 Welcome back everyone and this video we're going to show that e to the x is equal to its Maclaurin series the sum We're in ranges from n equals zero to infinity of x to the n over n factorial Now in a previous video we have already determined that this is the Maclaurin series for e to the x This is the Taylor series centered at zero and hopefully you should see a link on your screen right now to that video If you're curious on how one does that computation What we plan to do right now is actually prove that e to the x is equal to this Maclaurin series It's it's it's only one question to find the Maclaurin series But actually to ask to determine that these two things are equal to each other It's a separate thing and that's what we're talking about right now Now in order to show that e to the x is equal to its Maclaurin series We have to show that the remainder term r n of x Which is a reminder. What does this new patient mean here in this context that the function In question, which for us will be e to the x minus its Taylor polynomial sequence Right, so r n of x is f of x minus its Taylor polynomial We have to show that this thing converges towards zero As n goes to infinity So as the Taylor polynomials converge towards the Taylor series We want the difference of that with the function to go off towards zero And that would force that the function's equal to its Taylor Series because we know that t in the Taylor polynomials converge towards the Taylor series t of x And so if they're equal then they'll convert to that as well Right, but the only thing is since t of x is the limit of the Taylor polynomial sequence The only way it could get sufficiently close to f of x is if these two things are equal right A sequence can only have one limit and so if something else is is acting like a limit You know if it walks like a limit and it quacks like a limit That mean that it's the limit as well and that'll force f of x to equal the Taylor series That's what that's our that's our approach here And the goal here is to use Taylor's inequality Which we had learned about that in a previous video Taylor's inequality remember tells us that the remainder r n of x It is bounded above by m times x minus a to the n plus one over n plus one factorial And so this is going to be true for all x's That are sufficiently close to a and so what that means is we have some fixed finite value d So if you're no more than d units away from the center a is the center there This this applies only to that situation right there And likewise we also have that this number m is an upper bound for The n plus first derivative of the function at a Likewise where uh, not not of a sorry of x Likewise on this interval. So there's some stuff to unravel right there So now let's use some specifics about this function. We have in play So first of all our center is going to be zero because we're looking at in the chlorine series And so statements like this is that we need x to be less than or equal to some fixed value d Now I want to mention that this thing is an arbitrary number right now We will eventually allow for d to get bigger and bigger and bigger and bigger. But for right now We're gonna we're gonna keep d as a fixed number We're gonna allow it to get bigger in just a second So we have to consider x's whose absolute value is less than or equal to d great So the next part is we have to deal with this m right here. What do we do with the m? Well, that comes from this that comes from this n plus first derivative And so let's think about that. What are the derivatives of e to the x? Well, if f of x equals e to the x as we know already right the derivative e to the x is equal to e to x itself And in fact all derivatives of e to the x are equal to e to the x right and therefore the n plus first derivative Is likewise going to equal e to the x so if If x has absolute value less than d Let's think about the function for example the the n plus first derivative if we were to graph it It would look something like this. We know the graph of e to the x Let's just kind of sketching it out right here. We would get A graph that looks something like this. This is just our natural exponential And so what we've done is we've specified some specific interval right if you have d and d right? So here's d and here's negative d. What's the biggest that e to the x gets in this interval? Since it's an increasing function that's going to happen right here And this happens at the point d comma e to the d Like so and so what this tells us is that our m value m can be set to be e to the d Because if you are inside this interval negative d to d the function Never gets bigger than e to the d. So that's the biggest value we can use so we can use m to be that So now let's look at Taylor's inequality again the absolute value of our n times x This will be less than or equal to e to the d times the absolute value of x raised to the n plus one power over n plus one factorial So we get like that We get we get something like that now. Okay continuing on We get this inequality. Let's now consider Uh, let's consider what happens as we allow this thing to go off towards Infinity now first of all, I should mention that this this is going to be bounded below by zero, right? The error can never be less than zero. It has to be it's probably going to be a positive number But I mean it could be it could be zero. It could be that we're perfect in our calculation here, right? Um, so we're going to try to set up some type of squeeze argument right here, right a squeeze A squeeze going on right here for our n of x A squeeze so what happens to what happens to the term on the left as n goes towards infinity, right? That that's that's the squeeze we're trying to get right now So if we investigate that limit in a little bit more detail, right? The limit as n goes to infinity of e to the d times the absolute value of x to the n plus one over n plus one factorial Now e to the d notice it does not depend on n whatsoever So as a constant we can bring it out of the limit here So we get a constant e to the d times the limit as n goes to infinity You get this absolute value of x over n plus one over n plus one factorial Now this limit, um, I'm going to make a slight modification here because as n goes to infinity so will n plus one, right? So as n goes to infinity n plus one will also go to infinity So we can honestly get rid of the plus ones in the denominator right here It's not changing the limit calculation whatsoever We get e to the d times the limit as n goes to infinity of The absolute value of x to the n over Infectorial so a little bit simpler there and it turns out that this limit is equal to zero That's an important thing to mention here So we're going to get we're going to get e to the d times zero Which does equals zero right here. So that's an important value to get and how do we get that? How did we get that this limit is equal to zero? This one can be a little bit of a challenge to do Um, there's there's sort of like, uh, there's a couple ways one could try to do this limit Um, but one thing I like to mention is let's come up above right Let's look at this thing right here. Hmm. This looks surprisingly similar to the limit We're looking at just a moment ago x to the n over in factorial if this series is in fact convergent Right, uh, which it is right. We showed previously that this Maclaurin series has a radius of convergence of r equals infinity That is it's convergent for all real numbers negative infinity to infinity, right? x it'll be convergent for every real number positive negative whatever right and so this series is convergent now a series that's convergent that implies That the sequence Actually must converge towards zero That's that we usually use it the other way around right that's we call it the tests for divergence That the if the sequence doesn't go to zero that means the series is divergent But originally we proved that if the series is convergent then the sequence has to converge towards zero, right? And so that's the sequence that we have in play right now Let me erase this The sequence we have in play right now is x to the n over in factorial. Uh, so come coming back down here Right the only way the only way that the sequence here converges I mean it has to converge because the associated series converges So it has to go to zero. So we get this limits equal to zero. So that that's the first step So for a fixed So for a fixed d remember that's the assumption we had before so for a fixed d What we see is that this inequality right here Goes off towards zero This goes to zero on so then that's where the squeeze theorem comes into play right because r n of x It's it'll be greater than equal to zero and it'll be less than equal to zero as n goes to infinity And so by the squeeze theorem we get that R n of x will go off towards zero But we have to be careful for that We have to say that four For the absolute value of x is less than d. So for a finite interval we can guarantee We can guarantee that the remainder will the remainder will go to zero That means e the x is equal to its Taylor series at least as long as x is not bigger than d So let's relax d. Let's let let's let d get a little bit bigger, right? What if we pick pick a bigger d or even a bigger d, right? What happens to that situation? Well As d gets bigger and bigger and bigger bigger bigger bigger bigger bigger pick you pick the number in question as d gets bigger bigger bigger bigger This calculation here makes no difference whatsoever. It never changes and therefore as we take the limit Uh, so if we allow d to approach infinity in this discussion right here Then we still get that r n of x will converge towards zero And therefore that then gives us the proof that e to the x is equal to It's maclaurin series So that that actually confirms this observation here And so that's that's a little bit of a difficult argument for many students the first time they see it here So what we're saying is we have a fixed d. We show that taylor's inequality goes to zero Um, and then we allow d to relax with the statement that oh it doesn't really depend on d too much anyways And therefore we get all possibilities, right? And so what we're saying is when when x is two we can set d equal to two and that limit will go to zero So each square is equal to the maclaurin series if we want to do e to the 12th We allow d to be 12 and then we do that limit the 12 didn't really matter You're going to get the r n of x is going to go to zero And therefore e to the 12 is equal to the the the series and you just do this for every value of x And we can actually do all of them simultaneously And we see that e the x is equal to its own Its own maclaurin series Now one other comment I want to make a mention here before we stop this video Is that what happens if you take the derivative of this function? Take the derivative of this function. This is going to be kind of fun here. Well, the right hand side You're going to get the series will by the power of you're going to n x the n minus one Over n factorial and this is going to go from n equals one to infinity, right? Because you lose the constant term when you take the derivative But look at that you have an n on top you have an n factorial on the bottom the n will cancel with this guy leaving an n minus one factorial on the bottom In which case you end up with The sum of n equals one to infinity You get x to the n minus one over n minus one factorial Now notice that you have n minus one right here if you plug in the first term n equals one into n minus one You actually get a zero and so by an index shift We actually could rewrite this sum as the sum where n equals zero to infinity Of x to the n over n factorial Which this is just e to the x like we saw a moment ago And so this this I hope is a good thing to notice here that the maclaurin series for e to the x Is exactly a series Which if you take the derivative of the series using the derivative techniques We've learned for series in the past that the derivative of this series gets back itself It's kind of like this interesting thing where you have this like infinite line Of numbers, right? This you know, you know, there's this little game, right? There's there's nine in the bed and the little one said roll over roll over one fell out right You know this this little nursery rhyme that kids know Um, the things we have this infinite number of little bears in the bed And so when one falls out all of them roll over we still have an infinite amount Um, and so this maclaurin series has the property that we take its derivative You get back e to the x Which is a good thing because the derivative of e to the x is equal to e to the x and in fact Some people when it comes to this this function e to the x they actually define e to the x That is they might say something like oh, we're going to define a function exp Of x we're going to find this function by this maclaurin series And this maclaurin series has the property that it's equal to its own derivative So we don't start off with some exponential function and this weird number e We actually start off with the property that we want we want e to we want a function which is equal to its own derivative It's it's a solution to a very important differential equation And then once you have that like okay, you have this fact I'm going to clear out some of this stuff right here You now have this observation that I've created a function using power series So that e to the x prime is equal to e to the x Or I should say exp prime We have this function which is equal to its own derivative wonderful We've created a solution to a differential equation Then you can actually discover from this that it's exponential Because if you set exp of equal to one right, there's a number here And then you can call that number you'll find out it's rational You can call that number e it's a great number And then from there you can actually show that exp of x is equal to e to the x You can show that this function we've created acts like an exponential function You basically have to prove the exponential rules You want to show that you'd want to show that exp evaluated at a plus b Is the same thing as exp times a times exp To the b that basically proving that right there would prove that exp is an exponential function Which would necessarily have to be this natural exponential Now proving this right here does require doing some algebra with power series You have to factor it but You know and so that that that that takes some work to do But if you actually compare how we did it the other way around in calculus one There's there's some fruit to this perspective that we actually create the natural exponential Not as just some arbitrary number, but we choose it because we want it to be equal to its own derivative I cannot emphasize enough how important it is that an actual exponential is equal to its own derivative It's a fundamental principle for difference of equations the integrating factor We use for linear differential equations is based upon this observation right here And so we could have actually introduced this transcendental function by using power series We created a function which is equal to its own derivative And so if we can do that if we can sort of tailor make no pun intended there I'm jk that there's a pun intended there if we can tailor make A solution to a differential equation using tailor series Like we can do that for the differential equation y prime Equals y what stops us from doing that for other differential equations And although we will not see that in this series I hope this is starting to open up your minds the power that A power series can do again pun not intended jk. It really was intended We can there's a lot of power to power series We can use them to to tailor make solutions to differential equations much like y prime equals y right here with e to the x