Analytic Continuation of Generating Function of Dirichlet Series

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Theorem

Let $\displaystyle \map \lambda s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series

Let $c \in \R$ be greater than the abscissa of absolute convergence of $\lambda$ and greater than $0$.

Let $\displaystyle \map g z = \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k $ be the generating function of $\map \lambda {c k}$

Then the analytic continuation of $g$ to $\C$ is equal to

$\displaystyle \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}$


Proof

We will first show that the series is meromorphic functions on $\C$ with simple poles at $n^c$.

For $\cmod z < R$, pick M such that $M^c > 2 R$.

This is always possible, as $c > 0$.

Then for $n > M$:

\(\displaystyle \cmod {1 - \frac z {n^c} }\) \(>\) \(\displaystyle 1 - \cmod {\frac z {n^c} }\) Triangle Inequality
\(\displaystyle \) \(>\) \(\displaystyle 1 - \cmod {\frac R {M^c} }\)
\(\displaystyle \) \(>\) \(\displaystyle 1 - \cmod {\frac R {2 R} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2\)
\(\displaystyle \frac 1 {\cmod {1 - \frac z {n^c} } }\) \(<\) \(\displaystyle 2\)

Therefore, for $\cmod z= < R$, pick $M$ as described above.


For all $K > M$:

\(\displaystyle \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }\) \(=\) \(\displaystyle \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} \frac {n^c} {n^c - z} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } \cmod {\frac 1 {1 - \frac z {n^c} } } }\)
\(\displaystyle \) \(<\) \(\displaystyle \cmod {2 \sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } }\)
\((1):\quad\) \(\displaystyle \) \(<\) \(\displaystyle \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }\)


Because $c$ is chosen to be greater than the abscissa of absolute convergence of $\lambda$, the above series converges.

Therefore, for all $\epsilon > 0$ we can pick $N$ such that for $K > N$:

\(\displaystyle \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }\) \(<\) \(\displaystyle \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }\)
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\)

Because the summands are meromorphic functions, the series is an absolutely uniformly convergent series of analytic functions on the region $\cmod z < R, z \ne k^c \text{ for all } k \in \N$.


By Uniform Limit of Analytic Functions is Analytic, the series converges to an analytic function on the domain $z \in \C, z \ne k^c \text{ for all } k \in \N$.

Also, each $k^c$ is a pole at exactly one summand, where it is a simple pole.

Therefore, the series has simple poles at each $k^c$, meaning that the function is meromorphic on $\C$.


We now show that the Generating Function is equal to the series for $\cmod z < 1$:

\(\displaystyle \map g z\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty z^k \sum_{n \mathop = 1}^\infty \frac {a_n} {n^{c k} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k\)


By Fubini-Tonelli Theorem, if:

$\displaystyle \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}$

converges, then:

$\displaystyle \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k$

We see that for $\cmod z < 1$:

\(\displaystyle \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \cmod {a_n} \sum_{k \mathop = 1}^\infty \paren {\frac {\cmod z} {n^c} }^k\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\frac {\cmod z} {n^c} } {1 - \frac {\cmod z} {n^c} }\) Sum of Infinite Geometric Sequence
\((2):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {n^c - \cmod z}\)


The summands of $(2)$ are equal to the summands of:

$\displaystyle \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {\cmod {n^c - z} }$

for all but finitely n.

By $(1)$, the above series converges, and thus $(2)$ as well.

Thus we may write:

\(\displaystyle \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n \frac {\frac z {n^c} } {1 - \frac z {n^c} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}\)

Thus the Generating function is equal to the series on the unit circle.

Thus by Definition:Analytic Continuation, the series is the unique analytic continuation of the Generating Function.

$\blacksquare$