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.