Intereting Posts

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Number of isomorphism classes of a tree on n vertices

The following identity between generalized Dirichlet series, **both absolutely convergent in the whole complex plane**, occurs when investigating functions of the Selberg class of degree $d=0$:

\begin{equation}

\sum_{n=1}^\infty a(n)\left(\frac{Q^2}{n}\right)^s=\omega Q\sum_{n=1}^\infty\overline{a(n)}n^{s-1}.

\end{equation}

To prove that

\begin{equation}

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

\end{equation}

is actually a Dirichlet polynomial, using an **uniqueness theorem for generalized Dirichlet series** one derives from the above identity **that $Q^2/n$ is an integer for all n with $a(n)\neq 0$. But how?**

If we set $b(n):=\omega Q \overline{a(n)}$, $\lambda_n=\log(n/Q^2)$ and $\nu_n=\log(n)$ we got

\begin{equation}

f(s):=\sum_{n=1}^\infty a(n)e^{-\lambda_n s}=\sum_{n=1}^\infty b(n)e^{-\nu_n(1-s)}=:g(1-s)

\end{equation}

- Showing $\sum\frac{\sin(nx)}{n}$ converges pointwise
- Convergence of the Fourier Transform of the Prime $\zeta$ Functions
- Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$
- How can I prove my conjecture for the coefficients in $t(x)=\log(1+\exp(x)) $?
- Determining the coefficients of the reciprocal of a Dirichlet series
- An identity involving the Möbius function

From the uniqueness theorem I know we would get, that if $f(s)=g(s)$ in some half-plane, then $\lambda_n=\nu_n$ and $a(n)=b(n)$ for all $n$, which is exactly what I need. But in fact, we just got $f(1/2)=g(1/2)$.

- Theorem 11.14, Apostol, pg 238 - need explanation
- Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$
- Does the Abel sum 1 - 1 + 1 - 1 + … = 1/2 imply $\eta(0)=1/2$?
- Determining the coefficients of the reciprocal of a Dirichlet series
- Translations AND dilations of infinite series
- Is this Dirichlet series generating function of the von Mangoldt function matrix correct?
- On Dirichlet series and critical strips
- Derivative of Riemann zeta, is this inequality true?

We use the

Theorem:Let $(\lambda_k)$ a sequence of distinct real numbers without accumulation point, and $(\alpha_k)$ a sequence of complex numbers, such that the series $$f(s) = \sum_{k = 1}^{\infty} \alpha_k e^{\lambda_k s}\tag{1}$$ converges absolutely on the line $\operatorname{Re} s = c$, where $c\neq 0$, and $f(s) = 0$ for all $s$ with $\operatorname{Re} s = c$. Then $\alpha_k = 0$ for all $k$.

With that uniqueness theorem, we consider

$$\sum_{n = 1}^{\infty} a(n)\biggl(\frac{Q^2}{n}\biggr)^s – \sum_{n = 1}^{\infty} \frac{\omega Q \overline{a(n)}}{n} n^s \equiv 0,\tag{$\ast$}$$

and write the left hand side in the form of the theorem, letting $(\lambda_k)$ be an enumeration of

$$\{\log n : n \in \mathbb{N}\} \cup \{2\log Q – \log n : n \in \mathbb{N}\}$$

and setting

$$\alpha_k = \begin{cases}a(m) – \dfrac{\omega Q\overline{a(n)}}{n} &, \lambda_k = \log n = 2\log Q – \log m \\ \quad -\dfrac{\omega Q \overline{a(n)}}{n} &, \lambda_k = \log n \\ \qquad a(m) &, \lambda_k = 2\log Q – \log m\end{cases}$$

where we choose the first matching condition. The hypotheses of the theorem are satisfied, since the two series in $(\ast)$ converge absolutely on the entire plane by assumption, and since $Q^2/n \to 0$ and $n\to +\infty$, the sequence $(\lambda_k)$ has no accumulation point. Hence we have $\alpha_k = 0$ for all $k$, and by the definition of $\alpha_k$ this yields $a(m) \neq 0 \implies Q^2/m \in \mathbb{N}$.

**Proof** (of the theorem)**:** For $\lambda \in \mathbb{R}$, consider the integral

$$I(\lambda) := \frac{1}{2\pi i}\int_{c – i\infty}^{c + i\infty} \frac{e^{\lambda s}}{s^2}\,ds.$$

The integrand is an entire meromorphic function with only one pole at $0$, and the residue there is $\lambda$. By the residue theorem, we have

$$I(\lambda) = \begin{cases} \lambda &, \lambda \geqslant 0 \\ 0 &, \lambda < 0 \end{cases}\quad \text{for } c > 0,\text{ and}\quad I(\lambda) = \begin{cases}\; 0 &, \lambda \geqslant 0 \\ -\lambda &, \lambda < 0\end{cases}\quad\text{for } c < 0.$$

Since the series in $(1)$ converges absolutely we can interchange summation and integration, and since $f(s)$ vanishes identically on the integration contour we obtain

\begin{align}

0 &= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{f(s)e^{-\lambda s}}{s^2}\,ds \\

&= \sum_{k = 1}^{\infty} \frac{\alpha_k}{2\pi i}\int_{c – i\infty}^{c + i\infty} \frac{e^{(\lambda_k – \lambda)s}}{s^2}\,ds \\

&= \sum_{c(\lambda_k – \lambda) > 0} \alpha_k \lvert \lambda_k – \lambda\rvert

\end{align}

for every $\lambda \in \mathbb{R}$. Varying $\lambda$ between two adjacent (in order, not with respect to index) $\lambda_k$ shows that

$$\sum_{\lambda_k > \lambda}\alpha_k = 0\qquad\text{resp.}\qquad \sum_{\lambda_k < \lambda}\alpha_k = 0\tag{2}$$

for every $\lambda\in \mathbb{R}$. Choose $m\in \mathbb{N}$ and consider $(2)$ for a $\lambda$ slightly smaller than $\lambda_m$ and a $\lambda$ slightly greater. It follows that $\alpha_m = 0$. Since $m$ was arbitrary, the theorem is proved.

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