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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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    $\begingroup$ 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 $\frac{dA_n}{dx}=0\space\forall n$ $\endgroup$
    – Ross Millikan
    Commented Jul 10, 2011 at 1:01
  • $\begingroup$ Does not make much sense to me. You could, for example set $a_n = b (n+1)$ $\endgroup$
    – leonbloy
    Commented Jul 10, 2011 at 1:15
  • $\begingroup$ @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. $\endgroup$
    – JBSnorro
    Commented Jul 10, 2011 at 2:08
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    $\begingroup$ 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) $\endgroup$
    – leonbloy
    Commented Jul 10, 2011 at 3:56
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    $\begingroup$ 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 $0<a<x<b$ for some interval $(a,b)$ in the positive reals. $\endgroup$
    – Vladimir Sotirov
    Commented Jul 10, 2011 at 6:47

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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  • $\begingroup$ 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 $-x/y^2>-1>-\alpha e^{-\alpha(y-x)}$. $\endgroup$
    – Vladimir Sotirov
    Commented Jul 10, 2011 at 18:57
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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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  • $\begingroup$ 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. $\endgroup$
    – Did
    Commented Jul 10, 2011 at 3:12
  • $\begingroup$ @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. $\endgroup$
    – Ross Millikan
    Commented Jul 10, 2011 at 3:17
  • $\begingroup$ But, of course, this implies that $A_n$ is implicitly a function of $x$ $\endgroup$
    – leonbloy
    Commented Jul 10, 2011 at 3:53
  • $\begingroup$ @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. $\endgroup$
    – Did
    Commented Jul 10, 2011 at 9:26
  • $\begingroup$ @Ross: Not interested in clarifying the above? $\endgroup$
    – Did
    Commented Jul 21, 2011 at 21:41
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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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    $\begingroup$ 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. $\endgroup$
    – Carl Brannen
    Commented Jul 10, 2011 at 5:42
  • $\begingroup$ 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). $\endgroup$
    – Vladimir Sotirov
    Commented Jul 10, 2011 at 6:41

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