[Prof Frenkel] What is the most difficult way to earn a million dollars?

[Brady] Making Youtube videos. [Prof Frenkel] *laughs*

[Prof Frenkel] Well, you probably know much more about that than I do.

[Prof Frenkel] One of the most difficult ones is to solve one of the Millenium Problems in Mathematics,

which were set by the Clay Mathematical Institute in the year 2000.

One of these problems is called "The Riemann Hypothesis".

It refers to a work of a german mathematician, Bernard Riemann,

which he did in the year 1859.

This is just one of the problems. In fact, there are seven.

And one of them has been solved so far.

And interestingly enough, the person who solved the problem has declined the one million dollars.

So...

It just shows that mathematicians work on these problems, not because they want to make some money.

I think it is now the most famous problem in mathematics.

It took the place of Fermat's Last Theorem,

which was solved by Andrew Wiles and Richard Taylor in the mid-1990s.

[Brady] But that wasn't a Millenium Problem. [Prof Frenkel] That was not a Millenium Problem.

[Prof Frenkel] The most essential thing here is what we call the Riemann Zeta function.

And the Riemann Zeta function is a function, so ...

A function is a rule which assigns to every value some other number.

And the Riemann Zeta function assigns a certain number to any value of s,

and that number is given by the following series:

1 divided by 1 to the power of s,

plus 1 divided by 2 to the power of s,

plus 1 divided by 3 to the power of s,

4 to the s, and so on.

So, for example, if we set x = 2.

Zeta(2) is going to be 1 divided by 1 squared plus

1 divided by 2 squared,

plus 1 divided by 3 squared,

plus 1 divided by 4 squared,

and so on.

So, what is this?

This is one.

This is 1 over 4.

This is 1 over 9.

1 over 16...

So this is an example of what mathematicians call a convergent series,

which means that, if you sum up the first n terms,

you will get an answer which will get closer and closer to some number.

And that number to which it approximates is called the limit.

But the limit here is actually very interesting.

And it has been a famous problem in mathematics to find that limit.

It is called the Basel problem,

named after the city of Basel in Switzerland.

And this Basel problem was solved by a great mathematician: Leonhard Euler.

And the answer is very surprising.

What Euler showed is that this sums up to pi squared over 6.

So you may be wondering.

What does this sum has to do with a circle?

Why would pi squared show up?

But Euler came up with a beautiful proof.

I'm not going to explain it now, but it's something that you can easily find online.

This series is just one example of this Riemann Zeta function,

but you can try to do the same for any other value of s.

So, for example, if you take s=3, you will get the reciprocals of all the cubes,

and you sum them up, and so on.

So this will, again, be a convergent series, and you can wonder what that answer is.

That would be zeta(3).

You can also try to substitute negative numbers.

And this is very interesting, because if you substitute... if you just substitute —

If s = -1, then what are we going to get?

So you will get 1 divided by...

1 to the 1 to the -1,

plus 1 to the 2 to the -1,

plus 1 over 3 to the -1...

If you take the reciprocal of something, which is the inverse of something,

then you will get that thing.

So this will be 1,

this will be 2,

this will be 3,

this will be 4...

[Prof Frenkel] Does it look familiar? [Brady] Yes, I have seen that before.

[Prof Frenkel] We have arrived at the famous sum of all natural numbers:

1 + 2 + 3 + 4 ...

But, you see, now we obtained in the context of the zeta function.

So this is what we call a divergent series.

There is no obvious way how we could possibly assign a finite value to it.

This sum is infinite, it does not converge to any finite value.

But, this context...

If we put this value, this infinite sum, in the context of this function,

there is actually a way to assign a value to s=-1.

And this is what Riemann explained in his paper.

And so what Riemann said is that, actually, we should allow s to be, not just a natural number

— for example, 2, or 3, or 4, when the series is convergent —

but we should allow also all possible real numbers.

And not only real numbers, but also complex numbers.

The way you get complex numbers is by realizing that, within real numbers,

you cannot find the square root of -1.

Then what to do?

One way is to ban the square root of -1 and say,

"This doesn't exist, we cannot use it"

But, in mathematics, we have understood, a long time ago, that actually there is a much better way to treat this.

the square root of -1.

We can simply adjoin it to the real numbers.

Think of real numbers as points on a line.

Here is 0,

and here is 1,

and here is 2,

and then you can mark your favorite fractions.

For example, one half is exactly in the middle way between 0 and 1.

And say,

1 1/3 would be a third of the way between 1 and 2.

But then, you also have things like square root of 2...

For example, somewhere here.

And then, there is pi, which is just to the right of 3.

So all the real numbers live here.

Square root of -1 cannot be found anywhere on this line.

But we don't give up. We say, "You know what?"

"Let's actually draw a plane, let's draw another coordinate system"

"And let's mark square root of -1 on this new coordinate axis."

You see, if we do that, then every point on this plane becomes a number.

So that would be 2 times square root of -1,

3 times the square root of -1, ...

But more than that, let me find a number which is on the intersection of this line.

I can draw a vertical line which goes from 2

and I can go... can also draw a horizontal line.

Then there's this point of intersection.

So this point also would represent a number,

which would be 2 plus 3 times square root of -1.

So, in other words, a general number is going to have what we call a real part,

that is the projection onto this axis;

and the imaginary part, that's the projection on the vertical one.

The notation is a little bit clumsy.

Instead of square root of -1, they write i.

So then for example: instead of writing 2 + 3 square root of -1, we'll just write 2 plus 3i.

It's an imaginary number, we imagine it. We cannot find it on this real line.

So we have imagined it, and then we have adjoined it in our imagination.

Real numbers comprise all points on the real line, on this axis;

and complex numbers comprise all the points on this brown paper,

if you could extend the brown paper all the way to infinity.

Right?

So let's go back to Riemann.

What Riemann's insight was is he said "look, let's think of this argument of the Zeta function, this number s...

Initially, we thought that s could be 2, 3, 4, and so on.

But then we realised that actually any real number to the right of number 1...

not including number 1, because actually in this case you cannot assign a value, it's a divergent series,

so it goes to infinity.

But anything to the right, and then drawing and marking it with red...

For all of them, this function is actually well defined.

So... But then he said '"We can actually do more... We can think of s as being a complex number."

So instead of thinking of s as just being a point on this line, we can take s anywhere.

It will be convergent if it is to the right of this line.

So you see if this is the line, which sort of to the right of this line... live all the complex numbers

whose real part is greater than 1.

So, it turns out — and it's very easy to show — that anywhere in the shaded area, except for this line —

so to the right of this line.

Now, for any value of s in this area, this function converges to something.

—So if I put 6 + 9i into the Riemann Zeta function, I'll get a convergent series...?

—That's right. You get a convergent series. It will converge to something,

which is not going to be a real number.

It's goint to be a complex number, because you're going to add up infinitely many complex numbers.

But there will be a certain number to which — which will be a closer and closer approximated

as you go along summing up the series.

—So far, basically everything to the right of this line... —Gives us a bona fide value

—will come out to play. —will come out to play, and will give us a bona fide value.

—Can the imaginary part go in negative?

Yeh, but the imaginary part... yes. The imaginary part is okay — can be negative or positive.

But the real part has to be greater than 1.

But, now, you are in the context of a theory of... functions with complex arguments.

And it is what we call a holomorphic function.

So it has some very special, very nice properties.

So one of the properties that this kind of — what we call holomorphic — functions enjoy is what we call

analytic continuation. So we can extend the definition, i.e. the domain of definition of the function.

There are methods which allow — which enable us — to kind of push the boundary,