 Welcome back to our lecture series Math 1220 Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. This is the first video for lecture 45 in our series and it represents the first video from section 11.10 of James Stewart's calculus textbook. We'll actually spend three whole lectures in this section. It's a pretty big one and one of the most important sections of the entire book here in our entire series. So we want to spend some good time talking about that. And now this lecture will be a continuation of lecture 44, which we were talking about power series representations. And so you'll recall that in the previous section, we were able to represent certain functions as power series. In particular, we did this for rational functions, a polynomial divided by a polynomial, and then some things that were related to rational functions like the natural log of one plus X or tangent of X, given that those functions have derivatives, which are rational functions. But is it possible for us to represent any function as a power series, at least limited to the interval of convergence, right? And so what we want to do in this section is explore that question, can we represent general functions as power series? And so what we're going to do is we're going to begin with a slightly related question that to answer the first question, we're going to say, yes, we have a function which can be represented. What is that power series going to look like? So again, we're not saying that every function can be represented by power series, but we're saying that if it can, what would the power series representation look like? How would we find it? Because all of the previous examples utilize the geometric series formula. If we don't have that in the general case, how do we find this power series representation? So let's take for just hypothetically right now a function F and let's say that it can be represented by a power series. And if that's the case, then F of X will equal some power series. It will say the power series is centered at A so that F of X equals the sum as N ranges from zero to infinity of the sequence C N times X minus A to the N. Now we have some coefficient sequence. We don't know what that is right now. And actually our goal right now is how do we relate the coefficient sequence of the power series representation to the original function F? How do we do that if we specify what the center is supposed to be? Now, of course, an expanded form. This power series will look like C N plus C 1 times X minus A plus C 2 times X minus A squared plus C 3 times X minus A cubed, et cetera, et cetera. And we're only assuming that this equality holds on the interval of convergence. So as the distance between X and the center is less than some radius convergence are, this is where equality holds. So how could we how could we determine what these power series coefficients are going to be? Now, it turned out that when we worked on this problem previously, we had things like the natural log of 1 plus X. We determined that this thing was equal to C plus something else, right? What that something else was doesn't really matter. But I mean, if we reference what we saw last time, we ended up with like a X minus X squared over 2 plus X cubed over 3. Again, that's sort of beside the point right now. But the point is when we we were able to take the antiderivative of the function 1 over 1 plus X, right? And that's where all these other pieces came from. But the antiderivative had this unknown constant term. So what we did was like, hey, let's plug in the center of the power series. Let's plug in X equals 0 into the above equation. And then this gave us the natural log of 1 plus 0, which equals the natural log of 1, which itself is equal to 0. That was the left hand side of that expression. But then when we plug in the center of the power series here, every term will disappear, except for the constant term, for which case we got the constant term was 0. And then this helped us this helped us determine that the constant term is 0. We can do that same trick right here that if we plug in the center of the power series, so that's in this case, it could be X equals A. Let's plug that into the power series, right? Well, when we do that, you're going to get an A minus A, which goes to 0. You get an A minus A, which goes to 0. You get an A minus A, which goes to 0. And all of these terms will disappear with exception to the constant term, the exact same trick that we did before. So in summary, if we were to plug in the center of the power series, we end up with F of A equals C0. So this helps us out for the very first coefficient, or the zeroth coefficient, I should say, F of A equals C0. And so then we could substitute that into this expression right here. So we know the very, very initial term of the sequence. Well, what about C1? What could we do to obtain C1? Well, since this is a power series, we could take its derivative upon doing so. If we take the derivative of this, we would see that C0 would disappear, right? X minus A would go away. Then you get a 2, C2, X minus A. You get a 3, C3, X minus A squared. That is, just go through the usual power rule here. We end up with an expression that looks like the following. In particular, the original constant term disappears when we take the derivative, because it's a constant, the derivative of constant 0. And so now the new constant term of the derivative F prime of X, this will be C1. And so playing the same game that we did a moment ago, if we plug in X equals A into the derivative, because the derivative, if the original function has a power series, the derivative has a power series. And the center of the derivative will be the center of the original power series as well, which in this case is still X equals A. In which case, then you see that all of the X minus A's will go to 0, and the entire power series will vanish away with one exception, we get C1. And so that's if we plug in X equals A into this power series. If you plug it into just the derivative itself, you end up with F prime of A. And therefore, we have the very important relation that we just discovered right here, that C1 is equal to F prime of A. So we now know what C0 is, we now know what C1 is, how do we find C2? Well, the trick we did last time seems like it works pretty well. What if we take the derivative of this thing right here, because the derivative of the constant goes to 0, and then if we apply the power rule for everywhere else, bringing these coefficients out in front by the usual power rule calculation, we're going to get something that looks like the following. You'll see that right here, right? You'll end up with 2 times C2. You're going to get 2 times 3 times C3 times X minus A. You'll get 3 times 4 C4 times X minus A squared. You'll get 4 times 5 times C5 X minus A cubed. And you'll end up with 6 times 5 times C6 times X minus A to the fourth. You'll notice here that I'm not writing things like 2 times 3 is 6. 3 times 4 is 12. 4 times 5 is 20 or anything like that. Because it actually turns out that in this situation, I'm delaying multiplication that's in computer science. This is often referred to as a lazy computation. It seems like I'm being lazy, but I'm actually it's actually going to be more enlightening to us that we don't have any need to multiply it out yet. So we're not going to do that. If we plug in X equals A into the power series, this term, this term, this term, this term and infinitely many following terms will all vanish. You'll be left with 2 C2 and that's going to equal F double prime of X. The second derivative, I should say at A, right? If you plug in A in which case, then if you solve for C2 divide both sides by 2, we end up with the following equation, which I'll mark right here. We end up with the second derivative of A equals 2 C2. That's great. But now, unlike the first one, we see that we have these coefficients, right? 1, 1, 2, we can see that. OK, we can relate the higher derivatives of the function of value at the center A to the coefficient C, but there's this coefficient. What is it? 1, 1, 2, is that like the Fibonacci sequence? Will the next number be 3? Then 5, right? Or what's the pattern, right? We might need to see a little bit more investigation before we get that. If we look at the third derivative, right? We take all the derivatives like we did before. When you take the derivative, the constant term will disappear. X minus A will disappear. We're left with 6 C3. The two will bring come out in front. The power lowers by one. The three will come out in front and lowers by one. The four will come out in front. The power then lowers by one. And if we summarize all these things we get, we're going to end up with the coefficient of C3 is going to equal one times two times three, the coefficient of C4 will be two times three times four. The coefficient in front of C5 will be three times four times five and then four times five times six and then five times six times seven. Great. If we plug in X equals A into the power series, this term, this term, this term, this term and all following terms will disappear. We end up with one times two times three times C3. And then if we plug in A on the left hand side, we end up with the equation that you see in the box right here, F triple prime, the third derivative evaluated A is equal to six times C3. So you can see that now the pattern might be established in itself. Let's just go one more just to make sure if we take the derivative, if the constant term will disappear, the X minus A will disappear, bring out the two lower power, bring out the three lower power, bring out the four lower power, you're going to get a new factor in front each and every time. And now when you plug in, now when you look at the the power series for the fourth derivative, you plug in X equals A. Again, all of the X minus A's will cancel out. You're left with just one times two times three times four times C4. That's a 24 C4. And then if we plug in A into the fourth derivative, we get the following equation, the fourth derivative value at A is equal to 24 times C4. But where 24 came from, where did it come from? You had one times two times three times four, you had a one time two times three. And then predicting what's going to happen, right? The next term that's going to come down, this is the first power. You're going to bring over the and you'll get a one times one there. You'll bring this two out here. You'll bring this three down here by the time the coefficient Cn becomes the constant term. We're going to have the coefficients also one times two times three times four all the way up to N, aka n factorial. So generally speaking, we're going to get the following equation. The nth derivative of f value at A will equal n factorial times Cn. If you divide both sides by n factorial, we get the very important equation right here. Cn equals the nth derivative of the function value at A divided by n factorial. And so this gives us the formula for the coefficients of the coefficients of the power series representation if it has one. And so if we summarize what we've discovered right here, if so, we'll write this as a theorem here. So if a continuous function has a power series representation, that's a big if, right? We haven't yet said that it has a power series representation. All we're saying is that if the function has a power series representation, then the power series will look like this, f of x equals the sum as n ranges from zero to infinity of Cn times x minus a to the n. So this right here is just the general formula for a power series centered at A, but what we can then specify is that the coefficient sequence will be given by the coefficient sequence will be the nth derivative of A divided by n factorial and this this theorem right here is commonly referred to as Taylor's theorem, Taylor's theorem. And this right here gives us Taylor's formula, the formula for the coefficient sequence in this series and a definition that will then conclude this video with here is that if a power series has this form where it's a power series centered at A and then the coefficient sequence follows Taylor's equation, right? The nth derivative of A, the nth derivative of f evaluated A over n factorial. If those are the coefficients of the sequence, we refer to this as a Taylor series centered at A. And then a special case here is that if you centered your Taylor series at A equals zero, which is our favorite place to center it, this is commonly referred to as a Maclaurin series. And so section 11.10 of James Stewart's calculus textbook is a discussion about Taylor series and Maclaurin series. And so what we'll do in subsequent videos for our lecture right now, lecture 45, in the subsequent videos, we're going to calculate some Taylor and Maclaurin series for common functions that we know very well and love.