# Solution of an equation with Jacobi theta function

I have been struggling with this equation for some time and I do not seem to find any conclusive answer (it's from my research, not a homework).

It has to do with the real solutions $x$ to the following equation

$$x + x f(x) = 1 + f(1),$$ where $$f(x) = 2\sum_{n=1}^\infty \mathrm{e}^{(-ax^2-b) n^2}$$ with $a$ and $b$ strictly positive.

I know that $x=1$ solves the equation trivially; from simulations, I cannot find a contradiction to the fact that it should be the only solution. However, I cannot prove nor disprove that $x=1$ is the only solution.

I have tried using the upper bound $$f(x) \leq \frac{ax^2+b+1}{ax^2+b}\mathrm{e}^{-ax^2 -b},$$ and i have tried relating $f(x)$ with the Elliptic theta function $$f(x) = -1 + \theta_3(0,\mathrm{e}^{-ax^2 -b});$$ I have also tried to prove that $x = -x f(x) + 1 +f(1)$ is a contractive mapping; however, I have only found (quite restrictive) sufficient conditions on $a$ and $b$ for it to be true.

If someone manages to solve it or help me find a counterexample, I will gladly acknowledge their contribution in the paper i am writing.

EDIT: $x = -x f(x) +1 +f(1)$ is not a contractive mapping in general. I have found counterexamples where it is not (but the equation still only has one solution $x=1$).

• You just need consider the monotonicity of the function $g(x)=x+f(x)$ with $x>-b/a$. It is seems that $g^{''}(x)<0$ for all $x>-b/a$ and thus $g(x)-g(1)=0$ has at most two real solution.
– Zhou
May 22 '18 at 7:34
• Hi @Zhou, thank you for your input! While it makes a lot of sense, I had a typo in the problem: the function I am looking at is $g(x) = x + x f(x)$. May 22 '18 at 7:51

Let $\, g(x) := \theta_3(0,\mathrm{e}^{-ax^2 -b}).\,$ Your question about solutions to $\, x + x f(x) = 1 + f(1) \,$ is now about $\, x g(x) = g(1).\,$ Now $\,g(x)\,$ is a bell shaped curve with $\, g(x) > 0 \,$ and $\, g(-x) = g(x).\,$ If we can prove that $\,xg(x)\,$ is monotone increasing we are done. If it holds for $\,a=1, b=0\,$ then it holds in general.

In that special case, using Jacobi theta identities (mentioned by Mikhail Skopenkov in another answer), $\,xg(x) = g(\pi/x)\sqrt{\pi}$ and since $\,g(x)\,$ is monotone decreasing for $\,x>0,\,$ then $\,g(\pi/x)\,$ must be monotone increasing and we are done.

• Hi! Thank you for your input; could you give me a pointer as to how I can prove that $xg(x)$ is monotone increasing? I get that $$\frac{\mathrm d}{\mathrm d x} \{x g(x)\} = g(x) - 4 a x^2 \sum_{n=1}^\infty \mathrm{e}^{-a x^2 -b}n^2$$ which is not obviously positive. Or is there some clever trick? May 23 '18 at 7:03
• I cannot thank you enough. Please, send me an email with your full name and affiliation if you want an acknowledgment in my paper. Otherwise, thank you very much! May 25 '18 at 9:37

Have you tried Jacobi's identities for theta-functions? At least for $a=1, b=0$ the identity $$xg(x)=\sqrt{\pi}\,g(\pi/x)$$ implies that the function $xg(x):=xf(x)+x$ is monotone increasing.

Edit: the identity corrected, thanks to Somos.

• The equation should be $xg(x) = g(\pi/x)\sqrt{\pi}.$ May 24 '18 at 16:02
• @Somos: Surely, many thanx May 24 '18 at 17:55
• I cannot credit you both for the answer; if you would like and acknowledgment in my paper, please send me an email with your full name and affiliation. Otherwise, thank you very much! May 25 '18 at 9:36
• That was just a standard trick with teta-functions estimates May 25 '18 at 18:58