in this video we're gonna talk about three different applications of Taylor series three different things from back in calculus one in calculus two problems that either were challenging to do or that you couldn't do by the methods of calculus one in calculus two but now that we know Taylor series we can go back and do them the first of these I'm going to focus on is an integration problem in particular consider the integral of e to the minus x squared DX and it's actually a very important integral this kind of distribution and probability and statistics is called a normal distribution it comes up all the time so the issue of doing integrals of this form are actually pretty important but we don't know how to do that by the methods of calculus 1 or calculus 2 we get for example do a u sub it doesn't work out and you'd want to try an integration by parts it's not gonna be nice in fact it turns out that there isn't a nice answer in terms of functions like e to the X and polynomials and trigonometry and so on but we can use series so for example so for example we know that the Taylor series for e to the X is just this sum from 0 to infinity of X to the N over N factorial indeed this is the Taylor series centered about X equal to 0 now this is e to the X but in my integrand i have e to the minus x squared so let me just substitute that in and everywhere I have an X I'm gonna replace it with a minus x squared now I'm doing the integral of this thing so let me go and take the integral of both sides of these things DX and then I'm gonna do one final just a little bit of cleanup job you see I've got the minus x squared all to the power of n let me take the minus sign out of that and so if I transform this what I'm gonna have is a minus 1 to the N and then an X 2 to it now what I'm gonna do with this Taylor series is I'm going to integrate it term by term and that is in effect I'm integrating some constant times X to the 2n what happened to that X to the 2n will go to X to the 2n plus 1 divided by 2n plus 1 which is going to give me this particular series the sum of minus 1 to the N over N factorial and then in X to two n plus one than you do I don't buy the two n plus one and finally because it's an indefinite integral you have to add a plus C so that's a perfectly acceptable answer we now have a power series answer to this integral of e to the minus x squared if we wanted to we could give this collection a name whatever you prefer you could call it after yourself you so much the only reason it doesn't have a name is because as a group of mathematicians we haven't ended up agreeing on one the way we agreed on a name for the special function e to the X and a special function sine of X and then if I want to I can also do a definite interval here so everywhere I've got my indefinite and replace it with a zero up to one and indeed when I have my final series I evaluate it from 0 up to 1 plugging in 0 doesn't do anything x to the 2n plus 1 when x is 0 is just 0 so that's all 0 and if I plug in 1 the X portions are gonna go away 1 to the 2n plus 1 is 1 and then just leave me behind minus 1 to the N over the N factorial and over that 2n plus 1 and this is a perfectly acceptable answer if in your application you needed to approximate this integral as accurately as you so wished well you could go and take the first 10 or 100 or million terms however main terms that you needed for your level of accuracy all right so that's an integration example next up let's look at a limit example so in this example I have an indeterminate form if I just plug in X equal to 0 I have 0 on the top and 0 on the bottom and what we learned back in calculus 1 is that in those scenarios you can do Lappe tal's rule in this case we have to do a once and twice and Lobby tells rule often works and D it works in this particular example but I'm gonna show you how I can do it with a Taylor series so in this expression I've got this e to the X and the cosine of X and both of those have series expansions which is to say if I go and plug them in so my x squared is still at the front but in place of e to the X I put its series and in a place of cosine of X I put its series let me try to clean this up a little bit so the numerator I have an x squared multiplied by a whole series so let me just put those together and the denominator I've got a plus 1 and a minus 1 let's cancel that is I'm gonna have this expression an x squared and higher-order terms on the top and a minus x squared over two and higher order terms on the bottom the fact that there's higher order terms here is important because I'm setting the limit as X goes to zero so that means that every one of these terms is gonna go to zero but the higher order terms go to 0 faster the X cubed goes to zero faster the next square does the X to the fourth go to zero faster than x squared does so when I look at the top and the bottom I try to figure out what just going to dominate as X is gonna go to zero well it's those lowest order ones that do it when when X is being sent to zero that is the things I really want to focus on is the first term on the top and the first term on the bottom so everything else is going to go to zero and it's gonna go zero in a way that is faster than these but these two are gonna cancel x squared over x squared cancels and it just leaves the coefficient 1 over minus 1/2 in other words this is equal to value of minus 2 so via this kind of analysis you're able to solve a whole class of limits that we're challenging before as well final application for Taylor series it's actually going to be series themselves so for example this is the Taylor series for e to the X the one over zero factorial plus X over one factorial is x squared over two factorial plus X cubed over three factorial and so on well what if instead I was giving you this problem this expression where there is no X there's a lot of ones places but it's just some particular series now if this was earlier on in calculus 2 we would have said well does it converge or dies that diverge and you could have used a ratio test or maybe a comparison test through a bunch of options but you would have deduced that yes it converges by some particular test but what does it converge to that was an open problem or at least would have been an open problem before but now if I look at this example this example is nothing but e to the X where I've plugged in X equal to 1 everywhere those entire numerators are now all 1 that's me plugging in X equal to 1 so what is this series well it's just whatever e to the 1 is that is it is e or if I slightly change things up if I consider the same basic example but I put some plus minus plus minus things again if we were doing a convergent test maybe you'd he was I don't know the alternating series test and deduce that it converts but what did it converge to well this thing is exactly e to the X but I am plugging in not executable one now but X equal to minus one indeed for like the X and X cube and X to the fifth and odd power terms put it in minus one it's gonna stick out a negative and for all the even power terms putting in minus one is not going to do anything it's gonna become positive so this is just e to the power of minus one aka 1 over e so in this video we've seen these three different ways that I can apply Taylor series to problems from the past in our development of calculus