Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let me start with the following on elementary symmetric polynomials:

The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the identity $$ [f(\lambda)=]\prod_{j=1}^n ( \lambda-X_j)=\lambda^n-e_1(X_1,\ldots,X_n)\lambda^{n-1}+e_2(X_1,\ldots,X_n)\lambda^{n-2}-\cdots+(-1)^n e_n(X_1,\ldots,X_n). $$

Let $n$ be even. The $k$th derivative evaluated at $\lambda=0$ is given by $$ \frac{d^k f(\lambda) }{d\lambda^k}\Biggr|_{\lambda=0}=(-1)^ke_{n-k}(X_1,\ldots,X_n) $$

So far. Now to something completely different: Take the $k$th power of the truncated Prime Zeta function $$ P_x(s)^k=\left(\sum_{p_t\,\in\mathrm{\,primes}\leq x} p_t^{-s}\right)^k =\sum_{k_1+k_2+\cdots+k_m=k} {k \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}p_{t}^{k_{t}}\;, $$ where ${k \choose k_1, k_2, \ldots, k_m} = \frac{k!}{k_1!\, k_2! \cdots k_m!} $, according to the Multinomial theorem, e.g. by W|A.

Let me show you the relevant part of the example $$ (11^s+13^s+17^s+19^s+23^s)^4=\ldots+24\cdot( 46189^s+55913^s+62491^s+ 81719^s+96577^s)+\ldots $$ where $46198=11\cdot 13\cdot 17\cdot 19$ and so on, in fact this part is the elementary symmetric polynomial $e_4(X_1,\ldots,X_5)$ with $X_k=p_{k+4}$ (the offset is just to not mix up the multinomial coefficient with the almost-primes).

Ok, now one more thing: The wiki page on Incidence Algebras states:

Multiplying by μ is analogous to differentiation.

Fine, now if I could multiply or apply (where I neglect the multinomial coefficients for the moment) the standard Möbius function $\mathbf \mu$ by the $k\,$th power of the truncated Prime Zeta function the only terms that survive are those, where all prime powers shows up only once, like in $46189^s$. All others have at least a square or a higher power of at least one prime and are therefore nulled out. Or more general, given a series $p_n(s)=\sum_{k=1}^n a_k $. I'd like to get $\hat{p}_n(s)=\sum_{k=1}^n \mu(k)a_k $. We are left with the elementary symmetric polynomial mentioned before.

So here is my question (again):

Is the application of $\mu$ on $P_x(s)^k$ analogous to the differentiation $\frac{d^k f(\lambda) }{d\lambda^k}\biggr|_{\lambda=0}$ ?

share|cite|improve this question
I think you meant $19$ and $23$ instead of $23$ and $29$ in that example? – joriki Oct 30 '12 at 7:34
I don't understand what you mean by "multiply or apply ... $\mu$ by ... the function", and the link in the last line doesn't really clarify it -- are you talking about pointwise multiplication or convolution? And in which variable? – joriki Oct 30 '12 at 7:40
@joriki, yes thanks for spotting. and yes: a pointwise multiplication (no variable needed in this case, right?) or something like a scalar product... Maybe it would be better to write "apply $\mu$ on the function" – draks ... Oct 30 '12 at 8:02
Well, "pointwise" requires points, i.e. a variable. Are you multiplying by $\mu(k)$, $\mu(x)$ or $\mu(s)$? I think it would be easiest and clearest if you just write out the expression you're thinking of. (After all you've written out enough other, less important parts of the question in considerable detail. :-) – joriki Oct 30 '12 at 8:05
@joriki Ok, but that's how the last linked question starts: Given a series $p_n(s)=\sum_{k=1}^n a_k $. I'd like to get $\hat{p}_n(s)=\sum_{k=1}^n \mu(k)a_k $. – draks ... Oct 30 '12 at 8:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.