Monday Math 89

A divisor function σ_x(n) is an arithmetic function consisting of the sum of the xth powers of the positive divisors of n:
$\sigma_x(n)=\sum_{d|n}d^x$
Notable special cases include $d(x)=\sigma_0(n)=\sum_{d|n}1$ , known as the divisor function, which gives the number of positive divisors of n; and $\sigma(x)=\sigma_1(n)=\sum_{d|n}d$ , the sum-of-divisors function, which adds all divisors of n, including n.
[The aliquot sum, the sum of all proper divisors of n, is thus s(n)=σ(n)-n; a "perfect" number is one for which s(n)=n.]
For example, consider n=12. The divisors are 1,2,3,4,6,12, and so:
$d(12)=\sigma_0(12)=1^0+2^0+3^0+4^0+6^0+12^0=1+1+1+1+1+1=6$
$\sigma(12)=\sigma_1(12)=1+2+3+4+6+12=28$
$\sigma_2(12)=1^2+2^2+3^2+4^2+6^2+12^2=1+4+9+16+36+144=210$ .
One can in fact define this for any real, or even complex, x. Note that for real x>0, we can see by multiplying σ_-x(n) by n^x, that $\sigma_{-x}(n)=\frac{\sigma_x(n)}{n^x}$

If we have n=ab, gcd(a,b)=1, with a having divisors 1,a₁,a₂,…,a and b having divisors 1,b₁,b₂,…,b, then we see that every product a_ib_j is a divisor of n, and that every divisor of n is such a product. Thus, we can show that divisor functions σ_x(n) are multiplicative functions.
For powers of primes, we see p^k has k+1 factors: 1,p,p²,…,p^k-1,p^k. Thus
$d\left(p^k\right)=\sigma_0\left(p^k\right)=k+1$
$\sigma\left(p^k\right)=\sigma_1\left(p^k\right)=1+p+p^2+\ldots+p^k=\frac{p^{k+1}-1}{p-1}$
$\sigma_x\left(p^k\right)=1^x+p^x+p^{2x}+\ldots+p^{kx}=\frac{p^{x(k+1)}-1}{p^x-1}$ (for x>0).
Considering, then, the prime factorization $n=p_1^{k_1}p_2^{k_2}\cdots{p_r^{k_r}}$ , we get
$\sigma_0(n)=\prod_{i=1}^r(k_i+1)$
$\sigma_x(n)=\prod_{i=1}^r\left(\frac{p_i^{x(k_i+1)}-1}{p_i^x-1}\right)$ .
[Note that $\lim_{x\to0}\left(\frac{p^{x(k+1)}-1}{p_i^x-1}\right)=k+1$ , so that the x=0 case is the limit of the x≠0 general case].

Recall previouly that the Dirichlet series generating function for a sequence consisting of a multiplicative function can be represented as a product over primes:
$F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{f\left(p^k\right)}{p^{ks}}$ .
For a divisor function, we see that
$\sum_{n=1}^{\infty}\frac{\sigma_x(n)}{n^s}=\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{\sigma_x\left(p^k\right)}{p^{ks}}$ .
Now, we need to use the formula for the product of power series:
$\left(\sum_{k=0}^{\infty}a_kx^k\right)\left(\sum_{k=0}^{\infty}b_kx^k\right)=\sum_{k=0}^{\infty}c_kx^k$ , where $c_k=\sum_{i=0}^{k}a_ib_{k-i}$ .
Letting a_k=1, b_k=b^k for some positive b, then:
$\left(\sum_{k=0}^{\infty}x^k\right)\left(\sum_{k=0}^{\infty}b^kx^k\right)=\sum_{k=0}^{\infty}\left(1+b+\ldots+b^k\right)x^k$ . (Note that both series on the left are geometric.)
Now, we see that since $\sigma_x\left(p^k\right)=1^x+p^x+p^{2x}+\ldots+p^{kx}$ , the sum inside the product for our Dirichlet series is equivalent to the right-hand side of the above with $b=p^x$ and $x=p^{-s}$ . Thus,
$\begin{eqnarray}\sum_{k=0}^{\infty}\frac{\sigma_x\left(p^k\right)}{p^{ks}}&=&\sum_{k=0}^{\infty}\left(1+p^x+\ldots+p^{kx}\right)p^{-ks}\\&=&\left(\sum_{k=0}^{\infty}\left(p^{-s}\right)^k\right)\left(\sum_{k=0}^{\infty}\left(p^x\right)^k\left(p^{-s}\right)^k\right)\\&=&\frac{1}{1-p^{-s}}\frac{1}{1-p^xp^{-s}}\\&=&\frac{1}{(1-p^{-s})(1-p^{x-s})}\end{eqnarray}$ .
Taking the product over the primes, we thus see that
$\begin{eqnarray}\sum_{n=1}^{\infty}\frac{\sigma_x(n)}{n^s}&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{\sigma_x\left(p^k\right)}{p^{ks}}\\&=&\prod_{p\;\text{prime}}\frac{1}{(1-p^{-s})(1-p^{x-s})}\\&=&\left(\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}}\right)\left(\prod_{p\;\text{prime}}\frac{1}{1-p^{x-s}}\right)\\\sum_{n=1}^{\infty}\frac{\sigma_x(n)}{n^s}&=&\zeta(s)\zeta(s-x)\end{eqnarray}$ .
This means that
$\sum_{n=1}^{\infty}\frac{d(n)}{n^s}=\zeta^2(s)$
and
$\sum_{n=1}^{\infty}\frac{\sigma(n)}{n^s}=\zeta(s)\zeta(s-1)$ .

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Tags: Dirichlet Series Generating Function, Divisor Functions, Divisors, Math, Monday Math, Multiplicative Function, Riemann Zeta Function

This entry was posted on September 14, 2009 at 12:11 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Monday Math 89”

Monday Math 90 « Twisted One 151’s Weblog Says:
September 28, 2009 at 12:06 am | Reply
[...] Math 90 By twistedone151 Last week, I talked about the divisor functions σx(n), defined as the sum of the xth powers of the [...]
Monday Math 94 « Twisted One 151's Weblog Says:
September 21, 2010 at 9:51 am | Reply
[...] function of the divisor function σx(n) is becomes much more simple than the proof seen here: Since , we see ; since the Dirichlet series generating function of 1(n) is , and the Dirichlet [...]

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Monday Math 89

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