 In this video, we're gonna find the Taylor series for the exponential function f of x equals e to the x and we wanna identify its radius convergence. And we're gonna do this for two different values of a, right? So the a is the center of the Taylor series. Now, we're gonna start off by looking at the Taylor series centered at zero. This is what's referred to as the Maclaurin series for e to the x. And this is actually a pretty important one that we're gonna want to remember. Now, as we saw previously, when we have our Taylor series, we see that the coefficients of the Taylor series start from zero and go towards infinity. The coefficients of the Taylor series is gonna be the nth derivative of the function evaluated at the center a and then divide that by n factorial and then you'll times that by x minus a to the nth. And so this gives us Taylor's formula for our Taylor series. Now, let's apply this to e to the x. Now, e to the x is a really great function in that we can predict what its higher derivatives are gonna be. Cause notice that for e to the x, it's, you know, if f of x equals e to the x, then the first derivative of e to the x is also equal to e to the x. And in fact, all derivatives of e to the x will likewise be e to the x. And so if we evaluate these things, if we evaluate these derivatives at the center zero, we're gonna end up with e to the zero, which is always equal to one. And that's all there is to it. Therefore the Taylor series for, I should say the Maclaurin series for e to the x will look like it equals zero to infinity. We're gonna get one over in factorial times x minus zero to the n are in simpler form. We get n equals zero to infinity. We're gonna get x to the n over in factorial. So this is a power series where the denominator is just the in factorial. So in expanded form, this will look like one plus x plus x squared over two plus x cubed over six plus x to the fourth over 24, continue on in the pattern we've established there. So this gives us the, this gives us the Taylor series or I should say the Maclaurin series, right? I mean, every Maclaurin series is a Taylor series, but it just Maclaurin series, this means we've centered at zero, like we mentioned before. So we consider this thing at zero, we have the Maclaurin series, but what is the radius of convergence? When will this thing be convergent? So to determine the convergence here, we're gonna use the ratio test. So the ratio test we need to consider the ratio of consecutive terms, a n plus one over a n. And so we do this, we'll taking the terms here, we're gonna get x to the n plus one over n plus one factorial, that's the a n plus one term, then take the reciprocal of a n, we're gonna get n factorial over x to the n. And now we've seen things like this many times, the x to the n will cancel with all but one of the x's, so we are going to get an absolute value of x, and n factorial will cancel with all but the n plus one factor, we're using the fact that n plus one factorial factors as n plus one times n factorial. So it's that n factorial that's canceling out there. And so we end up with, oh, I give myself way more space than I needed. So we're gonna have the absolute value of x over n plus one. And so taking the limit here as n goes to infinity, we end up with just the absolute value of x over infinity. And it does not matter what real number you place in for x, there is no way that a real number divided by infinity could be anything other than zero. So this thing is gonna squish towards zero, and that's the limit of the ratios. And so by the ratio test, since this limit is less than one, this tells us that this series is absolutely convergent for all real numbers. And so we end up seeing that the radius of convergence is infinity. And this series is convergent from negative infinity to infinity. It's absolutely convergent. Now what we are not saying at this moment is that e to the x is equal to this power series. But if this, because all we know from Taylor's formulas that if, if a function has a power series representation, then that power series representation will be given by Taylor's formula. We haven't yet said that the function is equal to its power series representation. But for e to the x, we're in a pretty good position, right? We have a power series, we have a Taylor series and it's convergent for all real numbers. That's pretty cool. Now let's try to look at it this again, but this time let's set the center to equal to two, right? So we'll have the Taylor series centered at two. If we do that, well be aware that the derivatives of e to the x are still the same thing, right? That the nth derivative of e to the x is still e to the x. And so as we look at the nth derivative evaluated at two, the center, we end up with e squared. And so in this situation, the Taylor series would be the sum as n goes from zero to infinity. You're gonna get e squared over n factorial times x minus two to the n. And so in an expanded form, this thing would look like when you plug in zero, you're gonna end up with an e squared. Other than the next term, you're gonna get an e squared times x minus two. You're gonna get an e squared over two times x minus two squared. Whoops. You get an e squared over six times x minus two cubed. And this would continue on and on and on. This is the Taylor series centered at two. Now I want you to just be aware that if you go through the calculation, as you go through the calculation of the Taylor series for this situation, using the ratio test a n plus one over a n, basically everything's gonna look the same. This time, this will simplify as x minus two over n plus one factorial. That's the only distinction here. This thing will still approach zero, which is less than one. And so we can see that the radius of convergence is equal to infinity. So this Taylor series would likewise have this infinite radius of convergence, just like when we did the Maclaurin series. And this gives us the Maclaurin series for e to the x. You can see that there was nothing particularly special about two. You could use a different number and easily find the Taylor series for e to the x. And what made e to the x so nice in this situation is that its higher derivatives were predictable. In fact, e to the x is a function whose higher derivatives are always equal to the original function. And therefore it has a pretty tame Taylor series associated to it.