 This lecture is part of Maths 1.15 Berkeley, an introductory undergraduate course on number theory, and will be mostly about Dirichlet series. So these are named after Dirichlet, and Dirichlet actually is one of the most impressive names of mathematicians, so his full name looked like this. I'm not quite sure why he had so many names, but his family name was apparently this double-barreled thing, so Dirichlet is really an abbreviation of his name. Anyway, Dirichlet series are a special case of generating functions, so I'll just stop by reviewing the concept of a generating function. So the most common form of an example of a generating function is a generating function that's a power series. So if we've got some sort of series of numbers, c0, c1, c2, we can encode them as a power series, c0 plus c1x plus c2x squared, and so on. So you might write this as a function, then you try and convert properties of the series ci into properties of the function f of x. So I'll just give one example of this, of the Fibonacci numbers. So the Fibonacci numbers are the property that f of n plus 2 is equal to f of n plus 1 plus f of n, and we take f0 equals 0, and f1 equals 1. So the first few numbers go 0, 1, 1, 2, 3, 5, 8, 13, and so on. And the problem is to find an explicit formula for these. So what we do is, you first of all put f of x equal to f0 x to 0 plus f1 x to 1 and so on. Now we try and find this function f of x by turning this recursion relation for the coefficients into some sort of formula for f of x. And if we write this out, we notice that x times f of x is equal to f0 x plus f1 x squared and so on. So f2 x squared. And x squared times f of x is equal to f of 0 plus f1 and so on. And now you notice that the recursion formula shows that if we take 1 minus x minus x squared times big f of x, so we're taking this term minus this term minus this term, then you notice everything cancels out except for these terms here, which just give us the factor of x. So we have this formula here. And now, using this, we can work out the function f of x and just find out what the coefficients f are, and you sort of do it like this. So we know f of x is now equal to x over 1 minus x minus x squared, which is equal to x over 1 minus phi of x times 1 minus 1 minus phi of x, where we've just taken phi to be one of the roots of this. So phi is equal to root 5 plus 1 over 2. We can write this as partial fractions. So this becomes 1 over root 5 times 1 over 1 minus phi of x minus 1 over 1 minus 1 minus phi of x, straightforward calculation. And now we can expand both of these as power series, and we see this is equal to 1 over root 5 times sum over all n of x to the n times phi to the n minus 1 minus phi to the n. So we've now got an explicit formula for the Fibonacci numbers. f of n is just 1 over root 5 times phi to the n minus 1 minus phi to the n, if I've got everything right. So that's a typical use of generating functions. The idea is you start with some sequence of numbers, you convert it into a function, you identify what this function is, and then you can work back and find information about your original sequence of numbers. So we try doing this with arithmetical functions. For example, we can take Euler's function phi of n, which is the number of integers less than n coprime to n, and we can try and form a power series. So phi of 1 times x to the 1 plus phi of 2 times x squared plus phi of 3 times x cubed and so on. So this looks like x plus x squared plus 2x cubed and so on. And the problem is this function here just seems to be a complete mess, and it's rather difficult to say anything much about it. For example, phi of n has this multiplicativity property, phi of mn is equal to phi of m times phi of n whenever m and n are coprime. And it's very difficult to convert this multiplicativity property to a nice property of this power series. So this just doesn't seem to work very well. However, power series are not the only form of generating function. What we can do is we can form a Dirichlet series. So if we've got a sequence of numbers c1, c2, c3, then instead of forming the power series with these as coefficients, it often turns out to be a good idea to form the following series. We take c1 over 1 to the s plus c2 over 2 to the s plus c3 over 3 to the s and so on. So you've got this function which is you traditionally use s as the variable for a Dirichlet series. And this sometimes works much better for arithmetical functions. So let's do some a few examples of this. The first example is we just take the series 1, 1, 1, 1 and so on. Now for power series, this doesn't give us anything terribly interesting because 1 plus x plus x squared and so on is just 1 over 1 minus x. So it's a fairly elementary function. For Dirichlet series, what we get is the function 1 over 1 to the s plus 1 over 2 to the s plus 1 over 3 to the s and so on. And this is the notorious Riemann zeta function which has caused more headaches than any other function in mathematics. So let's just review some properties of the Riemann zeta function. First of all, this converges provided the real part of s is greater than 1 or provided s is greater than 1 if s is real. That follows easily by comparison with the integral of 1 over x to the s from 1 to infinity. So this is a decreasing function of x. So the integral converges exactly when the series converges. At s equals 1, it definitely diverges because we get the harmonic series, which is well known for actually being infinite. The other another really useful property of the Riemann zeta function is its infinite product expansion. So zeta of s is 1 over 1 minus 2 to the minus s times 1 over 1 minus 3 to the minus s times 1 over 1 minus 5 to the minus s and so on. So this is a product over all primes p of 1 over 1 minus p to the minus s. And this sort of product expansion turns up a lot when you're studying the Ricollet series. So if we just quickly recall the proof of this, all you do is that you expand 1 over 1 minus 2 to the minus s as being 1 plus 2 to the minus s plus 2 to the minus s squared plus 2 to the minus s cubed and so on. And 1 over 1 minus 3 to the minus s is similar, 1 plus 3 to the minus s plus 3 to the minus s squared. And you get the same thing for 5 and for all other primes. And now if you multiply these all together, then you take a factor of 1 from most of them, but a factor of, and one of the factors further along. So for example, one of the many factors in the power series expansion, you will take 2 to the minus s cubed times 3 to the minus s and maybe you have a 7 to the minus s. And the product of these three factors and all these ones will be 2 to the minus 3s times 3 to the minus s times 7 to the minus s, which is equal to 2 cubed times 3 times 7 to the minus s. And what you notice is that every integer can be written as a product of powers of primes in a unique, so every positive integer can be written as a product of powers of primes in a unique way by the fundamental theorem of arithmetic, which means that every integer turns up exactly once when you multiply all these together. So this just becomes 1 over 1 to the s plus 1 over 2 to the s and so on. So the infinite product formula for the Riemann-Zeta function is equivalent to the fundamental theorem of arithmetic. So now let's see some other examples of this. First of all, we notice that this thing about having an infinite product expansion works whenever the series is multiplicative, so suppose you've got a series c1 over 1 to the s plus c2 over 2 to the s and so on. And suppose that the ci's are multiplicative, so cmn is equal to cm times cn, never m and n a co-prime. Then you can do this same trick, so this is, so here we have c1 equals 1, so so what we can do is we can multiply out 1 over 1 to the s plus c2 over 2 to the s plus c4 over 2 to the 2s and so on times 1 over 1 to the s plus c3 over 3 to the s plus c9 over 3 to the 2s and so on. And if we multiply this out just as before, we will get a lot of terms that a typical term will look something like, we might have a c4 over 2 to the 2s times c3 over 3 to the s times maybe c5 over 5 to the 2s. And what we will be getting is terms like c2 squared times c3 times c5 5 squared divided by 2 squared to the s times 3 to the s times 5 squared to the s. And this is equal to c2 squared times 3 times 5 squared over 2 squared times 3 times 5 squared to the s because the c's are multiplicative. So these two expressions are equal exactly when the c i's are a multiplicative function. And we can use this to identify a lot of Dirichlet series. So let's have some examples. The first simple example is just n to the k. So this is obviously multiplicative. In fact, it's strictly multiplicative because mn to the k is equal to m to the k times n to the k. And if we look at the corresponding Dirichlet series, we get 1 over 1 to the s plus 2 to the k over 2 to the s plus 3 to the k over 3 to the s. And it's not very difficult to figure out what this is. This is just zeta of s minus k. So I just sort of shifted zeta by a constant. And of course the product formula for zeta gives you a product formula for zeta of k minus s. So that's not terribly interesting. This is equal to product over p 1 over 1 minus p to the k minus s. So now let's go back to Euler's function phi of n. Now we recall this as multiplicative. So in order to work it out, the easiest way is just to find out its Euler product. So what we try and do is we work out 1 over 1 to the s plus 5p over p to the s plus phi of p squared over p to the 2s and so on. So this is called the Euler factor at the prime p. So all we have to do is to work out what this is. And this isn't terribly difficult because we know phi of p is p minus 1 and phi of p squared is p times p minus 1 and so on. So we just get 1 over 1 to the s plus p minus 1 over p to the s plus p squared minus p over p to the 2s and so on. And this splits as a sum of geometric series and you can see what it is. We just get a factor of 1 minus p to the minus s divided by 1 minus p to the 1 minus s, just by summing the geometric series. So sum of phi n times n to the s is equal to a product over all primes of 1 minus p to minus s over 1 minus p to the 1 minus s. And now we can see that this is just the factor we were getting for zeta of s and this is just the factor we were getting for zeta of s minus 1. So this is equal to zeta of s minus 1 over zeta of s. So when we take Euler's phi function as the coefficients of a power series we just get a horrible mess. When we take it as the coefficients of Dirichlet's series we actually get a reasonably nice function out of it. So now let's do the function tau of n which is the number of divisors of n. So tau of 1 is equal to 1, tau of 2 is equal to 2, tau of 3 is equal to 2 because 3 is 2 divisors and tau of 4 is equal to 3 because 4 is 3 divisors 1, 2 and 4 and so on. And we recall that tau of mn is tau of m times tau of n whenever m and n prime. So we can work out the Dirichlet's series tau of 1 over 1 to the s plus tau of 2 to the s and so on by working out the Euler factors and multiplying them together. And now we notice that tau of p to the k is just k plus 1 because the factors are 1 p p squared up to p to the k. So the Euler factor is going to be 1 over 1 to the s plus tau of p over p to the s plus tau of p squared over p to the 2s and so on which is 1 over 1 to the s plus 2 over p to the s plus 3 over p to the 2s and so on. And how do we sum this? Well this is just not quite a geometric series but it's quite close. It's something of the form 1 plus 2x plus 3x squared and so on and you probably recall from calculus that this is just 1 over 1 minus x squared. So this becomes 1 over 1 minus p to the minus s all squared. So this sum here becomes the product over all p of 1 over 1 minus p to the minus s squared which is obviously just zeta of s squared. So now let's look at a slightly more complicated example. Let's look at sigma of n which is the sum of the divisors of n. So sigma 1 is just 1, sigma 2 is 1 plus 2 which is 3, sigma 3 is 1 plus 3 which is 4 and sigma 4 is 1 plus 2 plus 4 which is equal to 7 and so on. So we're going to look at sigma 1 over 1 to the s plus sigma 2 to the s plus sigma 3 over 3 to the s and so on. And just as before we recall that sigma is multiplicative, sigma mn, sigma m times sigma n whenever m and n are co-primed so we just work out the Euler factors. And now we need to know what sigma of p to the k is. Well this is just 1 plus p plus p squared all the way up to plus p to the k which is equal to p to the k plus 1 minus 1 divided by p minus 1. So we have the Euler factor is going to be 1 over 1 to the s plus 1 plus p over p to the s plus 1 plus p plus p squared over p to the 2s. And if we sum this up we get that it's 1 minus p to the minus s sorry 1 over 1 minus p to the minus s times 1 minus p to the 1 minus s. So if you multiply this out it's just the series 1 plus 1 over p to the s plus 1 over p to the 2s and so on times 1 plus p over p to the s plus p squared over p to the 2s and so on. And if you multiply this expression by this expression you see you're just getting this expression here. So sum of sigma n over n to the s is just the product over all p 1 over 1 minus p to the minus s times 1 minus p to the 1 minus s and this is just zeta of s times zeta of s minus 1. There's something very similar you can do if you if you define sigma k of n to be the sum overall d dividing n of the k's powers of d then we can also form sum of sigma k of n over n to the s and this turns out to be zeta of s times zeta of s minus k. And the proof of this is very similar to the case of sigma n so I'm not going to write it out in detail. We just noticed that sigma 0 of n was just the number of divisors and sigma 1 of n was just the sum of the divisors. So the two special cases of this here we get sigma of s times sigma of s minus 0 and here we get sigma of s times sigma of s minus 1 so the two results about sigma and tau that I gave are special cases of this more general result. Next we come to the Moebius function mu of n and it's defining this rather peculiar way so it's equal to 0 if n is divisible by a square greater than 1 and it's equal to minus 1 to the k if n is a product of k distinct primes. This seems to be a completely bizarre definition but you'll see why it's a useful definition in a moment. So let's just look at a few values of it so if n is equal to 1 2 3 4 5 6 7 8 then mu of n is equal to 1 minus 1 minus 1 0 minus 1 1 minus 1 0 and so on. So it sort of oscillates in a more or less random fashion between 1 and minus 1 except it's sometimes 0 as well. So where on earth does this funny looking function come from? Well the explanation comes if you look at sum of mu to the n over n to the s and now you can see mu is multiplicative so we can write this as a product of its oiler product so let's look at 1 plus mu of p over p to the s plus mu of p squared over p to the 2s and so on well this is incredibly easy to work out because most of these terms just vanish this is just equal to 1 minus 1 over p to the s. So this sum here is just product overall primes of 1 minus p to the minus s and now we see this is just 1 over zeta of s so this sort of explains where this funny definition comes from is just the Dirichlet series coefficients of 1 over zeta of s. Well you may think it would have been more natural to define mu lambda of n to be minus 1 to the number of prime factors of n so it's like mu except it's never zero so if n is equal to 1 2 3 4 5 6 7 8 then lambda of n is well here it's 1 minus 1 minus 1 now it's plus 1 instead of zero for mu of n minus 1 plus 1 minus 1 and now it's going to be minus 1 again so so it's a bit like mu except whenever n is divisible by square it's it's it's not zero and we can work out some of lambda n over n to the s in much the same way so the corresponding Euler factor is 1 plus minus 1 over p to the s plus plus 1 over p to the 2 s and so on which is equal to 1 minus 1 over p to the s plus 1 over p to the 2 s minus and so on this is just a geometric series which is equal to 1 over 1 plus p to the minus s so how do we work this out so we want the product overall primes of 1 over 1 plus p to the minus s well there's a neat trick for doing this we can write this as a product overall primes of 1 minus p to the minus s over 1 minus p to the minus 2 s so and now each of these can be each of these products can be written in terms of the Riemann zeta functions this just becomes zeta of 2 s over zeta of s if I've got it the right way round so we can also look at an example of Dirichlet characters and these will be coming up later so let's just give a very simple example as far as I put chi of n to be given as follows so it's plus 1 if n is congruent to 1 mod 4 minus 1 if n is congruent to 3 mod 4 and 0 if n is even this is about the simplest non-trivial example of a Dirichlet character so so the sum of chi of n over n to the s looked like this it's 1 over 1 to the s minus 1 over 3 to the s plus 1 over 5 to the s minus 1 over 7 to the s and so on and we we can write this as an Euler product as follows so what we need to work out is for each prime p we need to work out 1 over 1 to the s plus chi of p over p to the s and so on plus chi of p squared over p to the 2 s and this obviously depends on what p is modulo 4 so if p equals 2 we just get 1 because all these terms are 0 if p is congruent to 1 mod 4 we get 1 plus 1 over p to the s plus 1 over p to the 2 s and so on which is equal to 1 over 1 minus p to the minus s if p is congruent to 3 mod 4 we get 1 minus 1 over p to the s plus 1 over p to the 2 s and so on which is 1 over 1 plus p to the minus s and in all cases you see that the factor we get is actually 1 over 1 minus chi of p times p to the minus s so this Dirichlet series l of s is just the product over all primes of 1 over 1 minus chi of p times p to the minus s and we'll be seeing lots of examples of this coming up in a few lectures time now let's give an example finished by giving the example that is not multiplicative this is the the function sometimes denoted by lambda of n which has the following rather funny definition it's equal to log of p if n is equal to p to the k for k greater than or equal to 1 and p prime and it's equal to 0 otherwise and it's definitely not multiplicative so we can't write it as an Euler product and you may ask where does this rather bizarre looking function come from well it turns out that um sum of um so lambda 0 over so lambda 1 over 1 to the s plus lambda 2 over 2 to the s plus lambda 3 over 3 to the s and so on turns out to be the zeta prime of s over zeta of s which is actually the the derivative of the log of zeta of s and this this function turns out to be really important when you're trying to count primes because it enables you to turn the problem of counting primes into the problem of studying the Riemann zeta function and to see this it isn't very difficult so we start with zeta of s is the product overall primes of of not taking long yet of 1 over 1 minus p to the minus s so log of zeta of s is sum overall primes of log of 1 minus p to the minus s which is just the sum overall primes of minus sorry p to the minus s plus p to the minus 2s over 2 plus p to the minus 3s over 3 and so on so we we're just writing down the usual power series for log of 1 minus x and now if we differentiate it with respect to s we get the sum overall p of log of p times p to the minus s plus log of p times p to the minus 2s plus log of p times p to the minus 3s and so on and you notice when we differentiate p to the minus 3s we get a factor of 3 but that then cancels out with this 3 in the denominator so this this is a sum over all primes and sum over all n's of log p times p to the minus ns so this is just the sum overall n of lambda n times n to the minus s so this function will be turning up when we briefly discuss the prime number theorem in in in a couple of lectures time okay um so um the next video will be about converting various other properties of um arithmetical functions into properties of Dirichlet series