in this video we're going to talk about how to represent a function as a power series so this is going to be the first example let's say the function f of x is 1 over 1 plus x how can we write that as a power series well first you need to be familiar with a geometric series now this two forms that you might have seen it here's one form a times r to the n is equal to a over one minus r this is the sum of an infinite geometric series now sometimes you might see it in this form so rather than starting at n equals zero sometimes it may start at n equal one in that case it's going to be the form a times r to the n minus one where a is the first term of the series and the sum is still going to be the same now r is the common ratio and keep in mind a geometric series will converge if the absolute value of r is less than one in this lesson we're going to represent our power series starting from the n equals zero value as opposed to the n equal one value now a power series contains a variable x raised to the n power so a geometric power series will be in this form x to the n in this case a is 1 and so the sum for that is going to be 1 over 1 minus x notice that this formula looks very similar to that one so therefore this function should have a power series that looks somewhat like that but not exactly like that now this particular power series is centered at x equals zero if you want to write a power series centered at x equals c you can write it in this form so it's going to be x minus c raised to the n power where c is some number so just keep that in mind so let's start with this example right now we need to write this function in the form of a over one minus r so somehow we need to convert the positive sign to a negative sign how can we do that now it's important to know that a negative times a negative is a positive so we can write it like this 1 plus x is the same as 1 minus negative x and so now it's in this form so looking at this equation what is a and what is r so notice that 1 is equal to a and negative x is equal to r so now that we have the a value and the r value we can now write the geometric power series so it's going to be in this form starting from 0 going to infinity a times r raised to the n so this is going to be 1 times negative x raised to the n power now negative x is negative 1 times x so this is going to be negative 1 to the n times positive x to the n and so this is our power series for this particular example this is the answer now there's still a few more things that we need to do the next thing is we need to find the interval of convergence so how can we go about doing that so recall that the common ratio has to be less than one in order for the geometric series to converge now we know that the common ratio is negative x so therefore the absolute value of negative x has to be less than one now once you have it in this form the absolute value of x minus c it's less than r in this case c is zero but r is going to be this number if there's no coefficients in front of the absolute value symbol so therefore r is the radius of convergence and in this example it's equal to 1. now let's determine the interval of convergence the absolute value of negative x is the same as the absolute value of x because the absolute value of negative one is one and so now we can get rid of the absolute value symbol and say that x is between one and negative one and that is the interval of convergence so what does that mean before we get into the answer of that question let's plot the interval of convergence on a number line so it's centered at zero and it's from negative one to one so we have an open circle at the end because negative one and one are not included so when dealing with a geometric power series you do not need to check for the end points because it's always going to be parenthesis the endpoints will never be included and the reason for that is due to the fact that the common ratio has to be less than 1 in order for the geometric series to converge remember if it's equal to or greater than 1 then the geometric series will diverge so the interval of convergence for a geometric power series will never include the endpoints so you should always have parentheses and not a bracket so this is our center our x equals c value and the radius of convergence is basically one half of the interval so in this case the radius of convergence is one now let's talk about what it means so what does it mean that the interval of convergence is between negative 1 and 1. so recall that the function f of x which is one over one plus x it can be represented by the power series from zero to infinity negative one to the n x to the n now let's write out the first five or maybe six terms in a series so when n is zero negative one to the zero is one and x to the zero power is one so the first term is one when n is one we're going to get negative x when n is 2 negative 1 squared is 1 and so it's going to be positive x squared and then it's just going to repeat so you can write out as many terms as you want to and that's just going to go to it's going to go on forever now this function can be represented accurately by this series only when x is between negative one and one so let's calculate the value of f of point five so if you plug in into your calculator 1 over 1 plus 0.5 or 1 divided by 1.5 you should get 2 over 3 which is 0.6 repeating so i'm going to round that as 0.667 now let's use the power series to estimate the answer and you want to choose an even amount of terms since we have an alternate series so you want to use either like 6 or 8 or 10 terms not like 7 9 or 11 because your answer is going to be a little off so i'm going to use the first a terms so it's going to be 1 minus 0.5 plus 0.5 squared minus 0.5 cubed plus 0.5 to the fourth minus 0.5 to the fifth plus 0.5 to the sixth power and then minus 0.5 to the seventh power go ahead and take a minute to type that in your calculator the more terms that you add the more accurate your answer will be to the true value so stop it at negative 0.57 i got approximately 0.6641 so 0.6641 is a good approximation of 0.667 so we can use a geometric power series to represent almost any rational function in the form of a polynomial now let's work on some example problems so let's say that f of x is one over x go ahead and represent the function as a power series and find the interval of convergence so the first thing we need to do is we need to put it in this form a divided by one minus r so right now we're missing a one so how can we introduce a one into the denominator of the fraction so when you're adding new numbers into the picture you cannot change the value of what was there so in the bottom the value is equal to x now we can add 0 to x 0 plus x is still x but adding 0 won't help us however 1 plus negative 1 is equal to 0 and so that's what we're going to add to the bottom of the fraction because the value of the fraction will still be equal to x now i need a negative sign between 1 and r so i'm going to move this one to the left side so right now i have 1 plus x minus 1. now i need to turn the positive sign into a negative sign how can i do that the best way to do that is to factor out a negative one as your greatest common factor and so it's going to be if we take out a negative from positive x we're going to get a negative x and then if we take out the negative sign from negative one it changes to positive one and so this is what we have so now it's in the right form so notice that a is equal to 1 and r is equal to negative x plus one so now that we have a and r we can write the geometric power series so let me get rid of this so now let's use this expression and then all we need to do is replace a and r and then we can simplify it so a is one we can ignore one you don't have to write one so we're going to have the summation symbol and then r to the n which is negative x plus 1 to the n now can we leave the answer like this when representing a function as a power series you want to write it in this form x to the n or you want to write it in this form x minus c raised to the n so right now we have negative x and not positive x so therefore we're going to have to factor a negative 1 inside the parentheses so this is going to be let me use brackets so we're going to have brackets negative 1 times positive x and then positive 1 will change to negative 1. and this is all raised to the n power now i'm going to distribute n to the negative 1 on the outside and x minus one on the inside so this is going to be negative one to the n power times x minus one to the n starting from n equals zero going to infinity so this is the answer that's the power series that corresponds to the function one over x now we need to find the interval of convergence we need to know the x values where one over x can be represented by that particular power series so notice that this power series is centered at x equal one because if you plug in one for x you're going to get zero so we can say that c is equal to one now what else do we know we know that the common ratio of the geometric series is negative one times x minus one and we need to write that because it's going to help us to find the interval of convergence that's the next step so for any geometric series the absolute value of r has to be less than one in order for it to converge so in this case the absolute value of negative one times x minus one that has to be less than one so the absolute value of negative one is one so we could just ignore that number so now we have this expression which we can write this as x minus one has to be less than one but greater than negative one so if we add one to both sides x is less than two but greater than zero and so the interval of convergence is from zero to two and you could see that c is one one is the midpoint of zero and two the radius of convergence is basically this number that you see here when you have it in the form of the absolute value of x minus c is less than r so the radius of convergence is 1. so plotting it on a number line we have one at the center and zero and two at the end and keep in mind zero and two is not going to be included for geometric power series it converges between 0 and 2 but add 0 and 2 the series diverges so this is our c value and this is r now let's try another example so let's say we have the function one divided by one minus x cubed go ahead and represent the function as a power series so this function is already in the appropriate form so we could see that a is equal to one and the common ratio r is equal to x cubed so we can write the power series using this expression so we have the summation from 0 to infinity and this is going to be 1 times x to the third raised to the n so we can write the final answer like this it's going to be x to the 3n now if we decide to list out a few terms when n is zero x to the three times zero that's x to the zero so that's gonna be one and then when n is one it's gonna be x cubed when n is two two times three is 6 so x to the 6 and so forth and so this power series is equal to 1 over 1 minus x cubed for certain x values and those x values will be based on the interval of convergence to find the interval of convergence start with this expression the absolute value of the common ratio is less than one and we can see that the common ratio is x cubed so we could say that the absolute value of x cubed has to be less than one so that's x cubed has to be between one and negative one now if you take the cube root of each side of the inequality you're going to get x is between 1 and negative 1. therefore the interval of convergence is just negative one to one and that's it for this problem the next problem is going to be a little different than the last one in this example we're going to have the function 1 divided by three minus x and we're going to do two things let's write the representation of this function as a power series where it's centered at zero and when it's centered at 1. now keep in mind when it's centered at 0 you want to put it in this form x to the n but when it's centered as some other number you want to put it in the form x minus c raised to the n so let's start with c equals zero and let's put the fraction in this form so notice that we have a three somehow we need to change that to a one so what we need to do is multiply the top and the bottom by a third so we're gonna have one third divided by one minus x over three and so now it's in the appropriate form so we could see that a is equal to one-third and r is equal to x over three so writing our answer in this form it's going to be 1 3 times r which is x over 3 to the n now let's simplify the expression so one over three that's just one over three to the first power and now let's distribute the exponent so this is x to the n divided by 3 to the n and now we can combine these two into a single number so let's add the exponents 1 plus n so this is going to give us the power series x to the n divided by 3 to the n plus 1 from 0 to infinity so this is the answer let's find the interval of convergence for this problem and let's start with this expression now we said that the common ratio is x over three so therefore the absolute value of x over three has to be less than one and let's take out the three so this becomes one third absolute value of x is less than one and now our next step is to multiply both sides by three and so the absolute value of x is less than three now once you have it in this form you can identify the radius of convergence so the radius of convergence is equal to three and x is between three and negative three centered at zero so the interval of convergence is from negative three to three and so this is the second part of the answer now let's go back to the original function 1 divided by 3 minus x and let's write the power series representation when c is equal to one so we need to put it in this format that is x minus c raised to the n and so that's going to be x minus one raised to the n power some sort of variation of that so the first thing we need to do is get the x minus 1 term so how can we get an x minus 1 expression in the denominator how can we introduce it and notice that we need the negative sign to be in front of it as well so if we distribute the negative sign this is negative x plus one somehow we got to take a one from the three so let's break down three into two plus one so we have two plus one minus x which is the same as three minus x so whatever changes you do make sure that the value of the fraction remains the same so now i'm gonna write this as two minus x plus one so i haven't changed the value the only thing i did is i move the one to the right now the next thing i'm going to do is take out a negative 1 from negative x plus one if i do so negative x becomes positive x and positive one becomes negative one so now it's centered at x minus one and the value is still the same two minus x minus one this whole thing is still equal to three minus x now it's almost in this format the next thing we need to do is change the two into a one and so to do that we're going to multiply the top and the bottom by one half and so this is going to be one half divided by two times one half that's one and then this times one half so i'm going to write it like brackets one half and then parentheses x minus one do not distribute the one half to x minus one the x minus one part is your x minus c term now it's in the appropriate form so we could see that a is one half so let's write that down and what is the common ratio in this example the common ratio r is one half times x to the minus one so now that we have that let's get rid of this and let's determine the interval of convergence first so let's start with this expression so that means that the absolute value of one-half times x minus one has to be less than one so taking the one-half to the front so we have this one-half x minus one is less than one so we need to multiply both sides by two so the absolute value of x minus one is less than two so now it's in this form and so the radius of convergence we can clearly see that it's equal to two so now let's get rid of the absolute value expression so we're going to have that x minus 1 is less than 2 and greater than negative 2. so now let's add one to all three sides and so x is less than three but greater than negative one so the interval of convergence is negative one to three now let's plot it on a number line so it's from negative 1 to 3 and our c value is the midpoint of those two values so c is one so this is where x is equal to c and so the geometric power series will converge anywhere where x is between negative one and three so now let's go back to our original function and so this function centered at c equal 1 can be represented by this particular geometric power series right now this is the geometric series but once we replace r with x then it becomes a geometric power series so a is one half r is going to be one half times x minus one raised to the n power now this one half outside of it we can write it i'm not sure what just happened there but we can write it like this it's gonna be one over two to the first power and then the n we're going to distribute to the one half inside the brackets and the x minus one so it's going to be times one over two to the n times x minus one raised to the n power so we can write the final answer like this it's going to be from 0 to infinity x minus one raised to the n divided by and these two we can combine and write it as a single two so two to the n plus one so this is the series for one over three minus x where the series is centered at one and you could see the x minus one term now that you've seen how to do the last example you should be able to do the next one consider the function f of x is equal to eight over two x minus nine now what i want you to do is write our power series representation for this function and this time we're not going to center it at one but let's center it at three so go ahead and try this problem so if we're going to center it at 3 we need the expression x minus c or x minus 3. and notice that there's a 2 in front of the x so that's we're going to have 2 times x minus 3 and if we distribute the 2 it's going to be 2x minus 6. somehow we need to get a negative 6 into this expression there's two ways we can do this we can add negative six and add plus six to the bottom of the fraction or we could just break down negative nine as negative six and negative three so let's just do it this way for now so this is going to be equal to eight divided by two x minus six minus three now since i wanted to put it in the a1 over 1 minus r form i need x to be on the right side not on the left so i'm going to move the negative 3 to the left side now this becomes eight divided by negative three plus two x minus six so now the next thing i need to do is take out the 2 and get x minus 3. in addition i need to get rid of the positive sign so recall that a positive number is the multiplication of two negative numbers so what i now have is eight over negative three minus now i'm going to take out right now i'm just going to write it like this for now so i didn't take out the two yet but these two negative signs they will cancel to make a positive sign so these two expressions are equivalent as of now now i'm going to take out the 2 from 2x minus 6. so this is going to be negative 2 times x minus 3. now the next thing i need to do is change this number to put it in a over one minus r form i need this to be a one so i got to multiply the top and the bottom by what number i need to multiply the top and bottom by negative one-third and so this is going to be negative eight over three since i'm running out of space i'm just to put this at the top so it's negative 8 over 3 on the top and then we have negative 1 3 times negative 3 which is positive one and then minus let's keep this negative sign let's not change it and then negative two times negative one third that's going to be inside the brackets but positive two over three and then times x minus three so as you can see it's now in the appropriate form so we can clearly see that a the first term is 8 over 3 negative 8 over 3 that is and the common ratio is everything inside the brackets two-thirds x minus three as soon as you identify a and r you don't need this equation anymore so now let's write the geometric series so a is negative eight over three and r is two over three x minus three and we're gonna raise r to the n power so this is the answer but we need to clean it up a bit how can we simplify this expression now the first thing we have is a negative sign so i'm going to write that as negative 1 for now and then we have an 8. notice that 8 and 2 they share the same base a common base of 2. so i'm going to rewrite 8 as 2 to the 3rd power and the 3 on the bottom that's going to be 3 to the first power and then everything inside the brackets i'm going to distribute the exponent n to it so this will become 2 to the n power divided by 3 to the n power and then times x minus 3 to the n power so we can write the final answer from 0 to infinity and then negative here we have 2 to the third times 2 to the n so we're going to add 3 and n and just write that as n plus three times x minus three to the n divided by and here we have one plus n or n plus one so it's divided by three to the n plus one so this is the power series representation of our original function which was eight over two x minus nine and this is true when it's centered at three and we can see three in our expression x minus three which tells us that it's centered at three the only thing we need to do at this point is find the interval of convergence so as always let's start with this expression so r is two over three times x minus three now let's remove the two thirds from the absolute value expression to get rid of the 2 over 3 on the left side and we need to multiply both sides by 3 divided by 2. and so on the left the 3's will cancel and the 2's will cancel so now we're going to have the absolute value of x minus 3 is less than 3 over 2. so now it's in this form x minus c is less than r so the radius of convergence we can clearly see that it's 3 over 2. and now let's get rid of the absolute value expression so x minus 3 is less than 3 over 2 but greater than negative 3 over 2. so once you get rid of the absolute value expression and you write it like this all you need to do is solve for x so we're going to add 3 to both sides now 3 plus 3 over 2 we need to get common denominators 3 is the same as 6 over 2 so we're going to add 6 over 2 to all three sides so these will cancel three over two plus six over two three plus six is nine so this is going to be nine over two and negative three plus six is positive three so that's positive three over two so as a decimal you could say that the interval of convergence is between three over two and nine over two three divided by two is one point five and 9 divided by 2 is 4.5 so anywhere between 1.5 and 4.5 the power series will be a very good representation of the function in fact it's going to be an exact representation of the function anywhere outside of that the power series will diverge now let's plot the interval of convergence on a number line so we said it's between negative 1.5 i mean positive 1.5 and 4.5 and it's centered at c equals 3. the radius of convergence is one half of the difference between 4.5 and 1.5 and so we said r is 3 over 2 which is 1.5 and so that's the distance between 3 and 1.5 and it doesn't include the endpoints so the interval of convergence is between 1.5 and 4.5 but it does not include 1.5 and 4.5 itself and so that's it for this problem consider the next example f of x is equal to x cubed divided by x plus 2. so what do you think we need to do for this problem in order to write the power series representation for that function notice that it doesn't appear to be in this format a is a constant but we have a variable in the numerator so for this particular example what do you think we need to do what we need to do is separate the function into two parts so we can write this as x cubed times one over x plus two and so we're going to write the power series representation of this part first so let's focus on that now we need to put x on the right side so right now what we have is 1 over 2 plus x and we need a negative sign and we're going to write the function where it's centered at c equals zero if it's not specified always assume c equals zero now we need to convert the positive sign into two negative signs so 2 plus x is going to be 2 minus negative x now the last thing we need to do is multiply the top and the bottom by one half so this becomes one-half over one minus negative x times one-half that's going to be negative x divided by two so now this is in this form so we can see that a is one half and r is negative x over two so at this point we can write the power series representation for this expression so it's going to go from 0 to infinity and then it's going to be a which is one-half times r to the n power so this is 1 over 2 to the first power and then if we distribute the n it's going to be negative x to the n power divided by 2 to the n power so let's simplify that negative x to the n power we can write negative x as negative 1 times x and they're both going to be raised to the n power because the negative was on the inside of the parentheses and then we can combine these two so that's going to be 2 to the n plus one and so this is equal to one over x plus two now if one over x plus two is equal to this series then f of x which is x cubed times one over x plus two that's going to be x cubed times the power series that we have up there at this point all we need to do is multiply x cubed by x to the n and so we're going to add 3 plus n so we're going to get the power series from 0 to infinity of negative 1 to the n times x raised to the n plus 3 divided by 2 to the n plus 1. so this is the power series representation of x cubed divided by x plus 2. so sometimes you need to break down a function into two separate parts and then just multiply the other part if you can combine it with something in this expression the last thing we need for this particular problem is the interval of convergence and early in this problem we said that r is equal to negative x over 2. and we know the absolute value of r has to be less than 1 in order for the geometric series to converge so the absolute value of negative x over 2 has to be less than one now we can break this apart into two absolute value expressions negative x over two is basically negative one-half times x and the absolute value of negative one-half is positive one-half so we have one-half times the absolute value of x which is less than one now let's multiply both sides by two and so the absolute value of x is less than two and so this is the radius of convergence which is two and to get rid of the absolute value expression we could say that x is less than two but greater than negative two so the interval of convergence is from negative two to two and so that concludes this problem that's all we need to do let's work on one more problem so consider the function three divided by x squared plus x minus two what do we need to do for this example for this particular rational function the first thing we need to do is factor the trinomial in the denominator of the fraction so x squared plus x minus 2. if we find two numbers that multiply to negative two but add to the middle coefficient of one in front of x that's going to be positive two and negative one so it factors to x plus two times x minus one two times negative one is negative two two plus negative one is positive one now we need to split this fraction into two smaller fractions and so we need to use partial fraction decomposition so we're going to write this as a over x plus 2 plus b divided by x minus 1. now let's multiply both sides of the equation by the denominator on the left side so that is by x plus two and by x minus one so these will cancel therefore on the left side we're just going to get three and then once we multiply these two the x plus two terms will cancel leaving behind a times x minus one and then when we take b over x minus one multiplied by these two factors the x minus one factor will cancel giving us b times x plus two now let's see what happens when x is equal to one when x is equal to one a times x minus one that becomes zero so we have three is equal to b times one plus two which means three is equal to three b and so b is one and when x is negative two we have three equals a times negative two minus one plus b times zero and so this is going to be negative 3a equals 3 and so a is negative 1. so therefore f of x can be written as negative 1 over x plus 2. plus one over x minus one so we said that a is negative one and b is one now let's write the power series representation for each of these two fractions separately so let's start with one divided by x minus one so the first thing i would do is reverse x to negative one so this is one over negative one plus x so we need to get it in this form now what i would recommend doing is multiplying the top and bottom by negative one so it's negative one over one minus x and now it's in the appropriate form so we could see that a is negative one and r is equal to x so thus we can write the series as a which is negative 1 times r which is x raised to the n and so we can leave it like that now for the interval of convergence we know that r has to be less than 1 and r is x so x is between one and negative one so for this series the interval of convergence is between negative one and one now for the other one one over x plus two we did that in the last example before we multiplied it by x cubed and so for that example we had the series it was negative 1 to the n times x to the n over two to the n plus one so we don't need to redo this again and the interval of convergence for this one we found it to be negative two to two now when you combine in two series together if you're adding or subtracting them you need to find the intersection of these two intervals so at what x value will both intervals be true it's going to be true for the more restrictive interval that is from negative one to one so this is going to be the interval of convergence for the entire problem because let's say if you plotted both answers so here's the first one this is for one over x plus two and for the other fraction 1 over x minus 1 the interval of convergence is more restrictive so for the combined function we need to choose the region of the two number lines where it's both shaded notice that it's shaded between negative one and one for both number lines and so that is the intersection of the two intervals so therefore the interval of convergence for the original function will be from negative one to one it has to be true for each individual fraction now let's put this all together so the original function we broke it down into negative 1 over x plus 2. plus one over x minus one now the power series representation for one over x plus two we said that's negative one to the n times x to the n over two to the n plus one and for one over x minus one we find it to be negative one times x to the n now this negative i'm going to move it inside the expression so i'm going to move the negative one to the front so to speak so this is going to be negative 1 to the 1 power times negative 1 to the n the only reason why i chose to move it here is to combine it with negative 1 to the n now this negative i'm going to move it outside because it doesn't have an n power to it so this is just going to be minus x to the n now i'm going to combine these two power series into one power series so notice that i can factor out x to the n so let's take that out and then negative 1 to the 1 times negative 1 to the n you can combine that as negative 1 to the n plus 1 and then divided by 2 to the n plus 1. and since we factor out x to the n we're going to be left with a negative 1 out here and so this is going to be the power series representation for the original function so this is the answer and the interval of convergence is negative one to one which means the radius of convergence is going to be one-half the difference between the two extremes so one minus negative one that's two so half of two is one so the radius of convergence is one in this example and so now you know how to write a power series representation for a problem where you have to use partial fraction decomposition to split into two parts you