or cosine. The idea is to separate off one copy from the odd power that one copy becomes part of our D U. and the remaining even power gets converted using the Pythagorean identity. In this case, we convert sine squared into one minus cosine squared. This allows us to do the use substitution and evaluate the integral. In this video, we'll evaluate integrals of trig functions involving only even powers of sine and cosine. There are three trig identities that will come in handy here, the first ones that are familiar, the factory and identity. The second one's an identity that allows us to rewrite cosine squared of x in terms of cosine of 2x. And the third one is in nd that lets us write sine squared of x. Also in terms of cosine of 2x. The only difference between the second and third is the fact that one's cosine squared and one sine squared. But the other difference is the plus sign versus the minus sign here. To recap, we separated out one copy of cosine x to be part of our D U. And then we converted the rest of the cosines in the signs isn't a Pythagorean identity. This allowed us to do use of the tuition with u equal to sine x and evaluate the integral. This same technique with some modifications works on a lot of other problems. In this next problem, if we tried separating out one copy of cosine of x to be part of our do, we'd run into problems, because we just have one copy of cosine x left. And we can't use the Pythagorean identity can to convert a single cosine into signs. Well, I guess technically, we could do something like this. But then we'd have to introduce a plus or minus sign and a square root sign into our integrand, which would make things difficult to integrate. It's a lot easier if we can just use the cosine squared x is equal to one minus sine squared x identity to replace our cosines with signs. But this identity only applies if we have an even power of cosines leftover that we want to convert. So instead of trying to save out a copy of cosine x, let's say about a copy of sine x instead. So we'll rewrite our integral as the integral of sine to the fourth of x times a sine of x. And we'll keep the cosine squared x and the dx. Now we want the sine x to be part of our D U. So let's set u equal to cosine of x. That way d u is equal to negative sine of x dx. And so sine of x dx is equal to negative d U. since u is equal to cosine x, this time, we want to replace all of our stray signs with cosines. We know that sine squared of x is equal to one minus cosine squared of x by the Pythagorean identity. So let's rewrite this sine to the fourth as sine squared of x squared. That allows us to substitute in one minus cosine squared of x for sine squared of x And now we can replace everything with us and d is I'll bring out the negative sign and multiply things out. One minus u squared squared is one minus two u squared plus u to the fourth. And this multiplies out to us squared minus two u to the fourth plus u to the sixth. Now I can integrate, distribute my negative sign. And finally, plug in cosine of x for you. That completes this problem. In this video, we'll use use substitution and the Pythagorean identity. To evaluate integrals with at least one odd power of sine or cosine. The idea is to separate off one copy from the odd power that one copy becomes part of our D U. and the remaining even power gets converted using the Pythagorean identity. In this case, we convert sine squared into one minus cosine squared. This allows us to do the use substitution and evaluate the integral. In this video, we'll evaluate integrals of trig functions involving only even powers of sine and cosine. There are three trig identities that will come in handy here, the first ones that are familiar, the factory and identity. The second one's an identity that allows us to rewrite cosine squared of x in terms of cosine of 2x. And the third one is in nd that lets us write sine squared of x. Also in terms of cosine of 2x. The only difference between the second and third is the fact that one's cosine squared and one sine squared. But the other difference is the plus sign versus the minus sign here. Let's start with the simplest possible even power, the integral of cosine squared x dx. According to my calculator, this integral evaluates to sine x cosine x over two plus x over two. And if I look it up in the back of my book, I get the integral to be one half x plus one for a sine of 2x. So what's going on here? Are these two answers really the same? Well, yes, because sine of 2x is equal to two times sine x cosine x. So if I replace that, here, I get this second answer is equivalent to one half x plus 1/4 times two sine x cosine x, which is equivalent to the first answer. Notice that the calculator on the table in the back of the book omit the plus c the cost of integration, but you should always include them for indefinite integrals. Now let's see where these answers come from. So let's compute the integral of cosine squared x dx by hand that is, without the aid of a calculator or computer integral table, the hint is that cosine squared x is equal to one plus cosine 2x over two, or sometimes I like to write this as one half plus one half cosine of 2x. So I'll use that identity to rewrite my integral. Now I can split up the integral using rules about integration. The integral of one half is just one half x, and the integral of cosine of 2x is one half times sine of 2x. It's possible to get that answer by use substitution with u equal to x. Or as a shortcut, we can observe that the derivative of sine of 2x is cosine of 2x times two, the times two comes from the chain rule taking the derivative of the inside. So to get rid of that, or counteract that, times two, we have to multiply by one half. In any case, our final answer is then one half x plus 1/4 sine of 2x plus C. This is the same answer that we can find using the integral table in the back of the book. Please pause the video for a minute and try to compute the integral of sine squared x dx. Using a similar technique. You'll want to use the identity sine squared of x is equal to one half minus one Half cosine of 2x. If you rewrite the integral, you can integrate like before, and get an answer of one half x minus 1/4 sine of 2x plus C, the same technique can be used to evaluate a lot of other integrals involving even powers of sine and cosine, often with a lot more effort. Let's try evaluating the integral of sine to the sixth x dx. We know that sine squared of x is equal to one half minus one half cosine of 2x. And since six is an even power, we know we can rewrite sine to the sixth as sine squared to the x raised to a power, in this case, the power of three. Now let's substitute in our identity. I think I'm going to factor out the one half. One half cubed is 1/8. So I'll pull the one eight out of the integral side and multiply out. Now let me divide up my integral into pieces. And I'll try to handle each piece separately. The integral of one dx, that's easy, that's x, and the integral of cosine of 2x, that's going to be one half times sine of 2x, like before. Now to integrate cosine squared of 2x, we can use the same trick we used to integrate cosine squared of x, cosine squared of 2x is going to be one half, plus one half cosine of two times 2x. So that's 4x. Let's start with the simplest possible even power, the integral of cosine squared x dx. According to my calculator, this integral evaluates to sine x cosine x over two plus x over two. And if I look it up in the back of my book, I get the integral to be one half x plus one for a sine of 2x. So what's going on here? Are these two answers really the same? Well, yes, because sine of 2x is equal to two times sine x cosine x. So if I replace that, here, I get this second answer is equivalent to one half x plus 1/4 times two sine x cosine x, which is equivalent to the first answer. Notice that the calculator on the table in the back of the book omit the plus c the cost of integration, but you should always include them for indefinite integrals. Now let's see where these answers come from. So let's compute the integral of cosine squared x dx by hand that is, without the aid of a calculator or computer integral table, the hint is that cosine squared x is equal to one plus cosine 2x over two, or sometimes I like to write this as one half plus one half cosine of 2x. So I'll use that identity to rewrite my integral. Now I can split up the integral using rules about integration. The integral of one half is just one half x, and the integral of cosine of 2x is one half times sine of 2x. It's possible to get that answer by use substitution with u equal to x. Or as a shortcut, we can observe that the derivative of sine of 2x is cosine of 2x times two, the times two comes from the chain rule taking the derivative of the inside. So to get rid of that, or counteract that, times two, we have to multiply by one half. In any case, our final answer is then one half x plus 1/4 sine of 2x plus C. This is the same answer that we can find using the integral table in the back of the book. Please pause the video for a minute and try to compute the integral of sine squared x dx. Using a similar technique. You'll want to use the identity sine squared of x is equal to one half minus one Half cosine of 2x. If you rewrite the integral, you can integrate like before, and get an answer of one half x minus 1/4 sine of 2x plus C, the same technique can be used to evaluate a lot of other integrals involving even powers of sine and cosine, often with a lot more effort. Let's try evaluating the integral of sine to the sixth x dx. We know that sine squared of x is equal to one half minus one half cosine of 2x. And since six is an even power, we know we can rewrite sine to the sixth as sine squared to the x raised to a power, in this case, the power of three. Now let's substitute in our identity. I think I'm going to factor out the one half. One half cubed is 1/8. So I'll pull the one eight out of the integral side and multiply out. Now let me divide up my integral into pieces. And I'll try to handle each piece separately. The integral of one dx, that's easy, that's x, and the integral of cosine of 2x, that's going to be one half times sine of 2x, like before. Now to integrate cosine squared of 2x, we can use the same trick we used to integrate cosine squared of x, cosine squared of 2x is going to be one half, plus one half cosine of two times 2x. So that's 4x. And finally, to integrate cosine cubed of 2x, well, cosine cubed is an odd power, so we can use our odd power trick and save aside one copy of cosine, and turn the remaining cosine squared into one minus sine squared. So that looks like this. Notice we have one minus sine squared of 2x. Because we're replacing a cosine squared of 2x. I'll copy the first part down. Now integrate one half to get one half x, and I'll integrate one half cosine of forex. To get one half times 1/4 sine of forex, we need the 1/4. To counteract the four that we get by taking the derivative of sine of 4x. And using the chain rule, we can finish off the integration by using a use substitution here where u is sine of 2x. And so d u is two cosine of 2x. So we can replace this integral here with minus the integral of one minus us squared times, one half do I'll copy the first part down again. and integrate that last part, I'll pull out the negative a half, and I get you minus 1/3 u cubed, I need a plus C. Now. I'll plug back in for you and simplify a little bit. And I get a final answer of five sixteenths x minus one for sine of 2x plus 360 fourths sine of 4x plus 148, sine cubed of 2x. Oh, and we need a plus C here. That's a complicated answer. And it was a complicated computation. But it was all based on the same sort of tricks as before, using identities like this one, to handle even powers of sine and cosine, and using the Pythagorean identity to handle odd powers. In this video, we use these last two identities to rewrite even powers of cosine and sine in terms of lower powers of cosine. This trick can be used to compute the integrals of complicated even powers of sine and cosine, provided that you have a lot of time and patience on hand. This video is about some special trig integrals, namely, the integral of tangent squared of x and the integral of secant of x. these integrals are special only in the sense that there's some special tricks required to integrate them. When I come across the integral of tangent squared X, I find myself wishing that I could integrate secant squared x instead, because integrating secant squared x is easy. It's just tangent x plus c since the derivative of a tangent secant squared. But happily, I know how to rewrite tangent squared in terms of secant, because tangent squared is secant squared minus one. So I'll just rewrite this integral as the integral of secant squared minus one. And that integrates easily to tangent of x minus x plus C. The same sort of trick works to integrate cotangent squared of x, since there's a similar identity, relating cotangent and cosecant every now and then, you have to integrate secant of x. There's several possible tricks that can be used to do this. One of them is to multiply secant x by secant x plus tangent x in the numerator and denominator. Now distributing, we get seacon squared x plus seek index tangent x in the numerator, and secant x plus tangent x in the denominator. Since secant squared is the derivative of tangent, and secant tangent is the root of a secant. We can say u equal to secant x plus tangent x and have d u, sitting right where we want it in the integrand. And finally, to integrate cosine cubed of 2x, well, cosine cubed is an odd power, so we can use our odd power trick and save aside one copy of cosine, and turn the remaining cosine squared into one minus sine squared. So that looks like this. Notice we have one minus sine squared of 2x. Because we're replacing a cosine squared of 2x. I'll copy the first part down. Now integrate one half to get one half x, and I'll integrate one half cosine of forex. To get one half times 1/4 sine of forex, we need the 1/4. To counteract the four that we get by taking the derivative of sine of 4x. And using the chain rule, we can finish off the integration by using a use substitution here where u is sine of 2x. And so d u is two cosine of 2x. So we can replace this integral here with minus the integral of one minus us squared times, one half do I'll copy the first part down again. and integrate that last part, I'll pull out the negative a half, and I get you minus 1/3 u cubed, I need a plus C. Now. I'll plug back in for you and simplify a little bit. And I get a final answer of five sixteenths x minus one for sine of 2x plus 360 fourths sine of 4x plus 148, sine cubed of 2x. Oh, and we need a plus C here. That's a complicated answer. And it was a complicated computation. But it was all based on the same sort of tricks as before, using identities like this one, to handle even powers of sine and cosine, and using the Pythagorean identity to handle odd powers. In this video, we use these last two identities to rewrite even powers of cosine and sine in terms of lower powers of cosine. This trick can be used to compute the integrals of complicated even powers of sine and cosine, provided that you have a lot of time and patience on hand. This video is about some special trig integrals, namely, the integral of tangent squared of x and the integral of secant of x. these integrals are special only in the sense that there's some special tricks required to integrate them. When I come across the integral of tangent squared X, I find myself wishing that I could integrate secant squared x instead, because integrating secant squared x is easy. It's just tangent x plus c since the derivative of a tangent secant squared. But happily, I know how to rewrite tangent squared in terms of secant, because tangent squared is secant squared minus one. So I'll just rewrite this integral as the integral of secant squared minus one. And that integrates easily to tangent of x minus x plus C. The same sort of trick works to integrate cotangent squared of x, since there's a similar identity, relating cotangent and cosecant every now and then, you have to integrate secant of x. There's several possible tricks that can be used to do this. One of them is to multiply secant x by secant x plus tangent x in the numerator and denominator. Now distributing, we get seacon squared x plus seek index tangent x in the numerator, and secant x plus tangent x in the denominator. Since secant squared is the derivative of tangent, and secant tangent is the root of a secant. We can say u equal to secant x plus tangent x and have d u, sitting right where we want it in the integrand. So now we just have to integrate one over u d u, which is the ln of u plus C. And now plugging back in for you, we get that our original integral of seeking dx evaluates to the natural log of secant x plus tangent x plus a constant. A similar trick can be used to evaluate the integral of cosecant x. And that's all for the integral of tangent squared, and the integral of secant. This video introduces the technique of trig substitution to evaluate integrals involving square root signs like this one. There are a few trig identities that are especially useful for this technique. The first one's the Pythagorean identity. And the second is the related identity that involves tangent and secant. There's also a third related identity that involves cotangent cosecant. This one could also be used in the method of tree substitution, although it doesn't come up as often as the first two. As our first example, let's look at the integral of x squared over the square root of 49 minus x squared. According to Wolfram Alpha, this integral evaluates to this expression involving a square root expression kind of an as expected and a sine inverse, which just sort of come seems to come out of the blue here. Let's see where this answer comes from. Using a trig substitution. The inverse sine function in the answer gives us a hint that we may want to substitute in something related to sign, I'm going to substitute in x equals seven sine theta. If x is seven sine theta, then dx is going to be seven cosine theta d theta. Now I'll substitute in for x and dx in my integral to get the integral of seven sine theta squared over the square root of 49 minus seven sine theta squared times seven cosine theta d theta. let me simplify a little. I have a seven cubed in the numerator times sine squared theta cosine theta. And the nominator I have 49 minus 49 sine squared theta. I'll factor out the 49 here. And since the square root of 49 is seven, I can pull a seven out of the square root sign. Now here is where a little bit of magic occurs. I know that one minus sine squared theta is equal to cosine squared The data by the Pythagorean identity, and the square root of cosine squared theta is equal to cosine theta, well, actually, it's equal to the absolute value of cosine theta, which is the same thing as cosine theta, if cosine theta is positive. Or in other words, if theta is between pi over two, and negative pi over two, for example, I would really like to replace my square root of one minus sine squared theta by just cosine theta, not the absolute value of cosine theta. So I'm just going to assume that theta is between negative pi over two and pi over two when I make my substitution. This might seem like cheating, but it's actually legit. Because if you think about the unit circle, as theta ranges from negative pi over two to pi over two, well, cosine is always positive like we want it to be. But sine is ranging all the way from negative one to one. And so we can actually achieve all of the x values between negative seven and seven this way, which are the only x values in the domain of our function anyway. Now if we go back to our integral, we can actually replace the square root of one minus sine squared theta with a simple cosine theta, with success successfully gotten rid of this tricky square root sign, which is the whole point behind trig substitution. Now, if we simplify, we just have to integrate seven squared sine squared theta, d theta. And this is a familiar problem that we can solve using another trig identity. So now we just have to integrate one over u d u, which is the ln of u plus C. And now plugging back in for you, we get that our original integral of seeking dx evaluates to the natural log of secant x plus tangent x plus a constant. A similar trick can be used to evaluate the integral of cosecant x. And that's all for the integral of tangent squared, and the integral of secant. This video introduces the technique of trig substitution to evaluate integrals involving square root signs like this one. There are a few trig identities that are especially useful for this technique. The first one's the Pythagorean identity. And the second is the related identity that involves tangent and secant. There's also a third related identity that involves cotangent cosecant. This one could also be used in the method of tree substitution, although it doesn't come up as often as the first two. As our first example, let's look at the integral of x squared over the square root of 49 minus x squared. According to Wolfram Alpha, this integral evaluates to this expression involving a square root expression kind of an as expected and a sine inverse, which just sort of come seems to come out of the blue here. Let's see where this answer comes from. Using a trig substitution. The inverse sine function in the answer gives us a hint that we may want to substitute in something related to sign, I'm going to substitute in x equals seven sine theta. If x is seven sine theta, then dx is going to be seven cosine theta d theta. Now I'll substitute in for x and dx in my integral to get the integral of seven sine theta squared over the square root of 49 minus seven sine theta squared times seven cosine theta d theta. let me simplify a little. I have a seven cubed in the numerator times sine squared theta cosine theta. And the nominator I have 49 minus 49 sine squared theta. I'll factor out the 49 here. And since the square root of 49 is seven, I can pull a seven out of the square root sign. Now here is where a little bit of magic occurs. I know that one minus sine squared theta is equal to cosine squared The data by the Pythagorean identity, and the square root of cosine squared theta is equal to cosine theta, well, actually, it's equal to the absolute value of cosine theta, which is the same thing as cosine theta, if cosine theta is positive. Or in other words, if theta is between pi over two, and negative pi over two, for example, I would really like to replace my square root of one minus sine squared theta by just cosine theta, not the absolute value of cosine theta. So I'm just going to assume that theta is between negative pi over two and pi over two when I make my substitution. This might seem like cheating, but it's actually legit. Because if you think about the unit circle, as theta ranges from negative pi over two to pi over two, well, cosine is always positive like we want it to be. But sine is ranging all the way from negative one to one. And so we can actually achieve all of the x values between negative seven and seven this way, which are the only x values in the domain of our function anyway. Now if we go back to our integral, we can actually replace the square root of one minus sine squared theta with a simple cosine theta, with success successfully gotten rid of this tricky square root sign, which is the whole point behind trig substitution. Now, if we simplify, we just have to integrate seven squared sine squared theta, d theta. And this is a familiar problem that we can solve using another trig identity. Now we've got an integral we can compute. So the integral of one half is one half theta, and the integral of cosine of two theta is one half sine of two theta. So let me add my constant of integration. And I'm almost done, almost, but not quite. Because this answer is in terms of theta, and my original problem was in terms of x. Now I know that x and theta are related, according to this equation here. So if I solve for theta, I get sine theta is equal to x over seven. So theta is inverse sine or arc sine of x over seven. So I could just plug in for theta here and get seven squared, that's 49 times one half sine inverse x over seven minus one half times one half is 1/4. sine of two times sine inverse x over seven plus C. This is a correct answer. But it's very awkward looking sine of twice sine inverse x over seven, there's got to be a way to simplify this. Unfortunately, we can't just pull the two out and cancel the sign with a sine inverse, because sine does work that way. But instead, we can use the double angle formula. So we can rewrite sine of two theta as twice sine theta cosine theta. So let me ignore this line for now. And I'll rewrite the above line as 49 times one half theta, minus 1/4 times twice sine theta, cosine theta Now we've got an integral we can compute. So the integral of one half is one half theta, and the integral of cosine of two theta is one half sine of two theta. So let me add my constant of integration. And I'm almost done, almost, but not quite. Because this answer is in terms of theta, and my original problem was in terms of x. Now I know that x and theta are related, according to this equation here. So if I solve for theta, I get sine theta is equal to x over seven. So theta is inverse sine or arc sine of x over seven. So I could just plug in for theta here and get seven squared, that's 49 times one half sine inverse x over seven minus one half times one half is 1/4. sine of two times sine inverse x over seven plus C. This is a correct answer. But it's very awkward looking sine of twice sine inverse x over seven, there's got to be a way to simplify this. Unfortunately, we can't just pull the two out and cancel the sign with a sine inverse, because sine does work that way. But instead, we can use the double angle formula. So we can rewrite sine of two theta as twice sine theta cosine theta. So let me ignore this line for now. And I'll rewrite the above line as 49 times one half theta, minus 1/4 times twice sine theta, cosine theta plus C. plus C. Now there's one more trick we need to use. And that trick is to draw a right triangle. I'm going to label the sides of that triangle using the equation for the substitution we made, or this equivalent forms a little easier to use. So since sine theta is x over seven, if I label one of my angles as theta, then the opposite side needs to have a measure of x, and I have partners should have a measure of seven. You can check that by the Pythagorean Theorem, that means that this bottom side needs to have a measure of the square root of 49 minus x squared. Notice that that's exactly the same expression that was in our original integrand. Once we have the triangle labeled, we can use it to find expressions for sine of theta and cosine of theta in terms of x. So Sine of theta is opposite of our hypothesis. So that's x over seven. Well, we already knew that. But now cosine of theta is adjacent over hypotenuse. So that's the square root of 49 minus x squared over seven. We can use these two equations to substitute in for sine theta and cosine theta. And to get rid of the, this naked theta here, we'll still have to use the fact that theta is inverse sine of x over seven. Those substitutions lead us to this answer. And a little bit of simplification leads us to the same answer that Wolfram Alpha gave us. This was a complicated problem. But the key step was to use this trig substitution. And then, that allowed us to use a trig identity in order to rewrite our square root and get rid of the square root. After that, it was a fairly routine integration problem, until we got to the point where we had to substitute back in for theta, and then drawing the right triangle helped us figure out what to do. In this video, we use two trig substitution to evaluate an integral with a square root and that This video introduces the method of partial fractions as a way to integrate many rational functions. Recall that a rational function is a function of the form a polynomial divided by another polynomial. Let's work out the integral of 3x plus two divided by x squared plus 2x minus three. According to Wolfram Alpha, the integral evaluates to this expression involving the log of x minus one and the log of x plus three. So what do x minus one and x plus three have to do with our original rational function, I challenge you to pause the video for a moment and try to figure this out. You might have noticed that the denominator x squared plus 2x minus three factors into x minus one times x plus three. Let's look at what happens if I take two separate fractions, one with denominator x minus one and one with denominator x plus three and add them up. The common denominator of these two fractions is x minus one times x plus three. So when I add these fractions together using standard algebra, I should get a fraction with common denominator, x minus one times x plus three. Well, my original rational function was just such a fraction. So it seems plausible, that I might be able to find numbers, a and b, so that these two partial fractions add up to this original fraction. Let's figure out what numbers a and b might make this work. To solve for A and B, I'm going to clear the denominator by multiplying both sides of my equation by this least common denominator, x minus 1x plus three. When I distribute on the left side, I get x plus three times a plus x minus one times B. And on the right side, my x minus one and x plus three cancel out to give me 3x plus two. I'll distribute the left side a little further. And I'm going to group the terms that involve x and the terms that don't involve x. If I want this equation to hold true for all values of x, then I need the coefficients of x to be the same on the left and the right. Now there's one more trick we need to use. And that trick is to draw a right triangle. I'm going to label the sides of that triangle using the equation for the substitution we made, or this equivalent forms a little easier to use. So since sine theta is x over seven, if I label one of my angles as theta, then the opposite side needs to have a measure of x, and I have partners should have a measure of seven. You can check that by the Pythagorean Theorem, that means that this bottom side needs to have a measure of the square root of 49 minus x squared. Notice that that's exactly the same expression that was in our original integrand. Once we have the triangle labeled, we can use it to find expressions for sine of theta and cosine of theta in terms of x. So Sine of theta is opposite of our hypothesis. So that's x over seven. Well, we already knew that. But now cosine of theta is adjacent over hypotenuse. So that's the square root of 49 minus x squared over seven. We can use these two equations to substitute in for sine theta and cosine theta. And to get rid of the, this naked theta here, we'll still have to use the fact that theta is inverse sine of x over seven. Those substitutions lead us to this answer. And a little bit of simplification leads us to the same answer that Wolfram Alpha gave us. This was a complicated problem. But the key step was to use this trig substitution. And then, that allowed us to use a trig identity in order to rewrite our square root and get rid of the square root. After that, it was a fairly routine integration problem, until we got to the point where we had to substitute back in for theta, and then drawing the right triangle helped us figure out what to do. In this video, we use two trig substitution to evaluate an integral with a square root and that This video introduces the method of partial fractions as a way to integrate many rational functions. Recall that a rational function is a function of the form a polynomial divided by another polynomial. Let's work out the integral of 3x plus two divided by x squared plus 2x minus three. According to Wolfram Alpha, the integral evaluates to this expression involving the log of x minus one and the log of x plus three. So what do x minus one and x plus three have to do with our original rational function, I challenge you to pause the video for a moment and try to figure this out. You might have noticed that the denominator x squared plus 2x minus three factors into x minus one times x plus three. Let's look at what happens if I take two separate fractions, one with denominator x minus one and one with denominator x plus three and add them up. The common denominator of these two fractions is x minus one times x plus three. So when I add these fractions together using standard algebra, I should get a fraction with common denominator, x minus one times x plus three. Well, my original rational function was just such a fraction. So it seems plausible, that I might be able to find numbers, a and b, so that these two partial fractions add up to this original fraction. Let's figure out what numbers a and b might make this work. To solve for A and B, I'm going to clear the denominator by multiplying both sides of my equation by this least common denominator, x minus 1x plus three. When I distribute on the left side, I get x plus three times a plus x minus one times B. And on the right side, my x minus one and x plus three cancel out to give me 3x plus two. I'll distribute the left side a little further. And I'm going to group the terms that involve x and the terms that don't involve x. If I want this equation to hold true for all values of x, then I need the coefficients of x to be the same on the left and the right. So I need a plus b to equal three. Similarly, I need the constant term to be the same on the left and the right. So I need three A minus B to equal two. I now have two linear equations in the two unknowns, a and b. So I can use standard algebra to solve these equations for A and B. For example, if I add the two equations together, the B and negative b cancel out to give me four A equals five, and so a equals five fourths. And then I can substitute in this five fourths into either one of my equations and solve for b. So I was able to find an a value of a and b that let me rewrite the original fraction as to partial fractions. Let me fill those values of a and b in here. Now to calculate the integral of my original expression, I can calculate the integral of my partial fractions. Instead, I split up my integrals here. Now the integral of one over x minus one is natural log of absolute value of x minus one, you can check this by taking the derivative. Or you can do a simple use of the tuition where you as x minus one to compute this integral. Similarly, the integral of one over x plus three is natural log of absolute value of x plus three. This completes the computation of the integral using the method of partial fractions. Recall that after factoring our denominator, the key step in this process was finding numbers a and b that made this equation hold. And you might wonder, Well, yeah, it work this time. But can we always find numbers a and b that will work like that? What if our numerator had been something different? What if our numerator had been instead something like, say 7x minus 15? Could we have still found an A and B that worked? Well, yes, because we still would have gotten two linear equations in two unknowns down here, and we would have been able to solve them, we'd get different values of a and b, but we still would have been able to find a solution. Even if our numerator had just been a number, like six, we still couldn't use the same method, we could have thought of this as 0x plus six. And we could have used the same equations down here, where A plus B would have to equals zero, and three, A minus B would have to equal six. So we still have two equations and two unknowns to solve for. In fact, if you think about it for a while, you may be able to convince yourself that this method will always work if you have two conditions. First, the denominator factors into distinct linear factors. By a linear factor, I just mean that the factor can be written like a number times x plus another number, with no x squared or anything in it. And by distinct, I just mean that these two factors are different from each other. The second condition is that the degree of the numerator is lower than the degree of the denominator. The second condition guarantee is that we'll have the same number of unknowns here, A and B, as we do coefficients here, three, and two, so that we'll have equations that we'll be able to solve for our unknowns. As a technical note, that distinctness of the linear factors guarantees that we won't have contradictory equations. If we have these two conditions, then we can proceed to integrate, like we did in this example. In fact, this method even works if we have more than two linear factors, in our denominator, we can have three or four, any number of distinct linear factors, it'll just get a little bit more complicated, because we'll have more constants and more equations to solve for those constants. There are also several related methods that allow us to integrate rational functions, even if the denominator factors into linear factors that aren't necessarily distinct. Or even if we can't get all the way down to linear factors, and some of our factors have squares in them, or even if the degree of the numerator is too big. But I won't get into the details of those related methods. In this video, you'll have to read about them in the book or wait for class or wait for a future video to find out more. In this video, we integrated a rational function by splitting up the fraction into two partial fractions. So I need a plus b to equal three. Similarly, I need the constant term to be the same on the left and the right. So I need three A minus B to equal two. I now have two linear equations in the two unknowns, a and b. So I can use standard algebra to solve these equations for A and B. For example, if I add the two equations together, the B and negative b cancel out to give me four A equals five, and so a equals five fourths. And then I can substitute in this five fourths into either one of my equations and solve for b. So I was able to find an a value of a and b that let me rewrite the original fraction as to partial fractions. Let me fill those values of a and b in here. Now to calculate the integral of my original expression, I can calculate the integral of my partial fractions. Instead, I split up my integrals here. Now the integral of one over x minus one is natural log of absolute value of x minus one, you can check this by taking the derivative. Or you can do a simple use of the tuition where you as x minus one to compute this integral. Similarly, the integral of one over x plus three is natural log of absolute value of x plus three. This completes the computation of the integral using the method of partial fractions. Recall that after factoring our denominator, the key step in this process was finding numbers a and b that made this equation hold. And you might wonder, Well, yeah, it work this time. But can we always find numbers a and b that will work like that? What if our numerator had been something different? What if our numerator had been instead something like, say 7x minus 15? Could we have still found an A and B that worked? Well, yes, because we still would have gotten two linear equations in two unknowns down here, and we would have been able to solve them, we'd get different values of a and b, but we still would have been able to find a solution. Even if our numerator had just been a number, like six, we still couldn't use the same method, we could have thought of this as 0x plus six. And we could have used the same equations down here, where A plus B would have to equals zero, and three, A minus B would have to equal six. So we still have two equations and two unknowns to solve for. In fact, if you think about it for a while, you may be able to convince yourself that this method will always work if you have two conditions. First, the denominator factors into distinct linear factors. By a linear factor, I just mean that the factor can be written like a number times x plus another number, with no x squared or anything in it. And by distinct, I just mean that these two factors are different from each other. The second condition is that the degree of the numerator is lower than the degree of the denominator. The second condition guarantee is that we'll have the same number of unknowns here, A and B, as we do coefficients here, three, and two, so that we'll have equations that we'll be able to solve for our unknowns. As a technical note, that distinctness of the linear factors guarantees that we won't have contradictory equations. If we have these two conditions, then we can proceed to integrate, like we did in this example. In fact, this method even works if we have more than two linear factors, in our denominator, we can have three or four, any number of distinct linear factors, it'll just get a little bit more complicated, because we'll have more constants and more equations to solve for those constants. There are also several related methods that allow us to integrate rational functions, even if the denominator factors into linear factors that aren't necessarily distinct. Or even if we can't get all the way down to linear factors, and some of our factors have squares in them, or even if the degree of the numerator is too big. But I won't get into the details of those related methods. In this video, you'll have to read about them in the book or wait for class or wait for a future video to find out more. In this video, we integrated a rational function by splitting up the fraction into two partial fractions. This video is about improper integrals, especially the first type, which I'll define in a moment. Two examples of improper integrals are the integral from one to infinity of one over x squared dx, and the integral from zero to pi over two of tan x dx. What makes these integrals improper? Well, in the first example, it's this infinity in the bound of integration. And then the second example, is the fact that the function, tan of x itself goes to infinity on the interval from zero to pi over two where we're integrating. So an integral is called improper. If either of these two situations occur. It's called a type one improper integral if we're integrating over an infinite interval. In other words, there's an infinity or negative infinity somewhere in the bounds of integration. So the first example, is a type one improper integral. A type two improper integral occurs when the function that we're integrating itself has an infinite discontinuity on the interval. by an infinite discontinuity, I mean the function is going to infinity or negative infinity, this is also called a vertical asymptote. This vertical asymptote could occur in the interior of the interval we're integrating over. Or, as in this example, it could occur on the endpoint of that interval of integration. Now, it's possible that both of these situations could occur for the same integral. And that's also an improper integral. This video will focus on type one improper integrals. A type one improper integral, asks us to integrate over an infinite interval. To do this, we take the integral over larger and larger finite intervals and take the limit. So for example, to find the integral from one to infinity of one over x squared dx, we'll evaluate the integral from one to some finite number T, and then take the limit as t goes off to infinity. In symbols, we can write the limit as t goes to infinity of the integral from one to T of one over x squared dx. Since one over x squared is the same thing as x to the minus two, we can integrate it to get negative x to the minus one evaluated between one and T, and then take that limit or rewrite this as negative one over x and evaluate on the bounds of integration. as t goes to infinity, one over t goes to zero. So the limit just comes from this expression, which evaluates to one. If we think of it, the integral is representing area, this is a little surprising. Even though we're taking the area of an infinitely long region, the area still evaluates to a finite number of one. In this situation, we say that the improper integral converges. So in general, the improper integral from some finite number a to infinity of f of x dx is defined as the limit as t goes to infinity of the integral from a to T of f of x dx. We say that the integral converges if this limit exists as a finite number. And we say that the integral diverges if the limit is infinity, or negative infinity, or if it doesn't exist. Similarly, we evaluate the integral from negative infinity to some finite number by taking bigger and bigger intervals that extend off to negative infinity. That is, this integral is defined as the limit as the left endpoint t goes to negative infinity of the integral from t to b of f of x dx. We say that this integral converges if the limit exists as a finite number, and diverges otherwise. So to evaluate the integral from negative infinity to negative one of one over x dx, we take the limit as t goes to negative infinity This video is about improper integrals, especially the first type, which I'll define in a moment. Two examples of improper integrals are the integral from one to infinity of one over x squared dx, and the integral from zero to pi over two of tan x dx. What makes these integrals improper? Well, in the first example, it's this infinity in the bound of integration. And then the second example, is the fact that the function, tan of x itself goes to infinity on the interval from zero to pi over two where we're integrating. So an integral is called improper. If either of these two situations occur. It's called a type one improper integral if we're integrating over an infinite interval. In other words, there's an infinity or negative infinity somewhere in the bounds of integration. So the first example, is a type one improper integral. A type two improper integral occurs when the function that we're integrating itself has an infinite discontinuity on the interval. by an infinite discontinuity, I mean the function is going to infinity or negative infinity, this is also called a vertical asymptote. This vertical asymptote could occur in the interior of the interval we're integrating over. Or, as in this example, it could occur on the endpoint of that interval of integration. Now, it's possible that both of these situations could occur for the same integral. And that's also an improper integral. This video will focus on type one improper integrals. A type one improper integral, asks us to integrate over an infinite interval. To do this, we take the integral over larger and larger finite intervals and take the limit. So for example, to find the integral from one to infinity of one over x squared dx, we'll evaluate the integral from one to some finite number T, and then take the limit as t goes off to infinity. In symbols, we can write the limit as t goes to infinity of the integral from one to T of one over x squared dx. Since one over x squared is the same thing as x to the minus two, we can integrate it to get negative x to the minus one evaluated between one and T, and then take that limit or rewrite this as negative one over x and evaluate on the bounds of integration. as t goes to infinity, one over t goes to zero. So the limit just comes from this expression, which evaluates to one. If we think of it, the integral is representing area, this is a little surprising. Even though we're taking the area of an infinitely long region, the area still evaluates to a finite number of one. In this situation, we say that the improper integral converges. So in general, the improper integral from some finite number a to infinity of f of x dx is defined as the limit as t goes to infinity of the integral from a to T of f of x dx. We say that the integral converges if this limit exists as a finite number. And we say that the integral diverges if the limit is infinity, or negative infinity, or if it doesn't exist. Similarly, we evaluate the integral from negative infinity to some finite number by taking bigger and bigger intervals that extend off to negative infinity. That is, this integral is defined as the limit as the left endpoint t goes to negative infinity of the integral from t to b of f of x dx. We say that this integral converges if the limit exists as a finite number, and diverges otherwise. So to evaluate the integral from negative infinity to negative one of one over x dx, we take the limit as t goes to negative infinity of the integral from t to negative one of one over x dx. Now the integral of one over x is ln of the absolute value of x, which will need to integrate between t and negative one and take a limit. We evaluate here we know that ln of the absolute value of negative one, that's ln of one which is zero. A graph of ln is helpful for evaluating the rest of this expression. as t goes to negative infinity, the absolute value of t is going to infinity. And so ln is also going to infinity. Therefore, our limit is actually negative infinity, and so the integral diverges. In this video, we evaluated improper integrals Which the interval that we're integrating over is infinite by looking at the integrals over larger and larger, finite intervals and taking a limit. This video is about type two improper integrals. These are integrals for which the interval that we're integrating over is finite. But the function that we're integrating goes to infinity on that interval. To integrate a type to improper integral like this one, or this one, we integrate our function over a larger and larger sub intervals on which the function is finite. And then we take a limit. So in this first picture, where the function approaches infinity, as x goes to be from the left, we can call that moving endpoint T. and evaluate the integral as the limit as that right endpoint t approaches B from the left of the integral from a to T of f of x dx. The same definition works if f of x is going to negative infinity instead of infinity. In the second picture here, we again want to take a limit of integrals over sub intervals on which the function is finite. So in symbols, that says that if f goes to infinity or negative infinity, as x goes to a from the positive side, then the integral from a to b of f of x dx is going to be the limit as t goes to a from the right of the integral from t to b of f of x dx. As an example, let's find the area under the curve y equals x over the square root of x squared minus one between the lines of x equals one and x equals two. That area can be described as the integral from one to two of our function. But this is an improper integral, because the function is going to infinity, as x goes to one from the right. So we'll evaluate it by taking the limit as t goes to one from the right of the integral from t to two of the function. I'll compute this integral using u substitution, where u is x squared minus one, and d u is 2x dx. Since I see an x dx, in my integrand, here, I'll solve for x dx, and I get x dx is equal to one half deal. I'm also going to change my bounds of integration, when x is equal to t. That means u is equal to t squared minus one. And when x is equal to two, u is two squared minus one, that's three. So I'll rewrite my integral x dx is the same thing as one half d u, and my square root of x squared minus one becomes a square root of u in the denominator. Rewrite again, by pulling the one half out of the integral and rewriting the square root of u in the denominator as u to the negative one half. Now I can integrate, I get two times you to the one half evaluated between three and t squared minus one, my one half and two cancel, and I'll plug in my bounds of integration here. Now, as t goes to one from the right, t squared is also going to one, so t squared minus one is going to zero. of the integral from t to negative one of one over x dx. Now the integral of one over x is ln of the absolute value of x, which will need to integrate between t and negative one and take a limit. We evaluate here we know that ln of the absolute value of negative one, that's ln of one which is zero. A graph of ln is helpful for evaluating the rest of this expression. as t goes to negative infinity, the absolute value of t is going to infinity. And so ln is also going to infinity. Therefore, our limit is actually negative infinity, and so the integral diverges. In this video, we evaluated improper integrals Which the interval that we're integrating over is infinite by looking at the integrals over larger and larger, finite intervals and taking a limit. This video is about type two improper integrals. These are integrals for which the interval that we're integrating over is finite. But the function that we're integrating goes to infinity on that interval. To integrate a type to improper integral like this one, or this one, we integrate our function over a larger and larger sub intervals on which the function is finite. And then we take a limit. So in this first picture, where the function approaches infinity, as x goes to be from the left, we can call that moving endpoint T. and evaluate the integral as the limit as that right endpoint t approaches B from the left of the integral from a to T of f of x dx. The same definition works if f of x is going to negative infinity instead of infinity. In the second picture here, we again want to take a limit of integrals over sub intervals on which the function is finite. So in symbols, that says that if f goes to infinity or negative infinity, as x goes to a from the positive side, then the integral from a to b of f of x dx is going to be the limit as t goes to a from the right of the integral from t to b of f of x dx. As an example, let's find the area under the curve y equals x over the square root of x squared minus one between the lines of x equals one and x equals two. That area can be described as the integral from one to two of our function. But this is an improper integral, because the function is going to infinity, as x goes to one from the right. So we'll evaluate it by taking the limit as t goes to one from the right of the integral from t to two of the function. I'll compute this integral using u substitution, where u is x squared minus one, and d u is 2x dx. Since I see an x dx, in my integrand, here, I'll solve for x dx, and I get x dx is equal to one half deal. I'm also going to change my bounds of integration, when x is equal to t. That means u is equal to t squared minus one. And when x is equal to two, u is two squared minus one, that's three. So I'll rewrite my integral x dx is the same thing as one half d u, and my square root of x squared minus one becomes a square root of u in the denominator. Rewrite again, by pulling the one half out of the integral and rewriting the square root of u in the denominator as u to the negative one half. Now I can integrate, I get two times you to the one half evaluated between three and t squared minus one, my one half and two cancel, and I'll plug in my bounds of integration here. Now, as t goes to one from the right, t squared is also going to one, so t squared minus one is going to zero. Therefore, my limit is just three to the one half, or square root of three. And that's the area underneath my curve. This video was about type two improper integrals, and how to compute them as the limit of integrals over larger and larger sub intervals on which the function is finite. Sometimes we don't really care what the value of an improper integral is, we just want to know whether it's finite or infinite, whether it converges or diverges. In this situation, the comparison theorem can be very handy. The comparison theorem can allow us to determine if an integral converges or diverges without actually having to evaluate the integral instead by comparing it to the integral of a function that we know converges or diverges. So suppose that g of x and f of x are both positive valued functions. They're both greater than zero for all x's on the interval A B. And let's suppose also that g of x is less than f of x on that interval a, b. here, A or B could be infinity or negative infinity. So in the picture, we'll call this blue function, g of x, and the orange function, f of x. And let's consider the interval from one to infinity, where g of x is less than f of x, and both of them are bigger than zero. Now, if we already know that the integral of f of x on this interval converges to a finite number, then g of x, which is less than f of x also has to converge to a finite number. So if the integral of f of x converges, then g of x converges. If we turn this around, we can say that if g of x diverges, so it doesn't converge to a finite number, then f of x has to diverged also. So in the situation where the integral of the bigger function converges, or the integral of the smaller function diverges, then we can make some conclusions about the integral the other function, but you have to be a little careful about this. Because if instead, the integral of the bigger function diverges, then we really can't make any conclusions at all about the integral of the smaller function, it could also diverged or it could be small enough to converge as similarly, if the integral of the smaller function converges, then we really just don't know anything about the integral of the bigger function, it could converge, or it could divert. Let's look at an example. Suppose we want to find out if the integral from two to infinity of two plus sine x over square root of x dx converges or diverges. Instead of trying to evaluate it, which could get tricky, because of the sine x and the square root of x in here, I'm going to just try to compare it to something that I know converges or diverges. The first thing that I notice is that sine of x is bounded. It's always between one and negative one. And that means that the numerator two plus sine x is always going to be in between three and one. Therefore, the function two plus sine x over square root of x is going to be between three over square root of x and one over square root of x. Here's the general idea of the picture. Now the comparison theorem tells us that if our function is less than a function whose interval converges, then the integral of our current function will converge. And if our function is greater than a function whose integral diverges than the integral of our function will diverged. So which one of these two inequalities if we want to use depends on what happens to the integral of these functions on the ends? Now we know that the integral of one over the square root of x dx from two to infinity has to die verge. That's because this is a p function, where p is equal to one half, which is less than one, the integral from two to infinity of three over the square root of x dx also diverges since it's just three times the other integral. So if we want to compare our function to a function whose integral diverges, it It better be bigger than that divergent integral in order to get any useful information out of it being less than a function whose integral diverges doesn't tell us anything. So we need to focus on this inequality. Therefore, my limit is just three to the one half, or square root of three. And that's the area underneath my curve. This video was about type two improper integrals, and how to compute them as the limit of integrals over larger and larger sub intervals on which the function is finite. Sometimes we don't really care what the value of an improper integral is, we just want to know whether it's finite or infinite, whether it converges or diverges. In this situation, the comparison theorem can be very handy. The comparison theorem can allow us to determine if an integral converges or diverges without actually having to evaluate the integral instead by comparing it to the integral of a function that we know converges or diverges. So suppose that g of x and f of x are both positive valued functions. They're both greater than zero for all x's on the interval A B. And let's suppose also that g of x is less than f of x on that interval a, b. here, A or B could be infinity or negative infinity. So in the picture, we'll call this blue function, g of x, and the orange function, f of x. And let's consider the interval from one to infinity, where g of x is less than f of x, and both of them are bigger than zero. Now, if we already know that the integral of f of x on this interval converges to a finite number, then g of x, which is less than f of x also has to converge to a finite number. So if the integral of f of x converges, then g of x converges. If we turn this around, we can say that if g of x diverges, so it doesn't converge to a finite number, then f of x has to diverged also. So in the situation where the integral of the bigger function converges, or the integral of the smaller function diverges, then we can make some conclusions about the integral the other function, but you have to be a little careful about this. Because if instead, the integral of the bigger function diverges, then we really can't make any conclusions at all about the integral of the smaller function, it could also diverged or it could be small enough to converge as similarly, if the integral of the smaller function converges, then we really just don't know anything about the integral of the bigger function, it could converge, or it could divert. Let's look at an example. Suppose we want to find out if the integral from two to infinity of two plus sine x over square root of x dx converges or diverges. Instead of trying to evaluate it, which could get tricky, because of the sine x and the square root of x in here, I'm going to just try to compare it to something that I know converges or diverges. The first thing that I notice is that sine of x is bounded. It's always between one and negative one. And that means that the numerator two plus sine x is always going to be in between three and one. Therefore, the function two plus sine x over square root of x is going to be between three over square root of x and one over square root of x. Here's the general idea of the picture. Now the comparison theorem tells us that if our function is less than a function whose interval converges, then the integral of our current function will converge. And if our function is greater than a function whose integral diverges than the integral of our function will diverged. So which one of these two inequalities if we want to use depends on what happens to the integral of these functions on the ends? Now we know that the integral of one over the square root of x dx from two to infinity has to die verge. That's because this is a p function, where p is equal to one half, which is less than one, the integral from two to infinity of three over the square root of x dx also diverges since it's just three times the other integral. So if we want to compare our function to a function whose integral diverges, it It better be bigger than that divergent integral in order to get any useful information out of it being less than a function whose integral diverges doesn't tell us anything. So we need to focus on this inequality. And now we can say that, since the integral from two to infinity of one over the square root of x dx diverges by the P test, the integral of our function also diverges by the comparison test. In this video, we saw that if we have two positive functions, and one function is always less than or equal to the other function on an interval, then if the smaller functions integral diverges, the bigger functions integral also has to diverge. And if the Biggers functions integral converges, the smaller functions integral also has to converge. That's the comparison there. In this video, I'll give some definitions and notation for sequences. A sequence is a list of numbers in a particular order, the sequence 314159 and So on is a sequence that gives the digits of pi. A sequence is sometimes described abstractly, with letters in place of numbers, a sub one, a sub two, a sub three, and so on. Or more concisely, by writing a sub n with these curly brackets. Here, we're told that the index n ranges from one all the way up towards infinity. Sometimes the sequence is written just a sub n with curly brackets. Here, it's implied that n ranges through all positive integers. For these sequences, given by formulas, let's write out the first few terms. We start with n equals one, and we get a sub one is three times one plus one over one plus two factorial. That is, for over three factorial. Recall that three factorial means that we start with three and then multiply with consecutive numbers all the way down to one, this simplifies to four, six, or two thirds. To find the next term, a sub two, we plug in two for add, that's seven over four factorial, which is 720 fourths. Similarly, a sub three is 10, over five factorial, which is 10 over 120, or 112. So the first three terms of the sequence are two thirds, 720, fourths, and 112. For the second example, we're asked to start with K equals two. So I'll call the first term, a sub two, and just plug in two for K, which is five ninths. Since negative one squared is positive one. Plugging in K equals three, we get negative 6/27, or to negative two nights. For as of four, we again get a positive number, since negative one to the fourth is positive. And in fact, as we keep writing down terms, they're going to alternate between negative numbers and positive numbers, because of the negative one to the K in the definition. Sometimes, the nth term of a sequence is defined indirectly, in terms of previous terms. This is called a recursive formula. To write out the first few terms of this recursive sequence, we're told that a sub one equals two, to find a sub two, we just use the recursive formula for minus one over a sub one. Since a sub one is two, that's four minus a half, or seven half, defined a sub three, we just apply the recursive formula again, four minus one over seven has simplifies to 26/7. Sometimes as possible, describe a sequence with either a recursive formula, or a closed form non recursive formula. For example, if I look at the sequence 246 810, I can describe that recursively by saying a sub one is two, and each a sub n is equal to a sub n minus one plus two. Or I can describe as a closed form expression by saying the sequence is of the form two times n, where n starts at one. And now we can say that, since the integral from two to infinity of one over the square root of x dx diverges by the P test, the integral of our function also diverges by the comparison test. In this video, we saw that if we have two positive functions, and one function is always less than or equal to the other function on an interval, then if the smaller functions integral diverges, the bigger functions integral also has to diverge. And if the Biggers functions integral converges, the smaller functions integral also has to converge. That's the comparison there. In this video, I'll give some definitions and notation for sequences. A sequence is a list of numbers in a particular order, the sequence 314159 and So on is a sequence that gives the digits of pi. A sequence is sometimes described abstractly, with letters in place of numbers, a sub one, a sub two, a sub three, and so on. Or more concisely, by writing a sub n with these curly brackets. Here, we're told that the index n ranges from one all the way up towards infinity. Sometimes the sequence is written just a sub n with curly brackets. Here, it's implied that n ranges through all positive integers. For these sequences, given by formulas, let's write out the first few terms. We start with n equals one, and we get a sub one is three times one plus one over one plus two factorial. That is, for over three factorial. Recall that three factorial means that we start with three and then multiply with consecutive numbers all the way down to one, this simplifies to four, six, or two thirds. To find the next term, a sub two, we plug in two for add, that's seven over four factorial, which is 720 fourths. Similarly, a sub three is 10, over five factorial, which is 10 over 120, or 112. So the first three terms of the sequence are two thirds, 720, fourths, and 112. For the second example, we're asked to start with K equals two. So I'll call the first term, a sub two, and just plug in two for K, which is five ninths. Since negative one squared is positive one. Plugging in K equals three, we get negative 6/27, or to negative two nights. For as of four, we again get a positive number, since negative one to the fourth is positive. And in fact, as we keep writing down terms, they're going to alternate between negative numbers and positive numbers, because of the negative one to the K in the definition. Sometimes, the nth term of a sequence is defined indirectly, in terms of previous terms. This is called a recursive formula. To write out the first few terms of this recursive sequence, we're told that a sub one equals two, to find a sub two, we just use the recursive formula for minus one over a sub one. Since a sub one is two, that's four minus a half, or seven half, defined a sub three, we just apply the recursive formula again, four minus one over seven has simplifies to 26/7. Sometimes as possible, describe a sequence with either a recursive formula, or a closed form non recursive formula. For example, if I look at the sequence 246 810, I can describe that recursively by saying a sub one is two, and each a sub n is equal to a sub n minus one plus two. Or I can describe as a closed form expression by saying the sequence is of the form two times n, where n starts at one. Now let's practice writing out a formula for the general term, a sub n of a sequence. For this first sequence, notice that each term is three more than the previous term. So this is like a linear function with slope three, each time that n goes up by one, our ace events go up by three. And so I can write a sub n is three times n plus b, where B functions like my y intercept and a linear equation, to find B, I can plug in seven, four, a sub one that corresponds to an N value of one, and I get that B has to equal four. So my general formula is three times n plus four, where n starts at one. I can check this by plugging in a few values of n, like we did in the previous example, just to make sure it works. Notice that it would also be possible to write this as three times n plus seven If we're willing to start with n equals zero instead of one, if we let n start with zero, then our first term functions like our y intercept. This is an example of an arithmetic sequence, a sequence for which consecutive terms have the same common difference. And in general, if A is the first term, and d is the common difference than an arithmetic sequence has the form d times k plus a, if our index is K and starts at zero, like it did over here. Or if we'd rather start with an index of one, we can rewrite that as D times n minus one plus a. Notice these two expressions are exactly the same, if we just set k equal to n minus one, in particular, the starting value of n equals one here, if I plug in one for n, I get the equivalent starting value of K to be zero. Let's read some more formulas for sequences. In this first sequence, notice that consecutive terms always have the same ratio of 1/10. In other words, each time n increases by one, a sub n gets multiplied by 1/10. This is the same property that exponential functions have. And in fact, we can write a sub n in the form of an exponential function with base 1/10. But we need to multiply by the right initial value, so that when n is one, we'll get a first term of three, that correct initial value is 30. As usual, we can check our answer by plugging in a few values of n, n equals 123. And making sure we get back the terms in our sequence. If we prefer to start with our index at zero, we can rewrite this as three times 1/10 to the nth power. Since a value of zero for n in this formula, gives us three, just like a value of one for n in this formula gives us three. In the second example, we again have a common ratio, if I divide the second term by the first term, I get a ratio of five halves. And that's the same ratio as I get when I divide the third term by the second term. So if I use an index starting with zero, I can write this series as 15 halves, which is the first term times five halves to the nth power. If I prefer to start with my index at one, one way to do this is to let K equal n plus one and making a variable substitution. When n is 0k, will be one. And since K is n plus one, n is k minus one, so I can replace n with k minus one. This gives the following representation of the sequence. The third example has a common ratio of negative two thirds. so we can write it as the initial term of three times that ratio, negative two thirds raised to the nth power, where n starts at zero. Or, as above, we can write it as three times negative two thirds to the k minus one. Now let's practice writing out a formula for the general term, a sub n of a sequence. For this first sequence, notice that each term is three more than the previous term. So this is like a linear function with slope three, each time that n goes up by one, our ace events go up by three. And so I can write a sub n is three times n plus b, where B functions like my y intercept and a linear equation, to find B, I can plug in seven, four, a sub one that corresponds to an N value of one, and I get that B has to equal four. So my general formula is three times n plus four, where n starts at one. I can check this by plugging in a few values of n, like we did in the previous example, just to make sure it works. Notice that it would also be possible to write this as three times n plus seven If we're willing to start with n equals zero instead of one, if we let n start with zero, then our first term functions like our y intercept. This is an example of an arithmetic sequence, a sequence for which consecutive terms have the same common difference. And in general, if A is the first term, and d is the common difference than an arithmetic sequence has the form d times k plus a, if our index is K and starts at zero, like it did over here. Or if we'd rather start with an index of one, we can rewrite that as D times n minus one plus a. Notice these two expressions are exactly the same, if we just set k equal to n minus one, in particular, the starting value of n equals one here, if I plug in one for n, I get the equivalent starting value of K to be zero. Let's read some more formulas for sequences. In this first sequence, notice that consecutive terms always have the same ratio of 1/10. In other words, each time n increases by one, a sub n gets multiplied by 1/10. This is the same property that exponential functions have. And in fact, we can write a sub n in the form of an exponential function with base 1/10. But we need to multiply by the right initial value, so that when n is one, we'll get a first term of three, that correct initial value is 30. As usual, we can check our answer by plugging in a few values of n, n equals 123. And making sure we get back the terms in our sequence. If we prefer to start with our index at zero, we can rewrite this as three times 1/10 to the nth power. Since a value of zero for n in this formula, gives us three, just like a value of one for n in this formula gives us three. In the second example, we again have a common ratio, if I divide the second term by the first term, I get a ratio of five halves. And that's the same ratio as I get when I divide the third term by the second term. So if I use an index starting with zero, I can write this series as 15 halves, which is the first term times five halves to the nth power. If I prefer to start with my index at one, one way to do this is to let K equal n plus one and making a variable substitution. When n is 0k, will be one. And since K is n plus one, n is k minus one, so I can replace n with k minus one. This gives the following representation of the sequence. The third example has a common ratio of negative two thirds. so we can write it as the initial term of three times that ratio, negative two thirds raised to the nth power, where n starts at zero. Or, as above, we can write it as three times negative two thirds to the k minus one. Where k starts at one. Sometimes people like to write the negative one separately. The negative one to the power makes a series alternate between negative and positive terms. These three sequences are all examples of geometric sequences, which are key sequences where consecutive terms have the same common ratio. And in general, if A is the first term and RS, the common ratio than a geometric sequence can be written in the form of a times r to the n, where n starts at zero, or as a times r to the n minus one, where n starts at one. These next two sequences are neither arithmetic sequences nor geometric sequences. Since they're terms neither have common differences nor common ratios, but I can still figure out a formula for the nth term just by looking for the pattern. In this example, since the terms are alternating, if I start at n equals one, I know that 90 to negative one to the nth power to make it start with a negative and then alternate, positive, negative Again, the numerator looks like it's just twice n. And the denominator looks like it's always a perfect square, starting with the square of three, so I'll write that as n plus two squared. The next sequence doesn't have a simple closed form formula, but I can describe it recursively by saying that a one is negative six, a two is five. And in general, a n is equal to the sum of the two previous terms, a sub n minus one, plus a sub n minus two. This sequence is closely related to the standard Fibonacci sequence, which has the same recursive formula, but different initial values. This video gave an introduction to sequences, including arithmetic sequences, geometric sequences, and recursively defined sequences. This video introduces the idea of a series and how to find it sum. For any sequence, a sub n for n equals one to infinity, the sum of its terms, a sub one plus a sub two plus a sub three, and so on, it's called a series. Often this series is written in sigma notation, as a sum from n equals one to infinity of a sub n. Let's look at the sequence one over two to the n for n equals one to infinity. If we add together all the terms, we get the series, the sum from n equals one to infinity of one over two to the N, that is the sum of one over two to the one, that's one half, plus one over two squared, that's 1/4, plus one over two cubed 1/8 plus 1/16, plus 1/32, and so on. But what does it really mean to add have infinitely many numbers? How can we figure out what this infinite song equals? Well, the start out, we could add up finitely, many at a time, may write down the first half dozen or so terms, that is the ace of ends. And I'll keep adding more and more of them together one more at a time. But just add the first term, well, that's just one half. Now add the next term on, that gives me a song of three fourths. If I add 1/8 to that, my son goes up to seven eights. And then I'll add the next one. I get 15 sixteenths, and so on just adding one more term each time. This process of repeated addition gives me a new sequence down at the bottom. That's called the sequence of partial sums. And it's usually denoted by s sub n. So the first term in the sequence of partial sums as S sub one, just adding together the first term, here's the second partial sum S sub two, adding together the first two terms. So three means add together, the first three terms, and so on. Let me contrast the sequence of partial sums with the sequence of terms that we started out with. Those are denoted a sub n. So here's a sub one, the first term, a sub two, the second term, and so on. Where k starts at one. Sometimes people like to write the negative one separately. The negative one to the power makes a series alternate between negative and positive terms. These three sequences are all examples of geometric sequences, which are key sequences where consecutive terms have the same common ratio. And in general, if A is the first term and RS, the common ratio than a geometric sequence can be written in the form of a times r to the n, where n starts at zero, or as a times r to the n minus one, where n starts at one. These next two sequences are neither arithmetic sequences nor geometric sequences. Since they're terms neither have common differences nor common ratios, but I can still figure out a formula for the nth term just by looking for the pattern. In this example, since the terms are alternating, if I start at n equals one, I know that 90 to negative one to the nth power to make it start with a negative and then alternate, positive, negative Again, the numerator looks like it's just twice n. And the denominator looks like it's always a perfect square, starting with the square of three, so I'll write that as n plus two squared. The next sequence doesn't have a simple closed form formula, but I can describe it recursively by saying that a one is negative six, a two is five. And in general, a n is equal to the sum of the two previous terms, a sub n minus one, plus a sub n minus two. This sequence is closely related to the standard Fibonacci sequence, which has the same recursive formula, but different initial values. This video gave an introduction to sequences, including arithmetic sequences, geometric sequences, and recursively defined sequences. This video introduces the idea of a series and how to find it sum. For any sequence, a sub n for n equals one to infinity, the sum of its terms, a sub one plus a sub two plus a sub three, and so on, it's called a series. Often this series is written in sigma notation, as a sum from n equals one to infinity of a sub n. Let's look at the sequence one over two to the n for n equals one to infinity. If we add together all the terms, we get the series, the sum from n equals one to infinity of one over two to the N, that is the sum of one over two to the one, that's one half, plus one over two squared, that's 1/4, plus one over two cubed 1/8 plus 1/16, plus 1/32, and so on. But what does it really mean to add have infinitely many numbers? How can we figure out what this infinite song equals? Well, the start out, we could add up finitely, many at a time, may write down the first half dozen or so terms, that is the ace of ends. And I'll keep adding more and more of them together one more at a time. But just add the first term, well, that's just one half. Now add the next term on, that gives me a song of three fourths. If I add 1/8 to that, my son goes up to seven eights. And then I'll add the next one. I get 15 sixteenths, and so on just adding one more term each time. This process of repeated addition gives me a new sequence down at the bottom. That's called the sequence of partial sums. And it's usually denoted by s sub n. So the first term in the sequence of partial sums as S sub one, just adding together the first term, here's the second partial sum S sub two, adding together the first two terms. So three means add together, the first three terms, and so on. Let me contrast the sequence of partial sums with the sequence of terms that we started out with. Those are denoted a sub n. So here's a sub one, the first term, a sub two, the second term, and so on. Although I can't physically add up infinitely many numbers, I can observe that as I add up more and more numbers, my partial sums are approaching the number one. That is the limit, as the number of terms I add up and goes to infinity of the nth partial sum is equal to one. So it makes sense that if I could add up all infinitely many numbers, I should get an exact sum of one, the sum of this infinite series is one. In fact, for this particular series, there's a nice way to see that the sum is one using geometry. If I draw a square with side length one and fill in half of the square, that gives me an area of one half. Now if I draw a line here, that gives me an additional area of 1/4. Here's an area of 1/8 1/16 and I keep spiraling in here. I keep adding areas that exactly match the terms of this series. In the limit, I'll have filled in the entire R square, which has an area of one. In this example, which found the sum of the series by evaluating the limit of the partial sums. And in general, this is how we find the sum of any series. For any series, the partial sums of the series are defined as the sequence s sub n, where S sub one is equal to just the first term, a sub one, a sub two is equal to the sum of the first two terms is a one plus a sub two, a sub three is a sum of the first three terms. And in general, s sub n is the sum of the first n terms. I can also write this in sigma notation, as the sum of say, k equals one to n of a sub k. I'm using a different letter K here as the index, just because I'm already using n to represent the number of terms that I'm adding up. That sum of a sub n is said to converge if this sequence of partial sums converges as a sequence. That is, if the limit as n goes to infinity of the sub ends, exists as a finite number. Otherwise, if the limit does not exist, or the limit is infinity or negative infinity, then the series is said to diverged. I want to emphasize that we're talking about the limit of the partial sums here, not the limit of the original terms as a bad, it's important to keep in mind when you're working with series that for any series, they're actually two sequences of interest. There's the sequence of terms, the ace of ends. And then there's the sequence of partial sums, the S events, it's the sequence of partial sums is telling us what the sum of our series is. So if the sequence of partial sums converges to a number L, then we say the cert series converges to L. Or in other words, the sum of the series is out. Let's look at the series The sum of one over n squared plus n, please pause the video and take a moment to write out the first four terms and the first four partial sums. The first four terms are a sub one, a sub two, a sub three and a sub four. So plugging in one for n, one over one squared plus one, so that's one half, a sub two is one over two squared plus two, that's one six, Ace of three ends up being 1/12. And a sub four is 1/20. Now S sub one is the sum of the first one terms, that's the same as a sub one is just one half Although I can't physically add up infinitely many numbers, I can observe that as I add up more and more numbers, my partial sums are approaching the number one. That is the limit, as the number of terms I add up and goes to infinity of the nth partial sum is equal to one. So it makes sense that if I could add up all infinitely many numbers, I should get an exact sum of one, the sum of this infinite series is one. In fact, for this particular series, there's a nice way to see that the sum is one using geometry. If I draw a square with side length one and fill in half of the square, that gives me an area of one half. Now if I draw a line here, that gives me an additional area of 1/4. Here's an area of 1/8 1/16 and I keep spiraling in here. I keep adding areas that exactly match the terms of this series. In the limit, I'll have filled in the entire R square, which has an area of one. In this example, which found the sum of the series by evaluating the limit of the partial sums. And in general, this is how we find the sum of any series. For any series, the partial sums of the series are defined as the sequence s sub n, where S sub one is equal to just the first term, a sub one, a sub two is equal to the sum of the first two terms is a one plus a sub two, a sub three is a sum of the first three terms. And in general, s sub n is the sum of the first n terms. I can also write this in sigma notation, as the sum of say, k equals one to n of a sub k. I'm using a different letter K here as the index, just because I'm already using n to represent the number of terms that I'm adding up. That sum of a sub n is said to converge if this sequence of partial sums converges as a sequence. That is, if the limit as n goes to infinity of the sub ends, exists as a finite number. Otherwise, if the limit does not exist, or the limit is infinity or negative infinity, then the series is said to diverged. I want to emphasize that we're talking about the limit of the partial sums here, not the limit of the original terms as a bad, it's important to keep in mind when you're working with series that for any series, they're actually two sequences of interest. There's the sequence of terms, the ace of ends. And then there's the sequence of partial sums, the S events, it's the sequence of partial sums is telling us what the sum of our series is. So if the sequence of partial sums converges to a number L, then we say the cert series converges to L. Or in other words, the sum of the series is out. Let's look at the series The sum of one over n squared plus n, please pause the video and take a moment to write out the first four terms and the first four partial sums. The first four terms are a sub one, a sub two, a sub three and a sub four. So plugging in one for n, one over one squared plus one, so that's one half, a sub two is one over two squared plus two, that's one six, Ace of three ends up being 1/12. And a sub four is 1/20. Now S sub one is the sum of the first one terms, that's the same as a sub one is just one half S sub two, is what I get when I add one half plus one six, that's four, six, or two thirds. To get a sub three, I need to add on the next term, two thirds plus 112, is 912, or three fourths. And finally, as sub four, I need to add on the next term, which gives me four fifths. There's a nice pattern going on with the S events here. For the nth partial song, the numerator is just n, and the denominator is just n plus one. So the limit as n goes to infinity of the partial sums, is the limit as n goes to infinity of n over n plus one, which is just one. That means that the sum of the infinite series is equal to one, by coincidence the same sound as in the previous example. In this video, we saw that to find the sum of an infinite series, we have to calculate the partial sums. If we add together more and more finitely many terms, our partial sum will get close to our infinite sum. And in fact, the sum of our infinite sum is defined as the limit as n goes to infinity of the partial sum. This video gives some more definitions about sequences, including the definition of bounded and the definition of monotonic sequence is bounded above if all of its terms are less than or equal to some number in other Words, there's a number capital M, such that the term a sub n is less than or equal to capital M, for all values of the index n. A sequence is bounded below if all of its terms are greater than or equal to some number. In other words, there's a number lowercase m, such that a sub n is greater than or equal to lowercase m, for all and we say that a sequence is bounded if it's bounded both above and below. In other words, all of its terms are trapped between two numbers. Please pause the video and decide which of these three sequences are bounded. This first sequence is bounded above by three. Since all of its terms are less than or equal to three, of course, we could have also used four as an upper bound, or 100 is an upper bound, but three is the tightest upper bound that we can use. It's also bounded below by zero, since all of the terms are positive. So we say that this sequence is bounded. The second sequence is bounded below by one, but it's not bounded above, since its terms will eventually grow past any potential bound. The third sequence is bounded below and above. In fact, if we graph n on the x axis and a sub n on the y axis, then our terms bounce around between positive and negative values. But since we're always multiplying by negative two thirds to get from one term to the next, the oscillations are dying down in magnitude. And in fact, the terms can never get above three or below negative two. We say that a sequence is increasing if each term is less than the next term, that is, a sub n is less than a sub n plus one for all n. sequences called non decreasing if each term is less than or equal to the next term. So a sub n is less than or equal to a sub n plus one for all n. Non decreasing is like increasing, it's just we allow equality between 2/6 consecutive terms S sub two, is what I get when I add one half plus one six, that's four, six, or two thirds. To get a sub three, I need to add on the next term, two thirds plus 112, is 912, or three fourths. And finally, as sub four, I need to add on the next term, which gives me four fifths. There's a nice pattern going on with the S events here. For the nth partial song, the numerator is just n, and the denominator is just n plus one. So the limit as n goes to infinity of the partial sums, is the limit as n goes to infinity of n over n plus one, which is just one. That means that the sum of the infinite series is equal to one, by coincidence the same sound as in the previous example. In this video, we saw that to find the sum of an infinite series, we have to calculate the partial sums. If we add together more and more finitely many terms, our partial sum will get close to our infinite sum. And in fact, the sum of our infinite sum is defined as the limit as n goes to infinity of the partial sum. This video gives some more definitions about sequences, including the definition of bounded and the definition of monotonic sequence is bounded above if all of its terms are less than or equal to some number in other Words, there's a number capital M, such that the term a sub n is less than or equal to capital M, for all values of the index n. A sequence is bounded below if all of its terms are greater than or equal to some number. In other words, there's a number lowercase m, such that a sub n is greater than or equal to lowercase m, for all and we say that a sequence is bounded if it's bounded both above and below. In other words, all of its terms are trapped between two numbers. Please pause the video and decide which of these three sequences are bounded. This first sequence is bounded above by three. Since all of its terms are less than or equal to three, of course, we could have also used four as an upper bound, or 100 is an upper bound, but three is the tightest upper bound that we can use. It's also bounded below by zero, since all of the terms are positive. So we say that this sequence is bounded. The second sequence is bounded below by one, but it's not bounded above, since its terms will eventually grow past any potential bound. The third sequence is bounded below and above. In fact, if we graph n on the x axis and a sub n on the y axis, then our terms bounce around between positive and negative values. But since we're always multiplying by negative two thirds to get from one term to the next, the oscillations are dying down in magnitude. And in fact, the terms can never get above three or below negative two. We say that a sequence is increasing if each term is less than the next term, that is, a sub n is less than a sub n plus one for all n. sequences called non decreasing if each term is less than or equal to the next term. So a sub n is less than or equal to a sub n plus one for all n. Non decreasing is like increasing, it's just we allow equality between 2/6 consecutive terms a sequence is decreasing if each term is greater than the next one. So a sub n is greater than a sub n plus one for all n. And a sequence is not increasing, if a sub n is greater than or equal to a sub n plus one for all n. Again, not increasing is like decreasing, but we allow for equality between consecutive terms. If we were to graph and on the x axis and a sub n on the y axis, then increasing looks like this just like an increasing function, whereas decreasing would go down non decreasing would go up possibly with some flat patches and mnandi increasing would go down possibly with some flat patches, notice that increasing is a stronger condition than non decreasing since being strictly less than something is stronger than just being less than or equal to it. For that reason. If a sequence is increasing, it is also non decreasing. And similarly, if a sequence is decreasing, it is also not increasing. A sequence is called monotonic, if it is either non decreasing or non increasing. Please pause the video and try to decide which of the following sequences are monotonic. The first two sequences are monotonic. The first one is monotonically. Non increasing, since we never increase when we go from one term to the next. In fact, we could also say that it's monotonically decreasing. Since we always go strictly down as we go from one term to the next, we never have equality between terms. The second sequence is monotonically non decreasing. Since we never go down, as we go from one term to the next, we either go up or stay at the same level. In this case, however, we could not say that the sequence is monotonically increasing because of the equality between some pairs of consecutive terms. The third sequence is not monotonic, the numbers bounce around between positive and negative numbers, and therefore sometimes we are decreasing while other times we're increasing. And the fourth sequence is not monotonic, because of the first few terms. However, from the fifth term on the terms are monotonically, non decreasing. And we could also say monotonically increasing. In this video, we give definitions for bounded and monotonic sequences. We'll see later that these definitions can be important for determining when a sequence converges. In class, we talked briefly about the sequence n minus five over n squared, where n goes from one to infinity. We want to know if the sequence is monotonic. And if it's bounded. If we compute the first few terms, the sequence appears to be steadily increasing. But in this case, appearances are deceiving. And a better way to decide whether it's monotonic is to use calculus to decide if the associated function f of x equals x minus five over x squared, it is increasing for x values bigger than one. So let me take the derivative. Using the quotient rule and simplify, I get that f prime of x is equal to minus x squared plus 10x over x to the fourth. And now I need to decide is F prime of X greater than zero for x bigger than one, if so my function, and therefore my sequence will, will be increasing. To check if f prime of x is greater than zero, I'll first set f prime of x equal to zero. So I'll set my ratio here equal to zero, which means my numerator needs to be equal to zero. And if I factor that, I get that x equals zero, or x equals 10. a sequence is decreasing if each term is greater than the next one. So a sub n is greater than a sub n plus one for all n. And a sequence is not increasing, if a sub n is greater than or equal to a sub n plus one for all n. Again, not increasing is like decreasing, but we allow for equality between consecutive terms. If we were to graph and on the x axis and a sub n on the y axis, then increasing looks like this just like an increasing function, whereas decreasing would go down non decreasing would go up possibly with some flat patches and mnandi increasing would go down possibly with some flat patches, notice that increasing is a stronger condition than non decreasing since being strictly less than something is stronger than just being less than or equal to it. For that reason. If a sequence is increasing, it is also non decreasing. And similarly, if a sequence is decreasing, it is also not increasing. A sequence is called monotonic, if it is either non decreasing or non increasing. Please pause the video and try to decide which of the following sequences are monotonic. The first two sequences are monotonic. The first one is monotonically. Non increasing, since we never increase when we go from one term to the next. In fact, we could also say that it's monotonically decreasing. Since we always go strictly down as we go from one term to the next, we never have equality between terms. The second sequence is monotonically non decreasing. Since we never go down, as we go from one term to the next, we either go up or stay at the same level. In this case, however, we could not say that the sequence is monotonically increasing because of the equality between some pairs of consecutive terms. The third sequence is not monotonic, the numbers bounce around between positive and negative numbers, and therefore sometimes we are decreasing while other times we're increasing. And the fourth sequence is not monotonic, because of the first few terms. However, from the fifth term on the terms are monotonically, non decreasing. And we could also say monotonically increasing. In this video, we give definitions for bounded and monotonic sequences. We'll see later that these definitions can be important for determining when a sequence converges. In class, we talked briefly about the sequence n minus five over n squared, where n goes from one to infinity. We want to know if the sequence is monotonic. And if it's bounded. If we compute the first few terms, the sequence appears to be steadily increasing. But in this case, appearances are deceiving. And a better way to decide whether it's monotonic is to use calculus to decide if the associated function f of x equals x minus five over x squared, it is increasing for x values bigger than one. So let me take the derivative. Using the quotient rule and simplify, I get that f prime of x is equal to minus x squared plus 10x over x to the fourth. And now I need to decide is F prime of X greater than zero for x bigger than one, if so my function, and therefore my sequence will, will be increasing. To check if f prime of x is greater than zero, I'll first set f prime of x equal to zero. So I'll set my ratio here equal to zero, which means my numerator needs to be equal to zero. And if I factor that, I get that x equals zero, or x equals 10. Now if I draw my number line, since f prime is equal to zero at zero and 10, it'll be positive and negative in between these values. And by plugging in values, like x equals negative one, one and 11, I can see that f prime is negative for x less than zero, positive for x between zero and 10, and negative for x bigger than 10. In particular, f increases when x increases from one to 10. And then it decreases. And so the same thing is happening to our sequence. Therefore, the sequence is not monotonic. We can also use calculus to check if the sequence is bounded. Our first derivative test shows that our function f of x has a maximum at x equals 10. At least that's the maximum for x values, ranging from one to infinity. And that's all that's relevant for our sequence. Therefore, our sequence is bounded above by its value at 10, which is 10 minus five over 10 squared, or 1/20. Now notice that our sequence and minus five over n squared is always bigger than zero, for n bigger than five since the numerator and denominator are both positive in this situation. And since there are only finitely, many terms where n is less than five, we can just use the minimum of these terms and zero as a lower bound. The smallest of the first four terms is negative four, which is less than zero, so that negative four forms a lower bound. So the sequence is in fact bounded. And somewhat surprisingly, the calculus ideas of derivative and maximum and minimum. In the past, we've encountered limits, like the limit as x goes to two of x minus two over x squared minus four. We can't evaluate this limit just by plugging In two for x, because x minus two goes to zero, and x squared minus four goes to zero as x goes to two. This is known as a zero over zero indeterminate form. It's called indeterminate because we can't tell what the limits going to be just by the fact that the numerator goes to zero and the denominator goes to zero. It depends on how fast the numerator and the denominator are going to zero compared to each other. And the final limit of the quotient could be any number at all, or it could be infinity, or it could not even exist. In the past, we've been able to evaluate some limits in zero over zero indeterminate form by using algebraic tricks to rewrite the quotients. In this video, we'll introduce lopatok rule, which is a very powerful technique for evaluating limits and indeterminate forms. A limit of the form the limit as x goes to a of f of x over g of x is called a zero over zero indeterminate form. If the limit as x goes to a of f of x is equal to zero, and the limit as x goes to a of g of x is equal to zero. A limit in this form is called an infinity over infinity indeterminate form. If the limit as x goes to a of f of x is equal to infinity or minus infinity. And the limit as x goes to a pub g of x is equal to infinity or minus infinity. We saw an example of a zero over zero indeterminate form in the introductory slide. One example of a infinity over infinity indeterminate form is the limit as x goes to infinity of 3x squared minus 2x plus seven divided by negative 2x squared plus 16. Notice that as x goes to infinity, the numerator goes to infinity while the denominator goes to negative infinity. In these definitions of indeterminate form, it's possible for a to be a negative infinity or infinity, like it is in this example, but it doesn't have to be loopy. talls rule can be applied when f and g are differentiable functions. And the derivative of g is nonzero in some open interval around a except possibly in a Now if I draw my number line, since f prime is equal to zero at zero and 10, it'll be positive and negative in between these values. And by plugging in values, like x equals negative one, one and 11, I can see that f prime is negative for x less than zero, positive for x between zero and 10, and negative for x bigger than 10. In particular, f increases when x increases from one to 10. And then it decreases. And so the same thing is happening to our sequence. Therefore, the sequence is not monotonic. We can also use calculus to check if the sequence is bounded. Our first derivative test shows that our function f of x has a maximum at x equals 10. At least that's the maximum for x values, ranging from one to infinity. And that's all that's relevant for our sequence. Therefore, our sequence is bounded above by its value at 10, which is 10 minus five over 10 squared, or 1/20. Now notice that our sequence and minus five over n squared is always bigger than zero, for n bigger than five since the numerator and denominator are both positive in this situation. And since there are only finitely, many terms where n is less than five, we can just use the minimum of these terms and zero as a lower bound. The smallest of the first four terms is negative four, which is less than zero, so that negative four forms a lower bound. So the sequence is in fact bounded. And somewhat surprisingly, the calculus ideas of derivative and maximum and minimum. In the past, we've encountered limits, like the limit as x goes to two of x minus two over x squared minus four. We can't evaluate this limit just by plugging In two for x, because x minus two goes to zero, and x squared minus four goes to zero as x goes to two. This is known as a zero over zero indeterminate form. It's called indeterminate because we can't tell what the limits going to be just by the fact that the numerator goes to zero and the denominator goes to zero. It depends on how fast the numerator and the denominator are going to zero compared to each other. And the final limit of the quotient could be any number at all, or it could be infinity, or it could not even exist. In the past, we've been able to evaluate some limits in zero over zero indeterminate form by using algebraic tricks to rewrite the quotients. In this video, we'll introduce lopatok rule, which is a very powerful technique for evaluating limits and indeterminate forms. A limit of the form the limit as x goes to a of f of x over g of x is called a zero over zero indeterminate form. If the limit as x goes to a of f of x is equal to zero, and the limit as x goes to a of g of x is equal to zero. A limit in this form is called an infinity over infinity indeterminate form. If the limit as x goes to a of f of x is equal to infinity or minus infinity. And the limit as x goes to a pub g of x is equal to infinity or minus infinity. We saw an example of a zero over zero indeterminate form in the introductory slide. One example of a infinity over infinity indeterminate form is the limit as x goes to infinity of 3x squared minus 2x plus seven divided by negative 2x squared plus 16. Notice that as x goes to infinity, the numerator goes to infinity while the denominator goes to negative infinity. In these definitions of indeterminate form, it's possible for a to be a negative infinity or infinity, like it is in this example, but it doesn't have to be loopy. talls rule can be applied when f and g are differentiable functions. And the derivative of g is nonzero in some open interval around a except possibly in a under these conditions, if the limit as x goes to a of f of x over g of x is zero over zero or infinity over infinity indeterminate form, then the limit as x goes to a of f of x over g of x is the same thing as the limit as x goes to a of f prime of x over g prime of x, provided that the second limit exists, or as plus or minus infinity. Let's look at loopy towers rule in action. In this example, as x goes to infinity, the numerator x goes to infinity and the denominator three to the x also goes to infinity. So we have an infinity over infinity indeterminate form. So let's try applying lopi tos rule, our original limit should equal the limit as x goes to infinity of the derivative of the numerator, which is one divided by the derivative of the denominator, which is ln of three times three to the x, provided that the second limit exists or as infinity or negative infinity. In the second limit, the numerators just fixed at one. And the nominator goes to infinity as x goes to infinity. Therefore, the second limit is just zero. And so the original limit evaluates to zero as well. In this example, we have a zero over zero indeterminate form, because as x goes to zero, sine of x and x, both go to zero in the numerator, and sine of x cubed goes to zero in the denominator. So using loopy tells rule, I'll try to evaluate instead, the limit I get by taking the derivative of the numerator and the derivative of the denominator. The derivative of sine x minus x is cosine of x minus one, and the derivative of sine x cubed is three times sine x squared times cosine x using the chain rule. Now let me try to evaluate the limit again. As x goes to zero, cosine of x goes to one, so the numerator here goes to zero. As x goes to zero, sine of x goes to zero and cosine of x goes to one, so the denominator also goes to zero. So I still have a zero over zero and determinant form. And I might as well try applying leptitox rule again. But before I do, I want to point out that cosine of x is going to one. So the cosine of x here really isn't affecting my limit. And in fact, I could rewrite my limit of a product as a product of limits, where the second limit is just one and can be ignored from here on out. Now apply lopatok rule on this first limit, which is a little bit easier to take the derivatives. And so the derivative of the top is minus sine x. And the derivative of the bottom is six times sine x times cosine x. Now let's try to evaluate again, as x goes to zero, our numerator is going to zero and our denominator is also going to zero. But hang on, we don't have to apply lobby towels rule again, because we can actually just simplify our expression, the sine x on the top cancels with the sine x on the bottom. And we can just rewrite our limit as the limit of negative one over six cosine of x, which evaluates easily to negative one, six. In this example, I want to emphasize that it's a good idea to simplify after each application of lopi talls rule. If you don't simplify, like we did here, then you might be tempted to apply lopatok rule and additional time when you don't need to, which might make the problem more complicated. Instead of simpler to solve this video, we were able to evaluate zero over zero and infinity over infinity indeterminate forms by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x, provided that second limit exists. This trick is known as lopi tels rule. We've seen that lopatok rule can be used to evaluate limits of the form zero over zero, or infinity over infinity. In this video, we'll continue to use loopy towels rule to evaluate additional indeterminate forms, like zero times infinity, infinity to the 00 to the zero, and one to the infinity. under these conditions, if the limit as x goes to a of f of x over g of x is zero over zero or infinity over infinity indeterminate form, then the limit as x goes to a of f of x over g of x is the same thing as the limit as x goes to a of f prime of x over g prime of x, provided that the second limit exists, or as plus or minus infinity. Let's look at loopy towers rule in action. In this example, as x goes to infinity, the numerator x goes to infinity and the denominator three to the x also goes to infinity. So we have an infinity over infinity indeterminate form. So let's try applying lopi tos rule, our original limit should equal the limit as x goes to infinity of the derivative of the numerator, which is one divided by the derivative of the denominator, which is ln of three times three to the x, provided that the second limit exists or as infinity or negative infinity. In the second limit, the numerators just fixed at one. And the nominator goes to infinity as x goes to infinity. Therefore, the second limit is just zero. And so the original limit evaluates to zero as well. In this example, we have a zero over zero indeterminate form, because as x goes to zero, sine of x and x, both go to zero in the numerator, and sine of x cubed goes to zero in the denominator. So using loopy tells rule, I'll try to evaluate instead, the limit I get by taking the derivative of the numerator and the derivative of the denominator. The derivative of sine x minus x is cosine of x minus one, and the derivative of sine x cubed is three times sine x squared times cosine x using the chain rule. Now let me try to evaluate the limit again. As x goes to zero, cosine of x goes to one, so the numerator here goes to zero. As x goes to zero, sine of x goes to zero and cosine of x goes to one, so the denominator also goes to zero. So I still have a zero over zero and determinant form. And I might as well try applying leptitox rule again. But before I do, I want to point out that cosine of x is going to one. So the cosine of x here really isn't affecting my limit. And in fact, I could rewrite my limit of a product as a product of limits, where the second limit is just one and can be ignored from here on out. Now apply lopatok rule on this first limit, which is a little bit easier to take the derivatives. And so the derivative of the top is minus sine x. And the derivative of the bottom is six times sine x times cosine x. Now let's try to evaluate again, as x goes to zero, our numerator is going to zero and our denominator is also going to zero. But hang on, we don't have to apply lobby towels rule again, because we can actually just simplify our expression, the sine x on the top cancels with the sine x on the bottom. And we can just rewrite our limit as the limit of negative one over six cosine of x, which evaluates easily to negative one, six. In this example, I want to emphasize that it's a good idea to simplify after each application of lopi talls rule. If you don't simplify, like we did here, then you might be tempted to apply lopatok rule and additional time when you don't need to, which might make the problem more complicated. Instead of simpler to solve this video, we were able to evaluate zero over zero and infinity over infinity indeterminate forms by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x, provided that second limit exists. This trick is known as lopi tels rule. We've seen that lopatok rule can be used to evaluate limits of the form zero over zero, or infinity over infinity. In this video, we'll continue to use loopy towels rule to evaluate additional indeterminate forms, like zero times infinity, infinity to the 00 to the zero, and one to the infinity. In this example, we want to evaluate the limit of a product. Notice that as x goes to zero from the positive side, sine x goes to zero, and ln x goes to negative infinity. Remember the graph of y equals ln x. So this is actually a zero times infinity indeterminate form. Even though the second factor is going to negative infinity, we still call it a zero times infinity and indeterminate form, you can think of the Infinity here as standing for either positive or negative infinity. It's indeterminate. Because as x goes to zero, the sine x factor is pulling the product towards zero, while the ln x factor is pulling the product towards large negative numbers. And it's hard to predict what the limit of the product will actually be. But the great thing is, I can actually rewrite this product to look like an infinity over infinity and determinant form, or a zero over zero and determinant form. Instead of sine x times ln x, I can rewrite the limit as ln x divided by one over sine x. Now as x goes to zero, my numerator is going to negative infinity. And since sine x is going to zero through positive numbers, my denominator one over sine x is going to positive infinity. So I have an infinity over infinity indeterminate form. Now, I could instead choose to leave the sine x in the numerator, and instead, put a one over ln x in the denominator. If I do this, then as x goes to zero through positive numbers, sine x goes to zero. And since ln x goes to negative infinity, one over ln x goes to zero, and so I have a zero over zero indeterminate form. Sometimes it can be difficult to decide which of these two ways to rewrite a product as a quotient. One rule of thumb is to take the version that makes it easier to take the derivative of the numerator and denominator. Another trick is just to try one of the ways and if you get stuck, go back and try the other. I'm going to use the first method of rewriting it because I recognize that one over sine x can be written as cosecant of x and I know how to take the derivative of cosecant x using lopi towels rule On this infinity over infinity indeterminate form, I can rewrite my limit as the limit of what I get when I take the derivative of the numerator, that's one over x divided by the derivative of the denominator, that's negative cosecant x cotangent x. As always, I want to simplify my expression before going any further, I can rewrite my trig functions in the denominator in terms of sine and cosine. cosecant x is one over sine x. and cotangent x is cosine of x over sine of x. Now flipping and multiplying, I get the limit as x goes to zero plus of one over x times the sine squared of x over negative cosine of x. In other words, the limit of negative sine squared x over x cosine x, we know that cosine of x goes to one as x goes to zero. So I can rewrite this as the limit of negative sine squared x over x times the limit of something that goes to one. So I once again have a zero over zero and determinant form. And I can apply lopatok rule yet again, taking the derivative of the top, I get negative two sine x, cosine of x, and the derivative, the bottom is just one. Now I'm in a good position just to evaluate the limit by plugging in zero for x in the numerator, I get negative two times zero times one, the denominator is just one, so my final limit is zero. In this limit, we have a battle of forces. As x is going to infinity, one over x is going to zero. So one plus one over x is going to one, but the exponent x is going to infinity, it's hard to tell what's going to happen here. If we had one, to any finite number, that would be one. But anything slightly bigger than one, as we raise it to have bigger and bigger powers, we would expect to get infinity. In this example, we want to evaluate the limit of a product. Notice that as x goes to zero from the positive side, sine x goes to zero, and ln x goes to negative infinity. Remember the graph of y equals ln x. So this is actually a zero times infinity indeterminate form. Even though the second factor is going to negative infinity, we still call it a zero times infinity and indeterminate form, you can think of the Infinity here as standing for either positive or negative infinity. It's indeterminate. Because as x goes to zero, the sine x factor is pulling the product towards zero, while the ln x factor is pulling the product towards large negative numbers. And it's hard to predict what the limit of the product will actually be. But the great thing is, I can actually rewrite this product to look like an infinity over infinity and determinant form, or a zero over zero and determinant form. Instead of sine x times ln x, I can rewrite the limit as ln x divided by one over sine x. Now as x goes to zero, my numerator is going to negative infinity. And since sine x is going to zero through positive numbers, my denominator one over sine x is going to positive infinity. So I have an infinity over infinity indeterminate form. Now, I could instead choose to leave the sine x in the numerator, and instead, put a one over ln x in the denominator. If I do this, then as x goes to zero through positive numbers, sine x goes to zero. And since ln x goes to negative infinity, one over ln x goes to zero, and so I have a zero over zero indeterminate form. Sometimes it can be difficult to decide which of these two ways to rewrite a product as a quotient. One rule of thumb is to take the version that makes it easier to take the derivative of the numerator and denominator. Another trick is just to try one of the ways and if you get stuck, go back and try the other. I'm going to use the first method of rewriting it because I recognize that one over sine x can be written as cosecant of x and I know how to take the derivative of cosecant x using lopi towels rule On this infinity over infinity indeterminate form, I can rewrite my limit as the limit of what I get when I take the derivative of the numerator, that's one over x divided by the derivative of the denominator, that's negative cosecant x cotangent x. As always, I want to simplify my expression before going any further, I can rewrite my trig functions in the denominator in terms of sine and cosine. cosecant x is one over sine x. and cotangent x is cosine of x over sine of x. Now flipping and multiplying, I get the limit as x goes to zero plus of one over x times the sine squared of x over negative cosine of x. In other words, the limit of negative sine squared x over x cosine x, we know that cosine of x goes to one as x goes to zero. So I can rewrite this as the limit of negative sine squared x over x times the limit of something that goes to one. So I once again have a zero over zero and determinant form. And I can apply lopatok rule yet again, taking the derivative of the top, I get negative two sine x, cosine of x, and the derivative, the bottom is just one. Now I'm in a good position just to evaluate the limit by plugging in zero for x in the numerator, I get negative two times zero times one, the denominator is just one, so my final limit is zero. In this limit, we have a battle of forces. As x is going to infinity, one over x is going to zero. So one plus one over x is going to one, but the exponent x is going to infinity, it's hard to tell what's going to happen here. If we had one, to any finite number, that would be one. But anything slightly bigger than one, as we raise it to have bigger and bigger powers, we would expect to get infinity. So our limit has an independent permanent form, it's hard to tell whether the answer is going to be one infinity, or maybe something in between. Whenever I see an expression with a variable in the base, and a variable in the exponent, I'm tempted to use logarithms. If we set y equal to one plus one over x to the x, then if I take the natural log of both sides, I can use my log roles to rewrite that by multiplying by x in the front. Now, if I wanted to take the limit as x goes to infinity of ln y, that would be the limit of this product, x times ln one plus one over x. As x goes to infinity, the first factor x goes to infinity. One plus one over x goes to just one and ln one is going to zero. So we have infinity times zero indeterminate form, which we can try to rewrite as an infinity over infinity, or a zero over zero indeterminate form. Let's rewrite this as the limit of ln one plus one over x divided by one of our x. This is indeed a zero over zero indeterminate form. So we can use lobi tiles rule and take the derivative of the top and the bottom. The derivative of the top is one over one plus one over x times the derivative of the inside, which would be negative one over x squared. And the derivative on the bottom, the derivative one over x is negative one over x squared, we can actually cancel out these two factors, and rewrite our limit as the limit as x goes to infinity of one over one plus one over x, which is just equal to one since one over x is going to zero. So we found that the limit of ln y is equal to one, but we're really interested in finding the limit of y which we can think of as e to the ln y. Since ln y is going to one, e to the ln y must be going to e to the one. In other words, E. So we found that our original limit is equal to E. And in fact, you may recognize that this limit is one of the ways of defining E. So our limit has an independent permanent form, it's hard to tell whether the answer is going to be one infinity, or maybe something in between. Whenever I see an expression with a variable in the base, and a variable in the exponent, I'm tempted to use logarithms. If we set y equal to one plus one over x to the x, then if I take the natural log of both sides, I can use my log roles to rewrite that by multiplying by x in the front. Now, if I wanted to take the limit as x goes to infinity of ln y, that would be the limit of this product, x times ln one plus one over x. As x goes to infinity, the first factor x goes to infinity. One plus one over x goes to just one and ln one is going to zero. So we have infinity times zero indeterminate form, which we can try to rewrite as an infinity over infinity, or a zero over zero indeterminate form. Let's rewrite this as the limit of ln one plus one over x divided by one of our x. This is indeed a zero over zero indeterminate form. So we can use lobi tiles rule and take the derivative of the top and the bottom. The derivative of the top is one over one plus one over x times the derivative of the inside, which would be negative one over x squared. And the derivative on the bottom, the derivative one over x is negative one over x squared, we can actually cancel out these two factors, and rewrite our limit as the limit as x goes to infinity of one over one plus one over x, which is just equal to one since one over x is going to zero. So we found that the limit of ln y is equal to one, but we're really interested in finding the limit of y which we can think of as e to the ln y. Since ln y is going to one, e to the ln y must be going to e to the one. In other words, E. So we found that our original limit is equal to E. And in fact, you may recognize that this limit is one of the ways of defining E. In the previous example, we had a one to the infinity and determinant form. And we took logs and used log rules to write that as an infinity times zero indeterminate form. Well, the same thing can be done if we have an infinity to the zero indeterminate form, or a zero to the zero and determinant form. So one to the infinity, infinity to the zero, and zero to the zero are all indeterminate forms that can be handled using lobi TAs role. In this video, we saw that a zero times infinity indeterminate form could be converted to a zero over zero, or infinity over infinity indeterminate form by rewriting f of x times g of x as f of x divided by one over g of x, or as g of x divided by one over f of x. We also saw how to use lopi talls rule on these three sorts of indeterminate forms by taking the ln of y, where y is our f of x to the g of x that we want to take the limit of this video gives some tricks for deciding whether a sequence converges. We say that a sequence a sub n converges, if the limit as n goes to infinity of the terms, a sub n exists as a finite number. Otherwise, we say the sequence diverges. In other words, a sequence diverges if the limit is infinity, or negative infinity, or does not exist. More formally, we said the limit as n goes to infinity of a sub n equals L, if for any small number epsilon, there's a number capital N, such that when the index little n is bigger than or equal to capital N, a sub n is within distance epsilon of owl. Saying that a sedan is within distance epsilon of L is the same thing as saying that the absolute value of a sub n minus L is less than epsilon. Let me draw this as a picture. If we put in on the x axis and a sub n on the y axis, we can plot our terms a sub n like this, here, it looks like our a sub n are settling at a particular value, I'll draw the value on the y axis and call it L. We say that the limit of a sub n is equal to L. Because for any tiny number epsilon, here, I've tried to draw a distance epsilon above L and a distance epsilon below, we can trap or a sub ends within epsilon of L by requiring the index and to be big enough, for the epsilon I've chosen here, are a sub ends are trapped within epsilon of L, when little n is bigger than equal to three. So there is that big number n, which is here, like two or three, such that when little n is bigger than that the a sub n are always trapped within epsilon of L. And if I chosen a tinier epsilon, I would just have to go a little bit farther out to make sure that my sequence was trapped within epsilon of n about this is the epsilon definition of a limit. Formally, we say the limit as n goes to infinity of a sub n is infinity. If for any big number, omega, there's a number capital N, such that a sub n is bigger than capital N, for a little n bigger than or equal to capital N. In other words, no matter how big an Omega I originally pick, my terms a sub n are eventually true. Above omega. Let me draw a picture for this one too. My terms here again, are drawn in red. And now if I pick a certain height omega, eventually all my turns will be above omega. And if I pick a different, bigger value of omega, my terms will still eventually be bigger than omega, I might just need to go further out in my sequence. In the previous example, we had a one to the infinity and determinant form. And we took logs and used log rules to write that as an infinity times zero indeterminate form. Well, the same thing can be done if we have an infinity to the zero indeterminate form, or a zero to the zero and determinant form. So one to the infinity, infinity to the zero, and zero to the zero are all indeterminate forms that can be handled using lobi TAs role. In this video, we saw that a zero times infinity indeterminate form could be converted to a zero over zero, or infinity over infinity indeterminate form by rewriting f of x times g of x as f of x divided by one over g of x, or as g of x divided by one over f of x. We also saw how to use lopi talls rule on these three sorts of indeterminate forms by taking the ln of y, where y is our f of x to the g of x that we want to take the limit of this video gives some tricks for deciding whether a sequence converges. We say that a sequence a sub n converges, if the limit as n goes to infinity of the terms, a sub n exists as a finite number. Otherwise, we say the sequence diverges. In other words, a sequence diverges if the limit is infinity, or negative infinity, or does not exist. More formally, we said the limit as n goes to infinity of a sub n equals L, if for any small number epsilon, there's a number capital N, such that when the index little n is bigger than or equal to capital N, a sub n is within distance epsilon of owl. Saying that a sedan is within distance epsilon of L is the same thing as saying that the absolute value of a sub n minus L is less than epsilon. Let me draw this as a picture. If we put in on the x axis and a sub n on the y axis, we can plot our terms a sub n like this, here, it looks like our a sub n are settling at a particular value, I'll draw the value on the y axis and call it L. We say that the limit of a sub n is equal to L. Because for any tiny number epsilon, here, I've tried to draw a distance epsilon above L and a distance epsilon below, we can trap or a sub ends within epsilon of L by requiring the index and to be big enough, for the epsilon I've chosen here, are a sub ends are trapped within epsilon of L, when little n is bigger than equal to three. So there is that big number n, which is here, like two or three, such that when little n is bigger than that the a sub n are always trapped within epsilon of L. And if I chosen a tinier epsilon, I would just have to go a little bit farther out to make sure that my sequence was trapped within epsilon of n about this is the epsilon definition of a limit. Formally, we say the limit as n goes to infinity of a sub n is infinity. If for any big number, omega, there's a number capital N, such that a sub n is bigger than capital N, for a little n bigger than or equal to capital N. In other words, no matter how big an Omega I originally pick, my terms a sub n are eventually true. Above omega. Let me draw a picture for this one too. My terms here again, are drawn in red. And now if I pick a certain height omega, eventually all my turns will be above omega. And if I pick a different, bigger value of omega, my terms will still eventually be bigger than omega, I might just need to go further out in my sequence. For this first value of omega, I would just need to pick a capital N of about 123456. Once my little ends are all bigger than six, my Ace events are bigger than that omega, and for this bigger value of omega, I need to pick a value of capital N of about nine, once my little ends are bigger than about nine, all my as events would be bigger than this omega. There's a similar definition for the limit as n goes to infinity of a sub n equaling negative infinity. Now we just need to say that for any big negative number, omega, there's a number. And such that a sub n is less than omega for a little n bigger than or equal to capital N. Please take a moment and try to come up with an example of a sequence that converges, a sequence that diverges to infinity or negative infinity, and a sequence that's bounded but still diverges. One example of a convergent sequence is a sequence whenever n, this sequence converges to zero, since the limit as n goes to infinity of one over N is zero. A divergent sequence is two to the N. Since the limit as n goes to infinity of two to the n is infinity. One example of a sequence that's bounded, but still diverges is negative one to the n. The sequence alternates between negative one and one, depending on whether n is odd, or even. So it's bounded in between negative one and one. But it still diverges because the limit does not exist. Since the sequence doesn't settle at a single value. The rest of this video will give some techniques for proving that sequences converge. The first technique is to use standard calculus tricks for finding limits of functions. Even though a sequence is only defined on positive integers. Sometimes it's possible to find a function defined on all positive real numbers that agrees with our sequence on the integers. In other words, the terms as of n are equal to f of n for this function F. When this happens, then if the limit as x goes to infinity of f of x equals L, then the limit as n goes to infinity of a sub n also equals out, the red dots are converging to the same limit as the blue function. So a lot of times, we can figure out if a sequence converges by replacing the terms a sub n with F of X for some appropriate function, and then using lobaton ruler, other tricks from calculus one to show that the function limit exists. Let's try that for the following example. When the indices are missing, as in this example, we'll assume that n starts at one and goes to infinity. In order to prove that this sequence converges, let's instead look at the function f of x equals ln one plus two e to the x over x, where x is a real number. Now let's look at the limit as x goes to infinity of f of x. That's the limit as x goes to infinity of ln one plus two e to the x over x. And as x goes to infinity in the x goes to infinity, so one plus two times e to the x goes to infinity, which means ln of that goes to infinity. So the numerator is going to infinity as x goes to infinity. And so as the denominator, we have an infinity over an infinity indeterminate form. So we can apply lobby towels rule and take the derivative of the numerator over the derivative of the denominator, the derivative of the numerator one over one plus two times e to the x times two times e to the x using the chain rule, and the derivative of the denominator is just one. We can take derivatives here, because we're thinking of x as a real number, not just an integer. Simplifying, we still have an infinity over infinity indeterminate form. So let's take the derivatives again, For this first value of omega, I would just need to pick a capital N of about 123456. Once my little ends are all bigger than six, my Ace events are bigger than that omega, and for this bigger value of omega, I need to pick a value of capital N of about nine, once my little ends are bigger than about nine, all my as events would be bigger than this omega. There's a similar definition for the limit as n goes to infinity of a sub n equaling negative infinity. Now we just need to say that for any big negative number, omega, there's a number. And such that a sub n is less than omega for a little n bigger than or equal to capital N. Please take a moment and try to come up with an example of a sequence that converges, a sequence that diverges to infinity or negative infinity, and a sequence that's bounded but still diverges. One example of a convergent sequence is a sequence whenever n, this sequence converges to zero, since the limit as n goes to infinity of one over N is zero. A divergent sequence is two to the N. Since the limit as n goes to infinity of two to the n is infinity. One example of a sequence that's bounded, but still diverges is negative one to the n. The sequence alternates between negative one and one, depending on whether n is odd, or even. So it's bounded in between negative one and one. But it still diverges because the limit does not exist. Since the sequence doesn't settle at a single value. The rest of this video will give some techniques for proving that sequences converge. The first technique is to use standard calculus tricks for finding limits of functions. Even though a sequence is only defined on positive integers. Sometimes it's possible to find a function defined on all positive real numbers that agrees with our sequence on the integers. In other words, the terms as of n are equal to f of n for this function F. When this happens, then if the limit as x goes to infinity of f of x equals L, then the limit as n goes to infinity of a sub n also equals out, the red dots are converging to the same limit as the blue function. So a lot of times, we can figure out if a sequence converges by replacing the terms a sub n with F of X for some appropriate function, and then using lobaton ruler, other tricks from calculus one to show that the function limit exists. Let's try that for the following example. When the indices are missing, as in this example, we'll assume that n starts at one and goes to infinity. In order to prove that this sequence converges, let's instead look at the function f of x equals ln one plus two e to the x over x, where x is a real number. Now let's look at the limit as x goes to infinity of f of x. That's the limit as x goes to infinity of ln one plus two e to the x over x. And as x goes to infinity in the x goes to infinity, so one plus two times e to the x goes to infinity, which means ln of that goes to infinity. So the numerator is going to infinity as x goes to infinity. And so as the denominator, we have an infinity over an infinity indeterminate form. So we can apply lobby towels rule and take the derivative of the numerator over the derivative of the denominator, the derivative of the numerator one over one plus two times e to the x times two times e to the x using the chain rule, and the derivative of the denominator is just one. We can take derivatives here, because we're thinking of x as a real number, not just an integer. Simplifying, we still have an infinity over infinity indeterminate form. So let's take the derivatives again, the derivative of the numerator is now two times e to the x, and the derivative of the denominator is also two times e to the x. So their limit here is one. Since our function converges to one, our sequence also converges to one. So this original sequence converges to one. the derivative of the numerator is now two times e to the x, and the derivative of the denominator is also two times e to the x. So their limit here is one. Since our function converges to one, our sequence also converges to one. So this original sequence converges to one. Another technique for proving that sequences converge is to use the squeeze theorem and trap the sequence between two simpler sequences that converge to the same limit. In this example, since sine and cosine are both bounded in between one and negative one, we know that cosine of n plus sine of n can't be any bigger than two, certainly, or on a smaller than negative two. If I divide all sides of this inequality by n to the two thirds, we can see that the original sequence is bounded in between negative two over n to the two thirds and two over n to the two thirds. Notice that n to the two thirds has to be positive, since as usual, we're assuming that n starts as one and goes to infinity. And so dividing by n to the two thirds does not switch the inequality signs. Now it's easy to check that the limit as n goes to infinity of negative two over n to the two thirds is zero, since as n goes to infinity, and to the two thirds also goes to infinity. Similarly, the limit as n goes to infinity of two over n to the two thirds is zero. Since these two limits are the same, we know by the squeeze theorem, that the limit of our sequence in the middle has to exist at equals zero. Also, it's a remarkable fact that I won't prove here that if a sub n is bounded, and monotonic, then it has to converge. You might get some intuition for this fact, by looking at a graph. If the ace of ends are monotonically, increasing, for example, but are bounded, then there's no place for these events to go. And they can't oscillate up and down. Because they're monotonically increasing. So it makes intuitive sense that they have to settle on some limit. A fun example of this fact is a sequence that starts point 1.1 2.12 3.1234, and so on. Where we just keep stringing together the counting numbers as our decimal, the sequence is certainly monotonically increasing, and is bounded. Since every term of the sequence is greater than zero and less than, say, point five. So we have a bounded monotonic sequence. And so the sequence has to converge. Now, what it actually converges to is a little mysterious, since it doesn't converge to some number we're already familiar with like, like, point six, or pi over three or something like that. But it does converge to some real number. And that real number is called champion nouns constant. And it has some interesting properties. And of course, it has a decimal expansion that's easy to to come up with, since you get it just by stringing together the counting numbers. If you can recognize a sequence to be a geometric sequence, then it's pretty easy to decide whether it converges or diverges. Recall that a geometric sequence is a sequence of the form a times r to the n minus one where n runs from one to infinity. Or sometimes it's written as a times r to the n, where n runs from zero to infinity. Let's try to figure out for what values of our the sequence r to the n converges. It's not really important here whether and starts at zero or starts So one, since when we talk about convergence, we're talking about the behavior of the terms as n goes to infinity. So the first few terms really don't matter. if r is greater than one, then the sequence r to the n is an increasing sequence. In fact, for the sequence r to the n, if we replace n with x, and look at the function f of x equals r to the x, that's an exponential function. And if we're assuming r is greater than one, the base for our exponential functions is greater than one. So we know that the limit as x goes to infinity of our to the X is infinity, which means that our sequence r to the n also has to diverge to infinity. Another technique for proving that sequences converge is to use the squeeze theorem and trap the sequence between two simpler sequences that converge to the same limit. In this example, since sine and cosine are both bounded in between one and negative one, we know that cosine of n plus sine of n can't be any bigger than two, certainly, or on a smaller than negative two. If I divide all sides of this inequality by n to the two thirds, we can see that the original sequence is bounded in between negative two over n to the two thirds and two over n to the two thirds. Notice that n to the two thirds has to be positive, since as usual, we're assuming that n starts as one and goes to infinity. And so dividing by n to the two thirds does not switch the inequality signs. Now it's easy to check that the limit as n goes to infinity of negative two over n to the two thirds is zero, since as n goes to infinity, and to the two thirds also goes to infinity. Similarly, the limit as n goes to infinity of two over n to the two thirds is zero. Since these two limits are the same, we know by the squeeze theorem, that the limit of our sequence in the middle has to exist at equals zero. Also, it's a remarkable fact that I won't prove here that if a sub n is bounded, and monotonic, then it has to converge. You might get some intuition for this fact, by looking at a graph. If the ace of ends are monotonically, increasing, for example, but are bounded, then there's no place for these events to go. And they can't oscillate up and down. Because they're monotonically increasing. So it makes intuitive sense that they have to settle on some limit. A fun example of this fact is a sequence that starts point 1.1 2.12 3.1234, and so on. Where we just keep stringing together the counting numbers as our decimal, the sequence is certainly monotonically increasing, and is bounded. Since every term of the sequence is greater than zero and less than, say, point five. So we have a bounded monotonic sequence. And so the sequence has to converge. Now, what it actually converges to is a little mysterious, since it doesn't converge to some number we're already familiar with like, like, point six, or pi over three or something like that. But it does converge to some real number. And that real number is called champion nouns constant. And it has some interesting properties. And of course, it has a decimal expansion that's easy to to come up with, since you get it just by stringing together the counting numbers. If you can recognize a sequence to be a geometric sequence, then it's pretty easy to decide whether it converges or diverges. Recall that a geometric sequence is a sequence of the form a times r to the n minus one where n runs from one to infinity. Or sometimes it's written as a times r to the n, where n runs from zero to infinity. Let's try to figure out for what values of our the sequence r to the n converges. It's not really important here whether and starts at zero or starts So one, since when we talk about convergence, we're talking about the behavior of the terms as n goes to infinity. So the first few terms really don't matter. if r is greater than one, then the sequence r to the n is an increasing sequence. In fact, for the sequence r to the n, if we replace n with x, and look at the function f of x equals r to the x, that's an exponential function. And if we're assuming r is greater than one, the base for our exponential functions is greater than one. So we know that the limit as x goes to infinity of our to the X is infinity, which means that our sequence r to the n also has to diverge to infinity. If instead, r is equal to one, then r to the n is just one to the n, which is just a sequence of ones, so that converges to one. if r is between zero and one, then r to the n, is decreasing. This time, it's like the exponential function f of x equals r to the x with the base between zero and one. And so the limit of that exponential function, as x goes to infinity is going to be equals zero. Therefore, the sequence also converges to zero. Of course, if r equals exactly zero, and the sequence is just a sequence of zeros, so it also converges to zero. Next, let's look at the case when r is between negative one and zero. Now, the sequence of rd ends are going to oscillate between positive and negative numbers that get smaller and smaller and magnitude as n goes to infinity. That's because we get from one number to the next by multiplying by r, which is a negative number of magnitude less than one. So our limit as n goes to infinity of our A to the ends, is going to be zero again. Another way of thinking about this case, is by thinking about the squeeze theorem, since r to the n is always less than or equal to the absolute value of r to the n, which is o n is always greater than or equal to negative the absolute value of r to the n. Technically, r to the n is always exactly equal to the absolute value are the n when n is even, so that art of the N is positive. And it's always exactly equal to the negative of the absolute value of r, the N, when n is odd, so that are the n is, is negative. But in any case, the inequality still does technically hold. And so therefore, since by the squeeze theorem, the limit of the left terms is zero, and the limit of the right terms is zero. As we noted before, the limit of the middle term also has to be zero. So our sequence converges to zero. Now, if r is equal to negative one, then our sequence r to the n is just negative one to the n, which alternates between negative one and one. And so that sequence diverges. Finally, if r is less than negative one, then to get from one term to the next, we're multiplying by negative number that has magnitude bigger than one. And so our terms are going to oscillate between positive and negative numbers, but they're going to be going up in magnitude. And so the limit as n goes to infinity of a sub n, does not exist. And we see that our sequence diverges. If we look through all these cases, we see that the sequence r to the n converges to zero, when r is between negative one and one. it converges to one when r is exactly equal to one, and it diverges when r is bigger than one or less than negative one. All right, then Summary below. In fact, almost the same thing as true, when we look at the sequence a times r to the n, where A is any real number, the sequence a times r to the n converges to zero, when r is between negative one and one. it converges to a, when r is equal to one. And it diverges when r is less than negative one, or greater than one. This follows because multiplying all the terms in the sequence by a just multiplies the limit by a and zero times as zero, well, one times i is a. So anytime you encounter a geometric sequence, that is a sequence that can be written in the form of a times r to the n, If instead, r is equal to one, then r to the n is just one to the n, which is just a sequence of ones, so that converges to one. if r is between zero and one, then r to the n, is decreasing. This time, it's like the exponential function f of x equals r to the x with the base between zero and one. And so the limit of that exponential function, as x goes to infinity is going to be equals zero. Therefore, the sequence also converges to zero. Of course, if r equals exactly zero, and the sequence is just a sequence of zeros, so it also converges to zero. Next, let's look at the case when r is between negative one and zero. Now, the sequence of rd ends are going to oscillate between positive and negative numbers that get smaller and smaller and magnitude as n goes to infinity. That's because we get from one number to the next by multiplying by r, which is a negative number of magnitude less than one. So our limit as n goes to infinity of our A to the ends, is going to be zero again. Another way of thinking about this case, is by thinking about the squeeze theorem, since r to the n is always less than or equal to the absolute value of r to the n, which is o n is always greater than or equal to negative the absolute value of r to the n. Technically, r to the n is always exactly equal to the absolute value are the n when n is even, so that art of the N is positive. And it's always exactly equal to the negative of the absolute value of r, the N, when n is odd, so that are the n is, is negative. But in any case, the inequality still does technically hold. And so therefore, since by the squeeze theorem, the limit of the left terms is zero, and the limit of the right terms is zero. As we noted before, the limit of the middle term also has to be zero. So our sequence converges to zero. Now, if r is equal to negative one, then our sequence r to the n is just negative one to the n, which alternates between negative one and one. And so that sequence diverges. Finally, if r is less than negative one, then to get from one term to the next, we're multiplying by negative number that has magnitude bigger than one. And so our terms are going to oscillate between positive and negative numbers, but they're going to be going up in magnitude. And so the limit as n goes to infinity of a sub n, does not exist. And we see that our sequence diverges. If we look through all these cases, we see that the sequence r to the n converges to zero, when r is between negative one and one. it converges to one when r is exactly equal to one, and it diverges when r is bigger than one or less than negative one. All right, then Summary below. In fact, almost the same thing as true, when we look at the sequence a times r to the n, where A is any real number, the sequence a times r to the n converges to zero, when r is between negative one and one. it converges to a, when r is equal to one. And it diverges when r is less than negative one, or greater than one. This follows because multiplying all the terms in the sequence by a just multiplies the limit by a and zero times as zero, well, one times i is a. So anytime you encounter a geometric sequence, that is a sequence that can be written in the form of a times r to the n, you can know that it converges if r is bigger than negative one and less than or equal to one. This sequence here, although it looks really complicated, it's really a geometric sequence in disguise. One way to see this is by simplifying the form of the terms, so this is negative one to the t, e to the t times e to the minus one over three to the T times three squared. This is the same thing as negative e over three to the T times one over three squared times E. Now this is looking like the tail end of a geometric sequence, where A is one over three squared times E, and R, the common ratio is minus e over three, say the tail end, because we have T starting at three instead of zero. Now since he is less than three, the magnitude of our is got to be smaller than one. In other words, our as a negative number, that's between negative one and zero, and therefore, the tail end of this geometric sequence converges. It's kind of interesting to note that we could also rewrite this geometric sequence if we wanted to, using an index and going from zero to infinity. And one way to figure out how to do that the are the common ratio stays the same as negative e over three. But since this version starts at t equals three, the first term here is really minus e over three cubed times one over three squared E. And that becomes our value of A. Notice that when n is zero here, I get this value. And when t is three, here, I get the same value for this sequence. So these sequences are equivalent. But in any case, for either a sequence, the common ratio, R is negative B over three, and a sequence converges. Therefore, the final trick that I want to mention for deciding whether sequences converge or diverge is limit laws, the usual limit laws about addition, subtraction, and so on hold for sequences as well for functions. So for example, if the limit as n goes to infinity of a sub n is out, and the limit of b sub n is am than the limit of the sum, a sub n plus b sub n is going to be equal to L plus m. And the limit of a sub n times b sub n is L times M. And the limit of C times a sub n where C is some constant is going to be c times though similar roles holds for subtraction and division. I want to emphasize that these limit holds hold under the condition that the limits of the component sequences exist as finite numbers. I can use these limit laws to decide if this sequence converges, since the limit of the terms is equal to the difference of the limits, provided those limits exist. Now the first limit is zero, since the degree of the numerator is less than the degree of the denominator here, and the second limit is also zero. Since this is a geometric sequence, with ratio of four fifths, and four fifths is between zero and one. Therefore, the limit of our original sequence must be zero. In this video, we saw several ways to prove that a sequence converges. We saw that we could use calculus techniques like lopi taas rule. After replacing the sequence with its associated function defined on real numbers, we also saw that we could use the squeeze theorem. we noted that all sequences that are bounded and monotonic must converge. And we saw that geometric sequences always converge if r is bigger than negative one and less than or equal to one. Finally, we saw that we can use limit laws to handle sums and products and other conglomerations of sequences. This video is about geometric series, and when they converge, and when they diverged. you can know that it converges if r is bigger than negative one and less than or equal to one. This sequence here, although it looks really complicated, it's really a geometric sequence in disguise. One way to see this is by simplifying the form of the terms, so this is negative one to the t, e to the t times e to the minus one over three to the T times three squared. This is the same thing as negative e over three to the T times one over three squared times E. Now this is looking like the tail end of a geometric sequence, where A is one over three squared times E, and R, the common ratio is minus e over three, say the tail end, because we have T starting at three instead of zero. Now since he is less than three, the magnitude of our is got to be smaller than one. In other words, our as a negative number, that's between negative one and zero, and therefore, the tail end of this geometric sequence converges. It's kind of interesting to note that we could also rewrite this geometric sequence if we wanted to, using an index and going from zero to infinity. And one way to figure out how to do that the are the common ratio stays the same as negative e over three. But since this version starts at t equals three, the first term here is really minus e over three cubed times one over three squared E. And that becomes our value of A. Notice that when n is zero here, I get this value. And when t is three, here, I get the same value for this sequence. So these sequences are equivalent. But in any case, for either a sequence, the common ratio, R is negative B over three, and a sequence converges. Therefore, the final trick that I want to mention for deciding whether sequences converge or diverge is limit laws, the usual limit laws about addition, subtraction, and so on hold for sequences as well for functions. So for example, if the limit as n goes to infinity of a sub n is out, and the limit of b sub n is am than the limit of the sum, a sub n plus b sub n is going to be equal to L plus m. And the limit of a sub n times b sub n is L times M. And the limit of C times a sub n where C is some constant is going to be c times though similar roles holds for subtraction and division. I want to emphasize that these limit holds hold under the condition that the limits of the component sequences exist as finite numbers. I can use these limit laws to decide if this sequence converges, since the limit of the terms is equal to the difference of the limits, provided those limits exist. Now the first limit is zero, since the degree of the numerator is less than the degree of the denominator here, and the second limit is also zero. Since this is a geometric sequence, with ratio of four fifths, and four fifths is between zero and one. Therefore, the limit of our original sequence must be zero. In this video, we saw several ways to prove that a sequence converges. We saw that we could use calculus techniques like lopi taas rule. After replacing the sequence with its associated function defined on real numbers, we also saw that we could use the squeeze theorem. we noted that all sequences that are bounded and monotonic must converge. And we saw that geometric sequences always converge if r is bigger than negative one and less than or equal to one. Finally, we saw that we can use limit laws to handle sums and products and other conglomerations of sequences. This video is about geometric series, and when they converge, and when they diverged. A geometric sequence is a sequence of the form a a times r, a times r squared, a times r cubed, and so on. For some numbers a and r. this can be written in the wrapped up notation, a times r to the k, where k goes from zero to infinity. For example, if a is three, and r is one half, the sequence would be three, three halves, three fourths, three eighths, and so on, which could be written as three times one half to the k, where k ranges from zero to infinity. A geometric series is the series you get by adding all these numbers up. So that would be a plus a times r plus a times r squared, and so on, which can be written in summation notation is the sum from K equals zero to infinity of A times r to the K. For our example, our series would be the following, which could be written as the sum from K equals zero to infinity of three times one half to the K. Sometimes you might get a geometric series that's in disguise, like this one. If I write out the first few terms, notice that we're told to start with i equals two. So plugging in i equals two here, I get negative one, squared over three to the two times two minus three, that's one over three. If I plug in i equals three, I get negative one cubed, that's over three to the two times three minus three, or negative one over three cubed, negative 127. When I go four, I get one over 243. If I look at the ratio of consecutive terms, negative 127 divided by 1/3, is negative one night. Similarly, the ratio here is also negative one nights, suggesting that we might have a geometric series with ratio R of negative 1/9. And first term coming from here, of 1/3. But there's another way to see this that's less arithmetic intensive, and makes use of exponent rules. If we look at the basic term, and rewrite it using exponent rules, this is three to the two i times three to the minus three, I can rewrite the three to the two is three squared to the I and the three to the minus three and the denominator becomes a three cubed and the numerator. Now I can group together all the pieces that are raised to the nth power. And write this as negative 1/9 to the i times three cubed, or 27 times negative one night to the I reading like this, I can see that every time I increases by one, I get another factor of negative one nine multiplied in my expression. And therefore, the this is a geometric series with ratio negative one nine, as I saw before, now, you might think that the first term should be 27. But remember that it doesn't start at zero, it starts at two. So the first term is going to be what I get I get when I plug in the first value of i. So that's 27 times negative one nine squared, which works out to 1/3. as before. So we do have a geometric series with first term of 1/3, a common ratio of negative 1/9. In fact, we could rewrite it in a more standard form as the sum of 1/3 times negative 1/9 to the k, where k ranges from zero to infinity. Since k equals zero, and this expression corresponds to i equals two in this expression, k is equal to, can be thought of as equal to i minus two. And this is really a reindexing tugann simplification to get from this version to this version. A geometric sequence is a sequence of the form a a times r, a times r squared, a times r cubed, and so on. For some numbers a and r. this can be written in the wrapped up notation, a times r to the k, where k goes from zero to infinity. For example, if a is three, and r is one half, the sequence would be three, three halves, three fourths, three eighths, and so on, which could be written as three times one half to the k, where k ranges from zero to infinity. A geometric series is the series you get by adding all these numbers up. So that would be a plus a times r plus a times r squared, and so on, which can be written in summation notation is the sum from K equals zero to infinity of A times r to the K. For our example, our series would be the following, which could be written as the sum from K equals zero to infinity of three times one half to the K. Sometimes you might get a geometric series that's in disguise, like this one. If I write out the first few terms, notice that we're told to start with i equals two. So plugging in i equals two here, I get negative one, squared over three to the two times two minus three, that's one over three. If I plug in i equals three, I get negative one cubed, that's over three to the two times three minus three, or negative one over three cubed, negative 127. When I go four, I get one over 243. If I look at the ratio of consecutive terms, negative 127 divided by 1/3, is negative one night. Similarly, the ratio here is also negative one nights, suggesting that we might have a geometric series with ratio R of negative 1/9. And first term coming from here, of 1/3. But there's another way to see this that's less arithmetic intensive, and makes use of exponent rules. If we look at the basic term, and rewrite it using exponent rules, this is three to the two i times three to the minus three, I can rewrite the three to the two is three squared to the I and the three to the minus three and the denominator becomes a three cubed and the numerator. Now I can group together all the pieces that are raised to the nth power. And write this as negative 1/9 to the i times three cubed, or 27 times negative one night to the I reading like this, I can see that every time I increases by one, I get another factor of negative one nine multiplied in my expression. And therefore, the this is a geometric series with ratio negative one nine, as I saw before, now, you might think that the first term should be 27. But remember that it doesn't start at zero, it starts at two. So the first term is going to be what I get I get when I plug in the first value of i. So that's 27 times negative one nine squared, which works out to 1/3. as before. So we do have a geometric series with first term of 1/3, a common ratio of negative 1/9. In fact, we could rewrite it in a more standard form as the sum of 1/3 times negative 1/9 to the k, where k ranges from zero to infinity. Since k equals zero, and this expression corresponds to i equals two in this expression, k is equal to, can be thought of as equal to i minus two. And this is really a reindexing tugann simplification to get from this version to this version. As you may know, a geometric sequence converges to zero, when the common ratio R is between negative one and one, it converges to a, when r is equal to one, and it diverges. Otherwise, when r is less than or equal to negative one, or when r is greater than one. We're assuming here, that A is not equal zero, since otherwise, we'd have a very boring though convergent sequence of all zeros. I want to restate this convergence criteria in limit form, since we'll use it later. But it's saying is that when r is between negative one and one, the limit, as K goes to infinity of A times r to the K is zero, that's what it means to converge to zero. when r is equal to one, that limit is equal to A. And when r is less than or equal to negative one or greater than one, that limit does not exist as a finite number. Now I'd like to find similar rules in terms of r for when a geometric series converges or diverges. Again, we'll assume that a is not equal to zero. Since otherwise, I just have a sum of a bunch of zeros, a boring series that would converge to zero. Our strategy is going to be to find a formula for the nth partial sum of the series, and then take the limit of partial sums. Since by definition, that limit tells us whether the series converges or diverges. Before we carry out that strategy, I want to consider one special case, when r is equal to one, then the series is just a plus a plus a, and so that diverges to infinity if A is positive, or to negative infinity, if a is negative, remember, we're assuming that A is not zero. so far is one our series diverges. And from now on, we'll assume that R is not equal to one. Let's look at a few partial sums. The first partial sum S sub one, just the first term A, S sub two is a plus a times R, and so on. In general, the nth partial sum s sub n is a plus a times r plus a times r squared. And I continue until let's see, the last term will be a times r to the n minus one, the second last term will be a times r to the n minus two. Notice that the nth partial sum only goes up to A times r to the n minus one. Since we're starting at a which is a 10 to the zero. I'd like to write the partial sum in a nicer form, so I don't have to write all those dots. And to do that, I'm going to use a trick, I'm going to multiply both sides of this equation by R. So on the left, I get r times s sub n. And on the right, I'm going to multiply each term by R. So the first term becomes a times R, the second term A times r squared, a times r cubed, and so on, the second to last term becomes a times r to the n minus one, and the last term becomes a times r to the n. Notice that this equation, and the equation under it have a lot of terms in common, they match up along the diagonals. So if I subtract the second equation from the first, on the left side, I'm going to get s sub n minus r times s sub n. But on the right side, a lot of my terms will cancel, this term will cancel the next one, this one cancels with the previous one. And what I'm left with is just a minus a times r to the n, I get a minus sign here, because I'm subtracting the whole equation. Now I can solve for a sub n, I can factor it out. I'll factor out the A also just to keep things tidy. And then I get that s sub n is a times one minus r to the n over one minus R. I don't have to worry about dividing by one minus r that can't be zero, because remember, I'm assuming that R is not one. Now that I have a nice tidy formula for s sub n, I can proceed to take the limit as n goes to infinity and see if my sequence of partial sums converges or diverges sets the limit of my formula. And notice that the only part of this formula that depends on n is the art of the N, a one one minus are all constants as far as n is concerned. So we can use limit rurals. To take those out of the limit, I can rewrite the limit as a over one minus r times the limit of one minus r to the n or even better as a over one minus r times one minus the limit of r to the n. Now this limit I've seen before, right on this page, is the same as the limit I was considering when I was looking at convergence of sequences, where I use the same R, and my A is equal to one. So we know that this limit goes to zero, when r is between negative one and one, and does not exist as a finite number. when r is less than or equal to negative one, or r is greater than one. Therefore, the limit of the partial sums is going to equal a over one minus r times one minus zero, which is just a over one minus r, when r is between negative one and one, and that limit, will not exist as a finite number. when r is less than or equal to negative one, or r is greater than one. Since the limit also doesn't exist as a finite number, when r is one, I can add a little equality sign. And now I've got all this cases for our covered me write this down as a conclusion. The geometric series converges to a over one minus R. As you may know, a geometric sequence converges to zero, when the common ratio R is between negative one and one, it converges to a, when r is equal to one, and it diverges. Otherwise, when r is less than or equal to negative one, or when r is greater than one. We're assuming here, that A is not equal zero, since otherwise, we'd have a very boring though convergent sequence of all zeros. I want to restate this convergence criteria in limit form, since we'll use it later. But it's saying is that when r is between negative one and one, the limit, as K goes to infinity of A times r to the K is zero, that's what it means to converge to zero. when r is equal to one, that limit is equal to A. And when r is less than or equal to negative one or greater than one, that limit does not exist as a finite number. Now I'd like to find similar rules in terms of r for when a geometric series converges or diverges. Again, we'll assume that a is not equal to zero. Since otherwise, I just have a sum of a bunch of zeros, a boring series that would converge to zero. Our strategy is going to be to find a formula for the nth partial sum of the series, and then take the limit of partial sums. Since by definition, that limit tells us whether the series converges or diverges. Before we carry out that strategy, I want to consider one special case, when r is equal to one, then the series is just a plus a plus a, and so that diverges to infinity if A is positive, or to negative infinity, if a is negative, remember, we're assuming that A is not zero. so far is one our series diverges. And from now on, we'll assume that R is not equal to one. Let's look at a few partial sums. The first partial sum S sub one, just the first term A, S sub two is a plus a times R, and so on. In general, the nth partial sum s sub n is a plus a times r plus a times r squared. And I continue until let's see, the last term will be a times r to the n minus one, the second last term will be a times r to the n minus two. Notice that the nth partial sum only goes up to A times r to the n minus one. Since we're starting at a which is a 10 to the zero. I'd like to write the partial sum in a nicer form, so I don't have to write all those dots. And to do that, I'm going to use a trick, I'm going to multiply both sides of this equation by R. So on the left, I get r times s sub n. And on the right, I'm going to multiply each term by R. So the first term becomes a times R, the second term A times r squared, a times r cubed, and so on, the second to last term becomes a times r to the n minus one, and the last term becomes a times r to the n. Notice that this equation, and the equation under it have a lot of terms in common, they match up along the diagonals. So if I subtract the second equation from the first, on the left side, I'm going to get s sub n minus r times s sub n. But on the right side, a lot of my terms will cancel, this term will cancel the next one, this one cancels with the previous one. And what I'm left with is just a minus a times r to the n, I get a minus sign here, because I'm subtracting the whole equation. Now I can solve for a sub n, I can factor it out. I'll factor out the A also just to keep things tidy. And then I get that s sub n is a times one minus r to the n over one minus R. I don't have to worry about dividing by one minus r that can't be zero, because remember, I'm assuming that R is not one. Now that I have a nice tidy formula for s sub n, I can proceed to take the limit as n goes to infinity and see if my sequence of partial sums converges or diverges sets the limit of my formula. And notice that the only part of this formula that depends on n is the art of the N, a one one minus are all constants as far as n is concerned. So we can use limit rurals. To take those out of the limit, I can rewrite the limit as a over one minus r times the limit of one minus r to the n or even better as a over one minus r times one minus the limit of r to the n. Now this limit I've seen before, right on this page, is the same as the limit I was considering when I was looking at convergence of sequences, where I use the same R, and my A is equal to one. So we know that this limit goes to zero, when r is between negative one and one, and does not exist as a finite number. when r is less than or equal to negative one, or r is greater than one. Therefore, the limit of the partial sums is going to equal a over one minus r times one minus zero, which is just a over one minus r, when r is between negative one and one, and that limit, will not exist as a finite number. when r is less than or equal to negative one, or r is greater than one. Since the limit also doesn't exist as a finite number, when r is one, I can add a little equality sign. And now I've got all this cases for our covered me write this down as a conclusion. The geometric series converges to a over one minus R. For For r between negative one and one, I can also say that's the absolute value of r less than one, and it diverges for the absolute value of r greater than or equal to one. Let's use this fact in this example. Remember that we decided that this was a geometric series with a common ratio r equal to negative one night, and we found the first term a by plugging in i equals two, and I got that first term of 1/3. Since the absolute value of our sales value of negative 1/9 is 1/9, which is less than one, we know that the series converges, and it converges to a over one minus r, so that's 1/3 over one minus negative one nights, which is 1/3 over one plus one night, that's 1/3 over 10 nights, which simplifies to three tenths. In this video, we looked at geometric series, with first term a common ratio R, we saw that a geometric series will converge if the absolute value of r is less than one. And they'll diverged if the absolute value of r is greater than or equal to one. In the case that it converges, it converges to a over one minus r, where A is the first term, and r is the common ratio. This video explains how to determine whether a series converges or diverges using an integral. Let's start with the series, the sum of one over n squared. Please pause the video for a moment and think about why this might converge or diverge. The sum from n equals one to infinity of one over n squared is closely related to the integral from one to infinity of one over x squared dx. Let me show you what I mean. In this picture, I've graphed the function y equals one over x squared in blue. in green, I've drawn a bunch of rectangles, I've divided the x axis into sub intervals of length one, and so each rectangle has a base of length one, and a height given by my functions value on the right endpoint of the sub interval. So the first rectangle has a base of one and a height of one, so it has an area of one, the second rectangle has a base of one and a height of one over two squared, so that's 1/4. So the area here is 1/4. The next rectangle base of one again, and a height of 1/9, and so on. The area of each rectangle is just the same as its height, and its height is just given by one over n squared for the appropriate value of n In other words, if I write out the first few terms of this series, it's exactly the same as the areas of my rectangles. And the sum of my series is just going to be the total green area. Now my integral can also be thought of in terms of area, this integral represents the area from one to infinity. Under my blue curve, we know that this area is finite, because we know that this integral converges by the P test where p is two. Now if I just look at the rectangles, starting with the second rectangle on, all of those rectangles will lie below the blue curve, and to the right of the line x equals one, so they will have a smaller area, and therefore the area of these rectangles from the second one on has to converge to a finite number. That is, the sum from n equals two to infinity of one over n squared converges. We're interested in the sum from one to infinity. But that's just the area of this single rectangle plus the area of all these rectangles with just one more than this sum here. So it also converges to a finite number. To recap, the chain of logic goes like this. First, we solve the integral represents a finite area, because of the P test. From that, we can conclude that the sum from n equals two to infinity of one over n squared represents a smaller and also finite area. And from that, we can figure out that the sum from n equals one to infinity has to represent a finite area. So our series converges to a finite number. Let's look at another example. The sum from n equals one to infinity of one over the square root of n. r between negative one and one, I can also say that's the absolute value of r less than one, and it diverges for the absolute value of r greater than or equal to one. Let's use this fact in this example. Remember that we decided that this was a geometric series with a common ratio r equal to negative one night, and we found the first term a by plugging in i equals two, and I got that first term of 1/3. Since the absolute value of our sales value of negative 1/9 is 1/9, which is less than one, we know that the series converges, and it converges to a over one minus r, so that's 1/3 over one minus negative one nights, which is 1/3 over one plus one night, that's 1/3 over 10 nights, which simplifies to three tenths. In this video, we looked at geometric series, with first term a common ratio R, we saw that a geometric series will converge if the absolute value of r is less than one. And they'll diverged if the absolute value of r is greater than or equal to one. In the case that it converges, it converges to a over one minus r, where A is the first term, and r is the common ratio. This video explains how to determine whether a series converges or diverges using an integral. Let's start with the series, the sum of one over n squared. Please pause the video for a moment and think about why this might converge or diverge. The sum from n equals one to infinity of one over n squared is closely related to the integral from one to infinity of one over x squared dx. Let me show you what I mean. In this picture, I've graphed the function y equals one over x squared in blue. in green, I've drawn a bunch of rectangles, I've divided the x axis into sub intervals of length one, and so each rectangle has a base of length one, and a height given by my functions value on the right endpoint of the sub interval. So the first rectangle has a base of one and a height of one, so it has an area of one, the second rectangle has a base of one and a height of one over two squared, so that's 1/4. So the area here is 1/4. The next rectangle base of one again, and a height of 1/9, and so on. The area of each rectangle is just the same as its height, and its height is just given by one over n squared for the appropriate value of n In other words, if I write out the first few terms of this series, it's exactly the same as the areas of my rectangles. And the sum of my series is just going to be the total green area. Now my integral can also be thought of in terms of area, this integral represents the area from one to infinity. Under my blue curve, we know that this area is finite, because we know that this integral converges by the P test where p is two. Now if I just look at the rectangles, starting with the second rectangle on, all of those rectangles will lie below the blue curve, and to the right of the line x equals one, so they will have a smaller area, and therefore the area of these rectangles from the second one on has to converge to a finite number. That is, the sum from n equals two to infinity of one over n squared converges. We're interested in the sum from one to infinity. But that's just the area of this single rectangle plus the area of all these rectangles with just one more than this sum here. So it also converges to a finite number. To recap, the chain of logic goes like this. First, we solve the integral represents a finite area, because of the P test. From that, we can conclude that the sum from n equals two to infinity of one over n squared represents a smaller and also finite area. And from that, we can figure out that the sum from n equals one to infinity has to represent a finite area. So our series converges to a finite number. Let's look at another example. The sum from n equals one to infinity of one over the square root of n. Please pause the video for a moment and think about how you might use an integral to decide if this series converges or diverges. A natural integral to consider is the integral from one to infinity of one over the square root of x dx. This integral diverges by the P test, where p is equal to one half, which is less than one. Let's look at a picture to compare areas. The blue curve here is the graph of the function y equals one over the square root of x. And here, once again, I've drawn green rectangles using the right endpoints to get the heights of the rectangles. So the areas of my rectangles are the same as the terms in my series. As in the previous problem, if I ignore the first rectangle, then all the rest of the rectangles have an area that's less than the area under my curve from one to infinity. But there's a serious problem here, the integral from one to infinity of one over the square root of x dx diverges to infinity, the area of my rectangles from the second one on might be less than my area under the blue curve. But being less than something that diverges to infinity, it tells us nothing, this series could converge or it could diverged. But don't give up hope. If we just tweak this picture a little bit, we'll be able to get something that we can use here. In this second picture, I've used left endpoints instead of right endpoints for the heights of my rectangles. Because my function y equals one over the square root of x is a decreasing function. Using left endpoints, makes my rectangles have a larger area than the area under the corresponding section of the curve. let me label the rectangles with their areas. The areas of these rectangles correspond to the terms in my series. But now we have that the area of the green rectangles, that total area, that total area is bigger than the integral from one to infinity of one over x squared of x dx. Please pause the video for a moment and think about how you might use an integral to decide if this series converges or diverges. A natural integral to consider is the integral from one to infinity of one over the square root of x dx. This integral diverges by the P test, where p is equal to one half, which is less than one. Let's look at a picture to compare areas. The blue curve here is the graph of the function y equals one over the square root of x. And here, once again, I've drawn green rectangles using the right endpoints to get the heights of the rectangles. So the areas of my rectangles are the same as the terms in my series. As in the previous problem, if I ignore the first rectangle, then all the rest of the rectangles have an area that's less than the area under my curve from one to infinity. But there's a serious problem here, the integral from one to infinity of one over the square root of x dx diverges to infinity, the area of my rectangles from the second one on might be less than my area under the blue curve. But being less than something that diverges to infinity, it tells us nothing, this series could converge or it could diverged. But don't give up hope. If we just tweak this picture a little bit, we'll be able to get something that we can use here. In this second picture, I've used left endpoints instead of right endpoints for the heights of my rectangles. Because my function y equals one over the square root of x is a decreasing function. Using left endpoints, makes my rectangles have a larger area than the area under the corresponding section of the curve. let me label the rectangles with their areas. The areas of these rectangles correspond to the terms in my series. But now we have that the area of the green rectangles, that total area, that total area is bigger than the integral from one to infinity of one over x squared of x dx. Since this integral diverges, and this series is larger, it must diverged also. This method of comparing a series to an integral is a very general method for showing convergence. It's known as the interval test. The integral test says that if f is a continuous positive and decreasing function on the interval from one to infinity, and our terms a sub n are equal to f evaluated at n, then if the integral from one to infinity of f of x dx converges, the series from one to infinity of a sub n converges. And if the integral from one to infinity of f of x dx diverges, then the series diverges. Although I won't give a formal proof of this theorem, the logic behind is the same logic we used in the previous two examples. If the interval converges, we use the picture here, using write endpoints to draw rectangles. The area of each rectangle is the same as its height, since the rectangle has base one, and the height of each rectangle is just f sub n. So the area of the first rectangle is just f sub one, which is a sub one, the area the second rectangle is F sub two, which is a sub two, and so on. If we focus on the second rectangle onwards, then the combined area of those rectangles is less than the area represented by the integral. So we can say the sum from n equals two to infinity of a sub n is less than or equal to the integral. So the integral converges by assumption, this series here has to converge, and therefore, our original series from n equals one to infinity must also converge. If instead, our integral diverges, then we use the other picture, and we use left endpoints for our rectangles, the areas of the rectangles are still given by a sub one, a sub two, and so on. But this time, the combined area of the green rectangles, that area is now greater than or equal to the integral of our function. Since this integral diverges, the larger area must diverged as well. That's the idea behind the integral test. And to apply it, we just need to be able to integrate the function that corresponds to our terms, and check that that function is continuous, positive and decreasing. Actually, it's enough to check that the function is eventually continuous, positive and decreasing. By eventually I mean that it has these properties on some interval from R to infinity for some number R. This is good enough, because then I can always draw the same pictures, just starting with R instead of one in my picture, and get the integral converges if and only if the series starting at our converges by the same arguments we use before, but the series starting at our converges if and only if the series starting at one converges, since adding on Finally, many extra terms to my series doesn't affect whether it converges or not. And the integral from R to infinity converges if and only if the integral from one to infinity converges, because similarly, adding on a finite little piece of area from one to R, doesn't change the convergent status of the integral. So by this chain of logic, it's okay if our function starts out increasing for a while, as long as it's eventually positive, continuous and decreasing. Here's an example of the integral test and action. We want to know if the sum from n equals one to infinity of ln n over n, converges or diverges. So let's look instead at the integral from one to infinity of ln of x Since this integral diverges, and this series is larger, it must diverged also. This method of comparing a series to an integral is a very general method for showing convergence. It's known as the interval test. The integral test says that if f is a continuous positive and decreasing function on the interval from one to infinity, and our terms a sub n are equal to f evaluated at n, then if the integral from one to infinity of f of x dx converges, the series from one to infinity of a sub n converges. And if the integral from one to infinity of f of x dx diverges, then the series diverges. Although I won't give a formal proof of this theorem, the logic behind is the same logic we used in the previous two examples. If the interval converges, we use the picture here, using write endpoints to draw rectangles. The area of each rectangle is the same as its height, since the rectangle has base one, and the height of each rectangle is just f sub n. So the area of the first rectangle is just f sub one, which is a sub one, the area the second rectangle is F sub two, which is a sub two, and so on. If we focus on the second rectangle onwards, then the combined area of those rectangles is less than the area represented by the integral. So we can say the sum from n equals two to infinity of a sub n is less than or equal to the integral. So the integral converges by assumption, this series here has to converge, and therefore, our original series from n equals one to infinity must also converge. If instead, our integral diverges, then we use the other picture, and we use left endpoints for our rectangles, the areas of the rectangles are still given by a sub one, a sub two, and so on. But this time, the combined area of the green rectangles, that area is now greater than or equal to the integral of our function. Since this integral diverges, the larger area must diverged as well. That's the idea behind the integral test. And to apply it, we just need to be able to integrate the function that corresponds to our terms, and check that that function is continuous, positive and decreasing. Actually, it's enough to check that the function is eventually continuous, positive and decreasing. By eventually I mean that it has these properties on some interval from R to infinity for some number R. This is good enough, because then I can always draw the same pictures, just starting with R instead of one in my picture, and get the integral converges if and only if the series starting at our converges by the same arguments we use before, but the series starting at our converges if and only if the series starting at one converges, since adding on Finally, many extra terms to my series doesn't affect whether it converges or not. And the integral from R to infinity converges if and only if the integral from one to infinity converges, because similarly, adding on a finite little piece of area from one to R, doesn't change the convergent status of the integral. So by this chain of logic, it's okay if our function starts out increasing for a while, as long as it's eventually positive, continuous and decreasing. Here's an example of the integral test and action. We want to know if the sum from n equals one to infinity of ln n over n, converges or diverges. So let's look instead at the integral from one to infinity of ln of x over x. over x. This is a continuous function, because it's the quotient of two continuous functions, and we're starting at an x value of one, so we don't have to worry about the denominator being zero. It's also a positive function. Since we know that ln of x is greater than zero for x bigger than one, and therefore this quotient is greater than This is a continuous function, because it's the quotient of two continuous functions, and we're starting at an x value of one, so we don't have to worry about the denominator being zero. It's also a positive function. Since we know that ln of x is greater than zero for x bigger than one, and therefore this quotient is greater than zero also. zero also. Finally, let's check if our function is decreasing. One way to do that is to look at the derivative. If f of x is ln x over x, then f prime of x by the quotient rule is x times one over x minus ln x times one over x squared. This simplifies to one minus ln x over x squared, which is negative when one minus ln x is less than zero, that is one is less than our an x, that is E is less than x. So the function has a negative derivative and is decreasing whenever x is bigger than E, so it's eventually decreasing the three conditions Finally, let's check if our function is decreasing. One way to do that is to look at the derivative. If f of x is ln x over x, then f prime of x by the quotient rule is x times one over x minus ln x times one over x squared. This simplifies to one minus ln x over x squared, which is negative when one minus ln x is less than zero, that is one is less than our an x, that is E is less than x. So the function has a negative derivative and is decreasing whenever x is bigger than E, so it's eventually decreasing the three conditions They're met, so we can apply the integral test. Next, we need to figure out if this integral converges or diverges. This is an improper integral. So by definition, it's the limit as t goes to infinity of the integral from one to T of ln x over x. We can use use of the tuition to evaluate it, where u is equal to ln x, d u is equal to one over x dx. And when x is equal to one, u is equal to ln of one, that's zero, when x is equal to t, u is equal to ln of t. Substituting in, we get the integral from zero to ln T of you do this integrates to use squared over two evaluated between ln T and zero. Substituting in our bounds of integration, we get the limit as t goes to infinity of ln t squared over two minus zero. Now as t goes to infinity, ln of T also goes to infinity. So ln t squared over two goes to infinity, therefore, the integral diverges. And so by the integral test, the series also diverges. In this video, we saw that a series converges if and only if the corresponding integral converges, provided that the corresponding function is eventually continuous, positive and decreasing. In this video, we'll determine whether series converge or diverge by comparing them to more familiar series. Suppose that the sum of a sub n and the sum of b sub n are series. And suppose that the terms of the series are always greater than or equal to zero, and that a sub n is less than or equal to b sub n for all n. In the pictures below, the heights of the blue bars are supposed to represent the ACE events, and the heights of the green bars are supposed to represent the base events. If we put the pictures together, we see that the heights of the blue bars are less than the heights of the green bars. So we have the inequality, zero is less than or equal to a sub n is less than or equal to b sub n. Since the base of each bar has length one, the height of each bar is the same number as its area. So when we write the sum from n equals one to infinity, of a sub n, this represents the area of all the blue rectangles added up the total blue area. And when we write sum from n equals one to infinity of b sub n, this represents the area of all the green rectangles, the total green area. Because the blue bars have a smaller area than the green bars, we can make some conclusions. First of all, if the total green area is finite, then so is the total blue area. In other words, if the sum of the b sub n converges, then so does the sum of the A sub N. Furthermore, if the total blue area is infinite, then so is the total green area. So we can also say, if the sum of the ace of ends diverges, then so does the sum of the base events. These facts are known as the comparison test for series and are very useful in establishing convergence. But we have to be careful not to take the conclusions too far. In particular, if the smaller series of Ace events converges, then we really can't say anything about the larger series of these events. The some of the big events could converge, red could diverge. Also, if the larger series of B sub ends diverges, then we can't conclude anything about the smaller series of Ace events, the sum of the ACE events could converge or could diverged. When using the comparison test to establish convergence or divergence, it's handy to compare your unfamiliar series to a familiar series that you already know converges or diverges. The following series are especially handy when making these comparisons. First, the geometric series Sum of A times r to the n, which converges when the absolute value of r is less than one. And second, the P series, one over n to the P, which converges when p is greater than one. Let's use a comparison theorem to determine whether the sum of three to the n over five to the n plus n squared converges or diverges. What matters most as far as whether a series converges or diverges is the behavior of the terms, when n gets close to infinity, the behavior of the terms when n is small, the behavior the first few terms doesn't make any difference as far as whether that series converges or diverges. So I'm going to focus on what happens to these terms, as n goes to infinity, well, three to the n goes to infinity, and five to the n goes to infinity, and n squared also goes to infinity. But between five to the n and n squared, five to the n is going to infinity much faster. So I'm going to say that this five to the n term dominates the denominator, it's more important. And for that reason, the behavior of the series we're given should be similar to the behavior of the series, three to the n, over five to the n, where I've just left out the n squared term, which is insignificant compared to five to the n when n is large. So I'm going to compare our given series to this other series, which is a geometric series. In fact, the second series we know converges, because it has a common ratio of three fifths, and the absolute value of three fifths is less than one. In order to use the comparison theorem, I'm going to need to compare the terms of this series to the terms of this series. And I want to show that these terms are less than or equal to those terms, because being smaller than a convergent series will guarantee convergence is clear that everything's positive, so we don't have to worry about that. And it's also clear that five to the n plus n squared is bigger than or equal to five to the n. When you divide by a bigger number, you get a smaller ratio. So three to the n over five to the n plus n squared is therefore less than or equal to three to the n over five to the n, we've shown that the inequality we need holds. And so by the comparison theorem, since the sum of three to the n over five to the n converges, so does our original series. We've established convergence using the comparison test. This video was about the comparison test, the fact that if we have zero less than or equal to A ad less than or equal to bn, and the sum of the bands converges, then the sum of the smaller series A ends converges. And if the sum of the ends diverges, then the sum of the larger series diverges. The limit comparison test gives an alternative to the regular ordinary comparison test for series. In the previous video, we looked at the series the sum from n equals one to infinity of three to the n over five to the n plus n squared. And we use the ordinary comparison test and compare it to the series, the sum of three to the n over five to the n. This worked out pretty nicely, because the terms here are less than the terms here. And this series converges. being less than a convergent series ensures convergence. But if we change the problem very slightly, and look instead at the sum of three to the n over five to the n minus n squared, look how things start to go wrong. If we now try to compare to the same series, then we get the inequality five to the n minus n squared is less than or equal to five to the n, and therefore, three to the n over five to the n minus n squared is greater than or equal to three to the n over five to the n. Since dividing by a smaller number gives a larger fraction, but this inequality unfortunately, is not useful to us, being greater than a convergence series doesn't guarantee convergence or divergence. So we can't conclude anything Based on this inequality, the limit comparison test gives us one way around this. The limit comparison test says the following. Suppose that sum of a n and the sum of bn are series with positive terms. If the limit as n goes to infinity of the ratio of a n over bn, is a number L, where L is a finite number that's bigger than zero, then either a both series converge, or both diverge, so they have the same convergence status. Let's try the limit comparison test on the problem we were just working on. We still want to compare to the same series, three to the n over five to the n. But this time, we're going to try a limit comparison. So we're going to take the limit, as n goes to infinity of the ratio of terms, three to the n over five to the n, divided by three to the n over five to the n minus n squared, it doesn't actually matter which term goes on the top and which goes on the bottom, we could instead take the ratio the other way, whatever limit we get, when we do the ratio, this way, will just be the reciprocal of the limit, we get When do the ratio this way. So if this ratio is a finite number that's bigger than zero, this ratio is reciprocal will also be a finite number that's bigger than zero. So I'll just stick with the first computation. let me simplify by flipping and multiplying, I can cancel my three to the ends. And now I can actually rewrite this as the limit of one minus n squared over five to the n. By breaking out my limit, this is the same as one minus the limit of n squared over a five to the n. And the second limit is an infinity over infinity form. So using lopi talls rule, carry the one over, I get and get the limit of what I get when I take the derivative of the numerator and the derivative of the denominator, I've still got an infinity over infinity and determinant form. So I'll use Libby toss rule again, the derivative of two n is two, and the derivative of the denominator is ln five times ln five times five to the n. Now the numerator is fixed at two, while the denominator goes to infinity as n goes to infinity, therefore this fraction goes to zero. And my final limit is one. Since one is bigger than zero, and it's finite, it's less than infinity. The limit comparison test tells me that my original series and my comparison series, either both converge, or both diverged. But my comparison series is a geometric series with ratio three fifths, so it definitely converges. Therefore, by the limit comparison test, our given series also converge us. That's the Limit Comparison theorem in action. The limit comparison test tells us that for two series with positive terms, if the limit of the ratio of the terms is some number, which is bigger than zero and less than infinity, then the two theories have the same convergence status. That is, they either both converge or both diverged. The limit comparison test is especially handy when the ordinary comparison test doesn't seem to work when we know what we want to compare to, but we can't get the inequality to go the right direction. In this video, I'll prove that the limit comparison test works. The limit comparison test says that if the sum of the ACE events, and the sum of the B sub ends are two series with positive terms. And if the limit as n goes to infinity of the ratio, a sub n over B sub n is equal to L, where L is a finite number that's bigger than zero, then either both series converge or both series diverged. To prove this theorem, let's start by assuming that all the hypotheses are true. That is, we'll assume all the stuff in brackets here is true. The limit as n goes to infinity of a sub n over B sub n equals L means that if I plot the numbers, 123, and so on, on the x axis, so those are the values of n, and I plot the ratios, a sub n over B sub n, on the y axis, those ratios are going to settle down to a value of L, as n goes to infinity. Using more technical mathematical language, it means that for any small number, epsilon that's bigger than zero, we can trap the ratios within epsilon of L, as long as we go out far enough for our values of n. That is, there exists a number capital N, such that n over bn is between L plus epsilon. And l minus epsilon for little n bigger than or equal to capital N. In the picture here, value for capital N of three would work, because for all little ends bigger than or equal to three, my ratios are trapped in between r minus epsilon, which is right here, and l plus epsilon, which is that upper bound here. Let's pick a small enough epsilon. So this interval here doesn't extend all the way down to zero through zero on the y axis, it just extends through through positive numbers near owl. Recall that that l itself is a positive number, so it's possible to trap it in a little interval. That's all positive numbers. For example, we could pick epsilon to be equal to L over two, for example, that way, the interval that I'm drawing here would extend down to l minus l over two, which is equal to L over two, and our extend up to l plus l over two, which is three L over two. So we have that l over two is less than a n over bn is less than three L over two, four little n bigger than or equal to our capital N that works for that value of epsilon. Now, I'm going to multiply all three sides of this inequality by b sub n, recall that b sub n is a positive number, all the series terms are positive, so that doesn't change around the inequalities at all. So we get L over two times b sub n is less than a sub n is less than three L over two times b sub n. And remember that everything here is bigger than zero, since all the ACE events and B sub ends are positive. Let's think about what this inequality tells us. First of all, if the sum of the beast events converges, then so does the sum of three L over two times b sub n, because I'm just multiplying that series by a constant still convergent. But now my a sub n terms are less than the terms of a convergent series. So by the ordinary comparison test, we know that the sum of the ACE events converges also. Furthermore, if the sum of the base events diverges, then we can focus on this part of the inequality, we know that l over two times b sub n diverges also. And so here we have the ace of ends are bigger than the terms of a Divergent series, so the sum of the AC bands must diverged, just using the ordinary comparison test. Now, you might be worried about the fact that this inequality only holds for little n bigger than or equal to some capital N. So we could rewrite this argument a little bit to say, if the sum from n equals one to infinity of the base events converges, then so does the sum from n equals capital n to infinity of b sub n, because adding or subtracting finitely many terms off the front never changes the convergence status of a series. After that, we can say well, then so does the sum from n equals capital n to infinity. Three L over two times b sub n. And so then by the ordinary comparison test, since this inequality does hold for our little n bigger than or equal to capital N, we can conclude that the sum from little n equals capital n to infinity of a sub n converges. And so the sum from n equals one to infinity of a sub n converges Also, since again, adding or subtracting Finally, many terms from the beginning of a series doesn't change anything about convergence. We could similarly rewrite the second part of the argument using precise indices as well. We've shown that the sum of the B ends and the sum of the A ends either both converge, or both diverged. And that completes the proof of the theorem. In this video, we prove the limit comparison test using the ordinary comparison test. This video defines absolute convergence and how it's related to convergence for a series. A series is called absolutely convergent. If the series of absolute values of the terms converges. Please pause the video for a moment and try to decide which of the following series are convergent, and which ones are absolutely convergent. The first series is convergent, because it's a geometric series with ratio r equal to negative 0.8. It's also absolutely convergent, because if I take the series of absolute values, that's the same thing as the series, the geometric series with a ratio of 0.8, which is also convergent. The second series, the sum of one over the square root of k is not convergent, we can see this by the P test. Since p is equal to one half, which is less than one. It's also not absolutely convergent. In this case, the sum of the absolute values of the terms is just the same as the sum of the original terms, which we already said diverges. The third series is the sum of negative one to the J times one over j. This is a convergent series by the alternating series test. In fact, this is the alternating harmonic series. What about absolute convergence? If we look at the sum of the absolute values of the terms, that's just the same thing as the regular harmonic series, which diverges. So this series is not absolutely convergent. Here's a question for you. Is it possible to have a series that's convergent, but not absolutely convergent? Please pause the video for a moment and try to answer this question. The answer is yes. We just saw us an example of such a series, the alternating harmonic series. There are many other examples of such theories. And in fact, there's a special name for them, they're called conditionally convergent. A series is called conditionally convergent if it is convergent, but not absolutely convergent. In symbols, that is, the sum of the a ns converges, but the sum of the absolute values of the a ns diverges. Next question for you. Is it possible to have a series that's absolutely convergent, but not convergent? This is a little trickier. But please pause the video for a moment and think about your answer. The answer to this one is no. It's a fact that every absolutely convergent series is convergent. Let me prove to you why that's true. Let's suppose that we have a series that's absolutely convergent. That is the sum of the absolute value of the A ends convergence. We know that the A ends might be positive or negative, but they do have to lie in between the absolute value of a n and the negative absolute value of a n. Actually, it's true that a sub n is either equal to its absolute value or its negative absolute value, but I'm ready This way with inequalities to help set the mood for using the comparison test. Now, I can't quite use a comparison test here to prove that the series of Ace events converges, because even though the ace of ends are less than or equal to the terms of a convergent series, they're not necessarily positive or greater than equals zero. And the can parison test only applies to series whose terms are all greater than equal to or equal to zero. But there's a nice trick to get around that difficulty. And that trick is to add the absolute value of a sub n to all the sides of the inequality. So then I get zero is less than or equal to a sub n plus the absolute value of a sub n, which is less than or equal to twice the absolute value of a sub n. Now, since this sum of the absolute value of f sub n converges, so does the sum of twice the absolute value of a sub n, since it's just a constant multiple. Based on this inequality, which now does involve terms that are greater than or equal to zero, I can conclude that the sum of a sub n plus the absolute value of a sub n converges based on the ordinary comparison test. But now, the series that we want, the sum of the ace of ends can be written as a difference. And we know that a series formed by subtracting the terms of two convergent series itself converges. So this converges, in fact, to the difference of the sobs. We've proved that if our series is absolutely convergent, then it must be convergent. The fact that absolute convergence implies convergence can come in handy when you're trying to prove that a series converges, as in the following example. In this example, we want to prove that this series is convergent or divergent, but the cosine and sine make things a little bit tricky. My intuition here is that this series should converge since cosine and sine are bounded. And so this essentially should behave something like the sum of one over n cubed, which converges because of the P test. But we can't just compare our series to the sum of one over n cubed and use the ordinary comparison test like we've done in similar problems in the past. What's different here is that our terms are not always going to be positive, because sine and cosine can be positive and negative. And so can there's some to help us out of this pickle. Let's think about absolute convergence instead. If we look at the sum of the absolute values, which is the same as the sum of the absolute value of the numerator, divided by m cubed. Since cosine of n is between one and negative one, and sine of n is also in between one and negative one, we know that the som cosine n plus sine of n has to be less than or equal to two and bigger than or equal to negative two. In fact, it can't even get all the way to two or all the way down to negative two, we could find that tighter bound if we tried, but this is good enough for our purposes, this inequality can be rewritten as the absolute value of cosine n plus sine of n is less than or equal to two. And if I divide both sides here by n cubed, I have an inequality involving positive terms. So since we know the sum of one over n cubed converges by the P test, that implies that the sum of two over n cubed also converges. It's just a constant multiple. And that implies that the sum of my absolute value of cosine n plus sine n over n cubed converges by the comparison test. Therefore, we know that our series is absolutely convergent. And therefore, it must be convergent. Since an absolutely convergent series is always convergent. We've shown that our original series converges. In this video, we saw that if a series is absolutely convergent, then it has to be convergent But not vice versa. This video is about the ratio test, a test that can be used to prove that a series converges or diverges. The ratio test is all about looking at the ratio of consecutive terms. To figure out how it works, it can be helpful to think about geometric series first, recall it for a geometric series, the ratio of consecutive terms is given by this number R. And if R has absolute value less than one, the series converges. Why while if R has absolute value greater than one, it diverges. For more general series, the ratio of consecutive terms is not necessarily a constant. But if we look at the limit as n goes to infinity, of the absolute value of the ratio of consecutive terms, and if we get a limit of L, which is less than one, then the series converges, just like a geometric series. In fact, the series is absolutely convergent, meaning the sum of the absolute values of the a ns converges, and therefore, the sum of the a ns converges also, if instead, the limit as n goes to infinity of the absolute value of the ratio of consecutive terms, is a number l that's greater than one. Or if that limit is infinity, then just like a geometric series, the series diverges. Finally, if the limit of the absolute value of the ratio of consecutive terms is exactly equal to one, or if the limit doesn't exist, then the ratio test is inconclusive. That is, the sum of the ace of ends may converge. Or it may diverge. And to figure out, which will have to use a different test or a different argument, let's apply the ratio test to this series, we'll need to compute the limit as n goes to infinity of the absolute value of the ratio of consecutive terms. In this case, that's the limit as n goes to infinity, have the absolute value of n plus one squared times negative 10 to the n plus one divided by n plus one factorial, all of our n squared times negative 10 to the n over n factorial. I've just plugged in n plus one for n in this formula to get the a sub n plus one term. I can simplify this by flipping and multiplying. And I'm going to rearrange my factors. This expression is equivalent to the previous one, I've just arranged factors so that similar factors are on top of each other, this will make it easier to cancel things. Now, n factorial means n times n minus one times n minus two and so on. And n plus one factorial means n plus one, times n times n minus one, and so on. So if I divide n factorial by n plus one factorial, all my factors from the numerator will cancel with factors from the denominator. And I'll just be left with one over n plus one. Also, negative 10 to the n plus one over negative 10 to the n cancels out to just negative 10 to the One Power, so I can rewrite my limit as N plus one squared over n squared times negative 10 times one over n plus one, I'm going to divide my limit of a product into a product of limits. Now as n goes to infinity, n plus one over n goes to one. So this expression, which is equivalent to the square of n plus one over N, also goes to one. The limit of the absolute value of negative 10 is just 10. And the limit as n goes to infinity of one over n plus one is zero. Therefore our limit is one times 10 times zero, which is zero. And since zero is less than one by the ratio test, we know that our series converges absolutely. This video was about the ratio test, which focuses on the limit as n goes to infinity of the absolute value of the ratio of consecutive terms. Depending on whether this limit is less than one greater than one, or equal to one, we can determine whether the series converges or diverges. Or if we need to try another test. In this video, I prove the ratio test for convergence and divergence of series. The ratio test says that for a series, if the limit of the absolute value of the ratio of consecutive terms is equal to a number L, that's less than one, then the series is absolutely convergent, and therefore convergent. If however, the limit of the absolute value of the ratio of consecutive terms is a number L, that's bigger than one, or is equal to infinity, then the series is divergent. Although I didn't write it here, if the limit is equal to one exactly, or if the limit doesn't exist, then the ratio test is inconclusive and can't be used to establish convergence or divergence. To prove that the ratio test works, let's first assume that the limit is less than one. This means that if I graph n on the x axis, and my absolute value of the ratio of consecutive terms, on my Y axis, I get a bunch of dots that settle at a value of L. and this number L is less than one. Let's pick a tiny number epsilon, so that when I add epsilon to L, I'm still less than one. So pick epsilon greater than zero, such that l plus epsilon is less than one. By the definition of limit, if I go far enough to the right on the x axis, all of my red dots are going to be trapped in between this epsilon interval around L. In mathematical symbols, this means there exists a number capital N, such that the absolute value of ratios, the thing that's a limiting to L is between L plus epsilon. And minus epsilon, for all little n bigger than or equal to capital that I'm going to multiply all three sides of this inequality by the absolute value of a sub little m, I'm going to focus on the right end of the inequality, because ultimately, I want to prove that my series is absolutely convergent. So it's going to be more useful to show that my terms are smaller than things than to show that my terms are bigger than things. This inequality is true for all values of little n that are bigger than or equal to capital N. So in particular, it's true when little n equals capital N. The inequalities also true when little n is equal to capital N plus one. Here, I've just plugged in capital N plus one for a little n here. And when I plug it in here, I end up with capital N plus one plus one, which is capital N plus two. Now if I string these two inequalities together, essentially, I'm substituting in this inequality, right here, I get the following inequality, which simplifies to give me the absolute value of a sub capital N plus two is less than l plus epsilon squared times the absolute value of a sub capital N. Let's try this one more time. Going back to my original inequality here, I'm going to plug in capital N plus two for lowercase n. And stringing these two inequalities together, I get that this is less than l plus epsilon cube times the absolute value of a piece of capital N. And in general, the same reasoning shows that the absolute value of a sub capital N plus i is less than l plus epsilon to the i times the asset value of a sub capital N. For any I. With all these inequalities, I'm gradually building up to series. The first series is the series, the sum of the absolute value of a sub capital N plus i. Let's start that from i equals one to infinity. And the second series is the sum of L plus epsilon to the i times the absolute value of a so capital that again, let's start from equals one to infinity. Now the second series is a geometric series, where r is equal to L plus epsilon, which is less than one, because remember, we chose epsilon to make sure that was less than one. Therefore, this series converges. But that's great news. Because now looking at this inequality, we can show that series one converges, just using the ordinary comparison test, and recognizing that these terms are bigger than or equal to zero. So series one converges by the comparison test. But series one is just the tail end of the series, the sum from n equals one to infinity absolute value of a sub little n. So this series converges, because it's just finally many terms added on to a convergent series. And so we've shown that our original series converges absolutely. This proves the first part of the ratio test. Now let's prove the second part. And let's start by assuming that the limit is greater than one. This time, since l is bigger than one, we can pick a tiny number epsilon, so that when we go down from L by epsilon, we don't go as far as the number one. So we're going to pick an epsilon greater than zero, such that l minus epsilon is still greater than one. As before, we can use the definition of limit and a little algebra to get the same inequality as we got before. This time, I'm going to focus on the left side of the inequality though. Since l minus epsilon is greater than one. This tells me in particular, that the absolute value of a sub n plus one is always bigger than the absolute value of as little n for little n bigger than or equal to capital N. This means that the absolute value of Ace of capital n is less than the absolute value of a sub capital N plus one, which is less than the absolute value of capital A sub capital N plus two, and so on. And so my sequence of terms is actually ultimately an increasing sequence of positive numbers. So my sequence of terms cannot converge to zero, limit as little n goes to infinity as the value of a sub little n cannot be zero. And so the limit as little n goes to infinity of a sub little n cannot be zero either. But that means that my original series cannot converge, it has to diverged by the divergence test, we still have one detail that consider the possibility that the limit of the absolute value of ratios is infinity. In this case, the argument that we just used, works almost exactly the same. If we assume that the limit is infinity, then we can skip this part here. And here, and just note that there exists some capital n such that the ratios are always bigger than say to for all little n bigger than or equal to capital N. Since if the ratios are heading towards infinity, they're certainly going to be bigger than two, eventually. This gives us the same inequality that we need, I just read a two there. And now as before, we can conclude that a sub n plus one is absolute value is strictly greater than a sub n is absolute value. In fact, it's greater than twice of it for all little n bigger than equal to capital N. And we can make the same conclusion about an increasing sequence of positive terms. That so that the terms can't go to zero, and the series has to diverge by the divergence test. This concludes the proof of the ratio test. In this video, we prove the convergent part of the ratio test by comparing our series to a convergent geometric series, and we prove the divergent part of the ratio test using the divergence test. This video is a review of all the convergence tests we've talked about in class. I'll list the test roughly in the order that I would try to apply them. I like to start with the divergence test. Usually it's pretty easy to check if the limit as n goes to infinity of the terms is equal to zero. And if not, you're done because the series diverges. Be careful though. The divergence test can only be used to check for divergence, it cannot be used to prove convergence, because if the limits of the terms is equal to zero, the series may converge, but it may still diverge. The next thing I do is to check if the series is a simple p series, or a geometric series. Remember, a p series is a series of the form one over n to the p n is our indexing variable P is some number like two or 5.8. And this is easy to test for convergence since it converges. If p is greater than one and it diverges, otherwise, a geometric series this is the kind of the form a times r to the n, where A is the first term, and they start at zero here. So it's the really the first term and r is the common ratio. And this one's easy to check to because it converges if the absolute value of r is less than one and diverges otherwise. If the series happens to be alternating, than the alternating test is a good one to apply next. Be careful, this test can only be used to prove convergence. If the series is actually alternating. And this what I call this step size, that's the absolute value of the terms is going to zero and decreasing, then we can conclude that the series converges. But if some of those conditions are not satisfied, we can't automatically assume that the series diverges. Well, not at least by the alternating test, if the step size doesn't go to zero, then we should have already figured out the series diverges by the divergence test were really, really applying the divergence test there. And if the step size is not decreasing, not even ultimately decreasing, then the alternating test is just inconclusive. It doesn't apply, we don't know yet, whether it diverges or converges, and we have to look for another test. Now if the series is not one of these nice p series geometric or alternating series, my go to test is going to be the ratio test. The ratio test is especially good for series with n factorials in them or two to the ends and things with sort of geometric pieces that are not strictly geometric are good candidates for the ratio test. But be careful, the ratio test will be inconclusive for like all p like series, so series that just have you know things like ends and, and maybe the square roots of ends and things that can be easily compared to a p series are not good candidates for the ratio test. So if you happen to remember that you can save some time by not trying the ratio test on those. If the ratio test is not a good candidate or ends up being inconclusive, what I might try next is one of the comparison tests. So that would be like what I call the ordinary or the limit comparison test. We generally want to compare two series that we know a lot about that we know the convergence status of so we would generally want to compare to either p series or geometric series. The comparison tests are especially good for p like series that the ratio test is inconclusive for. And to figure out what to compare to it's a good rule of thumb to consider the dominant or highest power terms. One thing you need to be aware of when applying the comparison test is that it only applies to series with positive terms. Of course, the first few terms never matter for series convergence. So it's okay to apply it if you have eventually positive terms. But if the terms never become always positive, and they're not strictly alternating, so the alternating test doesn't apply. We don't have to give up hope we can use the fact that absolute convergence implies convergence. That's what I try and sort of mine since my seventh test to try. So you can just take the absolute values of your terms and then maybe use the comparison test. And if that works to prove convergence, then your original series will converge also. Another method I haven't mentioned before is to use limit laws to split up the series. So if you have the sum of two series, say a p series and a geometric series, then a natural thing to do is to split this up as the sum of two series and use a different method for each piece. If both pieces converge, like they do in this situation, then the psalm also converges. Also, if one piece happens to converge on the other diverges, then the sun will diverged. The only thing to be careful of is that if both pieces diverged, then the psalm may still diverged. But it may converge because there might be cancellation, one piece might be diverted into infinity and one piece may be diverging to negative infinity. And that's an indeterminate kind of form. If none of this stuff has worked so far, I might look to try the integral test and compare my series to an integral. This is especially handy in my experience four series with logs in them. So something and also it has to be a series where the integral of easy compute so you know, something like ln n over n, if you instead look at the integral of ln x over x, that's pretty easy to compute using use substitution. And so that would be a good candidate for the integral test. Be aware that the integral test can only be used when the series terms can be thought of as the functions values at integers for a function that is positive, continuous, and decreasing. Last on my list is the method of telescoping series. I put it last only because it's kind of a hassle to work with telescoping series, but it does have some good points. First of all, using the method of telescoping series, you can actually compute the sum rather than just tell if it converges or diverges. The only other tool on this list that will actually compute the sum of an infinite series is the geometric series test where we have a formula for the sum provided it converges. Another reason to use the telescoping series is if you happen to notice that your terms are the difference of related expressions. So something like the sum of e to the one over n plus one minus e to the one over N might be a good candidate for telescoping series. Something like the sum of one over n squared minus one would also be a reasonable candidate, because you can rewrite it using the method of partial fractions. And using that method well, and the telescoping series stuff will help you find an actual song. But if you just want to know convergence, it'll be a lot easier just to use comparison to a p series whenever n squared for this one. So that's pretty much everything I know about convergence tests for series. If you want to keep watching, I'll take a look at some examples on the next page. Before you keep watching, please take a moment to look at these six examples and decide which convergence or divergence test you might try. Please be aware that for many of these series, there are lots of tests that will work. So just because you pick a different one than I do doesn't mean that yours is wrong. I think this first example can be conquered using the divergence test. My hunch is if we took that limit and use lopi talls rule, we get a limit of infinity not zero. An alternative though would be to use the ratio test. Because this is a term that is has a geometric piece as well as some other stuff. The second example is an alternating series. So my first try is going to be the alternating series test. And my recollection is it does work to prove convergence in this case. For this third one, this is the kind that I call a p like series, because if I just look at the the dominant terms, the highest power terms, I could compare to the P series, which is one over and squared cubed rooted or in other words, one over n to the two thirds power. Since this one diverges, I'm gonna expect my original one also diverged, I probably need to use the limit comparison test because I don't think the inequalities will go the right direction for the ordinary comparison test. This next one's a perfect candidate for splitting up into two pieces. This one, the second piece, I could use the geometric series test to show convergence. And this first one says it has an n factorial that's a great candidate for the ratio test. This next one's kind of tricky for me. At first glance, I almost thought it would be a candidate for the integral test. If this had just been an N instead of an n squared, I might be able to integrate using use substitution. But because it's an n squared, the integral test would be more tricky to do or I might have to do integration by parts or something. So I'm gonna stay away from that. And I'm going to start by trying the ratio test because this does have a geometric kind of like piece to it. And this last example, I'm going to use the integral test here because I know like To integrate one over x ln x dx, using the use substitution u equals ln x, apply and converge, the divergence test is something that takes practice, the more you do it, the better you'll be at recognizing which tests might apply. But a lot of times, there's no substitute for just trying the test. And if it doesn't work, it's where it's inconclusive. Just try something else. Good luck, and see you in class. This video introduces some of the ideas and key formulas of Taylor series. One of the main ideas behind Taylor series, the idea of approximating functions with polynomials. So suppose we have some function f of x, we want to approximate this function with a polynomial, and we'd like the approximation to be good near x equals zero. And we're going to assume that f prime of zero exists, and f double prime of zero exists, and F third derivative exists at zero, all of its derivatives, we're gonna assume exists at x equals zero. Now, if we just want to get F value, right at zero, we can approximate F with the constant function, y equals f of zero, we can think of this constant function as being a degrees zero polynomial approximation. But it's a pretty lousy approximation, we can do much better than that, in fact, we can do a lot better even if we just use a degree one polynomial, that's a linear function. As you know, the tangent line at x equals zero is a linear function that provides a pretty good approximation for the actual function when x is near zero least that's the best approximation you can hope for out of a linear function. The equation for the tangent line is given by y equals f of zero, plus f prime of zero times x. This comes straight from the point slope form for a line where m is the slope of the tangent line, that's the derivative at zero, and the point x one y one is just the point zero, f of zero. So we get y minus f of zero is f prime of zero times x minus zero, which simplifies to that equation for the tangent line right there. Notice that the tangent line has the same value as f of x at x equals zero, and it has the same slope as f of x at x equals zero. But I'd like to do a little better than this, I'd like to approximate my function f of x with a polynomial that has the same value, the same slope or first derivative, and the same second derivative as f of x at x equals zero. It turns out, I can do this with a degree two polynomial. And I'll show you how, in general, a degree two polynomial, also called a quadratic is a polynomial of the form P of x equals c sub zero, plus c sub one times x, plus c sub two times x squared. And I just have to figure out what values of the constants C sub zero, c sub one, c sub two, will make my polynomial, agree with my function in its value, first derivative and second derivative. Well, if I want P of zero to equal f of zero, and that means I want c sub zero plus c sub one times zero plus c sub two times zero squared to equal s value at zero. In other words, c sub zero had better be equal to f of zero. But now I also want p prime of zero to equal f prime of zero. P prime of x is equal to c sub one plus two times c sub 2x. I'm just using my derivative rules on the equation for p remembering that my C's are constants. Therefore, to evaluate p prime of zero, I just plug in zero for x, and I get c sub one, but this needs to equal f prime of zero, and therefore c sub one is equal to f prime of zero. Finally, I want p double prime of zero to equal f double prime at zero, I can find p double prime of x by taking the derivative of my derivative, so that gives me two times C two, but that needs to equal f double prime of zero means that C two had better equal f double prime of zero divided by two. So now let's look what happened, requiring that P has the same value as zero as f forces c sub zero to have the value of f of zero, requiring that the polynomial and F had the same first derivative zero, forces the value of CS of one, two equal f prime of zero, and requiring that the polynomial and the function have the same second derivative at zero forces the value of c two to be f double prime of zero divided by two. So we've shown that there is a secondary polynomial that has the same value first derivative, the second derivative as F at X equals zero, and there's a unique such polynomial, and it's given by p of x equals f of zero, plus f prime of zero times x, plus f double prime of zero over two times x squared. Visually, that second degree polynomials going to look like a parabola, it might look something like this. Let's play this game again. But this time, I want to find a degree three polynomial p of x, such that P of zero is the same as f of zero, p prime of zero is the same thing as f prime of zero, p double prime of zero is equal to f double prime of zero, and peas, third derivative at zero is the same as F, third derivative at zero. Graphically, that's going to be a cubic polynomial that approximates my function, and it's going to be a pretty close approximation. We know that a three three polynomial in general has the form c sub zero plus c sub 1x, plus c sub 2x squared plus c sub 3x cubed. And we need to find the values of all the constant C's. Please pause the video and see if you can figure out the values of those constants, especially the value of c sub three in terms of the value of f and its derivatives at zero. If we write down the derivatives of P, we get the following expressions. And if we evaluate these expressions at x equals zero, all the terms with x's in them vanish. So we get these expressions. Now, because we want our polynomials value, and its derivatives to match the value and derivatives of our function, we get these equations from which we can solve for our constants, c sub zero is f of zero, c sub one is f prime of zero, c sub two is f double prime of zero over two, and c sub three is the third derivative of f at zero divided by six. Notice that that number six came from multiplying three times two, and that three and that two came from bringing down my success of exponents, when I took derivatives, we can keep track of this process by rewriting c sub three is the third derivative of f of zero divided by three times two or three factorial. So the third degree polynomial that approximates our function is p of x, which is f of zero, plus f prime of zero times x plus f double prime of zero over two times x squared, plus f third derivative at zero over three factorial times x cubed. I'll write this as P sub three to remember it's the third degree polynomial. We can repeat this process to get a fourth degree polynomial whose value add zero is the same as f of zero, and whose first four derivatives at zero are the same as f first for derivatives at zero. Please pause the video and either work out expressions for the coefficients of peace and four, or else make an educated guess what those coefficients should be based on the patterns you say. You should get the fourth degree polynomial has the same first terms as a third degree polynomial and has a final term of f, fourth derivative at zero over four factorial times x to the fourth. If we continue this process forever Finding polynomials of higher and higher degree that match more and more derivatives of f. Then in the limit, we'll have infinitely many terms that look like this. This is an infinite series, and it can be written in summation notation as the sum from n equals zero to infinity of the nth derivative of f at zero, divided by n factorial times x to the nth power. This works as long as we use the conventions, that the zeroeth derivative means just the function that zero factorial is equal to one, and that x to the zero is just equal to one, even if x is zero. This infinite series is called the Maclaurin series for f of x. And it's also called the Taylor series for f of x centered at x equals zero. Now, so far, we've been focusing on the value of f and its derivatives at x equals zero. What if we wanted to approximate f of x near x equals a, please pause the video and write down what you think the Taylor series centered at x equals A should look like. This is a series I'll call it T of x, that we want to match F's value at A. and we want all of its derivatives to match f derivatives at a, it makes sense that this series should have a formula similar to the formula we just found. But it should involve derivatives at a instead of derivatives at zero. In addition, the formula needs to involve powers of x minus a, instead of powers of x. I'll leave you to think about the details. And to verify that this series really does have the derivatives that we wanted to have. In this video, we tried to approximate a function by polynomials that had the same value and derivatives at x equals zero. And we ended up with a formula for a Taylor series at x equals zero, which we generalized to a Taylor series centered at x equals a. This video defines power series. informally, a power series is a series with a variable in it, often the letter X, and it looks like a polynomial with infinitely many terms. For example, if we look at the series, the sum from n equals zero to infinity of two n plus one times x to the n over three to the n minus one, that's a power series with variable x. If we expand that out by plugging in values of n, we get one n equals 01 times x to the zero over three to the minus 1x to the zero is one and three to the minus one on the denominator is the same as three on the numerator. So we can rewrite this term as just three. The next term, when n equals one, is three times x to the one over three to the zero, we can rewrite this as 3x, since three to the zero is one. The next term is 5x squared over three, and we can continue like this. I want to point out that when working with power series, x to the zero is always taken to be one, even though there's a possibility that x could end up being zero, and zero to the zero is considered undefined in other contexts. When working with power series, x to the zero for any value of x is one. The next series expands out to one plus five times x minus six, and so on. This is an example of a power series centered at six because of all the factors of x minus six. In general, a power series centered at a is a series of the form the sum from n equals zero to infinity of c sub n times x minus a to the n, where x is the variable. The cease events are real numbers, they're constants, called the coefficients, and a is also a real number a constant that's called the center. If I expand out the power series and read out the first few terms that looks like this, where c sub zero is the constant term, notice that x minus A to the zero is taken to be one, even when x equals a. If the power series is centered at zero, then we just set a equal to zero, we can write this a little more efficiently in the following form. Sometimes you might see a power series that starts with index of one instead of zero. That's perfectly legit. It just means there's no constant term. Or if you prefer, you can think of the constant term as being zero. It's also fun for the index to start at some other positive number. But it's not considered a power series if the index starts at a negative number, resulting in x's in the denominator. That's all for the definition of power series. Recall that a power series is a series with a variable in it. In this example, the C sub n and the A are supposed to be real numbers that are held constant, so the only variable is x. That's the only place where I can plug in different values at different times. This video explores the question of for what values of the variable x does the power series converge, and for what values of x does it diverged. Let's look at a few examples. First, for what values of x does this power series converge. There's one value of x that I know for sure it converges for, please pause the video for a moment and try to guess which value of x I'm thinking of the series definitely converges when x equals three. If I expand out the first few terms of the series, I get this expression. then plugging in three 4x, all of my terms vanish to zero, except for my constant term of one. So at x equals three, the series converges to its constant turn. And in fact, this is true of any power series all power series converge at their center. But let's see what other values of exit converges for. Although we have many tests for convergence in our toolkit, the ratio test is usually the best test to use to determine where of power series converges. For the ratio test, we need to take the limit as n goes to infinity of the absolute value of the ratio of consecutive terms. For our example, this is n plus one factorial times x minus three to the n plus one divided by n factorial times x minus three to the N. Let's simplify by canceling things. We get n plus one, times x minus three. Now x minus three is some number. And I'll assume it's a nonzero number, since I already dealt with the case when x equals three, so I have a nonzero number that stays fixed as n goes to infinity times a number that's going to infinity. So the absolute value of the product has to go to infinity, no matter what x value we have, other than the x value of three. The ratio test says that if this limit is infinity, the series diverges. Therefore, the power series diverges for all values of x except for three. The only place where it converges is at the center of three. In this next example, the center of this series is negative four. So the series definitely converges when x equals negative four. Let's use the ratio test to figure out what other values of x make it converge. So we'll take the limit as n goes to infinity of the absolute value of a sub n plus one over a sub n. This works out to the limit the absolute value of negative two to the n plus one times x plus four to the n plus one divided by n plus one factorial, all over negative two to the n x plus four to the n over n factorial. Let's simplify by flipping and multiplying and rearranging a little. After canceling, we get the limit of the opposite value of negative two times x plus four divided by n plus one. The numerator of this expression doesn't depend on an n, so it stays fixed as n goes to infinity, but the denominator goes to infinity. Therefore, we're dividing some fixed constant by larger and larger numbers, and so this limit is equal to zero. Once again, the limit doesn't depend on x as value It's always zero no matter what axis. So by the ratio test zero is less than one, the series converges for all values of x. Here's our third and last example, in this example, it's a little trickier to figure out what the center is, one thing we can do is to rewrite the series in a more standard form by factoring out the negative five, then we get negative five times x minus two fifths, all raised to the nth power over N, I can rewrite this again as negative five to the n times x minus two fifths to the n over n. And in this more standard form, it's easy to recognize that the center is two fifths. Another way to find the center is just to figure out the value of x, that makes terms go to zero. So in our case, if we want negative 5x plus two, two equals zero, we need X to equal two fifths. And therefore, two fifths must be the center, like we found before. In any case, our series converges for x equals two fifths, for sure, but it might converge for other values of x. So let's use the ratio test. To find other values of x that make the series converge. We start off the same way as usual, by taking a limit of a ratio of the n plus one than n terms. And then we simplify by flipping and multiplying. And after canceling, we get the limit of negative 5x plus two times n over n plus one as n goes to infinity, and over n plus one goes to one. And negative 5x plus two doesn't depend on n. So this final limit is just the absolute value of negative 5x plus two. So by the ratio test, our series converges when this limit is less than one, and it diverges when the limit is greater than one, the ratio test is inconclusive when the absolute value of negative 5x plus two is exactly equal to one, so we'll worry about that case later. Let's solve the first absolute value inequality. When the absolute value of something is less than one, that means that the quantity inside the absolute value sign has to be between one and negative one. So we can rewrite the absolute value inequality as negative one is less than negative 5x plus two, which is less than one, we can solve this for x by subtracting two and dividing by negative five. dividing by a negative number reverses the direction of the inequality signs. So we have that our series converges for these values of x. Now let's solve the second absolute value and equality. The one that tells us where the power series diverges, the series diverges when the absolute value of negative 5x plus two is greater than one. When the absolute value of something is greater than one, that means that whatever is inside the absolute value sign has to either be less than negative one or greater than one. So we can replace our absolute value in equality with two inequalities. Negative 5x plus two is less than negative one, or negative 5x plus two is greater than one. Let's solve these inequalities by subtracting two and dividing by negative five. And similarly on the other side. So our series diverges when x is greater than three fifths or less than 1/5. That makes sense, it's kind of the opposite of where it converges because we're solving an absolute value inequality with the opposite inequality side. Putting all this information together, we see that our series converges when x is between 1/5 and three fifths and divert On either side of this interval, we still don't know what happens when x is exactly equal to 1/5, or exactly equal to three fifths, since those values correspond to the case when the ratio test is inconclusive. So let's turn our attention to the x values of 1/5 and three fifths Next, let me write down our original power series, it was the sum from n equals one to infinity of negative 5x plus two to the n over n. If we want to know if this power series converges at x equals 1/5, let's just plug in x equals 1/5. This simplifies to the sum of one to the n over n, which is just the regular harmonic series, which diverges. If we plug in x equals three fifths, we get this series, which simplifies to the alternating harmonic series. So it converges. So now we know that the series converges when x equals three fifths and diverges when x equals 1/5. And our final answer is that the power series converges on the interval from 1/5 to three fifths, where we use an open bracket, to denote that we exclude the endpoint 1/5, because the series diverges there, and the closed bracket, square bracket means that we include the endpoint of three fifths where the series does converge. I want to make one more observation. Before we leave this example. Notice that the midpoint of this interval is the number 2/5. Remember, at the beginning of the problem, we calculated the center of the power series, and it was also two fifths. We'll see in a moment that this is no coincidence. In fact, the interval of convergence is always centered at the center of the power series. And we could in fact, describe the interior of this interval of convergence as the x values for which x minus that center is less than 1/5. Bet is all x values within a distance of 1/5. From the interval center. We've seen three examples of power series, and each one converge in a very different way. In general, it turns out that there are only these three different types of convergence that we've already seen. It's possible, like we saw on the first example, that a series might converge only at its center. It's also possible that a power series could converge for all values of x. This is what happened in our second example. But if neither of these two cases hold, then the only other possibility is that there exists a number R, such that R is series converges anytime, where within our units from the center a and the power series diverges for any x values that are more than our units from the center array. In symbols, I can right there exists a number are such that the power series converges when the absolute value of x minus a is less than R, and diverges when the absolute value of x minus a is greater than R, since the absolute value of x minus a represents the distance between x and a. This was a situation we saw in our third example. Now in the first case, we say that the radius of convergence of the power series is zero. In the second example, we say the radius of convergence is infinity. And in the third example, we say the radius of convergence is our since our represents the distance from the center of the interval, sort of like the radius of a circle represents the circles distance from the center. Now the interval of convergence is the interval of all x values for which the power series converges. So in our first situation, the interval of convergence is just the number a it's not really an interval, just a single number, but we call it the interval of convergence anyway, in the second situation, our interval of convergence is the interval from now negative infinity to infinity. And then the third situation, the interval of convergence includes this entire interval here that extends from the number A minus r to the number A plus R. So our interval could be the open interval, a minus r to a plus R. But it could also include one or more endpoints, so it could be the closed interval, or it could just include the left endpoint, or just the right endpoint. In this video, I worked out some examples using the ratio test to figure out what x values make a power series converge. I also stated the fact that there are only three options for convergence of a power series, convergence at the center only convergence for all real numbers and convergence on some finite interval centered at the center of the power series. This video gives an example of computing the interval of convergence and the radius of convergence for a power series. To compute the radius of convergence and interval of convergence for this power series, we start by using the ratio test. So we need to find the limit as n goes to infinity of the absolute value of a sub n plus one over a sub n, where the a sub ns are the terms for this power series, we can compute a sub n plus one, by just plugging in n plus one everywhere we see an N in this expression. So that's negative four to the n plus one, times x minus eight to the two times quantity and plus one divided by n plus one. We divide all that by the A sub N term, which is just negative four to the n x minus eight to the to n over n, which I've just copied from the formula here. Now I'm going to simplify, flip and multiply. And now I'm going to rearrange terms so that corresponding terms are on top of each other. So I'm going to write negative four to the n plus one over negative four to the n times x minus eight to the to quantity n plus one, that's the same thing as two n plus two, over x minus eight to the two n. And then I'll write the N and N plus one. So that's n over n plus one. Once I cancel terms, I get the limit of negative four times x minus eight squared times n over n plus one. Now, as n goes to infinity, n over n plus one is going to one, and the absolute value of negative four is just four. And the absolute value of x minus eight squared is just going to be x minus eight squared, since this expression is always positive, so we have our limit. And the ratio test says the series will converge where this limit is less than one. So next, let's set four times x minus eight squared to be less than one and solve for x. In other words, x minus a squared is less than 1/4. Now we can't solve this quadratic inequality by taking the square root of both sides, that would lead to something like x minus eight is less than plus or minus the square root of 1/4, which doesn't even make any sense. And it's not true. What we can do instead is solve the quadratic equation, x minus eight squared is equal to 1/4. And then use some logic to figure out the inequality. So now that we have an equation sign, we can take the square root of both sides to get x minus eight is is equal to plus or minus the square root of 1/4. In other words, x minus eight is plus or minus one half. So in other words, x is equal to eight plus one half, or eight minus one half. That's either 17 halves, or 15 halves. Now let's go back to the inequality that we're interested in. Since we know that x minus eight squared is equal to 1/4. At x values of 15 halves and 17 halves, we can test to see whether x minus eight squared is bigger or smaller than 1/4. by plugging in values in between these numbers, so when x is less than 15, half just by plugging in a sample value like zero, we can see that x minus eight squared will be bigger than 1/4. Well, when x is between 15 halves and 17 halves, say the value of eight, the plug in eight and 4x, and see that eight minus eight squared, which is zero, is going to be less than 1/4. And finally, plugging in a value of x over here, maybe something like 10 for x, we're going to get a value of x minus eight squared, that is, again, bigger than 1/4. So putting this together, we can see that x minus eight squared is less than 1/4. When x is between 15 halves, and 17 halves. So by the ratio test, our series converges. For x between 15 halves, and 17 halves the ratio test also tells us the series diverges. When this expression is greater than one in order other words, x minus eight squared is greater than 1/4. In other words, for x values less than 15 halves are greater than 17 halves. Now the only thing we still need to figure out is what happens at the endpoints of the interval 15 halves to 17 half's recall that our series was given by this formula. So when x is equal to 15 halves, we have negative four to the n 15 halves minus eight to the to n over n, which is equal to negative four to the n, negative one half to the two n over n, which I can rewrite as negative one to the n times four to the n times negative one to the two N over two to the to n divided by n, which is the same thing as negative one to the n four to the n, negative one to the two n divided by n times two squared to the N. Since two n is always even negative one to the two, n is always equal to one. And since two squared is four, this four to the n on the denominator cancels with the four to the n on the numerator. So I'm left with the alternating harmonic series, which we know converges. So the series converges for x equal to 15 halves. Now at x equals 17 halves, we can go through the same computation, just using 17 halves in place of 15 halves, that changes the negative one half here to a positive one half. And now we have a positive one to the two n, which is still always one, everything else works the same. And so we still have a conversion series. So going back up to the top, we know that the series actually converges for x greater than or equal to 15 halves and less than or equal to 17 halves. That is our interval of convergence. closed bracket 50 and a half to 17 halves close bracket. That is that the interval of convergence has length one, because 17 halves minus 15 halves, the difference of the two endpoints is two halves, which is one also, the interval of convergence has center of eight because the average of the endpoints 17 halves plus 15 halves over two is equal to eight. This should come as no surprise, because our original series was centered at eight. So if we draw our interval of convergence on the number line, it's centered at eight, and extends out a total distance of one unit. In other words, it extends out by half a unit on either side. And so the radius of convergence is the length of the interval divided by two, or one half. So we found the radius of convergence. And we found the interval of convergence, which was this closed interval here. And so that completes the problem. In this video, I'll prove some of the key facts about convergence of power series. My ultimate goal is to prove that there are only three possible options for convergence, a power series could converge only at its center, it could converge for all real numbers. And if these two options don't hold, then there must exist a number are such that the power series converges for all x, within our units away from the center a and the power series diverges. for all x, his distance from a is greater than our first I'll prove some preliminary facts. I'll start with this one. If a power series converges when x is equal to b, for some nonzero number B, then it also converges for any x whose absolute value is less than the absolute value of b. To prove this fact, let's assume that the power series converges when x is equal to b, that is, the sum of c sub n times beats to the n converges. If a series converges, then the limit of its terms has to equal zero. Since if the limit of the terms is not equal to zero, the series would have to diverged by the divergence test. Therefore, by the definition of limit, for any epsilon, there exists a number capital N, such that c sub n times b to the n is between zero plus epsilon, and zero minus epsilon for little n bigger than or equal to capital N. In particular, if we pick epsilon equal to one, this says there exists a capital n such that negative one is less than c sub n times b to the n is less than one for little n bigger than or equal to capital N, I can rewrite this statement as the absolute value of c sub n times b to the n is less than one for little n bigger than or equal to capital N. Now, if x is any number, with absolute value less than the absolute value of b, we can write the absolute value of c sub n times x to the n as the absolute value of c sub n times b to the n times x over a b to the n, just using algebra, I can rewrite this as c sub as the value of c sub n times b to the n times the absolute value of x over b to the n. For little n bigger than or equal to capital N, we know that the opposite value of c sub n times b to the n is less than one. So this expression has to be less than the absolute value of x over b to the n. Now if the absolute value of x is less than the absolute value of b, this means that the absolute value of x over b is less than one. So the series The sum of X absolute value of x over b to the n is a geometric series, whose ratio has absolute value less than one, so it's a convergent series. Now, the ordinary comparison test tells us that the sum of the absolute value of c sub n, x to the n also converges, because we know that the terms of that series are less than the terms of our convergent geometric series. Therefore, our original series The sum of c sub n, x to the n converges absolutely, and therefore converges. So we've proved the first statement. The second statement says that if the power series diverges when x is equal to d for some nonzero number D, then it also diverges. Whenever we have an x whose absolute value is greater than the absolute value of D. The statement follows directly from the first statement, because suppose we have the sum of c sub n times d to the n diverges. If the absolute value of x is bigger than the absolute value of d, and the sum of c sub n x to the n converged, then by part one, the sum of c sub n, d to the n would have to converge, since the absolute value of d is less than the absolute value of x. But this contradicts the assumption that the sum of c sub n times d to the n diverges. And therefore, we know that the sum of c sub n times x to the n must diverged after all. That's all for the proof of facts about convergence of series. In this video, we'll talk about power series as functions. Consider the function defined by f of x is equal to the sum from n equals zero to infinity of x to the n. This can also be written as one plus x plus x squared, and so on. When we think of this as a function, the variable x becomes our independent variable, or input variable. So to figure out the values of f, we can plug in numbers for x and evaluate. Please pause the video and calculate F of 1/3. f of 1/3 is equal to one plus a third plus a third squared, and so on, this is a geometric series. So it adds up to the first term one divided by one minus the ratio of 1/3. And that simplifies to three halfs. Next question, what's the domain of f of x, we can think of the domain of a function as the values of the input variable x, that give us a real number as output? Please pause the video to write down your answer for the domain of f of x. The series, the sum of x to the n converges when x is between negative one and one. So for these values of x, we get a finite real number as our answer for f of x. Also, the series diverges for other values of x. So we don't get a real number answer when we plug in values of x less than or equal to negative one or greater than or equal to one. Therefore, the domain of f of x is the set of X values for which negative one is less than x is less than one, or an interval notation, we can write this as the interval from negative one to one. And in general, the domain of a power series is exactly the set of values where it converges. By a closed form expression for f of x, I mean an expression that doesn't involve a summation sign, I can write the sum of X to the N, without the summation sign by using the geometric series formula, first term is one divided by one minus the common ratio of x. This holds for all x values between negative one and one as usual, where the geometric series converges. Therefore, I can write f of x as one over one minus x for x values in that interval. Notice that if I just looked at the function, g of x equals one over one minus x out of context, it would have domain spanning from negative infinity to one together with the interval one to infinity. Because this function g of x is defined for all x values that are not equal to one. So f of x and g of x are not exactly the same function. They have different domains, but they are exactly equal on the interval from negative one to one. And we say that the function g of x is represented by the power series, the sum from n equals zero to infinity of x to the n. If we want to be more precise, we can say it's represented by this power series 4x between negative one and one. We can think of the partial sums of this series sum of x to the n as a way to approximate the function, one over one minus x with polynomials. Please pause the video and write out the first few partial sums, your answers should have Xs. S sub zero is just the zeroeth term one, S sub one is one plus x, and so on. s sub n is one plus x plus all the way up through x to the n, an nth degree polynomial. In this figure, I've drawn the function one over one minus x in blue. And I've drawn the first partial sum the linear function one plus x in orange. Notice that these two functions are close to each other when x equals zero, but get farther away from each other when x is far from zero. In this graph, I've added the next partial sum s two, which is a degree two polynomial, a quadratic. Here I've got partial sums through s four. And here I've got partial sums through about s 12. The original function one over one minus x is here in blue on mark over Ed. And you can see that these partial sums are becoming very good approximations to that original function on the interval from negative one to one. Now outside that interval, for example, for x values below negative one, our partial sums deviate wildly from our original function. I want to show one more graph. In this graph, the function one over one minus x is shown in blue. And the graph of the power series, the sum of X to the N is shown in orange, the blue function is actually obscured by the orange function because the two functions are identical for values of x between negative one and one. The only difference between these two functions, as discussed before, is that the function one over one minus x is defined for all x values except for the x value of one. And that's why we can see the blue graph even when x values are less than negative one. However, the function given by this power series has domain in between negative one and one, and so it only exists here in between those x values. In this video, I talked about representing the function one over one minus x with the power series, the sum from n equals zero to infinity of x to the n. These two progressions are equal for x values between negative one where the power series converges. I also used a graph to give an idea of what this equation means. The partial sums drawn in various colors give excellent approximations to the original function drawn in blue on the interval of X values between negative one and one. Since the partial sums are polynomials, this gives us a way to approximate this rational function with simple polynomial equations. The idea of approximating functions with polynomials is a very important idea that we'll see again and again. This video is about rewriting functions in terms of power series. All the examples that we'll do in this section will be based on the formula for the geometric series, the fact that the sum from n equals zero to infinity of x to the n is equal to one over one minus x. For x between negative one and one. We want to express the function two over x minus three as a power series. And we want to do this by using the geometric sum formula. The trick here is going to be to rewrite to over x minus three, so it looks more like one over one minus something, then we can treat whatever that something is as x and plug into the formula to get a power series. So that's the idea. Now I'm starting with two over x minus three. And I don't really like the X minus three, I'd rather This was three minus x because that reminds me more of one minus x. So I could rewrite it like this, but my two expressions now are equal. This one's the negative of this one, so I can fix that by just sticking a negative sign out in front. Now my two expressions are equal, because I've just multiplied my first expression by negative one over negative one. To get this expression, but I still don't really like the three minus x, I wish that were one minus x, it'd be nice if I could just divide the three by three to get one. But in order to leave the expression on change, I'm going to need to divide everything by three, both the top and the bottom. This gives me negative two thirds, divided by three minus x over three, which I can also write as negative two thirds times one minus x over three. Now if I bring the negative two thirds out front, I have negative two thirds times one over one minus x over three. Using my geometric sum formula, this is the same as negative two thirds times the sum from n equals zero to infinity of x over three to the n, I'm just plugging in x over three 4x. In this formula, I've now found a power series representation. But I'm going to clean it up a little bit and make it look more standard, bring the negative two thirds into the summation sign and distributed my exponent to get x to the n over three to the N. And now I can rewrite this as negative two over three to the n plus one, times x to the n. To figure out the interval of convergence for this power series, there are two different approaches that I could take. First, I could do a standard computation using the ratio test. I'll let you work out the details yourself. But you should get that the radius of convergence is three, and the interval of convergence is from negative three to three. A second approach to finding the interval of convergence is to look at the history of how we made the power series. Our basic template power series was the sum from n equals zero to infinity of x to the n, which converges when x is between negative one and one. We then plugged in x over three for x. Well, this good should converge when x over three is between one and negative one. In other words, when x is between three and negative three, finally, we multiplied that series by negative two thirds. This doesn't change the interval of convergence. So the interval of convergence for our final power series is the interval between negative three and three, just like you could have gotten from the ratio test. As a second example, let's find a power series representation of x over one plus 5x squared. Again, we want to use the geometric series summation formula. So we want to make this expression look more like one over one minus something. Well, one plus 5x squared is the same thing as one minus minus 5x squared. So if I just wanted a power series for one over one plus 5x squared, I could do that easily by using the geometric sum formula, and getting the sum from n equals zero to infinity of minus 5x squared to the N. All I'm doing here is plugging in negative 5x squared for x in this formula. Since I want a power series for x over one plus 5x squared, instead, I can just multiply everything by x. Again, I'd like to clean things up and make this expression look more like a standard power series. So I'll drag the x inside the summation sign, I can do that because the summation is over n, which has nothing to do with x. Now I can use my laws of exponents to rewrite this as negative one to the n five to the n times x to the two n. Now, x times x to the two n is equal to x to the two n plus one. And so this gives me a good power series representation for my function. Although the problem didn't explicitly ask for it, it's a good idea to compute the interval of convergence to see for what values of x this equation actually holds. I'll use the history approach. We started with our old familiar power series, which converges when x is between one and negative one with plugged in net Get a 5x squared for x. So that converges when negative one is less than negative 5x squared is less than one, which is equivalent to the inequality 1/5 is greater than x squared, which is greater than negative 1/5. Notice that I had to flip around the inequality signs when i divided by negative five. Looking at a graph of y equals x squared, I can see that x squared is between negative 1/5 and 1/5. For x values corresponding to this section of the graph that I'm drawing here in green, or this interval of X values on the x axis that I'm drawing in pink. To find the endpoints of this pink interval, I just need to find where x squared is exactly equal to 1/5, which is when x is equal to plus or minus the square root of 1/5. So the x values that satisfy me and equality are the x values in between these two values, I have negative the square root of 1/5 is less than x is less than the square root of 1/5. Now the last step in my history is when I multiplied everything by x, this doesn't change my interval of convergence, which remains this interval here. In this video, I represented several functions with power series using the geometric series summation formula. Although only a limited class of functions can be handled using this formula, some of the techniques that I used in the process, like multiplying my whole power series by x or plugging in an expression for x, some of those techniques can be used in a much broader context to represent many different kinds of functions as power series. As we'll see. Up to now, we've only been able to calculate the sum of a few various specific series like geometric series and some telescoping series. In this video, we'll see how to use Taylor series to find the sums of other series. Let's start by finding the Taylor series for arc tan of x centered at x equals zero. Since arc tan of x is equal to the integral of one over one plus x squared, one easy way to find the Taylor series for arc tan of x is to build it up, starting with the formula for a geometric series, one over one minus x is the sum from n equals zero to infinity of x to the n. Now, one over one plus x squared is equal to one over one minus negative x squared. So I can plug negative x squared in for x in this power series. And I get the sum from n equals zero to infinity of negative x squared to the n, which simplifies to the sum from n equals zero to infinity, negative one to the n x to the two n. Therefore, arc tan of x, which is the integral of one over one plus x squared dx is going to be equal to the integral of this power series, at least up to a constant, I can integrate this power series term by term to get the sum of negative one to the n, x to the two n plus one divided by two n plus one plus a constant. To figure out the constant C, I can plug zero in for x on both sides of my equation. Since all of my powers of x involve a positive power of x, even when n equals zero, I've got x to the one, so there's always at least one copy of x there. So if I plug in x equals zero, all of these terms go to zero, and arc tan of zero is also zero. So plugging in x equals zero gives me zero equals the sum of a bunch of zeros plus C. In other words, the constant is zero. Therefore, this expression right here gives me a Taylor series representation of arc tan. Let's take a moment to figure out which x values this equation is actually true for. We know that the geometric series formula holds for x values between negative one and one not including the endpoints. Therefore, when I plug in negative x squared for x, I get In the equation that holds for negative x squared between one and negative one, which is equivalent to saying that x is between one and negative one. And when I take the integral of both sides, I still get an equation that holds true for x between negative one and one. So I'm guaranteed that this equation up here holds for x values between negative one and one. But in fact, it's not hard to check that this series actually converges at the endpoints of negative one and one. This follows from the alternating series test sets when we plug in x equals negative one, or x equals one. Either way, we get an alternating series that converges. So this series converges on the closed interval from negative one to one, and it's equal to arc tan on that open interval. In fact, it turns out that is equal to our tan even on the closed interval. In particular, the equation holds for x equal to one. When x is equal to one, that I plug into the equation to get the arc tan of one is equal to the sum from n equals zero to infinity, negative one to the n times one to the two n plus one, that's just one over two n plus one, I can rewrite that. Now, arc tan of one is the angle whose tangent is one. So arc tan of one is going to be pi over four. In other words, I now have a series that sums to pi over four, let's write out the first few terms of this series and see what it looks like. The first term is one, the next term minus a third, plus a fifth minus a seventh plus a ninth, and so on. In other words, multiplying both sides by four, we get that pi is equal to four minus four thirds plus four fifths minus four sevenths, plus four ninths, minus 4/11, and so on, if you've ever wondered how to generate digits for pi, here's one way we found the sum of kind of a natural series to look at. And we found a beautiful formula for pi. For the next example, let's start by finding the Taylor series for f of x equals ln x centered at x equals one, we can write out the pattern of the function and its derivatives. And we soon notice that the nth derivative will have negative one to the n minus one times n minus one factorial times x to the n. Since we're centering at x equals one, we'll plug in one and get the nth derivative of f at one is equal to negative one to the n minus one times n minus one factorial. Therefore, the Taylor series for ln of x will be the sum from n equals zero to infinity of f to the n at one over n factorial, times x minus one to the n. Since this pattern for the nth derivative of f really only works, starting with the first derivative, not with the zeroeth derivative, I'm going to split out the first term, which is just going to be ln of one, which is actually zero. And then all the other terms follow the same pattern. And we have negative one to the n minus one, times n minus one factorial divided by the n factorial times x minus one to the n. This simplifies to ln of x is equal to the sum from n equals one to infinity of negative one to the n minus one times x minus one to the n over n. Since the n minus one factorial cancels with almost all the n factorial leaving just the factor and in the denominator, so I now have a formula for the Taylor series for ln of x. It's easy to check using the ratio test that this power series has a radius of a convergence of one and so it converges when x minus one is between one and negative one. It other words when x is between zero, and two. And although I won't prove it here, it turns out that this Taylor series really does converge to its function ln of x. And in fact, it converges to ln x on the interval for x greater than zero and less than or equal to two. Now, if I plug in x equal to two into my equation, I get something interesting. I get that ln of two is equal to the sum from n equals one to infinity of negative one to the n minus one of two minus one to the n, well, that's just one to the n, which is just one. And so I get ln f two is equal to the sum of negative one to the n minus one over n. That should be looking familiar to you. And yes, it's true. This is just the alternating harmonic series, one minus one half, plus 1/3, minus 1/4, and so on. So Taylor series has given us the sum of the alternating harmonic series, and it is ln of two. In this video, we use Taylor series to find the sum of the alternating harmonic series. We also used Taylor series for arc tangent, to find that the sum of a different series is actually equal to pi. As you get more familiar with Taylor series, you'll be able to calculate the sum of other series by recognizing them as the series that you get by plugging in a certain value of x into the Taylor series of a particular function. For example, if you see the series one, plus one over two, plus one over three factorial, plus one over four factorial, and so on, you might recognize that as the number one plugged into the formula for the Taylor series of either dx, in other words, this series is equal to either the one which is e. for any function f of x, whose derivatives all exist, we can write down the Taylor series for f of x centered at x equals a, by using this formula. But just because we can write the Taylor series down, doesn't guarantee that the Taylor series actually converges to the function we started with. In this video, we'll address the questions of When can we be sure that the Taylor series converges to its function? And when the Taylor series does converge? How good is the approximation by taylor polynomials? or partial sums? In other words, how big is the remainder? The answer to the question, does the Taylor series always converge to the function that's made from is unfortunately, no. Sometimes the radius of convergence is zero. And sometimes, even though the radius of convergence is large, or even infinite, the Taylor series converges, but it converges to the wrong function. Here's an example of the second situation. If we look at this piecewise defined function, g of x is defined as e to the minus one over x squared, if x is not zero, and is defined so that it's continuous at zero to be zero, when x equals zero, it's possible to work out the value of g prime is zero, using the limit definition of derivative, g prime of zero is the limit as h goes to zero of g of zero plus h minus g of zero over h, which is the limit as h goes to zero of e to the minus one over h squared minus zero over h. I'll rewrite this as the limit as h goes to zero of one over h divided by e to the one over h squared. as h goes to zero from the positive side, this is an infinity over infinity indeterminate form. And as h goes to zero from the negative side, is a negative infinity over infinity indeterminate form. In either case, we can use loopy talls rule to replace this limit with the limit of the derivatives, which simplifies to a zero over infinity kind of limit, which is equal to zero. In a similar way, it's possible to prove that the second derivative of g zero is also zero, and so is the third derivative, and so are all the derivatives of G at zero. Therefore, if we write out The Taylor series is just the sum of a bunch of zeros or the zero function. Certainly this Taylor series converges for all x, but it converges to the constant zero function. And that's different from the function that we started with. In fact, the function that we started with g of x is not zero for any x, except x equals zero. So the Taylor series only matches the function at the single point x equals zero and nowhere else. We found an example where the Taylor series converges, but not to its function g of x. Fortunately, this behavior doesn't happen for most of the functions that we typically deal with. To understand better, which Taylor series are guaranteed to converge to their functions, let's take a look at the idea of remainders. For a function f of x as Taylor series T of x, the remainder is written r sub n of x equals f of x minus T sub n of x, where T sub n of x is the nth degree Taylor polynomial. This can be expanded out as follows. Previously, when we looked at remainders for series, we wrote that the remainder was the infinite sum, which I'll call s infinity, minus the nth partial sum. The analogous expression for Taylor series might be the entire Taylor series, minus the first terms up through the degree and term. But that's not what we define the remainder to be for Taylor series. Instead, the remainder for Taylor series is the difference between the function and the first terms up to the degree and term. The reason it's defined a little bit differently is because for Taylor series, we're super interested in the Taylor series converging to its function. And it's of less interest whether or not the Taylor series happens to converge to its infinite sum. Because we define the remainder as the difference between the function and its Taylor polynomial, it follows directly that the Taylor series for f of x converges to f of x and an interval around a if and only if the limit of the remainders is zero in this interval, to see this, just note that the limit as n goes to infinity of the Taylor series equals f of x, that's what it means to converge to f of x. If and only if f of x minus this limit is equal to zero. We can rewrite this as the limit as n goes to infinity of f of x minus the limit as n goes to infinity of T n of x equals zero. Since the limit as n goes to infinity of f of x is just f of x, there's no ends in this expression. Using limit laws, we can rewrite this again, as the limit of the quantity f of x minus t and f of x equals zero. But that's just the same thing as saying that the limit of the remainders is zero by definition of remainders. So we restated the question about when does a Taylor series converge to its function? As a question about when does the limit of the remainders equals zero? tailors in equality gives us a bound on these remainders that can help us answer the question of when the remainders limit to zero. This bound can also be a useful way to answer the question of how close is the approximation when we use taylor polynomials to approximate a function. Here's some details about when this bound holds. Suppose there's a number capital N, such that the n plus one derivative of X has magnitude less than or equal to capital N, for all X's within a distance d of the center a as a graph, this means that there's a number capital M. And for all X's within a distance d of the center a, the graph of the n plus one derivative of f lies between negative m and m. So if such a number M exists, then the remainder r sub n of x of the Taylor series satisfies the inequality that the magnitude of r sub n of x is less than or equal to this bound capital M divided by n one plus one factorial times the absolute value of x minus a to the n plus one to power for all x's in this interval. of length two D that we're talking about. Now, in the statement of Taylor's inequality, the number m can be chosen just to work for a particular derivative. But really nice things can happen if we are able to choose the same number of M to work for all derivatives for our values of little n. In fact, if all derivatives are bounded by the same value capital M, then we can guarantee that the Taylor series converges to the function. That gives us a nice practical convergence criterion that I'll show you on the next slide. The practical convergence condition says that if there's a number capital M, such that the magnitude of the nth derivative at x is less than capital M, for all numbers, x and a certain interval around the center. And for all numbers, and then the Taylor series for f of x converges to f of x 4x is in that interval. I represent the convergence condition visually. This time, we're agreeing that all the derivatives are within this bound. So the original function lies within this bound, and its derivative lies within this bound, and the second derivative lies within this bound. These are not necessarily accurate representations of the derivatives. But that's the idea. So as long as the bound holds for all the derivatives, then the Taylor series converges to the function. And it's not too hard to prove this practical convergence criteria. From Taylor's inequality. Remember that Taylor's inequality says that, because of this bound, the nth remainder is bounded by M over n plus one factorial times the absolute value of x minus a to the n plus one. But it's a fact that the limit as n goes to infinity of M over n plus one factorial, times the absolute value of x minus a to the n plus one is equal to zero. And it's not hard to prove this fact, by looking at the series. And using the ratio test, to show that the series converges, I'll leave that as an exercise for the viewer. Therefore, by the divergence test, we know that the limit of the terms has to be zero, which is what we want. Now, because the limit of this expression is zero, by the squeeze theorem, the limit of the our ends has to be zero as well, which means that the Taylor series converge to the function. This practical convergence criterion is a very good way to show that Taylor series converges to their function. But even if it doesn't hold, it's still possible that the Taylor series may converge to its function, or in some cases, it may not. Let's use this practical convergence condition to prove the Taylor series for sine x converges to sine x. Recall the Taylor series for sine x is given by this equation. And recall also that any nth derivative of x for f of x equals sine x will have to be of the form sine x, or negative sine x, or cosine of x, or negative cosine of x. That's simply because when we take repeated derivatives of sine and cosine, the values cycle around in between those four possible answers. Now, since the absolute value of sine x is always less than one, or equal to one and same thing for cosine, we know that the nth derivative of our function has to be bounded by one, so we'll let m be one. And this bound holds for all x values, all real numbers. Therefore, we know that the Taylor series converges to the function sine of x. for all values of x. We've used the practical convergence condition with M equals one to prove this. In this video, we defined the nth remainder for a Taylor series as a difference between the function and its nth Taylor polynomial. We also gave a bound on the size of the nth remainder. It's always less than or equal to M over n plus one factorial times the value of x minus a to the n plus one where M is a bound on the size of the n plus ones. derivative of x for x within some distance d of a. Because of this formula for the remainder known as Taylor's inequality, we can show that if the nth derivative of x is always bounded by the same M for all x within a certain interval around a, and for all values of n, then the Taylor series converge to its function. This video introduces the idea of parametric equations, instead of describing a curve as y equals f of x, we can describe the x coordinates and y coordinates separately in terms of a third variable t, usually thought of as time. so we can write x as a function of t and y as a separate function of t. This is especially useful as a way to describe curves that don't satisfy the vertical line test, and therefore can't be described traditionally as functions of y in terms of x. A Cartesian equation for a curve is an equation in terms of x and y only. parametric equations for a curve give both x and y as functions of a third variable, usually T. The third variable is called the parameter. That's our first example, let's graph the parametric equations given here on an x y coordinate axis. We'll do this by finding x and y coordinates that correspond to the same value of t. For example, when t is negative two, you can calculate that x, I plug in negative two for t gives you five and y, when you plug in negative two for t gives you eight. Please pause the video for a moment and fill in some additional values of x and y. For some additional values of t. Your chart should look like this. And when we plot the XY pairs and connect the dots, we get something like this. It says this point over here corresponds to a T value of negative two. And this point over here corresponds to the t value of two. So if we think of t as time, we're traversing the curve in this direction, to find a Cartesian equation for this curve, we need to eliminate the variable t from these equations. One way to do this is to solve for t and one equation, say the first equation. So two t is equal to one minus x, which means that t is one half minus x over two, then we can plug that expression for t into the second equation and get y equals one half minus x over two squared plus four, which simplifies to the quadratic equation, y equals 1/4 X squared minus one half x plus 17 fourths. Let's try some more examples. A table of values for the first example helps us draw the familiar graph of a circle of radius one. This should come as no surprise, since the equations x equals cosine t and y equals sine t, are familiar from trig as a way of describing the x and y coordinates of a point on the unit circle. Notice that when t equals zero, our curve lies on the positive x axis. And as t increases from zero to two pi, we traverse the curve once in the counterclockwise direction. A Cartesian equation for this unit circle is given by the equation x squared plus y squared equals one. This follows from the trig identity cosine squared t plus sine squared t equals one by substituting in X for cosine t, and y for sine t. Please pause the video for a moment to graph the second curve and rewrite it as a Cartesian equation. The table of values should help you see that the graph is again a unit circle. But this time, as t increases from zero to two pi, we actually traverse the circle twice in the clockwise direction. I'll draw this with a double arrow going clockwise The Cartesian equation for this graph is still x squared plus y squared equals one. And so we found two different parameterizations. For the same graph on the x, y axis. Let's take a look at the third equation. There's no interval value specified for t here. So let's just assume that t can be any real number. Now as T ranges from negative infinity to infinity, our Y values, which are given by cosine t, oscillate between one and negative one, our x values are always the square of our Y values. So the graph of this curve has to lie on the graph of x equals y squared, which is a sideways parabola. But a parametrically defined curve doesn't cover this whole parabola. Remember that y is given by cosine of t. So y can only range between negative one and one. And so we're only getting the portion of the parabola that I shade in here. As t varies from say, zero to pi, I traverse this parabola one time. And then as t goes from pi to two pi, I go back again in the other direction. And as T continues to increase, I traverse this parabola infinitely many times. The Cartesian equation for this curve is the equation x equals y squared, with the restriction that y is between negative one and one. We've seen several examples where we went from parametric equations to Cartesian equations. Now let's start with a Cartesian equation and rewrite it as a parametric equation. In this example, y is already given as a function of x. So an easy way to parameterize. This curve is to just let x equal t. And then y is equal to the square root of t squared minus t, substituting in T for x, the domain restriction in terms of x just translates into a restriction in terms of t. I call this the copycat parameterization. Since we have successfully introduced the new variable t, but T just copies, whatever x does. In the second example, we could try setting x equal to t, then we get 25 t squared plus 36, y squared is equal to 900. and solving for y, we'd have y squared equals 900 minus 25 t squared over 36. So y is plus or minus the square root of this quantity. This is a very awkward looking expression. And fact, why is that even a function of t here because of the plus and minus signs. So let's look for a better way to parameterize this curve. Because of the x squared and the y squared, this equation is a good candidate for parameterizing using sine and cosine. In fact, if we divide both sides of the equation by 900, we get 25x squared over 900 plus 36, y squared over 900 is equal to one, which simplifies to x squared over 36 plus y squared over 25 is equal to one. If I rewrite this as x over six squared plus Y over five squared equals one, then I can set x over six equal to cosine of t, and y over five equal to sine of t. And I can see that for any value of t x over six and y over five will satisfy this equation simply because cosine squared plus sine squared equals one. This gives me the parameterization x equals six cosine of t, y equals five sine of t, which is a handy way to describe an ellipse. As a final example, let's describe a general circle of radius r, and center HK. For any point, x, y on the circle, we know that the distance from that point x y to the center of the circle is equal to r. So using the distance formula, we know that the square root of x minus h squared plus y minus k squared has to equal Are squaring both sides. This gives us the equation for the circle in Cartesian coordinates. So for example, if our circle has radius five, and has Center at the point negative 317, then its equation would be x minus negative three, that's x plus three squared plus y minus 17 squared is equal to 25. One way to find the equation of a general circle in parametric equations, is to start with the unit circle and work our way up. We know that the unit circle with radius one centered at the origin is given by the equation x equals cosine t, and y equals sine t. If we want a circle of radius r centered around the origin instead, then we need to expand everything by a factor of R. So we multiply our x and y coordinates by R. If we now want the center to be at HK instead of at the origin, then we need to add h to our x coordinates and add K to all our Y coordinates. This gives us the general equation for a circle in parametric equations. to match the Cartesian equation above, we can write our same example circle and parametric equations as x equals five cosine t minus three, y equals five sine t plus 17. In this video, we translated back and forth in between Cartesian equations and parametric equations with a special emphasis on the equations for circles. This video is about finding the slopes of tangent lines, for curves to find parametrically. To find the slope of the tangent line, for a curve y equals p of x, given an ordinary Cartesian coordinates, we just take dydx or equivalently, we calculate p prime of x. If the curve is defined parametrically by the equations, x equals f of t, y equals g of t. To find the slope of the tangent line, we still want to find dydx. But since our curve is given parametrically, we don't have ready access to the y dx. Instead, we'll need to calculate it from the Y DT and dx dt, which are easy to get from our parametric equations. to relate dydx to d y DT and dx dt, we just need to use the chain rule. Recall that the chain rule says that d y dt is equal to d y dx times dx dt. So rearranging, we know that dou y dx is equal to d y DT divided by dx dt. And that's how we'll calculate the slope of our tangent line, we can write this formula equivalently as d y dx is equal to g prime of t over f prime of t. Now let's use these formulas in an example. For the list as you figure given by these equations, and drawn below, let's find the slopes of the tangent lines at the center point with x&y coordinates of zero. And let's find the way the tangent line is horizontal. The slope of the tangent line is given by dy dx, which is d y DT divided by dx dt. Now d y dt is going to be cosine of two t times two and dx dt is going to be negative sine of t. Taking the ratio, we see that dydx is twice cosine of two t over negative sine of t. Now we want to calculate this slope not when t is zero, but when x and y are 0x is 01. cosine of t is zero, which is when t is pi over two and three pi over two. Those are the only two values that work in the interval of t values that we're interested in. And it's easy to check that when T has these values, then y, which is sine of two t, is also going to be zero. So we want to calculate d y DT at t equals pi over two, and at t equals three pi over two. plugging into our formula for dydx, we get twice cosine of pi divided by negative sine of pi over two, which simplifies to positive two. And when t is three pi over two, we get twice cosine of three Pi over negative sine three pi over two, which simplifies to negative two. So our tangent lines at the origin have slopes positive two, and negative two. Next, let's find where the tangent line is horizontal. From the figure, there should be four places. If we set d y dx equals zero, we get that two cosine of two t over negative sine of t needs to be zero, which means that we need cosine of two t t equals zero, or two t, two equal pi over two, plus some multiple of pi. Solving for T, we get that T has to equal pi over four, plus some multiple of pi over two. There are indeed four such values of t in the interval from zero to two pi. And those are pi over four, three Pi over four, five pi over four, and seven pi over four, we can find the x and y coordinates of these points simply by plugging in these values of t into our original equations. Here are the XY coordinates of those four points. In this video, we saw that for a curve given by parametric equations, the slope of the tangent line is given by D y dX, which is d y d t, divided by dx dt. In this video, we'll find the area under a curve defined parametrically. Recall that the area under a curve y equals p of x defined in usual Cartesian coordinates is just the integral from x equals a to x equals b of y dx. Or, in other words, the integral from a to b of p of x dx. If instead, the curve is given by the parametric equations, x equals f of t and y equals g of t, the area is still going to be the integral of y dx. But now y can be written as g of t, and dx is going to be f prime of t dt using differential notation. Therefore, the area is going to be the integral of g of t, f prime of t, dt. And since we're integrating with respect to t, now, our bounds of integration have to also be t values, I'll still call them a and b, but I want them to represent t values here. It's important to note that this is still the area under a curve. In other words, in between the curve and the x axis. Let's use this formula to find the area enclosed by this list as you figure given by these equations. by symmetry, it's enough to compute the area under one segment of the lisu curve, and then multiply that area, which I'll call a bye for now, a is equal to the integral of y dx. And using our parametric equations, we know that y is sine of two t, and dx is the derivative of cosine. So that's negative sine of t dt. the rightmost point of the section of curve that we're interested in Then right here happens when x is one and y equals zero, it's easy to check that that occurs when t equals zero. The leftmost point of the section of curve occurs when x and y are both zero, setting both our equations equal to zero and solving for T, we see this happens when t is pi over two, plus any multiple of pi. So the first time we reach this point, after the t value of zero is when t is simply pi over two. So we'll set our bounds of integration as from t equals zero to t equals pi over two. plugging this information into our equation, we get that the area is the integral from zero to pi over two of sine of two t times negative sine of t dt. Let's pull the negative sign out and use the double angle formula to rewrite sine of two t as two sine t cosine t, multiply that by the sine t dt, we can pull the two out and rewrite this as sine squared of t cosine t dt. And then I use substitution will allow us to compute the integral we get negative two times the integral from u equals zero to one of u squared d u, which integrates to negative two u cubed over three, evaluate between one and zero, which is negative two thirds. Notice I get a negative answer here. And that's because I accidentally integrated from right and point to the left endpoint instead of from the left to the right, I accidentally followed the t values in increasing order when I should have integrated in the other order in order to make the x values in increasing order. In any case, I can correct this by switching my bounds of integration, or more quickly just by sticking a negative sign in front and changing my sign. Now I can figure out the total area inside the list of geo figure just by multiplying by four. In this video, we saw that the area underneath the curve given in parametric equations, is given by the integral of y dx, which if x is f of t and y is g of t, this is just the integral of g of t times f prime of t dt, our bounds of integration need to be t values. In this video, we'll calculate the length of curves given by parametric equations. As a warm up, let's calculate the length of this curve graph below. Since it's piecewise, linear, we can just calculate the length of each linear piece using the distance formula. For example, the length of the first segment is given by the square root of y two minus y one squared, that's three minus two squared plus x two minus x one squared, that's two minus one squared, which gives us the square root of two. Similarly, the second piece has length given by the square root of five minus three squared plus three minus two squared, which is the square root of five, we can make similar calculations for the length of the third segment and the fourth segment. adding these together, we get a total length of twice the square root of five plus the square root of two plus two. We can use the same process to approximate the length of any curve by dividing it up into and small pieces, approximating each piece with a straight line and using the distance formula to find the length of the line segments. If the curve is given by the parametric equations, x equals f of t and y equals g of t. Then we can write each of these points along the line in terms of f and g. For example, P of i minus one, we can write as the x coordinate F. Have t minus one and the y coordinate g of t minus one. And similarly, p II is going to be x coordinate f of t AI, y coordinate g of t AI. If we think of t as time, then this is just saying that T sub i is the time at which we get to point a piece of AI. Now, the distance formula tells us that the length of the line segment from P sub i minus one to P sub i is going to be given by the square root of x two minus x one squared. That's f of t sub i minus f of t sub i minus one squared plus y two minus y one, that's g of t sub i minus g of t sub i minus one squared. Now, the total length of the curve, which is called the arc length, is going to be approximately equal to the sum of the length of these segments. This is starting to look a bit like a Riemann sum, but it's missing the delta t. So I'll fix that by multiplying the numerator and the denominator by delta t. If I suck the delta t from the denominator, inside the square root sign, it needs to become a delta t squared, and I get the following expression. If I break up my fractions, I can rewrite this. Now there's something kind of exciting going on this quotient here looks a lot like a slope. In fact, it's exactly the expression for the slope of the secant line for the function f that we get if we were calculating the derivative of f with respect to t. And similarly, this expression, and here is exactly the expression for the slope of the secant line, we'd get if we were calculating the derivative of g with respect to t. Because these quotients here, are approximately equal to f prime and g prime, or more rigorously, because of the mean value theorem, I can replace my expression with f prime of t i star squared, and g prime of t star squared, where ti star is some time in the if time interval. Now exact arc length is going to be the limit of this expression, as the number of intervals goes to infinity. As usual, I can replace the limit of this Riemann sum with an integral where the bounds of integration are the t values that get me from the start of the curve to the end of the curve. This arc length formula has an alternative form, which is the integral from a to b of the square root of dx dt squared plus d y dt squared, dt. And these are my two versions of this very useful formula for arc length. Now let's use this formula to set up an integral to express the arc length of this list as you figure since dx dt is given by negative sine of t. And the Y dt is given by two cosine of two t, our Clank is given by the integral of the square root of sine of t squared plus two cosine of two t squared dt, we do still need to figure out the bounds of integration in terms of t. That will make us wrap around this curve exactly one time. It's easy to check that when t equals 0x is equal to one and y is equal to zero. So we're at this point right here. The next time that we get to this point, with an x coordinate of one, we need x equal cosine t to equal one. So the next time will be when t equals two pi. Therefore, our bounds of integration are going to be from zero to two pi. It's easy to set up This integral, but it would be very difficult to actually integrate it. And that's often the case with our clients. But we could use a calculator or a computer to get a numerical approximation of about 9.4. In this video, we derive an equation for arc length. This video introduces the idea of polar coordinates. Polar Coordinates give an alternative way of describing the location of points on the plane. Instead of describing a point, in terms of its x and y coordinates, those are the Cartesian coordinates of the point. When using polar coordinates, we instead describe the point in terms of radius r, and an angle theta, r is the distance of the point from the origin, and theta is the angle that radius line makes with the positive x axis. Let's plot these points given in polar coordinates. So the eight here is the value of the radius, and the negative two thirds pi is the value of the angle theta. The negative angle means that I need to go clockwise from the positive x axis, instead of counterclockwise like I normally would for a positive angle. So here, a negative two thirds pi means that I need to go to this line right here. And the eight of for the radius means I need to go eight lines out from the origin. So my point should be around right here. The next point has a radius of five and an angle of three pi. The angle of positive three pi means that I go counterclockwise starting at the positive x axis, here, I've gone around by two pi. And here, I've got an extra pi to make three pi. Now the radius of five means I need to go five units out from the origin. So that puts me about right here. Notice that I could have also labeled this point with the polar coordinates of five Pi, there's more than one way to assign polar coordinates to a point. The next point has an angle of pi over four, and a radius of negative 12. The negative radius means that I need to jump to the other side of the circle before I plot the point. In other words, instead of plotting the point at an angle of pi over four and a radius of 12, which would be about right here, I go to the opposite side of the circle, and plotted at the same distance from the origin, but 180 degrees or pi radians around the circle over here. Now I could have also labeled this point using a positive radius of 12. And using an angle of pi over four plus pi, or five pi over four. And in general, a point with polar coordinates of negative r theta means the same point as the point with polar coordinates r, theta plus pi. Adding pi just makes us jump around to the opposite side of the circle. To convert between polar and Cartesian coordinates, it's handy to use the following equations. First, x is equal to r cosine theta, y is equal to r sine theta r squared is equal to x squared plus y squared, which means that R is plus or minus the square root of x squared plus y squared. And tangent theta is equal to y divided by x. Let's see where these equations come from. If we draw a point with coordinates, x, y, and draw lines to make a right triangle, the height of that triangle is y. The length of the base is x. And the high partners has length R. Theta is the measure of this interior angle. From trig, we know that cosine theta is equal to adjacent over hypotenuse, so that's x over r, which means that x is equal to r cosine theta. Similarly, sine theta is opposite over hypotenuse. That's why over R, which means that y is equal to r sine theta. That gives us the first two equations the batang orien theorem tells us that x squared plus y squared is equal to r squared. And that gives us the third equation. Finally, tangent theta is opposite over adjacent. So that's y over x, which is the fourth equation. To convert five, negative pi over six, from polar to Cartesian coordinates, we just use the fact that x equals r cosine theta, and y equals r sine theta. So in this case, x is equal to five times cosine of negative pi over six, that's five times square root of three over two, and y is equal to five, sine negative pi over six. So that's equal to negative five halves, did convert negative one negative one from Cartesian to polar coordinates, we know that negative one and negative one are x and y values. So we need to use the fact that r squared is x squared plus y squared, that is r squared is negative one squared plus negative one squared, or two. Also, tangent theta is y over x, so that's negative one over negative one, or one. Now there's several values of r and theta that satisfy these equations are could be squared of two, or negative the square root of two, and theta could be Pi over four, or five pi over four. Or we could add multiples of two pi to either of these answers. But not all combinations of r and theta, get us to the right point. The point with Cartesian coordinates negative one negative one lies in the third quadrant. But if we use a theta value of say pi over four and an R value of square root of two, that would get us to the first quadrant. So instead, we need to use the polar coordinates of square root of two and five pi over four. Or if we prefer, negative square root of two, and pi over four. We could also add any multiple of two pi to either of these values of theta, and get yet another way of representing the point and polar coordinates. This video talked about polar coordinates and converting in between Cartesian coordinates and polar coordinates using some familiar equations from trig.