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Can we express $x$ as a Dirichlet series ? Thus $x = a_0 + a_1 2^{-x} + a_2 3^{-x} + a_3 4^{-x} + ...$ where the $a_i$ are real numbers ? I do not know how to get a nondivergent solution.

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mick: It's unclear whether the $a_k$ are considered already given and you wish to solve for $x$. If the $a_k$ can be chosen after $x$ is fixed, then using $x=2$ and all $a_k$ taken as $2/(\(pi^2/6))$, then your equation works, since with all $a_k=1$ the right side is $1+1/4+1/9+...=\pi^2/6.$ – coffeemath Jan 23 '13 at 12:05
@coffeemath The equation must hold for all $x$. Otherwise I would have given a specific value for $x$. Say $x$ was $2$ I could have used $a_0 = 2$ and all other $a_i = 0$. In other words we do not solve for $x$ and I think that is quite clear otherwise the question is quite silly. However you have a point. Maybe the question can be clarified somehow. However asking to express a function as a series is usually considered as I wrote , compare with $floor(x)$ in fourrier series. – mick Jan 23 '13 at 18:00
Consider $x\to\infty$. – anon Jan 24 '13 at 21:43
Yes ! thanks anon. – mick Jan 24 '13 at 23:08

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