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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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    You want to solve for $A_n$? That seems to be one equation in an infinite number of unknowns. To have convergence, you need the $A_n$ to increase fast enough, but once you do, you can bump $A_1$ up and $A_2$ down and maintain the equality. I'm not sure to make of the condition $\frac{dA_n}{dx}=0\space\forall n$2011-07-10
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    Does not make much sense to me. You could, for example set $a_n = b (n+1)$2011-07-10
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    @Ross, the condition $\frac{dA_n}{dx}=0$ only means $A_n$ does not depend on $x$. I state this for completeness, otherwise a trivial solution could be found, e.g $A_n(x)=-x(log(x)+e^{π2/6n^2})$. @leonbloy: That $A_n$ is not a solution of the equation.2011-07-10
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    I deleted my answer because I misinterpreted the question to mean the equality held for all $ x $.2011-07-10
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    But what does `$A_n$ does not depend on $x$` mean? Given a certain solution sequence $A_n$, either it satisfies the equation for a fixed $x$ (and then it is implicitly a function of $x$ : different $x$ give different $A_n$), either it satisfies the equation for all $x$ (I'd like to see that)2011-07-10
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    The limit of the rhs is infinity as $x\to-0$.2011-07-10
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    Clearly the right-hand side is always positive, so this can only happen for $x>0$. @Ross, you answer shows that this can always be arranged so that the series $\sum_{n=0}^\infty e^{-A_m/x}$ converges to a function $x$ for at least one point; I think the interesting question that's hiding in here is whether it's possible for the series to converge to $x$ for $0 for some interval $(a,b)$ in the positive reals.2011-07-10
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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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    Two remarks on your post. First, a necessary condition for the RHS to be finite is that $A_n\to+\infty$. Second, more importantly, the RHS is nondecreasing.2011-07-10
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    @Didier: I remarked on the need for $A_n$ to increase in my comment on the original question. I was more focused on the fact that there are so many solutions the problem is not too interesting.2011-07-10
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    But, of course, this implies that $A_n$ is implicitly a function of $x$2011-07-10
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    @Ross: My remarks addressed your post and your post only, and specifically your *As $x$ grows, the right side decreases monotonically* and *for any series $A_n$ that is convergent*. I can understand neither of them. First, as $x$ grows, the RHS increases. Second, for any sequence $(A_n)$ such that $A_n\not\to+\infty$ and any fixed positive $x$, $\mathrm{e}^{-A_n/x}\not\to0$ when $n\to+\infty$, hence the RHS is infinite.2011-07-10
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    @Ross: Not interested in clarifying the above?2011-07-21
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    @Didier: I thought the problem was to find a set of $A_n$ given a particular $x$, not to find a set of $A_n$ that works for all $x$. This was because "$x$ is given and real". It feels to me like a solution over an interval as Vladimir Sotirov suggests will not work because it is trying to fit uncountably many points with only countably many $A_n$. But that is not a proof.2011-07-21
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    @Ross: Sorry but this does not address either one of the two specific points I made.2011-07-21
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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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    It seems that what's going on here is that we're trying to write $1/x$ as a Fourier transform but where the transform is restricted to involve only a countable number of frequencies $A_n$. That's clearly impossible.2011-07-10
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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