# An interesting doubly infinite series

Let $0<\mu<1$ and $\alpha:=1-\mu^2$. Consider the function $$f(x):=x\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}x}-\frac{1}{x}\sum_{k=-\infty}^\infty\mu^{4k}e^{-\alpha\mu^{4k}/x},$$ defined for all $x>0$. Three properties are easy to check: $f(\mu^{2n})=0$ for every integer $n$, $f(x)=-f(1/x)$, $f(x)$ vanishes at $x=\mu^2$ and $x=1$ and $f(x)=f(\mu^4x)$.

I want to show that $f(x)<0$ for $\mu^2<x<1$, but I have not been able to prove it. Has anybody seen anything like this?

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Did you finish what you intended to write? I don't see a question. –  Greg Martin Sep 15 '11 at 21:17
@Greg: there was some error/bug with the LaTeX formatting. I've cleaned it up. –  Yemon Choi Sep 15 '11 at 21:36
Take a look at mathoverflow.net/questions/61350/… before getting too deeply involved –  Will Jagy Sep 15 '11 at 21:45
Where does this arise? The desired inequality fails for many choices of $\mu$, such as $\mu=1/4$ as detailed in my answer below; but perhaps the source of the problem might point to an inequality that does hold (or at least comes closer...) and is still tractable by the same method. –  Noam D. Elkies Sep 16 '11 at 2:52

I've seen sums like this, and they can get quite amusing, e.g. the Fourier coefficients of $f(x)$ as a periodic function of $\log(x)$ involve values of the Gamma function at complex arguments (see below); but it seems that this is overkill for the question at hand: there are several ranges of $\mu$ for which $f(\mu) > 0$, e.g. $\mu = 1/4$ works, giving $f(1/4) = 0.0892157+ > 0$. Are you sure this is what you meant?

If I computed everything correctly (and gp corroborates numerically), the following sine-Fourier expansion holds: write $\mu = \exp(-\lambda)$ and $x = \mu^t = \exp(-\lambda t)$; then $$f(x) = \sum_{n=1}^\infty \phantom. c_n \sin \frac{\pi n t}{2}$$ where $$c_n = \frac1\lambda \mathop{\rm Im} \left( \Gamma\bigl(1 + \frac{\pi i n}{2\lambda}\bigr) \Bigl/ \alpha^{1 + \frac{\pi i n}{2\lambda}} \right).$$ This does not depend on the choice $\alpha = 1 - \mu^2$.

P.S. See this Mathoverflow answer where such a sum (and its Fourier expansion with complex-Gamma coefficients) arises naturally.

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Noam, take a look at the edit history of the OP's other question as well. In short, others do all the work. –  Will Jagy Sep 16 '11 at 3:21

Thanks very much, you're right, it is not true for $\mu=1/4$, meaning that in general the real zeros of the function $f(x)$ are not just $\mu^{2n}$, $n$ an integer. I thought that this perhaps was the case, implying what I wanted to prove. I'll explain more:

Define $$\chi_q(t):=t^2\int_0^\infty \mu^{-4\langle\log_{\mu^4}(s)+q\rangle }s e^{-(\mu^{-1}-\mu) ts}ds ,\quad \Re(t)>0$$ where $\mu$ and $q$ are constants with $0<\mu<1$, $0\leq q < 1$, and $\langle x\rangle$ denotes the fractional part of $x$.

The function I am looking at originally is $$g_q(t)=\chi_q(t)-\chi_q(\mu^2/t),\quad \Re(t)>0.$$ This functions satisfies $g_q(\mu^4t)=g_q(t)$. What I thought would be the case is that $g_q(t)$ has $\mu$ as its only zero in the region $\Re(t)>\mu^2R/2$, $|t-1/R|<1/R$. Here $R=\mu^{-1}+\mu$. This region is its own reflection about the circle of radius $\mu$, and the intersection of this region with the real axis is the interval $(\mu^2R/2, 2/R)$.

The function $s\mu^{-4\langle\log_{\mu^4}(s)+q\rangle }$ is a step function, and by integrating we can write $g_q(t)$ as the doubly infinite series (up to a multiplicative constant):$$t\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}t}-\frac{\mu^2}{t}\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}(\mu^2/t)}$$

For $q=0$, this is essentially the function in the original question since $f(x)=\mu^{-1} g_0(\mu x)$.

However, computational experiments with Mathematica indicate that for small values of $\mu$, g_0(t) has at least three real zeros in the interval $(\mu^2R/2, 2/R)$, and this seems to be the case as well for $\mu=1/5$ and $q=1/1000$, which disproves my conjecture. Below is the Mathematica code. Here m is $\mu$, and I truncate the series at $k=M$.

M = 300;

m = 1/5;

q = 0;

N[m]

R = m + 1/m;

g[x_] := x*(Sum[ m^(4*(k - q))Exp[-(1/m - m) m^(4*(k - q))x], {k, 0, M}] + Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))(m^2/x)], {k, 1, M}]) - (m^2/ x)(Sum[m^(4*(k - q))*Exp[-(1/m - m)m^(4(k - q))(m^2/x)], {k, 0, M}] + Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))*(m^2/x)], {k, 1, M}]);

Plot[g[x], {x, m^2*R/2, 2/R}, PlotRange -> All]

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This should be merged with the question, not added as an answer. –  S. Carnahan Nov 28 '11 at 2:15