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How can I show that if $\Phi$ is a Cyclotomic polynomial, $$\Phi_n(x)=\prod_{\substack{1\leq k\leq n\\(n,k)=1}}(x-\zeta_n^k)$$ With $\frac{d}{dx}\Phi_n(x)=\Phi'_n(x)$

Then, $$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$

$$\sum_{n=1}^\infty\frac{1}{n^8}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{4725\zeta(7)}{\pi^8e^{2\pi}}+\frac{19}{12\pi e^{2\pi}}$$

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What is the source of the problem? It is interesting! +$1$ – Clayton May 1 '13 at 5:09
@Clayton Just my own stuff – Ethan May 1 '13 at 5:32
Dear Ethan: those are very nice. If you would like to indicate the line of reasoning which led you to them, maybe we could figure out a proof. I believe your identities are related to the theory of modular forms. All the best, – Bruno Joyal May 23 '13 at 20:41
up vote 9 down vote accepted

$\Phi'_n(x)/\Phi_n(x) = \sum 1/(x - \zeta_n^k) = x^{-1} \sum 1/(1 - \zeta_n^kx^{-1}) = x^{-1} \sum_{d \ge 0} \zeta_n^{kd} x^{-d}$.

Let $S(n,d) = \sum \zeta_n^{kd}$. We easily have $S(1,d) = 1, S(n,0) = \phi(n)$ and $S(n,1) = \mu(n)$ (the Möbius function).
If $\gcd(d,n) = 1$ then $\zeta_n^k \mapsto \zeta_n^{kd}$ is a permutation of the roots, so $S(n,d) = S(n,1) = \mu(n)$.
If not, then write $d = gd'$ and $n = gn'$ where $g = \gcd(d,n)$. We have $\zeta_n^d = \zeta_{n'}^{d'}$, and so $S(n,d) = \phi(n)S(n',d')/\phi(n')$. Then, since $\gcd(n',d') = 1$, $S(n,d) = \phi(n)\mu(n',d')/\phi(n')$.

Now, $\Psi(s,x) = \sum n^{-s}\Phi'_n(x)/\Phi_n(x) = x^{-1}\sum_{d \ge 0} (\sum_{n \ge 1} S(n,d)n^{-s})x^{-d}$.

Since $\phi$ and $\mu$ are multiplicative functions, so is $S(n,d)$ when $d$ is fixed.
If $v_p(d) = m$, with the convention that $v_p(0) = \infty$ and $p^{-\infty} = 0$, we see that $$\sum_{k \ge 0} S(p^k,d)p^{-ks} = \sum_{0 \le k \le m} \phi(p^k)p^{-ks} + \sum_{k \gt m} \phi(p^k)\mu(p^{k-m})p^{-ks}/\phi(p^{k-m}) \\ = 1 + \sum_{1 \le k \le m} (p-1)p^{-1+k(1-s)} - p^{-s+m(1-s)} = (1 - p^{-s})\sum_{0 \le k \le m} p^{k(1-s)} \\ = (1 - p^{-s})\sigma_{1-s}(p^m)$$

Hence, $$\sum_{n \ge 1} S(n,d)n^{-s} = \prod (1-p^{-s}) \sigma_{1-s}(p^{v_p(d)}) = \sigma_{1-s}(d)/\zeta(s)$$

Notice that if $d=0$, then $\sigma_{1-s}(0) = \zeta(s-1)$, and if $d>0$, then $\sigma_{1-s}(d) = \sigma_{s-1}(d)/d^{s-1}$, which proves that : $$\Psi(s,x) = \frac 1 {x\zeta(s)}\left(\zeta(s-1) + \sum_{d\ge 1} \frac{\sigma_{s-1}(d)}{d^{s-1} x^d} \right)$$

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If I am not mistaken, your $S(n,d)$ is Ramanujan's sum $c_d(n)$ – მამუკა ჯიბლაძე Feb 22 at 8:54

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