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The Riemann Zeta Function was actually first introduced first by Leonhard Euler,

who used it in the study of prime numbers.

He didn't really use it for complex arguments.

He used it for real arguments.

We'll see really what Euler's work was next class.

In particular, Euler used the properties of the zeta function to show

that if you take 1 over every prime number and add up all these numbers.

So that will be one-half because 2's a prime + one-third because 3 is a prime

+ one-fifth because 5 is prime + one-seventh + one-eleventh and so forth.

He used the properties of the zeta function to show that this sum diverges,

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so it is a little bit similar to the harmonic series but

we're missing quite a few terms.

And so, one could think that we're missing enough terms for

the series to converge but he showed the series diverges,

which in particular shows that they must be infinitely many prime numbers.

But it also shows something about the distribution, the gaps don't get too big,

the series still diverges.

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Bernhard Riemann used the zeta function a century after Euler

to obtain results on the asymptotic distribution of prime numbers.

Bernhard Riemann's work was seminal in making

progress under distribution of prime numbers.

And again, we'll talk more about that next class.

Today, we'll study the Riemann's zeta function which wasn't always called

the Riemann's zeta function.

When Riemann talked about it in Euler, they didn't call it the Riemann zeta

function, they just called it the zeta function.

But this function has gained such importance that it is now called

the Riemann zeta function.

We'll also talk about the Riemann hypothesis,

one of the most famous open problems in mathematics.

And next lecture, which is also our last lecture, we'll study the relation

between the Riemann zeta function and the distribution of prime numbers.

So, in order to introduce the Riemann zeta function,

recall that the harmonic series, which is the sum of all 1 over n's, diverges.

But, if you raise this n and the denominator to a power just slightly

bigger than one, then all of a sudden that series converges.

I showed you a proven how one can see that the harmonic series diverges.

Then we quickly show you a little trick on how to see that this series with an s

that's slightly bigger than 1, any value of s bigger than the 1 in fact, converges.

You can compare this series to an integral.

Here's how you do that.

The sum 1 over n to the s can be broken up until

the first term 1 which is 1 over 1 to the s plus

the sum n from 2 to infinity 1 over m to the s.

For m = 2, I get 1 over 2 to the s.

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And so, the area of this next rectangle is 1 over 3 to the s.

Similarly, the area of the next rectangle is 1 over 4 to the s, 1 over 5 to the s,

1 over 6 to the s, and so forth.

In other words, the value of this sum from 2 to infinity of 1 over n

to the s is equal to the area of all these rectangles that I drew.

In all these rectangles are underneath the curve because I drew right hand sums for

this curve and this is a decreasing function.

And so therefore, that sum is bounded above by the actual area under the curve,

which is given by the integral from 1 to infinity of 1 over x to the s tx.

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Because when that differentiate, x to the -s + 1,

I get -s + 1 times x to the -s + 1- 1 which is x to the -s.

If I wanted to rewrite that again,

this is 1 over 1- s times 1 over x to the s- 1.

And that's what you see right here.

So, here's the anti-derivative and

I need to evaluate that anti-derivative at the two bounds, 1 and infinity.

Well, I can't really plug in infinity.

What this really means, I would plug in a really large value,

say r, and then later on, let r go to infinity.

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But when I plug in a really, really large value of r for

x, I'm looking at this exponent s- 1 because s is bigger than 1.

s- 1 is still bigger than 0.

If I raise a really, really, really large number to something bigger than 0,

it's still really, really large and goes to infinity.

So, this denominator, x to the s- 1 goes to infinity

as x goes to infinity whereas r goes to infinity.

In other words, 1 over that denominator goes to 0.

So, at the upper bound, my anti-derivative goes to 0.

If I let the upper bound go to infinity.

And therefore, in the limit,

what's left is only the anti-derivative at the lower bound.

When I plug in the lower bound, namely 1, I just find 1 over 1- s.

Because it is the lower bound, I need to subtract that value.

So, I end up with 1 minus 1 over 1- s, and

I can make that into a 1 + 1 over s- 1 by just flipping the signs,

and then find the common denominator which is s- 1.

In the end, the value of the right hand side is s over s- 1,

which is a finite number.

So this shows that this series, 1 over n to the s,

is bounded above by s over s- 1.

The partial sums therefore, form an increasing sequence.

I keep adding more, and more, and more but they're bounded.

So, the partial sums keep getting bigger and

bigger but they are bounded by s over s- 1.

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In order to consider the Riemann zeta function, we now let s be a complex number

instead of just letting it be a real number as on the previous page.

So, for s being a complex number, whose real part is bigger than 1,

we define the zeta function as a function zeta(s) Equals

the sum from n = 1 to infinity of 1 over n to the s.

This is a zeta, a Greek zeta.

It's a little funny but it's traditional to call the complex variable s,

instead of z.

Normally we would want to called that z but we're calling it s, and

it's just a traditional way because Riemann did that.

What do we even mean by n to the s, if s is a complex number?

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Let's quickly figure this out.

Remember that for real values of s,

we know that n to the s is the same as e to the natural log of n to the s.

Because e and natural log, they just cancel each other out.

But natural log of n to the s, by the rules of

logarithm, it's s times natural log of n.

So, n to the s is therefore the same as e to the s times natural log on n and

we use that to define for complex s what n to the s equals.

We say n to the s = e to the s log n, where logarithm is the complex

logarithm but because we're talking about n just being a real number,

we can actually just use the natural logarithm right here.

One can actually more generally define complex powers where

n doesn't even have to be an integer or real number.

So, if we have the base being a complex number,

we would use this definition e to the s log n.

But because n is actually a natural number, we go back to e to the sl and

n and that is the definition of n to the s for complex values of s.

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So, does this series still converge when real product s is bigger than 1?

Now, we're plugging in complex values for s.

Does the series still converge?

Well, n to the s again is e to the s times natural log of n.

Let's find out if this series converges absolutely, so

if you put absolute values right there.

So, absolute value of n to the s is the absolute value of e to

the s natural log of n, and we know the absolute value of e

to the something is e to the real part of the exponent.

But the real part of s times natural log of n since natural log of n is

a real number.

The real part of s times natural log of n is simply the real part of s times

the natural log of n.

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But now, everything is back to being real and so

this expression is the same as n to the real part of s.

Therefore, the series of absolute values of 1 over n to the s

is the same as the series of 1 over n to the real part of s.

Real part of s is some real number bigger than 1.

We showed that if you raise n to any real number bigger than 1,

the series converges.

Since, the real part of s bigger than 1 by assumption,

the series on the right converges.

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Therefore, we have absolute convergence of the original series, and

we know that absolute convergence implies regular convergence and therefore,

the series 1 over n to the s converges.

In fact, one can show that the convergence is uniform so equally fast in

portions of the plane where the real part is strictly bigger than 1.

So, here's 1, if you choose an r that is strictly bigger than 1,

then the convergence is going to be uniform in

the half plane that starts at real part equals r.

One can use that, the fact that uniform convergence and

the fact that the function s goes to 1 over n to the s is

analytic to show that the zeta function is actually

analytic in the entire half plane that starts at 1.

Riemann's seminal contribution to the theory of the zeta function,

and the study of prime numbers is the following theorem.

The zeta function has an analytic continuation into the entire complex

plane with the exception of the number 1.

And this continuation satisfies that zeta of s goes to infinity as s goes to 1.

So far, the zeta function's only defined for a real part of s greater than 1.

So, over here, we have a definition.

Riemann showed that you can extend this

definition to everywhere except the point 1.

The point 1 forms a so-called pole of the zeta function s.

s approaches 1, the zeta function goes off to infinity.

The proof is not easy but a slightly easier construction is to just

construct an extension to the entire right half plane.

So, here this picture, the zeta function s of now is defined for s greater than 1,

and we can actually give the construction that shows you can define it for

s greater than 0 again, except at the point 1.

So, that construction is slightly easier and we'll outline it here.

Again, a motivation in R first.

The sum from n = 1 to some bound uppercase N of 1 over n to

the s again can be viewed as false.

The first term is 1 over 1 to the s.

1 over 1 to the s is this height right here, so

the first term is the area of this first rectangle.

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1 over 2 to the s is the area of the next rectangle.

1 over 3 to the s, 1 over 4 to the s, and so forth.

So, this entire sum here is the area of all these rectangles up

to the rectangle that starts at n and goes to n + 1, so this one right there.

If I, on the other hand, look at the area under the curve,

which is given by the integral from 1 to n + 1, 1 over x to the s,

I'm missing out on all these little pieces.

I'm missing this piece.

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So, I call each of these pieces delta.

This is delta 1 of s.

That's delta 2 of s and so forth.

So, if I add all these extra delta n's of s's in there,

then I get an equality right here for all these areas.

And delta n of s is just given by the correct difference here.

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This integral here converges to 1 over s- 1 which

is just the integral from 1 to infinity of 1 over x to the s dx.

And then, there's that sum in the end.

Now, observe that that sum is an analytic function, and

it is analytic in all s is real part of s is greater than 0.

We don't need it to be greater than 1,

it's enough to be greater than 0 because we're only looking at 1

over n to the s minus the integral from n to n + 1, 1 over x to the s dx.

So one can actually see that that is an analytic function and one can moreover

show that the sum of all these delta n of s, because we're only talking about little

pieces of area Converges, as n goes to infinity to an analytic function h(s).

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except at the point one.

So in the limit we find that zeta of s, which is the limit of the left hand side,

is equal to one over s minus one plus h of x.

This definitely holds for

a real part of s greater than one because that's where the zeta function is defined.

But the function h is actually analytic in a real part of s greater than zero, and

this guy here is analytic in the whole complex plane,

minus the point 1, so the right hand side

lives in a much larger region than what left hand side has defined.

So here again is this equation zeta of S is one over

S minus one plus h(s) holds for real part of s greater than one.

But h(s) is actually analytic in real part of s greater than zero.

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To define the zeta function using the right hand side.

In all of the right half plane except for the point one.

And that definition agrees with the original definition for

real part of s greater than 1 because we have this equality.

But now, we can define the zeta function in the entire right half plane except for

the point 1.

As s approaches 1, h(s) is well behaved and the only thing that's not so

well behaved is the term 1 over s-1 which goes to infinity.

Riemann was actually able to extend the zeta function to an analytic function in

all of C- the point 1, not just in the right half plane- point 1.

Of much interest are the zeros of the zeta function.

So those s's in the complex plane for which zeta of s = 0.

One can show the following.

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Those are often called the trivial zeros because once you know the analytic

continuation of the zeta function is to the whole complex plane,

it's not hard to see that these are zeros of the zeta function.

So the region to be studied remains the strip,

real part of s is between zero and one.

A key result is that the zeta function has no zero's on the line,

real part of s equals one.

This is the key result and the proof of the prime number theorem.

Again, we'll talk about that during the next lecture.

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From the fact that there are no 0's on the line where the real part is equal to 1,

one can use a functional equation for

the zeta function to show that there's also no 0's on the real part of s = 0.

And therefore, the only region that is

unknown is the region between zero and one.

So, in this region we still don't really know what happens for the zeroes or

the zeta function.

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At least one third of the zeroes in the critical strip lie on that critical line.

That is known, trillions of zeros of the zeta function have been calculated.

So far all of them lie on the critical line.

But that's not a proof.

It could be that once you get to a few higher powers,

all of a sudden you find a zero that's not on the critical line.

So none of these are proofs of the Riemann hypothesis,

though they strongly point toward this veracity.

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The Riemann hypothesis has strong implications on the distribution of

prime numbers and on the growth of many other important functions.

It would greatly shock many results in number theory.

Next class, we'll discuss the prime number theorem,

the relation of the Riemann hypothesis and the Riemann zeta function to the prime

number theorem, and we'll finish up this course.