Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
    
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
add comment

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]

share|improve this answer
1  
This should be merged with the question, not added as an answer. –  S. Carnahan Nov 28 '11 at 2:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.