 In this final video for lecture 44, we actually are gonna show you why someone might actually want a power series representation in the first place. We've learned a lot of how one could find a power series representation for rational functions and things whose derivatives are rational functions. But what's the point? Why do we want a power series representation? Why might it be more useful to use a power series than the function itself? Now imagine we are trying to compute the area under the curve that is we wanna integrate from zero to one half, one over one plus x to the seventh dx. Like maybe we wanna find the area in the curve or maybe this integral is measuring work or hydrostatic force or arc length. There's some other application of integrals, right? There's good reasons why one would do that. Now if we were to try to compute this the usual way using the fundamental theorem of calculus, it would require that we evaluate the indefinite integral one over one plus x to the seventh. But as hard as we try, we're gonna really struggle finding the anti-derivative of this function because maybe the anti-derivative is non-elementary or even if we could, it might just be really, really, really hard. So we sort of at some point might abandon our efforts to find an anti-derivative. And so therefore we might try to approximate the area under the curve. We could try Simpson's rule. That's a pretty good technique that we've learned before. But let me now demonstrate how one could approach this problem from a perspective of power series. Why might that be useful? Because with a power series, you're gonna get f of x equals one over one plus x to the seventh. With the right perspective, we could write this as one over one minus negative x to the seventh, right? We use the constant ratio r equals negative x to the seventh. This tells us that our power series representation, we can actually get a very quick power series representation. This is going to be the sum as n goes from zero to infinity of negative x. I'm gonna write negative one to the n, x to the seventh n. Like so, in expanded form, this would look like one minus x to the seventh plus x to the 14th minus x to the 21st and then continue on and on and on and on, right? And so we can get a power series representation of our function right here very quickly. Be aware that because this is a geometric series, we need to have that the absolute value of negative x to the seventh is less than one. This translates to tell us that the absolute value of x needs to be less than one. Or in other words, x needs to be between one and negative one. This is our interval of convergence from negative one to one. I want you to be aware that the domain that we're trying to integrate across zero to one half, this lives inside of our interval of convergence. Therefore, we could equate this function, right? This function with the power series, there's no difference between them. But with the power series, the anti-derivative is substantially, substantially easier to do. The power series anti-derivative would look like c plus x minus x to the eight over eight plus x to the 15th over 15 minus x to the 22nd over 22. And this continues on and on and on and on. If we want a closed formula for this, we can get something like c plus the sum. We're gonna go from n equals zero to infinity. It's an ultramain sum, so we're gonna negative one to the n. We're gonna get x to the seven n plus one all over seven n plus one. We could do something like that. We don't necessarily need this close formula at this moment, but we could have this closed formula. And therefore, as we go from zero to one half, because after all, that's our target range right here. As we go from zero to one half, we evaluate this function at zero, which we'll just give you c, and then we evaluate at one half. And admittedly, for the fundamental theorem of calculus, we don't need to have a plus c, right? Because it just cancels out anyways. So what I'm saying is this integral from zero to one half of f of x dx, this will equal the sum where n goes from zero to infinity. We get this ultramain series, negative one to the n. We're gonna get one half to the seven n plus one over seven n plus one. And so we see that this integral is equal to a geometric, not geometric, into an ultramain series. That's the word we're looking for right here. We're rewritten in one more fashion. We get this ultramain series where n goes from zero, it goes from n equals zero to infinity, negative one to n over, we're gonna get this seven n plus one still, but we're also gonna get two to the seven n plus one power. That shows up in the denominator. This is an ultramain series. You can see that from the negative one to the n. The sequence of absolute values does decrease towards zero quite rapidly, in fact. And therefore, we can approximate the integral using a partial sum of this series, right? Which partial sum should we use? Well, it depends on how accurate we wanna be, right? If we wanna be accurate, it doesn't specify, the original problem didn't specify any level of accuracy, right? What if we were just to use the first four terms? Cause I mean, that's what I see on the screen right here, right? I noticed I stopped at, I stopped here at the x to the 22. So if we just take these four terms right here, what we're gonna see is that our integral is approximately the same thing as one half minus one half to the eighth over eight plus one half to the 15th over 15 minus one half to the 22nd over 22. And so this is the first four terms of our power series. And if we approximate this, you can put this in a calculator, it's not too difficult to do. But in a calculator, you'll get 0.49951374242 dot dot dot. It's an irrational, it's not irrational. Well, I mean, this number over here, although it's irrational, this one right here is a rational number, but it'll repeat over and over and over again, that's fine. But we're good with this approximation right here. And so this is our approximation. It's about one half itself, right? And so what is, how good, how good of an estimate is this? Well, since it's an alternating series, right? Since this is an alternating series where notice this is the n equals zero term, this is the n equals one term, this is the n equals two term, this is the n equals three term. If we look at the next term in the sequence, right? How big is that term? The next term would look like a positive x to the 29th power over 29. But of course we're not using x here, we're using one half. So we get two to the 29th, like so. How big is that number? Cause this gives us an error bound. The error is gonna be bounded above by the next term in the alternating sequence. So we get one over 29 times two to the 29th, which that number is approximately a 64, 6.4 times 10 to the negative 11th right there, right? So you see that this number right here is accurate to at least 10 decimal places. And you'll notice one, two, three, four, five, six, seven, eight, nine, 10, right? There may be some wiggle room on that last number there, but this is actually fairly accurate right here. Now, if we were to do the same calculation using Simpson's rule, right? In order to get this same level of accuracy, we would have to be using Simpson's rule with 50 subdivisions. That is S sub 50, right? Now, admittedly, admittedly on a calculator, it's not too difficult to get S 50, right? We've seen a calculator in a previous lecture, right? It could run that really, really quickly, right? But the point is Simpson's rule takes 50 calculations to get this level of accuracy while the power series only took four. It took four calculations to get the same thing. So even though a calculator can do 50 calculations really quickly, think of how fast it can do four calculations, right? I mean, it's a fraction of the time. And this advantage will only compound the more and more sophisticated the problem gets. And so when a power series representation is at hand, it's gonna be one of the most accurate, one of the most efficient ways of estimating integrals compared to all the techniques we've done before. It's a much more sophisticated technique. And so this is meant to sort of illustrate not just how one uses a power series to estimate an integral, but why one would be interested in it. It's so much more effective than even Simpson's rule, which before this moment was our best technique. So in this game of King of the Hill, power series representations are now our most accurate, most efficient way of computing a numerical integral.