# About the asymptotic behavior of specific Jacobi $\theta$ function $\operatorname{\vartheta}_3\left(0;x\right)$ when $x\to{1-}$.

Since $$\displaystyle\sum_{n=1}^\infty{x^{n^2}}=\dfrac{\operatorname{\vartheta}_3\left(0,x\right)-1}2$$ for $$x\in\left(0,1\right)$$ (just in case), it suffices to consider the former below. (Another relevant identity is that $$\displaystyle\sum_{n=1}^\infty{x^{n^2}}=\left(1-x\right)\mspace{-2mu}\sum_{n=1}^\infty{\!\left\lfloor\mspace{-1mu}\sqrt{n}\right\rfloor\!x^n}$$.)
Many a post (e.g.,

has shown that $$\lim_{\mspace{-1mu}x\to1^-}{\sqrt{1-x\mspace{2mu}}\sum_{n=1}^\infty{\mspace{-1.5mu}x^{n^2}}}=\frac{\mspace{-1.5mu}\sqrt\pi\mspace{1mu}}2\text{.}$$ Nevertheless, none of them evaluated the limit $$\lim_{\mspace{-1mu}x\to1^-}{\mspace{-3mu}\left({\sqrt{\mspace{-0.5mu}\frac\pi{1-x}}-2\mspace{-0.75mu}\sum_{n=1}^\infty{\mspace{-1.75mu}x^{n^2}}}\right)\mspace{-1mu}}\text{,}$$ which is equal to $$1$$. how to find the asymptotic expansion of the following sum: sketched out a possible approach to finding this kind of limit. Some other related posts are as follows:

In Analysis of Series and Products. Part 1: The Euler–Maclaurin Formula and Analysis of Series and Products. Part 2: The Trapezoidal Rule, the authors show that $${{\sum_{j=1}^\infty\mathrm{e}^{-j^2/x^2}-\frac{x\sqrt\pi}2+\frac12}\to0}\quad\text{as}\quad{x\to{+\infty}}\text{,}$$ and the magnitude of the left-hand side decreases exponentially as $$x$$ increases. (Then set $$u=\exp\left(-x^{-2}\right)$$. How about the Abel–Plana formula?)

In addition, in On the asymptotics of some partial theta functions and Some new asymptotic expansions of certain partial theta functions, each author showes that $$\dfrac{\operatorname{\vartheta}_3\left(0,x\right)+1}2=\sum_{n\in\mathbb{N}}{x^{n^2}}=\frac12\sqrt{\frac{-\pi}{\ln{x}}}+\frac12+\operatorname{\mathcal{O}}\left(\ln^q{\!x}\right)\text{,}\quad\forall{q\in\mathbb{N_+}}\text{,}$$ as $$x\to1^-$$ (after making some substitutions).

It is generally known that $$\ln{x}\sim{x-1}$$ as $$x\to1$$; hence, as $$x\to1^-$$, $$\displaystyle\sum_{n=1}^\infty{x^{n^2}}\sim\frac12\sqrt{\frac{\pi}{1-x}}$$ just means that $$\displaystyle\sum_{n=1}^\infty{x^{n^2}}\sim\frac12\sqrt{\frac{-\pi}{\ln{x}}}$$. Yet, if one uses the former in the asymptotic expansion, what about the remainder term?

Via a number of numerical experiments, it seems that $$\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}=\frac1{\sqrt{\pi\!\left(1-x\right)}}\sum_{n\in\mathbb{Z}}{x^{n^2}}=\frac1{1-x}-\frac14+\operatorname{\mathcal{o}}\left(1\right)\quad\text{as}\quad{x\to1^-}\text{.}$$ Unfortunately, I have no idea how to prove it. Any idea? Many thanks!

• Updated for much better. Sep 28, 2021 at 9:55
• $$\sum_{n\in\mathbb{Z}}{x^{n^2}}=\sum_{n\in\mathbb{Z}}{e^{-n^2\ln\frac{1}{x}}}=\theta(\frac{1}{\pi}\ln\frac{1}{x})$$ Then you can use the functional equation for theta-function (which is $\mathbf{exact}$; for the reference - math.columbia.edu/~woit/fourier-analysis/theta-zeta.pdf , page 4): $$\theta (x)=\sum_{n\in\mathbb{Z}}{e^{-\pi x n^2}}=\frac{1}{\sqrt x}\theta\big(\frac{1}{x}\big)=\frac{1}{\sqrt x}\sum_{n\in\mathbb{Z}}{e^{-\frac{\pi n^2}{x}}}$$ $$\sum_{n\in\mathbb{Z}}{x^{n^2}}=\frac{\sqrt\pi}{\sqrt{|\ln x|}}\sum_{n\in\mathbb{Z}}{e^{-\frac{\pi^2 n^2}{|\ln x|}}}$$ Oct 13, 2021 at 2:18
• At $x=1-\epsilon; \epsilon\to 0$$\ln (1-\epsilon)=-(\epsilon+\frac{1}{2}\epsilon^2+\frac{1}{3}\epsilon^3+ ...)$$ All the terms, except for$n=0$, are exponentially small; therefore we leave the only term with$n=0$$$S=\sum_{n\in\mathbb{Z}}{x^{n^2}}\sim\frac{\sqrt\pi}{\sqrt{\epsilon}}\frac{1}{\sqrt{1+\frac{1}{2}\epsilon+\frac{1}{3}\epsilon^2+... }}$$ Then you can easily get the series, expanding the square root: $$S=\frac{\sqrt\pi}{\sqrt{\epsilon}}\Big(1-\frac{\epsilon}{4}-\frac{1}{2}\frac{\epsilon^2}{3}+\frac{3}{4}\frac{1}{2}\big(\frac{\epsilon}{2}\big)^2+ O(\epsilon^3)\Big)$$ Oct 13, 2021 at 9:09 • corrected the typo Oct 13, 2021 at 9:10 ## 2 Answers Considering$$f(x)=\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}$$ Adaptating an approximation nwork done in my group more than thirty years ago, ws have $$g(x)=-\frac{1+2 \left(t+t^4+t^9\right)}{x-1}-\frac{1}{4} \left(1+2 \left(t+t^4+t^9\right)\right)+$$ $$\frac 1{96}\left(7+14 t+96 t^4+14 t^9\right) (x-1)-\frac{5}{128} \left(1+2 \left(t+t^4+t^9\right)\right) (x-1)^2+O\left((x-1)^3\right)$$ where $$t=\exp\Bigg[\pi^2\left(\frac{1}{x-1}+\frac{1}{2}-\frac{x-1}{12}+\frac{1}{24} (x-1)^2-\frac{19}{720} (x-1)^3 \right) \Bigg]$$ is very small; for $$x=\frac 12$$, $$t=\exp\Big[ -\frac{8321 }{5760}\pi ^2\Big]=6.43\times 10^{-7}$$. So, using $$t=0$$, we have as another approximation $$h(x)=-\frac{1}{x-1}-\frac{1}{4}+\frac{7 }{96}(x-1)-\frac{5}{128} (x-1)^2+O\left((x-1)^3\right)$$ Using $$t=0$$, we could extend the expansion and have $$h(x)=-\frac{1}{x-1}-\frac{1}{4}+\frac{7 }{96}(x-1)-\frac{5}{128} (x-1)^2+\frac{309 }{10240}(x-1)^3-$$ $$\frac{763 }{40960}(x-1)^4+\frac{893209 }{61931520}(x-1)^5+O\left((x-1)^6\right)$$ Using the last $$h(x)$$, consider the norm $$\Phi=\int_{\frac 12}^{1} \Big[f(x)-h(x)\Big]^2\,dx=9.80 \times 10^{-9}$$ Edit I have been in touch with my former PhD student who continued working in this area. His latest results are $$\operatorname{\vartheta}_3\left(0,x\right)=\sqrt{\frac \pi t}\Bigg[1-\frac 14\sum_{n=1}^\infty a_n\,t^n\Bigg]\qquad \text{with}\qquad t=1-x$$ where the $$a_n$$ form the sequence $$\left\{1,\frac{7}{24},\frac{5}{32},\frac{787}{7680},\frac{763}{10240},\frac{893209}{15482880},\frac{2885597}{61931520},\frac{1153151299}{29727129600},\frac{261937547}{7927234560},\frac{3997632829}{139519328256},\frac{30141297349}{1195879956480},\frac{4101 190700056349}{182826127746662400},\cdots\right\}$$ This makes $$f(x)=\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}=\frac 1{1-x} -\frac 14\sum_{n=1}^\infty a_n\,(1-x)^{n-1}$$ Using the above coefficients $$\Phi=\int_{\frac 12}^{1} \Big[f(x)-h(x)\Big]^2\,dx=7.83 \times 10^{-15}$$ • Thanks, but what is$g(x)$? Sep 26, 2021 at 14:38 • @user688486.$g(x)$is the first expansion;$h(x)$is$g(x)$with$t=0$. Plot the original function and the last$h(x)$for$0 \leq x \leq 1\$ and tell me what you think. Sep 26, 2021 at 14:41
• Are there any research papers on 'an approximation (n)work done in my group more than thirty years ago'? Sep 26, 2021 at 20:33
• @user688486. This was part of a thesis work. If I remember, it was presented in a condefece around 1977 in the US. By yhe way, if this is an answer for you, may be you could accept it. Cheers :-) Sep 27, 2021 at 1:41

We use the fact that the Mellintransform $$M$$ of a Gaussian is given by

$$M(e^{- x^2})=\frac{\Gamma(s/2)}{2}\,\,\,\, (*)$$

In following we are using a method discovered by Flajolet and co-workers. First of all we can see that the sum in question ($$-a^2=\log(x)$$) $$\mathcal{S}(a)=\sum_n e^{-a^2 n^2}$$ is a harmonic sum with frequencies $$f_n=n$$ , amplitudes $$A_n=1$$ and base function $$g(x)=e^{-x^2}$$. Therefore by the above reference we have (using (*) and $$\Lambda(s)=\sum_{n\geq1 } f_n^{-s}$$): $$M(\mathcal{S})=\Lambda(s)g^*(s)=\zeta(s)\frac{\Gamma(s/2)}{2}$$

now we can use the "Mellin Summation Formula" stated on page 25 of the reference ($$H$$ is the set of residues left of the fundamental strip $$(0,\infty)$$ ). $$\mathcal{S}(a)\sim_{a\rightarrow 0} \sum_{s \in H}\text{res}(M(\mathcal S)a^{-s/2})$$

we have $$\text{res}(M(\mathcal S)a^{-s/2})|_{s=1}=\sqrt{\pi}/(2a^{1/2})$$ and $$\text{res}(M(\mathcal S)a^{-s/2})|_{s=0}=-1/2$$ and therefore $$\mathcal{S}(a)\sim_{a\rightarrow 0} \frac{\sqrt{\pi}}{2a^{1/2}}-\frac12$$

Note the corrections are expontially small since $$\zeta(-2n)=0$$ so no other residue has finite value and no further correction of $$O(x^n)$$ exists.