 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and will be about products of Dirichlet series. So to start off with, I first recall about products of power series. So suppose I've got a power series A0 plus A1x plus A2x squared and so on, and I multiply it by another power series B0 plus B1x and so on. Then I get the power series C0 plus C1x and so on, where the number Ci, given as follows. So C0 is just A0B0, C1 is A1B0 plus A0B1, C2 is A2B0 plus A1B1 plus A0B2, and so on. And in general, Cn is just the sum over all i of Ai times Bn minus i. This expression here is sometimes called a convolution of the series A and B. In the theory of Fourier integrals, you get a rather similar convolution, where if you take two functions f and g, the convolution is the integral of f of y minus x times g of x dx. And you can see this is rather similar to this expression here, except you've replaced integration over the reals by summation over the integers, or maybe the positive integers. So taking a product of power series corresponds to taking a convolution of the corresponding arithmetic functions. So what happens for Dirichlet series? So let's take a Dirichlet series f of s, which is f1 over 1 to the s plus f2 over 2 to the s and so on. And g of s is going to be g1 over 1 to the s plus g2 over 2 to the s. And let's just multiply them together. So f of s times g of s. Well, the first coefficient is just going to be f1 times g1 over 1 to the s. And the next one is f1 times g2 plus f2g1 over 2 to the s. So so far it's rather similar to power series. The next coefficient is f1g3 plus f3g1 over 3 to the s. And now you notice there's no, nothing like f2g2 in the numerator that we might expect by analogy with power series. And the next one is f1g4 plus f2g2 plus f4g1 over 4 to the s. And now you notice that what's going on is the pairs of subscripts 1, 4, 2, 2, 4, 1 are always of writing 4 as a product of two numbers. In other words, this is really a sum over divisors of 4. So f of s times g of s is a sum over all n of 1 over n to the s times the coefficient, which is a sum over all divisors of n of f of d times g of n over d. And now this little bit here, sum over d divides n of f of dg n over d is really a sort of multiplicative convolution. You see, it's rather like the additive convolution we had earlier, except instead of adding, instead of summing over all pairs of integers that sum up to n, we're summing over all pairs of integers whose product is n. So products of Dirichlet series correspond to taking a sort of multiplicative convolution of two arithmetic functions. So let's look at an example of this. Suppose we take f of s to be just the usual Riemann-Zeta function. So that's 1 over 1 to the s plus 1 over 2 to the s and so on. So the numbers, the function f of n is just given by f of n equals 1 for all n. And let's take g of s to be g of 1 over 1 to the s plus g of 2 over 2 to the s and so on, and work out what Zeta of s times g of s looks like. Well, if we write its coefficients as h of n over n to the s, then we see that h of n is just equal to sum over d divides n of g of d times the function 1 taking an argument n over d, which is just 1. So this is just sum over d divides n of g of d. We can see an example of this. If we take g of n to be, say, Euler's function phi of n, the number of numbers less than n that are co-prime to n. Then we recall we had the identity, the sum over d divides n of phi of d was just equal to n. And you remember, we used this identity, improving that numbers had primitive roots by carefully counting numbers of elements of various orders. So if we take f of s, so g of s to be the function phi of 1 over 1 to the s plus phi of 2 over 2 to the s and so on, then Zeta of s times g of s is equal to 1 over 1 to the s plus 2 over 2 to the s plus 3 over 3 to the s and so on, where the numerators 1, 2, 3 are just the values of this function f of n is n up here. And from this, well, so this series here is just Zeta of s minus 1 as we saw earlier. So from this, we can work out what g of s is. g of s is just equal to Zeta of s minus 1 divided by Zeta of s. So that's the function whose coefficients are Euler's phi function. So we found that in previous lecture, different method using Euler products, but here we've found it by taking products of Dirichlet series. We can also get something called the Mobius inversion formula. So the Mobius inversion formula says this, if g of s is equal to Zeta of s times f of s, then obviously f of s is equal to Zeta of s minus 1 times g of s. And if you write it like this, it looks completely and utterly trivial. If we write it in terms of the coefficients of the functions f and g, it looks rather more subtle. So the first equation says that g of n is equal to sum of d divides n of f of d. And the second equation says that f of n is equal to the sum of d divides n of g of d times the Mobius function of n over d. So the Mobius function comes because it's the Fourier coefficients of Zeta of s. So Zeta of s minus 1 is just sum of mu n over n to the s. So this pair of equations is called the Mobius inversion formula. Yeah, it's the same guy who invented the Mobius band with a twist in it, but he also did other things. So an example of this, suppose we recall that n is the sum of d divides n by 5n. So in this, we're taking g of n to be n and f of n to be phi of n. So we can apply the Mobius inversion formula to this and we find that phi of n is equal to the sum of d divides n of d times mu of n over d. So for instance, this is the phi of 6 is equal to 6 minus 3 minus 2 plus 1. So another example of a product of Dirichlet series, we can take Zeta of s times L of s, where L of s was the series 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 this series turns out to be equal to a quarter of sum over n of r of n over n to the s, where r of n is the number of solutions to n equals a squared plus b squared. So you remember we discussed the number of solutions to this before and we found that r of n was actually multiplicative and depended on the prime factorization of n and whether prime factors were 1 or 3 modulo 4. In particular, we see the series r1 over 1 to the s plus rp over p to the s plus r of p squared over p to the 2s and so on. The sum of this depends on the value of p. So for p equals 2, it's 1 over 1 minus 2 to the s because these numerators rule 1. For p1 mod 4, we get it's 1 over 1 minus p to the minus s squared as you recall from a few lectures ago. But for p3 mod 4, the number of representations of p to the n is 0 unless n is even. So we just get this factor of 1 over 1 minus p to the minus 2s and we can write this as 1 over 1 minus p to the minus s and 1 over 1 plus p to the minus s. And now if you compare all these Euler factors with the Euler factors of zeta of s. So the Euler factor of zeta of s was 1 over 1 minus p to the minus s and the Euler factor of l of s was 1 over 1 minus chi of p times p to the minus s. You see that zeta of s times l of s has exactly the same Euler factors as a quarter of sum of rn over n to the s. Quarter there I guess. So this expression here is actually something called the zeta function of the Gaussian integers. So the usual Riemann zeta function is just the zeta function of the ordinary integers. And there are analogs of this for various other rings and about the simplest example is the ring of Gaussian integers whose zeta function in fact turns out to be this. You can sort of understand the connection because the coefficients of zeta of s or the coefficient of 1 over n to the s of zeta of s is up to a factor of 2, the number of ordinary integers of absolute value n. And here the coefficient of 1 over n to the s is the number of Gaussian integers such that the square of the absolute value is n, again up to a factor of 4 or so. So the zeta function you can think of as being something like counting the number of Gaussian integers of given norm. Another thing we can do with Dirichlet series is differentiate them. So suppose I've got a Dirichlet series, f of s which is f of 1 over 1 to the s plus f of 2 over 2 to the s and so on. And what happens if I differentiate it with respect to s? Well what I get, well the derivative of n to the minus s is minus log n times n to the minus s. So I essentially multiply all the coefficients by log of n. If I put a minus sign in front there I get log of 1 times f of 1 over 1 to the s plus log of 2 times f of 2 over 2 to the s and so on. Log of 1 is 0 of course. I'm just putting it in so that you can see what the pattern is. So an example of this is the function lambda n which is equal to log of p if n is equal to p to the k and 0 otherwise. So you remember last time that sum of lambda n times n minus s was just the derivative with respect to s of log of zeta of s. So here we've got an example of taking the derivative of a Dirichlet series. As an example of all this we can prove an identity due to Selberg. So this is one version of Selberg's identity. So Selberg's identity was a rather remarkable identity found by Selberg that he used to give an elementary proof of the prime number theorem. And his identity isn't that he proved the prime number theorem isn't exactly the following identity but is rather closely related to it. So his identity says that lambda n times log of n plus sum over d divides n of lambda d times lambda n over d is equal to sum over d divides n of mu of d times log of n over d or squared. And if you were asked to prove this identity it would look like a complete nightmare. I mean this looks like this just completely bizarre complicated identity and you've no idea how you should even approach it. Well if you approach it using generating functions for Dirichlet series it becomes essentially trivial to prove. So all we do is we find the functions whose coefficients are given by each of these functions here. So for lambda n you remember these are the coefficients of this corresponds to the Dirichlet series zeta prime of s divided by zeta of s. And you remember if we multiply the coefficients by log of n this corresponds to differentiating the function so we get zeta prime of s over zeta of s differentiated possibly with a minus sign in there. I always get muddled up about the minus signs. And here the expression we've got with summing over d divides n of some function of d times some function of n over d. So this corresponds to the Dirichlet series whose coefficients are lambda of d. You remember this is just zeta of prime of s over zeta of s times the function whose coefficients are lambda d again. So we just take zeta prime of s over zeta of s again. Now here the function log of n squared, so this bit here would just correspond to the function where we take the second derivative of zeta of s. Now this whole thing here where we sum over mu of d times log of n squared you remember when we sum over d divides n of mu of d times something this corresponds to multiplying by 1 over zeta of s. So this term on the right corresponds to the Dirichlet series zeta double prime of s over zeta of s. And now all we've got is the identity zeta prime of s over zeta of s differentiated is equal to zeta prime of s squared over zeta of s squared times zeta double prime of plus zeta double prime of s over zeta of s. And this is essentially just Leibniz's rule for differentiating so it should be a plus there for differentiating the quotient of two functions. So this completely horrendous looking identity between arithmetical functions just turns out to be a elementary piece of calculus where you're just differentiating a product or quotient of functions. Okay, so next lecture I'll be talking a little bit about the prime number theorem and not giving a complete proof of it because that's rather too complicated but sketching the main ideas of a proof of it.