2
$\begingroup$

Can someone come up with a proof for this little theorem?

Suppose that $F_a(s)$ is a Dirichlet series and $a(n)$ is its associated arithmetic function, that is:

$$F_a(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^s}$$

Then the $a(n)$ are given by:

\begin{equation} \label{eq:a(n)} \nonumber a(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \end{equation}

The advantage of this formula is that if you know $F_a(s)$ at the even integers, you know the coefficients of its series expansion.


Let me give an example, before questions rain down asking for clarification.

We know that: $$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$ where $\mu(n)$ is the Mobius function. Therefore: $$\mu(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-1}}{(2i+1-2j)!}$$

Now, one good application of this formula is the generalization of the Mobius function. Whatever the $k$, real or complex, positive or negative, we have by definition that: $$\zeta(s)^{-k}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$$ and you may ask, what is the $\mu_k(n)$ for each $n$ that satisfies the above equation?

Well, they are given by: $$\mu_k(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-k}}{(2i+1-2j)!}$$

An example:

$\sqrt{\zeta(s)}=1+\frac{1}{2\cdot 2^s}+\frac{1}{2\cdot 3^s}+\frac{3}{8\cdot 4^s}+\frac{1}{2\cdot 5^s}+\frac{1}{4\cdot 6^s}+\frac{1}{2\cdot 7^s} +\frac{5}{16\cdot 8^s} +\frac{3}{8\cdot 9^s}+\frac{1}{4\cdot 10^s}+\frac{1}{2\cdot 11^s}+\cdots\\$

Another example, the Von Mangoldt function divided by the log is given by:

$$\frac{\Lambda(n)}{\log n}=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\log\zeta(2j)}{(2i+1-2j)!}\\$$

$\endgroup$
  • $\begingroup$ This should be made precise first. (What is $F_a(0)$ for $a(n)\equiv 1$?..) $\endgroup$ – metamorphy May 18 at 5:14
  • 1
    $\begingroup$ Then you're talking about $F_a(s)$ as the analytic continuation of the series, and thus require it to exist at $s\in\Bbb{Z}_{\geqslant 0}$ (it may not, as these results on natural boundaries indicate). That's exactly what I'm asking for - what are the premises to prove the formula under? $\endgroup$ – metamorphy May 18 at 6:13
  • 1
    $\begingroup$ Not really, if you drop i=0 and j=0 the formula still holds. To make it more interesting, I will edit the OP to show a generalization of the Mobius function. $\endgroup$ – user630964 May 18 at 15:56
  • 1
    $\begingroup$ I upvoted your question, it fits perfectly the requirements of MSE. It is still unobvious to me how you came with that formula without knowing the proof ($\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{-2j}}{(2i+1-2j)!}$ are the Taylor series coefficients of $\frac{1}{1-x^2} \frac{\sin(2\pi x/m)}{2\pi x/m}$) $\endgroup$ – reuns May 18 at 17:23
  • 1
    $\begingroup$ @metamorphy A last comment, your formula for $\mu_k(n)$ can't possibly work for complex $n$, right? It seems to require integer $n$. $\endgroup$ – user630964 May 19 at 5:49
1
$\begingroup$

First let's prove this for $$a(n)=\delta_{m,n}=\begin{cases}1,&n=m\\0,&n\neq m\end{cases}$$ with a (fixed) positive integer $m$. Consider the RHS of the formula, $$A_{m,n}=-2\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{-2j}}{(2i+1-2j)!}.$$ Replacing $j$ by $i-j$ in the inner sum, we get $$A_{m,n}=-2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}.\tag{1}\label{eq1}$$ Since $\displaystyle\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=\frac{\sin 2\pi m}{2\pi m}=0$, this implies $$A_{m,n}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i+1}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}$$ which, after replacing $j$ with $i+j$, gives $$A_{m,n}=2\sum_{i,j=0}^{\infty}\frac{(-1)^{i+j}(2\pi)^{2i+2j}n^{2i}m^{2j}}{(2i+2j+1)!}.\tag{2}\label{eq2}$$ In particular, $A_{m,n}=A_{n,m}$. Now if $n<m$ then we can change the order of summation in $\eqref{eq1}$: $$A_{m,n}=-2\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}\sum_{i=j}^{\infty}\Big(\frac{n}{m}\Big)^{2i}=-2\Big(1-\frac{n^2}{m^2}\Big)^{-1}\frac{\sin 2\pi n}{2\pi n}=0,$$ and if $n=m$ then, collecting terms in $\eqref{eq2}$ with $i+j=k$, we obtain $$A_{n,n}=2\sum_{k=0}^{\infty}\frac{(-1)^k(2\pi n)^{2k}}{(2k+1)!}(k+1)=\cos 2\pi n+\frac{\sin 2\pi n}{2\pi n}=1.$$ Concluding, we have $A_{m,n}=\delta_{m,n}$, and this finishes the proof for $a(n)=\delta_{m,n}$.


Essentially, we've proven the formula for functions $a(n)$ such that $\{n:a(n)\neq 0\}$ is finite. It's also easy to see that the formula holds when $\sum_{n=1}^{\infty}a(n)$ converges absolutely: the triple sum $$\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi)^{-2j}}{(2i+1-2j)!}\sum_{\color{blue}{m=n+1}}^{\infty}\frac{a(m)}{m^{2j}}$$ converges absolutely in this case, so we can change the order of summation, but each of the sums with a fixed value of $m$ equals $0$ as already proven; so the entire sum is $0$, and we're left with $\sum_{m=1}^{n}$, which again has been treated already.

Further results require analytic continuation. This can be done as usual (beginning with application of the above to $\bar{a}(n)=n^{-s}a(n)$ with $\Re s>\sigma$ the abscissa of absolute convergence of $F_a(s)$), but behaviour of the series in the RHS of the formula needs to be explored separately.

$\endgroup$
0
$\begingroup$

This proof is simple:

\begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \Rightarrow \end{equation} \begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}}{(2i+1-2j)!}\sum_{k=1}^{\infty}\frac{a(k)}{k^{2j}} \Rightarrow \end{equation} \begin{equation} \nonumber \sum_{k=1}^{\infty}a(k)\left(-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}k^{-2j}}{(2i+1-2j)!}\right)\\ \end{equation} The theorem then follows from the following equation: \begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi k)^{-2j}}{(2i+1-2j)!}=\begin{cases} 1, & \text{if}\ n=k\\ 0, & \text{otherwise}\\ \end{cases} \end{equation} And the above equation is justified for being the convolution of $\mu_0(n)$ (which is the same as $\delta_{n1}$) and the associated function of the series $k^{-s}$, $b(n)$, since from the convolution formula: \begin{equation} \nonumber c(n)=(\mu_0*b)(n)=\sum_{d|n}\mu_0(d) b\left(\frac{n}{d}\right)=b(n)=\begin{cases} 1, & \text{if}\ n=k\\ 0, & \text{otherwise} \end{cases} \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy