 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. We are getting near the end, right? We're near at the end of our series, right? So in this video, we're going to start lecture 47, which is a continuation of our discussion of Taylor's and McLaurin series that we've been discussing from section 11.10 of James Stewart's Calculus textbook. So I want to summarize some Taylor series that we have learned about so far. So this discussion began with a talk about geometric series. So the function 1 over 1 minus x, it can be represented as a power series using the geometric series. You're going to get the sum n equals 0 to infinity of x to the n, an expanded form that looks like 1 plus x plus x squared plus x cubed plus x to the fourth, et cetera, et cetera, et cetera. The geometric series formula, it has a radius convergence of 1. Now of course, if you were to substitute in x something like y over 4, you can of course get a larger radius. Your radius would be something like 4. It's adaptable because again, this variable could be replaced with something a little bit more complicated. And we'll see some examples of that in a little bit. But this geometric series is one you're going to want to know. And now from this geometric series formula, we can actually handle every rational function because combining, well, because polynomials, first of all, are their own Taylor series. That's something you should be aware of. So if you have ever a rational function, you can do a partial factor decomposition, start with division, divide out the polynomial part, your whole number part, and then you have a proper rational function. Do your partial fractions on the proper fraction. You're going to get things that look like this maybe up to some type of substitution maybe. You could have a repeated power, right? 1 over 1 minus x squared. In that situation, you can take the derivative of the geometric series to find the power series in that situation. We saw examples like that in section 11.9 of James Stewart's calculus textbook. If you have an irreducible quadratic in the bottom, complete the square, and you might get something like the following, in which case then, again, by substitution, you can compare that to this and get your power series representation. But then also from that, like if you take the anti-derivative of 1 over 1 plus x squared, we found a power series representation for arc tangent, right, arc tangent by taking the, if the anti-derivative of 1 over 1 plus x squared, we see that arc tangent has a power series representation where n equals 0 to infinity. This is an alternating sum, negative 1 to the n. You get x to the 2n plus 1, you only get odd powers of x, and then you divide by 2n plus 1. There's no factorial there, it's just 2n plus 1. That's a consequence from the anti-derivative power rule. And so in expanded form, this will look like x minus x cubed over 3 plus x to the fifth over 5 minus x to the seven over seven plus x to the ninth over nine, right? You get only odd powers of x, the denominator is just the exponent and it's alternating. Because this is an integral of a geometric series, its radius convergence will likewise be one. The derivatives of integrals of power series will retain the same radius of convergence. That doesn't change when you take derivatives or anti-derivatives. If you take, similarly, if you take the natural log of 1 plus x, because it's the anti-derivative of 1 plus x, which is a geometric series, if you take a double negative in that situation. Taking the anti-derivative of that, we see that its Maclaurin series will be the sum or n equals 1 to infinity of negative 1 to the n minus 1. You get x to the n over n. This is very similar to arc tangent where you don't get odd powers. You get all the powers except for the zeroth power, right? So you're gonna get x minus x squared over 2 plus x cubed over 3 minus x to the fourth over 4. Very similar to arc tangent, but natural log of 1 plus x, it only involves all the powers. So one, two, three, four, five. The denominator is just the exponent and it's alternating as well. So they both start at x. So there's some similarities there, but there's some important differences. This one only has the odd powers. This one has all the powers. And it likewise, its radius of convergence will likewise be one. So we did that all, we found all of those just by manipulating this observation with the geometric series. Now for some other functions, it was a little bit less difficult, it was more difficult to figure out what, we didn't have a formula beforehand, but we were able to use, we found formulas using Taylor's inequality and Taylor's equation, right? So we found the Maclaurin series for e to the x, it is in, it's the sum where n equals zero to infinity of x to the n divided by n factorial, right? This is a formula, which if you take the derivative, you'll get back to the exact same Maclaurin series, it's pretty awesome. An expanded formula looks like 1 plus x squared plus x squared over 2 factorial, which is 2 plus x cubed over 3 factorial, that's a six, then you get x to the fourth over four factorial. Unlike the other ones, this one has a radius of convergence of infinity, e to the x is equal to its Maclaurin series for every choice of x. These ones are only equal to their Maclaurin series, when x is strictly between one and negative one. Similarly, we found Maclaurin series for sine and cosine. The Maclaurin series for sine only alternating, you have n equals zero to infinity, it's an alternating factor. Take odd powers of x, it's like one, three, five, six and seven, and then you divide it by the same degree factorial, 2n plus one factorial. So you get x minus x cubed over three factorial plus five x to the fifth over five factorial minus x to the seventh over seven factorial. It's radius of convergence will be infinity, just like e to the x. Cosine is similar, you get the sum from n equals zero to infinity of negative one to the n, x times x to the two n over two n factorial. That looks like one minus x squared over two factorial plus x to the fourth over four factorial minus x to the sixth over six factorial. It's radius of convergence is infinity. All of these guys are, it's radius of convergence is infinity, right? Sine and cosine look very similar to each other. The main difference is that sine will have odd powers, three, four, seven, three, five, seven, excuse me, and cosine has even powers, two, four, six. And that's because sine is an odd function and cosine is an even function. They are equal to the Maclaurin series for any choice of x here. And then most recently, we also discovered another family of Maclaurin series, which we call the binomial series. This is useful when you have a series of the form one plus x to the k where k is any real number we want. In which case then the series will look like the sum of n equals zero to infinity of the binomial coefficient k choose n times x to the n. Remember the binomial coefficient has the definition k over n. This looks like k times k minus one times k minus two all the way up to k minus n plus one over n factorial, like so. And so in expanded form, you see things like this. Much like the geometric series, its radius of convergence is equal to one. And so these series that you see on the screen right now, these Maclaurin series are ones you're gonna wanna memorize. I mean, you could redrive them over and over and over again if you want to, but these ones are important significance that you will want to know. You wanna memorize them at the very least, put them on a note card or a sheet of paper or something in your notebook, someplace you can reference and come back to it over and over and over again. And so we can actually use these Maclaurin series to help us compute sums of series that otherwise we had difficult time doing it before. We did this for geometric series, but now it turns out we can do this for other types of series as well. So consider this infinite sum where we take one over one times two minus one over two times two squared plus one over three times two cubed and minus one over four times two to the fourth, right? Continue on and on and on and on. This sum looks like something like the following. It looks like you're taking a power of two in the denominator, which really is the same thing as taking a power of one half. So then you have like a one half and you're times in that by one over one, right? Then you subtracting from it one over two times one half squared. Then you're gonna be adding to that one over three times one half cubed. And then you subtract from that one over four times one half to the fourth. And so using the observation, you'll keep on going here, right? It's like, well, what, you know, what if we replace one half with an X? This would instead look like X over one minus X squared over two plus X cubed over three, right? Minus X to the fourth over four. And if we come back to our table that we see right here, that seems to be describing exactly this sequence right here. X minus X squared over two plus X cubed over three minus X squared over four. That's one of these Maclaurin series that we wanna know. And so this thing looks like the natural log of one plus X. But using the fact that X here is one half, this tells us that we actually are looking at the natural log of one plus one half for which one plus one half is three halves. So we get the natural log of three halves or if you prefer the natural log of three minus the natural log of two, either one works. This is an exact answer, right? And we can estimate this with our calculator and we get something like, I mean, for your homework, you should have an exact answer. But if we need an estimate, we can estimate this as 0.405565108, that's enough decimals for me, I guess. And so we can actually recognize a lot of Maclaurin or a lot of series as a power series that's been evaluated as a specific number. In which case we can then represent that series as the function and then you compute it in that like manner.