Welcome to Calculus. I'm Professor Ghrist. We're about to begin Lecture 57, Calculus Redux. We'd now come to the end of our story. It's been a long and difficult journey. And I'm guessing you probably felt like giving up more than once along the way. But you didn't, you persevered and you made it to the end. And now, it's time to enjoy the fruits of that hard labor. In this lesson, we're going to slow down and take a reflective look at what we can and cannot do with Calculus. If we take a moment and think about all of the things that we have learned in this course, we see that there's quite a lot that we can do. We understand limits, derivatives, integrals, ordinary differential equations, and Taylor series. However, there are a few things that we can't do. And in this lesson, we're going to examine the boundary between what is possible and is just beyond our reach. Let's begin with an example of a difficult integral. It is a fact that the integral from minus infinity to infinity of e to minus x squared over 2 dx is the square root of 2 pi. Now you may ask, well, who cares about such a result? Well, you care about this result if you want to do anything in probability or statistics because this is the result that says that a standard Gaussian is in fact a probability density function. So, let's see if we can solve this using the tools that we've learned in this course. And we know from the standard expansion for e to the e x that e to the minus x squared over 2 is the sum over n 1 over n factorial times quantity negative x squared over 2 to the nth. We know that this converges absolutely. And we know that because of that absolute convergence, we can integrate term by term. So let's try that. We can take this sum and integrate the individual terms. Reversing the integral and the sum, what do we get? Well, we get the sum over n of negative 1 to the n times 1 over n factorial 2 to the n x to the 2n plus 1 over 2n plus 1. That's great. But you have to evaluate that integral from negative infinity to positive infinity? That seems nonsensical, and in fact, that's not going to work. We have failed. Doing things term by term is not going to help for this improper integral. Here's a different strategy. Let's show that e to the minus x squared over 2 equals the limit as n goes to infinity of cosine raised to the nth power of x over square root of n. This is not so well-known of a result, but it's very suggestive. It's as if you're taking the first lobe of the cosine function and stretching the base out to infinity while squeezing the tail down. Well, let's see. How would we prove this, if we denote by L this limit then we could take the log of both sides. Assuming that we can reverse the limit and the log, then we pull down the exponent of n and we consider log of cosine of x of a root n. In the limit as n is going to infinity, then x over root n is getting small and we can perform a Taylor expansion of cosine about 0 that gives us the dominant terms 1 minus 1 half x squared over n, all other terms are big O of x to the 4th over n squared, and hence ignorable in the limit. So what do we get? We get that the log of L is equal to negative 1 half x squared. And that means that L is indeed e to the minus 1 half x squared. Now, why might this be useful? Well, if we want to integrate this going from negative infinity to infinity, well, it's a little bit better. I claim that it's possible to evaluate the limit of this integral, but it gets very delicate because we have to worry about the bounds. And the fact that it's an improper integral, and there's a limit there's not enough room on this slide to give a proper argument. You have to learn a little bit more about how limits and integrals interact. So we've failed, but this integral is easy when you use the techniques that you'll learn in multi-variable calculus. I guarantee you that you will do this integral, and it will take just a minute. Let's turn to a different problem that we can almost do. This one, coming from ordinary differential equations. Recall the result with which we began this course, namely Euler's theorem, that e to the it equals cosine t plus i times sine of t. Now, we took this as a given, but we didn't prove it. How might you give a proof of something like this? Well, one obvious thing to do is write everything out in terms of Taylor series and show that the Taylor expansion for e to the it is equal to that of cosine t plus i times sine of t. However, you may recall that we use this result to derive the series expansions for cosine and sine so that's a little bit of circular logic. What else could we do to try? Well, consider the following. If we let z be equal to e to the it, we're going to think of z as a function of t. Then, there's an approach using ordinary differential equations, because the one differential equation that you know for sure is that e to the constant times t is the solution to the differential equation z prime equals that constant times z. In this case, the constant is i square root of negative 1. Now this shouldn't be too weird. z of t is just a function. It now has a real and an imaginary part. Now let's write out the real and the imaginary parts of z as follows. z of t equals x of t plus i times y of t, where x and y are real functions. Now what happens when we multiply this by i? Well, i times e equals i times x plus i squared times y. The i squared becomes a negative 1, and we can reverse the order so that we keep it real part an n imaginary. Now we know that z prime equals i times z. So what is z prime? Well, I can use the linearity of the derivative and say that z prime equals x prime plus i times y prime. And now my differential equation z prime equals iz really turns into a system of two differential equations. One for the real part that says x prime equals negative y and one for the imaginary part that says y prime equals x. Now this is a system of two differential equations. There's no imaginary numbers in here. These are both real functions but they are not independent, they are coupled. The x prime depends on y, the y prime depends on x. We have not learned how to solve systems of coupled ordinary differential equations. And you might look at this and say, well, if x were cosine of t and y was sine of t, then this would work since the derivative of cosine is minus sine and the derivative of sine is cosine. That's fine. But this is not a principled or systematic approach, it's just a guess. When you do take Multi-variable Calculus, this will be an easy result. You will learn methods for solving systems of coupled linear ordinary differential equations from which will follow easily Euler's Theorem. Let's turn to one last problem that we can't do, this one involving series. It is a result that we've mentioned several times that the sum over n of 1 over n squared equals pi squared over 6. You know that the series converges, you know how to bound the error for finite approximation. But how do you show that the exact result is pi squared over 6? Well, let's give it a try. We're going to show as much of the proof of this as we can on one slide. Let's begin with the function u equals arc sine of x and in a somewhat unmotivated step, we're going to integrate u du as u goes from 0 to pi over 2. When we do so, we get, of course, u squared over 2 evaluated at the limits yielding pi square over 8. Note the presence of a pi squared. That is a critical piece. Now when we substitute in arc sine of x for u, we get the integral of arc sine dx over square root of 1 minus x squared. Changing the limits, this becomes the integral as x goes from 0 to 1. Now, we don't want to evaluate this integral, but we already know the answer. It's pi squared over 8. What we want to do is substitute in the Taylor series for arc sine of x. We've run across this once or twice before. It's an unusual looking series. The coefficients involve the products of odd numbers and the numerator, the products of even numbers and the denominator, and an x to the n over 2n plus 1. Now, this is looking rather complicated. We can integrate this series term by term, but integrating that x to the n over square root of 1 minus x squared is highly non-trivial. That is doable by the methods of this class. You can do it with integration by parts and a reduction formula. I'm not going to show you all of those steps and it wouldn't exactly fit on this slide. But trust me that when you do so, you will get after a lot of simplification. The sum then goes from 1 to infinity of 1 over quantity 2n minus 1 squared. that's so close to what we were looking for. This is the sum of the odd numbers in the denominator squared. Well, again, with just a little bit more of an argument involving a geometric series, one can show that that sum is 3 4th of the sum over n of 1 over n squared. And that sum of 1 over n squared is what we were looking for knowing that this is really 4 3rds times pi squared over 8 yields pi squared over 6. Now, you could certainly ask for a clearer or more complete proof. The one that I have sketched is due to Euler, the master himself. And it is, if my research is correct, his fourth published proof of that result. But it's certainly not easy. Now, you are going to expect that I'm going to tell you that this is easy when you learn Multi-variable Calculus. Well, you may see a proof of this using Multi-variable and it will be easier to follow than the one that I have sketched out that I am not aware of any proof of this result that I would call easy. Some mathematical truths are deep. They are difficult, and they require an extraordinary amount of effort to ascertain. Some truths are right at the boundary between what we can do and what we can't, and they are worth striving for. And that is the end. You made it. Congratulations, you got all the way to the end. It was a hard, and long, class, but you learned a lot. Take a moment, relax, prepare for the day of judgment, otherwise known as the final exam, and enjoy the fruits of your hard work. >> Cut. >> Oh, is that it? Are we done? Yes. I'm so happy. Oh, thank God. Oh, good job, Jordan. >> Thanks. You too. >> I'm going to take a nap.
