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Usually when working with indefinite sums, I want to work out the sum or whether it convergence. But now I encountered a problem they other way around and I'm clueless...

Is there even a general solution of $ x=\sum_{n=0}^\infty e^{-A_n/x}$ for $A_n$, where $x$ is given and real, $A_n >0\space\forall n$ and $\frac{dA_n}{dx}=0\space\forall n$?

Thank you

EDIT:

To make my question clearer for the commenters and others, I'm searching for a systematic sequence $A_n$ which, when entered in the equation above, yields $x$ and this should hold for all (real) $x$.

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    Quantifier abuse police is here!2013-08-30

3 Answers 3

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It cannot be done. For the proof write $x:={1\over y}$. Then we should have ${1\over y}\ \equiv\ \sum_{n=0}^\infty e^{-A_n y}\qquad(*)\ ,$ say for all $y\geq1$. In particular $\sum_{n=0}^\infty e^{-A_n}=1$, so necessarily $\lim_{n\to\infty} A_n=\infty$. It follows that $\alpha:=\inf_n A_n>0$ and therefore $\sum_{n=0}^\infty e^{-A_n y}=\sum_{n=0}^\infty e^{-A_n} \ e^{-A_n(y-1)} \leq e^{-\alpha(y-1)} \qquad (y\geq1)\ .$ This shows that $(*)$ cannot hold for all $y\geq 1$.

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    Brilliant. This in fact shows that if $\frac 1y=\sum_{n=0}^\infty e^{-A_ny}$ and $\frac1x=\sum_{n=0}^\infty e^{-A_nx}$ with some y>x, then $\frac xy\leq e^{-\alpha(y-x)}$, limiting the intervals on which we could have convergence (because when $x/y\to0$ slower than $e^{-\alpha(y-x)}$ as $y\to\infty)$. Some numerical exploration with Mathematica suggests that for $\alpha\geq 1$, $x\geq 1$ the above system has no solution other than $x=y$, which seems to follow by computing derivatives -x/y^2>-1>-\alpha e^{-\alpha(y-x)}.2011-07-10
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Certainly there are many solutions. As $x$ grows, the right side decreases monotonically, so for any series $A_n$ that is convergent there will be an $x$ that solves the equation. So pick any $x$ and series $A_n$ that solve the problem. Given a different $x$, just change your favorite $A_n$(s) to make it work.

For a specific example, take $x=1, A_n=\ln 2^{n+1}$. If you want a solution for $x=2$, just decrease any set of $A_n$'s to add enough to the RHS.

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    @Didier: Oops, you are right about the RHS increasing with increasing $x$. As $x$ grows, the exponent becomes less negative, so the RHS does increase. I did not mean to assert that the series converges for all choices of $A_n$, just that you needed a set of $A_n$ that would make the sum converge. As you say, they must increase to infinity, but I did not think about how fast they must.2011-07-21
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Divide both sides by $x$ to get:
$1 = \Sigma\frac{e^{-A_n/x}}{x},$ As $x\to \infty$, the left hand side is 1 while the right hand side goes to zero for each $n$.

This might give a hint on what $A_n$ won't work.

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    you're right that what's happening is writing $-1/x$ as a series in $\{e^{mx}\}$ for positive real $m$. I agree that this is impossible *when we require the series to converge absolutely*, but I'm not sure that we have impossibility under conditional convergence (to a non-holomorphic function).2011-07-10