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For any Dirichlet series, $$f(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ is the sequence, $a_n$, always unique to $f(s)$? In other words, is it possible to show that $a_n$ is the only sequence that will ever satisfy the series being equal to $f(s)$?

If this is not true, could someone try to provide a counter example if possible?

tyobrien
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  • (Assuming you mean $x(s)$) I would try to assume $a_n\not\equiv b_n$ give rise to the same function. Then if $c_n=a_n-b_n$ we have $g(s) = \sum \frac{c_n}{n^s} = 0$. As $s\to\infty$ the lowest non-zero term $\frac{c_k}{k^s}$ will dominate the sum. Then try to prove that $c_k$ has to be zero which would contradict it being the lowest non-zero term so we must have $c_n\equiv 0$. I don't know if there are any subtleties here that would obstruct such a proof. – Winther Sep 01 '15 at 17:05
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    What is $s$? Is it a fixed number? – Wojowu Sep 01 '15 at 17:05
  • Interesting @Winther. I'm not quite sure how to go about proving that $c_k$ has to be zero... and thank you. I will fix that. – tyobrien Sep 01 '15 at 17:09
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    You should clarify if you mean $x$ as just a number or $x(s)$ as a function. In the first case the statement is certainly not true. – Winther Sep 01 '15 at 17:10
  • $s$ is a fixed number @Wojowu. – tyobrien Sep 01 '15 at 17:11
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    If we take $s=1$, this isn't true. Example: first take $a_1=1,a_n=0$ otherwise, second take $a_2=a_3=a_6=1,a_n=0$ otherwise. They both give $x=1$. (edit: now you changed the question, so this invalidates my comment). – Wojowu Sep 01 '15 at 17:12
  • See here for another proof using Perron's formula – Bart Michels Jan 07 '18 at 19:32
  • This is 11.3 in Apostol's book. – Watson Nov 23 '18 at 13:57

1 Answers1

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If we assume that your series converges absolutely for all large $s$, then, yes, expansion of $f(s)$ as Dirichlet series is unique. See Theorem 4.8 at http://www.math.illinois.edu/~ajh/ant/main4.pdf for proof.

Otherwise, you should be able to provide a counterexample.

user0
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    Then what do we say about things like convergence acceleration methods? For instance for apery's constant $\zeta(3)=\sum_{n=1}^\infty\frac{a_n}{n^3}$ we can let $a_n=1$ or $a_n=\frac 52 (-1)^{n+1} \frac 1{\binom {2n}n}$ – tyobrien Jun 11 '17 at 17:49
  • @Ty O'Brien For all $s > 1$, $$\zeta(s) \neq \frac{5}{2} \sum_{n = 1}^\infty \frac{(-1)^{n+1}}{\binom{2n}{n} n^s}.$$ I plotted those two functions and saw that they have different graphs, but happen to intersect at $s = 3$. I see now in one of your comments that you want $s$ to be a fixed number, so my answer is not appropriate. – user0 Jun 12 '17 at 17:46
  • That's not what I'm saying. My comment says that there are two different values of $a_n$ such that the sum of $\frac{a_n}{n^3}$ equals $\zeta(3)$. In fact there are others as well such as $a_n=\frac 43 (-1)^{n+1}$ – tyobrien Jun 12 '17 at 18:17
  • @Ty O'Brien I was referring to your comment on Sep 1 '15 at 17:11.

    To be clear, if one is considering $f(s)$ for all values of $s$ where the Dirichlet series converges absolutely (i.e., the entire function for those large $s$), then the $a_n$ are unique. That is what my answer means. But as Winther notes in his second comment, if one is considering just one value of the function, then the $a_n$ are not unique. Both Wojowu and you give examples.

    To visualize, note that even though two different functions have different graphs, they can still share a common point of intersection.

     – user0 Jun 13 '17 at 16:04
  • I got it! Thanks a lot! – tyobrien Jun 13 '17 at 16:10
  • I know I've accepted your answer but I guess I'm not satisfied about one thing. What do we say about the zeta function when we define $a_n$ in terms of $s$? Because $a_n$ can equal $1$ or $\frac{(-1)^{n+1}}{1-2^{1-s}}$ and they are equal for all $s\gt 1$ – tyobrien Jun 16 '17 at 14:49
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    @Ty O'Brien By the definition of a Dirichlet series, $a_n$ is not a function of $s$. Your second case, therefore, is not a Dirichlet series and so does not contradict my answer or comments. – user0 Jun 16 '17 at 16:36