1:51

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

5:24

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.