 This lecture is part of an online course on the theory of numbers and will be about Dirichlet series. So, last lecture we were looking at an arithmetic function fn which is just a function defined for positive integers n. And there's a very powerful method for dealing with arithmetic functions which is called the method of generating functions. There are two sorts of generating functions. One is to form a sort of power series whose coefficients are fn. So you would take f0 plus f1x plus f2x squared plus f3x cubed and so on. We're not going to discuss power series very much, at least not in this lecture, although these quite often turn up in combinatorics. What we're going to do is a different sort of generating function which is a Dirichlet series. Here, instead of taking powers of x, we take f1 over 1 to the s plus f2 over 2 to the s plus f3 over 3 to the s and so on. So, here the numbers 1, 2 and 3 are in the exponents and here they're not, they're in the thing you raised to the power of s. So we could call this function big f of s. And the idea is to relate properties of the generating function little f to the Dirichlet series big f of s. So, let's start by looking at some examples. So, here's a function f of n and here's the corresponding Dirichlet series f of s. So let's start just by taking f of n equals 1 for all n. Well, here what we get is the series 1 over 1 to the s plus 1 over 2 to the s plus 1 over 3 to the s and so on, which is of course just the Riemann zeta function. So, even the trivial function 1 here turns into quite a complicated function in the Riemann zeta function. It's one of the most puzzling functions in mathematics, what with the Riemann hypothesis and so on. So, what happens if we take f of n equals n? Well, that's easy. That gives 1 over 1 to the s plus 2 over 2 to the s plus 3 over 3 to the s. You can see this is just zeta of s minus 1. And similarly, if we take f of n equals n to the k, you can see we're going to get zeta of s minus k. So, polynomials on the arithmetic function side can all be written in terms of zeta of s on the Dirichlet series side. So far, that's not terribly exciting. But now let's look at a more interesting function. Let's take f of n to be sigma 0 of n. So, you remember this is equal to the number of divisors of n. So, let's look at its Dirichlet series. It's 1 over 1 to the s because 1 is 1 divisor plus 2 over 2 to the s plus 2 over 3 to the s plus 3 over 4 to the s plus 2 over 5 to the s plus 4 over 6 to the s and so on. So, if you look at this, you can see it's probably not at all obvious what this series is. So, we want to identify this series. And the key point in identifying these is to use the fact that sigma 0 is multiplicative. So, you remember this means sigma 0 of mn is sigma 0 of m times sigma 0 of n whenever m and n are co-prime. And this isn't actually true in general. Now, if we've got a multiplicative function f, then the corresponding Dirichlet series splits up as a so-called Euler product. And let's see why. So, we take 1 plus f of 2 over 2 to the s plus f of 4 over 4 to the s plus f of 8 over 8 to the s. And then we're going to multiply this by 1 plus f of 3 over 3 to the s plus f of 3 squared over 3 to the 2s plus f of 3 cubed over 3 to the 3s and so on. And then we're going to do the same for all primes. 1 plus f of 7 over 7 to the s and so on. And the convention is when we have an infinite product like this, it means to form the infinite product, we pick up one term from each factor with the convention that all but a finite number of factors we have to pick the term 1. If you do this analytically, you can make more sense of this but we're just doing things formally. And now the key point is that if you take this infinite product over all primes, this is just equal to sum of fn over n to the s. And you can see this if you look at the prime factorization of n. So this is going to be f of p1 to the n1 times f of p2 to the n2 and so on, all divided by p1 to the n1, p2 to the n2 and so on, where n is equal to p1 to the n1 times p2 to the n2 and so on. So first of all we've got the fundamental theorem arithmetic which says that each denominator n to the s occurs exactly once because it can be written in one way as a product of prime powers. So we get the correspondence to these following from the fundamental theorem of arithmetic. And the numerator is equal to the product of these terms. You see we're taking one term from each line here. So the numerator looks like that because f is multiplicative. So whenever we've got a multiplicative function f, this makes it much easier to work out what the Dirichlet series is. So let's apply this to sigma zero. So we've got to work out the Euler factor. I should say each of these things here is called an Euler factor. This is because Euler found the first example of this. You remember he wrote the Riemann Zeta function as a kind of product like this. It's called the Euler product and these are called the Euler factors. So let's work out the Euler factors for sigma zero. Well, what we need to do is we just form sigma zero of one over one plus sigma zero of p over p to the s plus sigma zero of p squared over p to the 2s and so on. And we should remember what sigma zero is. This is just the number of factors. So this becomes one over one to the s plus two over p to the s plus three over p to the 2s and so on. So now this is looking much more reasonable because the numerators are just going one, two, three instead of going wildly all over the place as sigma zero does. And you in fact notice that the denominators are now essentially just making this into a power series. So this is just a power series in one over p to the s. So if we just restrict the terms that involve a power of p then Dirichlet series just become power series. So all the factors turn out to be power series which are much easier to evaluate. Well now we recall that one plus two x plus three x squared and so on is equal to one over one minus x all squared. So this power series here is just equal to one over one minus p to the minus s all squared. So now we can work out the Dirichlet series for sigma zero. So sum of sigma zero of n over n to the s is a product overall primes of the Euler factor. And we figured out what the Euler factor was. So this is just a product overall primes of one minus one over one minus p to the minus s all squared. Well this should look vaguely familiar because we recall that zeta of s is a product over p of one minus one minus p to the minus s. And if you compare this expression with this expression who see they're pretty similar apart from the fact that this is squared. So this is just zeta of s all squared. So there we found the Dirichlet series for sigma zero. It turns out to be a rather simple expression. So now that we've done sigma zero let's try sigma one and this is going to be very similar. So sigma one is also multiplicative. So we just have to work out the Euler factor. So what we want to know is the Euler factor which is going to be sigma one of one over one to the s plus sigma one of p over p to the s and so on. And this is going to be one over one to the s plus well the sum of the devices of p is one plus p over p to the s. And then we have one plus p plus p squared over p to the two s and one plus p plus p squared plus p cubed over p to the three s and so on. And this takes a little more thought to figure out what this is. And then you'll notice that this can actually be written as a product. It's a product of one over one to the s plus one over p to the s plus one over p to the two s plus one over p to the three s and so on. Times one over one to the s plus p over p to the s plus p squared over p to the two s plus p cubed over p to the three s and so on. And now we can work out each of these because they're just geometric series so this is equal to 1 over 1 minus p to the minus s times 1 over 1 minus p to the 1 minus s. So we can work out the Dirichlet series for sigma 1. We find sum of sigma 1 of n over n to the s is a product overall primes of 1 over 1 minus p to the minus s times 1 over 1 minus p to the 1 minus s. And now this bit here is giving you zeta of s and this bit here is just giving you zeta of s minus 1. So this Dirichlet series is just zeta of s times zeta of s minus 1. So again if we looked at the power series sum of sigma 1 n times x to the n this would be something really complicated and horrible but the Dirichlet series turns out to be something really nice. Well the next we can sort of generalize this example. So you can see for example that sum of sigma k of n over n to the s is now going to be sigma of s times sigma of s minus k by the same argument. And we can generalize this a little bit more because we know that sigma of k of n is the sum over all divisors of n of d to the k. And more generally you quite often have some arithmetic function g of n that's the sum of all over all the divisors of n of some other arithmetic function. And now suppose suppose we we form the Dirichlet series of these. So a form g of s is sum of g n over n to the s and f of s is sum over f of n over n to the s. And suppose g and f are related by this. How are big f and big g related? Well if we take f of s so this is 1 over 1 to the s plus 1 over sorry f of 1 over 1 to the s plus f of 2 over 2 to the s plus f of 3 over 3 to the s and so on. And let's multiply it by zeta of s which is 1 over 1 to the s plus 1 over 2 to the s plus 1 over 3 to the s and so on. And the product zeta of s times f of s let's multiply it out. Well it's going to be f of 1 over 1 to the s and then we get f of 1 plus f of 2 over 2 to the s. So this is so far this is just like multiplying a couple of power series. But then we get f of 1 plus f of 3 over 3 to the s plus f of 1 plus f of 2 plus f of 4 over 4 to the s. Because you can see we can get a 4 to the s by taking f of 4 or we can take f of 2 times 1 or we can take f of 1 times 1. And you see what's going on here is this is just summing over the divisors of 4 and this is summing over the divisors of 3 and so on. So here the coefficients are given by sum over d divides n of f of d which is this expression here. So we see that g of s is just equal to zeta of s times f of s. So if you multiply Dirichlet series by zeta of s what you're doing to the coefficients is this funny sort of summing over divisors. And as special cases of this we can we can now get the Dirichlet series for sigma k of n slightly more quickly. So if we look at sigma k of n well this is equal to sum over d divides n of d to the k. Now the Dirichlet series sum of d to the k over n to the s sorry that would be n to the k over n to the s is just zeta of n minus k. So the coefficients of the Dirichlet series sum of sigma k of n over n to the s when you take zeta of n minus k that comes from the d to the k and then you have to multiply this times zeta of so that's not n minus k so that should be s minus k. You should multiply this by zeta of s and the zeta of s comes because you're doing this sum over d divides n thing. So that's a slightly quicker way of finding the Dirichlet series of sigma k of s. There's a more general version of this. Sometimes we do what is known as a Dirichlet convolution. So if I take an arithmetic function f of n and g of n I might define a new arithmetic function h of n to be sum over d divides n of f of d g n over d. So these sorts of things turn up quite a lot so this is a Dirichlet convolution. And what I want to do is to find the relation between the corresponding Dirichlet series and this is quite easy because if we write f of s is f of 1 over 1 to the s plus f of 2 over 2 to the s plus f of 3, 3 to the s plus f of 4 over 4 to the s plus f of 5, 5 to the s plus f of 6 over 6 to the s and so on. And you write g of s to be the same g of 1 over 1 to the s plus g of 2, 2 to the s plus g of 3 over 3 to the s. Right, now I want to work out the coefficients of f of s times g of s and let's work out the coefficient of say 1 over 6 to the s. So what's the coefficient going to be here? Well so how can we get 6 to the s? Well we can get it as 1 to the s times 6 to the s so we get the product of these two terms and then we can get 2 to the s times 3 to the s so we get these two terms and then we get 2 to the s times 3 to the s so we get these two terms. I don't know if yellow shows up, well not very well. And then we get 1 to the s times 6 to the s so we get these two terms. So the coefficient of 1 over 6 to the s is going to be f 1 times g of 6 plus f of 2 g 3 plus f of 3 g 2 plus f of 6 g 1. And you can see this coefficient will typically be sum over d divides n of f of d times g of n over d. So if you've got two arithmetic functions you take this Dirichlet convolution which is really rather complicated from the point of view of Dirichlet series all you're doing is multiplying the corresponding Dirichlet series together. So we would have h of s equals f of s times g of s where h of s is sum over h of n over n to the s and so on. So now let's look at another Dirichlet series. Let's look at the series sum of phi of n over n to the s. As usual we notice that phi is multiplicative so we want to work out the Euler factor. So the Euler factor is going to be 1 plus phi of p over p to the s plus phi of p squared over p to the 2s and so on. And you can write this as 1 plus p minus 1 over p to the s plus p squared minus p over p to the 2s and so on. And you can split the numerator into two bits. So first of all we have a 1 plus 1 over p to the s minus 1 plus 1 over p to the 2s minus 2 and so on. So this bit is coming from these terms and then we've got these other terms so we get minus 1 over p to the s times 1 plus p over p to the s plus p squared over p to the 2s and so on. So this bit here is coming from these terms here and we can add this off easily because this is 1 over 1 minus p to the 1 minus s because it's a geometric series and this bit here is 1 over p to the s times 1 over p to the 1 minus s. So all together this is equal to 1 minus p to the minus s over 1 minus p to the 1 minus s. Now that we found the Euler factors we can work out the Dirichlet series for some of 5n over n to the s so this is a product overall primes of the Euler factor 1 minus p to the minus s over 1 minus p to the 1 minus s and this sort of looks suspiciously similar to zeta of s which is product 1 over p to 1 minus p to the minus s. So this turns out to be zeta of s minus 1 over zeta of s and this is because the zeta of s minus 1 comes from these bits here and the 1 over zeta of s comes from these bits here. So here we found the Dirichlet series for 5n and now you see there's actually a sort of dictionary between arithmetic functions and Dirichlet series. So let's write out what we've got. So if you've got an arithmetic function f of n it corresponds to a Dirichlet series f of s which is sum of f of n over n to the s and we've seen that 1 corresponds to the zeta function of s and n corresponds to zeta of s minus 1 and more generally if we've got n to the k times f of n this corresponds to the Dirichlet series f of s minus k so multiplying by power of n corresponds to shifting the Dirichlet series by constant. If the arithmetic function is multiplicative this corresponds to the Dirichlet series having an Euler product so it's a product of overall primes of some power series in p to the minus s and we saw that if we've got something like g of n equals sum of d divides n of f of d this just says that g of s equals zeta of s times f of s and more generally if h of n is sum over d divides n of f d g n over d this corresponds to h equals f times g so you see that this side is tends to be rather simpler than that side and then we had some special cases sigma nought corresponds to zeta of s squared sigma 1 corresponds to zeta of s times zeta of s minus 1 and phi Euler's phi function corresponds to zeta of s minus 1 divided by zeta of s well what's the point of all this well the point is we can look at the right-hand side and find various identities and translate these into identities for arithmetic functions so let's do one example so what we're going to start off with is looking at this expression and this expression so let's start off with the following identity zeta of s times zeta of s minus 1 divided by zeta of s is equal to zeta of s minus 1 so this identity is completely utterly trivial and so trivial it may wonder what is the point of writing it down well let's convert this into a statement about arithmetic functions so first of all you remember that zeta of s minus 1 over zeta of s is the Dirichlet series corresponding to Euler's phi function so let's take phi right phi of n here and then we saw that multiplying by zeta of s does this funny sum over divisors so let's change this to a d and do sum over d divides n phi of d and zeta of s minus n corresponds to n so this absolutely trivial identity here if we translate it into arithmetic functions it becomes this sort of non-trivial relation for Euler's phi function for example if we take n equals 12 it says that phi of 1 plus 5 2 plus 5 3 plus 5 4 plus 5 6 plus 5 12 equals 12 and you can check this these numbers are 1 1 2 2 2 and 4 now that doesn't add up 1 1 4 5 6 7 8 9 2 yes it does add up sorry well there's actually a different way to prove this identity what we can have a look is look at the numbers from 1 up to 12 so it's the greatest common divisor with 12 is 12 6 4 3 2 1 and you can see at the number whose greatest common divisor with 12 is 12 the number of these is just 1 because it has to be a multiple of 12 the greatest common divisor has to be a multiple of 6 and must be multiple of 6 that's it can't be a multiple of 6 that's something dividing dividing 12 and so on so these numbers here can also be thought of as the number of numbers from between 1 and 12 whose greatest common divisor with 12 is exactly one of these devices of 12 and so this is actually another way of seeing the sum of seeing this expression here you're just sort of summing over all numbers from 1 to n whose greatest common divisor is equal to something in fact you can draw this pictorially if you draw the numbers from 1 to 12 as a sort of clock face as we do in first grade you can identify the numbers so here 5 of 1 let's try these numbers 1 2 3 4 5 6 7 8 9 miss one out miss one out there 8 9 10 11 12 I was obviously asleep during the lecturing kindergarten where we learned how to draw clock faces and anyway what you see is this one here corresponds to this number here because there's only one number whose greatest common divisor is 12 once whose greatest common divisor is 6 is just 6 and then we get 3 and 9 corresponds to that and let me use a different color so I don't get too confused sorry go to that one there and this one here corresponds to these 2 and 6 corresponds to these ones here and the ones left over are 115 and 7 which are the four numbers co-prime to 12 so for another example of an identity let's try and prove the following identity suppose you want to prove that sum of d divides n or sigma 1 of d times phi of n over d is equal to n times sigma 0 of n okay well if you try and prove that in terms of arithmetic functions it's really a bit of a mess I mean it's not exactly difficult but it's absolutely trivial to do if we convert this into Dirichlet series so let's do this so what Dirichlet series does phi can correspond to well that's just zeta of s minus 1 over zeta of s and what does sigma 1 correspond to well sigma 1 corresponds to zeta of s times zeta of s minus 1 and what does the sum summing over d divides n of something of d times something of n over d well that just corresponds to multiplying these together and sigma 0 of n corresponds to zeta of s all squared now if we multiply it by n that just corresponds to shifting n by 1 so we should change that to zeta of s minus 1 or squared so this shift this minus 1 comes because we're we're multiplying by n there so this identity turns out to be equivalent to proving this identity here zeta of s times zeta of s minus 1 times zeta of s minus 1 over zeta of s is zeta of s minus 1 squared and I think you will agree that this identity here is completely and utterly trivial and so as the last example I will just talk about the Moebius inversion formula well to define the Moebius inversion formula we first need the Moebius function and this is the arithmetic function such that sum of mu n over n to the s is equal to 1 over zeta of s so let's try and work at what mu of n is well this is equal to product of 1 minus p to minus s so that's 1 minus 2 to minus s times 1 minus 3 to minus s 1 minus 5 to minus s and so on and if you multiply this out you can sort of see immediately what mu of n is so mu of n is equal to not if n is divisible by a square of some prime and it's minus 1 to the k if n is a product of k distinct primes so let's just have a little table of its values so if n is 0 1 2 3 4 5 6 7 8 9 10 then you can see that mu of n is equal to 1 minus 1 minus 1 0 minus 1 1 minus 1 0 0 1 and so on so it's sort of oscillates randomly between minus 1 and 1 except that it's sometimes 0 and now the Moebius function turns up with the following problem suppose that g of n is sum of d divides n of f of d so as we've seen this sort of relation between functions turns up fairly often and the problem suppose we're given g find f so we want to sort of invert this if we're given f it's easy to find g but you know what happens if we've got g and want to find f well this is easy to answer if we write it in terms of the reclase series so if this just says that g of s equals zeta of s times f of s where as usual g of s is sum of g of n over n to the s and f of s is sum of f of n over n to the s well how about we just divide both sides by zeta of s so we find f of s is g of s times 1 over zeta of s that was trivial now we convert it back into arithmetic function so this says f of n is sum over d divides n of mu of d times g of n over d so the mu of d are just turning up because they're the coefficients of 1 over zeta of s and of course the g's just turn up because they're the coefficients of big g yes this sorry this yellow really does seem to be a legible right it in blue which is now turning into a funny green because of the yellow who will never mind and the multiplication of the reclase series is turning into this funny sum over d divides n so this gives us a formula for f in terms of g and this is the moebius inversion formula yeah this is the same guy who invented the moebius band he did do other things as well so let's just have an example of the moebius inversion formula so we know so we showed earlier that sum over d divides n 5n is equal to n and let's just use this to find a formula for phi and this is kind of cheating a bit because we're really just reversing the calculation that we did for that well moebius the moebius formula just says that phi of n sorry that should be d of course phi of n is just equal to sum over d divides n of mu of d times n over d or equivalent you can also write this as sum of d divides n of d times mu of n over d doesn't really matter which way around you put these for instance it just says that phi of 12 is equal to 12 minus 6 minus 4 plus 2 because here we're taking d to be 12 6 4 and 2 and n over d is going to be 1 2 3 and 6 for which the moebius function is 1 minus 1 minus 1 and 1