Euler’s Pi Prime Product and Riemann’s Zeta Function

What has pi to do with the prime numbers, how can you calculate pi from the licence plate numbers you encounter on your way to work, and what does all this have to do with Riemann's zeta function and the most important unsolved problem in math? Well, Euler knew most of the answers, long before Riemann was born.

I got this week's pi t-shirt from here: https://shirt.woot.com/offers/beautiful-pi

As usual thank you very much to Marty and Danil for their feedback on an earlier version of this video.

Here are a few interesting references to check out if you can handle more maths: J.E. Nymann, On the probability that k positive integers are relatively prime, Journal of number theory 4, 469--473 (1972) http://www.sciencedirect.com/science/article/pii/0022314X72900388 (contains a link to a pdf file of the article).

Enjoy!

Burkard

Welcome to another Mathologer video. Last time I showed you how the mathematical

superstar Euler discovered this stunning
identity up there: PI squared over 6 is

equal to the sum of the reciprocals of
the squares. Today I'll introduce you to

the mathematical magic that allowed him
to morph this infinite sum into an

infinite product. And this infinite
product established as a connection

between PI and all the prime numbers
there on the right.

In fact, we'll see that this identity is
just a special case of the main bridge

that connects the famous Riemann
zeta function

to the prime numbers. Along the way we'll
come across many other beautiful

identities involving pi, a seriously
crazy way to calculate pi using random

numbers, wait for it, and a couple of
nifty ways to prove some mathematical

all-time classics. So buckle your
mathematical seat belts, it's going to be

a wild ride. We'll warm up by
tricking Euler's identity into giving us

a couple of other beautiful identities
involving pi. First we make a copy,

here we go. Now we'll line up things
like this and multiply everything at the

bottom by 1/2 squared. Expand term
by term and so we get 1/2 squared

times 1/1 squared equals well, of
course, 1/2 squared, 1/2 squared

times 1/2 squared equals 1/4
squared, and so on. Now we'll subtract the

bottom from the top. On the right side,
notice how nicely things line up there.

Beautiful! Anyway so when we subtract
every second term on the top gets wiped

out. On the left, we have 1 times the
fraction minus 1/2 squared times

the fraction and so we get this. And there
you have it -- two more beautiful identities for

pi pretty much for free. There's one more
very important identity hiding here

which we'll need later. So let's just
step back to the previous slide and

let's subtract the bottom from the top
one more time. On the right, that fills

the gaps on top with the
negatives of what was there originally.

There and there, etc. And on the left side
we get this which we

can also write like that. Okay, three new
PI identities just around the corner

from the original one. To be able to
go further let's switch back to the

unsimplified left sides. Now, at the top,
if we replace the 2s in the exponents

by an arbitrary number z we're now
looking at a function in the variable

z, the famous Riemann zeta function. The
extra three identities that we derived

for the special case z=2
actually work in general and to get

these zeta function identities we'll just
replace the PI fraction by zetas and all

the 2 exponents by z's. There you go.
And you can convince yourself that these

new identities hold in exactly the same
way as I showed you in the special case

z=2. Really quite
straightforward, so maybe give it a try.

Now as a first application of these
identities let's evaluate zeta at 1,

that's a special value. The resulting
infinite sum at the top is called the

harmonic series and is one of the most
important infinite sums ever. Of course

quite a few of you will know a lot about
this infinite sum but bear with me,

there's some nice stuff coming up here.
As usual, to evaluate this infinite sum

we just start adding: so 1 plus 1/2
plus 1/3, and so on.

Since the terms we add are all positive,
we get larger and larger partial sums, right?

This means that either our
partial sums explore to positive

infinity, in which case it makes sense to
say that the sum is plus infinity, or the

partial sums sneak up to a finite overall
sum. So which one is it? Do we get a

finite sum like in the case of zeta at
2 where the infinite sum adds to PI

squared over 6, or do we get an infinite
sum? Well, let's

first assume that the sum at the top is
finite. If this is the case, then we can

be absolutely sure that everything we
did to get these additional identities

stays valid. Okay
now let's compare those identities: one

at the top is greater than 1/2 at
the bottom 1/3 is greater than 1/4

at the bottom, and so on. The top is
always greater than the bottom and this

means that the sum at the top is greater
than the sum at the bottom, right? However

this contradicts what we get out of the
left sides. Here we've got 1 minus 1/2

which is 1/2 which means that the sums
should be equal. So what that means is

that our assumption that our original
sum is finite implies a contradiction

and this means the assumption was wrong
and therefore the sum has to be infinite.

In fact, from what we just said it
follows that all 3 sums have to be

infinite. Anyway, for later just remember
that zeta evaluated at 1 equals infinity. Now

what's important about the zeta function
is first and foremost its connection to

the prime numbers. Euler managed to pin
down this connection by pushing the

simple trick that got us this second
odd power identity here to its absolute

limit. You'll see what I mean by this.
Here's what he did ok. As earlier, we

start by making a copy. Then we multiply
the bottom by 1/3^z. Okay, so let's

just do it, here we go.
Subtract the bottom for the top and then

on the right all the fractions on the
top that have denominators divisible by

3 get wiped out. And, on the left,
well what have we got, we've got 1 times

something minus 1/3 to the power of
z times the same something which

gives this guy here. Now just rinse and
repeat. So we make a copy, times the

second term on the right and subtract the bottom

from the top which wipes out what? Well
all the terms with denominators

divisible by 5 this time. And we just
keep repeating this and in the limit we

get this. So all the terms on the right
except for the first one have been wiped

out and the numbers in the denominators
on the left are exactly the prime numbers.

Now, just in case you know a little bit
more, can you see the famous prime number

sieve of Eratosthenes in action in this
derivation? Now the right side, well

that's just 1. So now we can solve
for zeta and that gives Euler's famous

product formula for the Riemann zeta
function. Now this identity is one of the

biggest deals in mathematics and it's
the point of departure for the famous

paper in which Bernhard Riemann states
the Riemann hypothesis. So let's have a

quick look at this. There it is, all in
German. Let's zoom in a bit. There

it is, alright, that's exactly what we
have there, just written in a little bit

more compact way. So what is this paper
about? Well the title says it all, if you

happen to speak German: "Ueber die Anzahl der Primzahlen under einer gegebenen Groesse."

(That was perfect :) which translates to
"About the number of primes less than a

given value". Now what Riemann manages to do in this paper is to derive a formula

that allows to calculate the number of
primes less than a given value without

actually having to calculate all those
primes. That sounds like magic, right?

For example, recently mathematicians used this formula to figure out the exact

number of primes less than 10 to the
power of 25 which pans out to be this

monster number here and Riemann's magic
formula would be extra magical and the

prime numbers would be distributed in
the nicest imaginable way if the famous

Riemann hypothesis that's also part of
this paper was true. So that's what the

big deal is all about. Ok now I won't
prove Riemann's heavy-duty prime number

results for you
but what I would really like to do is to

show you some really amazing and
accessible results about prime numbers

that follow from Euler's product formula.
Okay,

so let's just go for this special value
again z is equal to 1. Then we know that

the left side is infinity. Now wait for
it...

This actually implies that there are
infinitely many prime numbers! Why?

Because if there was only finitely many
prime numbers, right, maybe just up to 7,

then the product on the right would
evaluate to the finite number. But that's

not possible. I'm pretty sure you didn't
see that one coming, right? Okay

next trick. Set z equal
to 2. Then we're back to where we started

from on the left there and actually this

shows again that there must be
infinitely many primes. Why because if

there were only finitely many the
expression on the right would be a

rational number. But this is impossible
because pi squared divided by 6 is

irrational. Well, of course, proving that PI squared

over 6 is irrational is much much much
harder and took more than 2,000 years

longer than proving that there's
infinitely many primes.

So our second proof of the infinitude of
the prime numbers is really similar to

killing a fly with a bazooka. Still a lot
of fun, of course, for people who are wired

like me, both the killing of the fly and
proving this. Now let's look at the

reciprocal of this identity. This is Euler's product connecting the primes with pi that I promised

you at the beginning. Now this stunning
identity also amounts to a proof of the

following very curious fact: What we
do is we pick two natural numbers

randomly. Then the probability that these
two numbers are relatively prime, so have

no common factors except for
1, that probability is equal to 6 over

PI squared which is about 61%. So how on
earth is this identity a proof of this

fact? Well let's have a look. The
probability of a randomly picked natural number

to be even is what? Well 1/2, obviously.
What about the probability of two

randomly picked numbers to be both
divisible by two? Well, they don't have

anything to do with each other. So
it's just 1/2 times 1/2 which is equal

to 1 over 2 squared. How about the
probability that not both are divisible

by 2? Well, that's simply 1 minus 1/2

squared and you can see something
happening, right? It's just our first

factor up there. Great, now we can play
the same game for all the other prime

numbers. So, for example, the
probability that not both numbers are

divisible by 3 is just 1-1/3^2 which is equal to the

second factor, and so on, which shows that
the probability of both numbers to have

no common prime factors is equal to the
infinite product. And this implies that

the probability of both numbers to be
relatively prime is equal to 6 over PI

squared. Well, actually, at least two of
the ingredients of this proof need a

little bit more justification and, well,
can you tell which? In any case, this

result really is true and can be turned
into a very very strange way to

approximate pi. What you do is you
randomly pick say a million pairs of

natural numbers and calculate how many
of these pairs are relatively prime. And

I've actually run a simulation on
Mathematica and that spat out

six hundred and eight thousand
three hundred twenty three and then the

probability is approximately, well, just
this fraction here, which means that pi

is approximately this expression here
which pans out to be 3.1405

Well, it's not great, but
it's not bad either. Now a number of

people actually got quite a bit of
mileage out of this insight by choosing

the random numbers from fun data set.
For example, astronomical data (sort of

pi in the sky), the digits of pi (sort of
pi from pi) chop pi into blocks and

interpret these as
random numbers, or license-plate numbers

that you come across on your way to work
(sort of pi from (pi)les of cars). Lame

joke but had to be done :)
Okay, so here's another way of writing

this formula. 1/zeta(2) is equal to this probability and

this actually does generalise in a
very straightforward way. Just change 2

to 3 and you get the
probability that three randomly chosen

numbers are relatively prime. And you can
play this game for any natural number.

Here's another fun question for you to
puzzle over. How good an approximation to

pi do you get if you use this identity
not indirectly, as we've just done, but

directly say by truncating the product
at the factor featuring 97, the largest

prime less than 100? Leave your
answers to this and all the other

teasers that I mentioned along the way
in the comments. And that's it for today.

Now how did this work for you? Hope you
all liked it. Well, actually, let me give

you a bit of a preview of what I'd like
to do next time (unless I get sidetracked

again). The point of departure for the
next video will be this third beautiful

identity that I derived for you at the
beginning of this video. It amounts to a

second way of defining the Riemann zeta
function. The plus/minus alternating sum

at the core of this definition is in
many ways much much better behaved than

the original pluses only one. I'll use
it and some of Euler's other ingenious

ideas to give a really accessible
description of the mysterious analytic

continuation of the zeta function that
many of you will have heard of. And this

will include a new take on the whole 1+2+3+...=-1/12

business, as well as chasing
down those elusive zeros that the

Riemann hypothesis is all about. Stay tuned.