# Convergence to $0$ of Jacobi theta function

I'm trying to prove that a function $$f(y) = \sum_{k=-\infty}^{+\infty}{(-1)^ke^{-k^2y}}$$ is $O(y)$ while $y$ tends to $+0$. I have observed that $f(y) = \vartheta(0.5,\frac{iy}{\Pi})$ where $\vartheta$ is a Jacobi theta function. It seems that these functions are very well studied but I am not too familiar with this area.

Any useful links or suggestions are very appreciated.

UPD: the previous version contained a square which shouldn't be there.

Actually, your function is even more simply expressed in terms of $\vartheta_4$-function. Also, I prefer this notation in which $$f(y)=\vartheta_4(0,e^{-y})=\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr).$$ I.e. I use the convention $\vartheta_k(z,q)=\vartheta_k(z|\tau)$.

Then, to obtain the asymptotics as $y\rightarrow 0^+$, we need two things:

• Jacobi's imaginary transformation, after which the transformed nome and half-period behave as $q'\rightarrow0$, $\tau'\rightarrow i\infty$ (instead of $q\rightarrow1$, $\tau\rightarrow0$): $$\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr)=\sqrt{\frac{\pi}{y}}\vartheta_2\Bigl(0\Bigr|\Bigl.\frac{i\pi}{y}\Bigr).$$

• Series representations for theta functions (e.g. the formula (8) by the first link), which implies that $$\vartheta_2(0,q')\sim 2(q')^{\frac14}$$ as $q'\rightarrow 0$. Note that you can also obtain an arbitrary number of terms in the asymptotic expansion if you want.

Taking into account the two things above, we obtain that the leading asymptotic term is given by $$f(y\rightarrow0)\sim 2\sqrt{\frac{\pi}{y}} \exp\left\{-\frac{\pi^2}{4y}\right\}.$$

• This is a VERY precise answer. I wonder if a cheaper method could be used to get a more basic estimate. – Mikhail Katz Jul 2 '13 at 19:00
• @user72694 This is a good question - I have no idea. But actually, in order to obtain the above estimate, one doesn't need to know anything about theta functions. Basically what was used is the Poisson summation formula which has transformed a series with a non-obvious asymptotics into a series with obvious asymptotics. But I am not sure this counts as a cheaper method. – Start wearing purple Jul 2 '13 at 21:42
• Many thanks @O.L.! I'll add the reference to these arguments to my paper (if it will be used there). – Mikhail Berlinkov Jul 3 '13 at 2:44
• @Misha You are welcome. Feel free to use the arguments without any reference, they are very well-known. – Start wearing purple Jul 3 '13 at 7:08

This does not appear to be correct. If $y>0$ then each summand is greater than 1 in absolute value, so there is no change the series will converge. In particular, one cannot claim that it is $O(y)$ in any obvious sense.

• Sorry, I have lost a minus in the exponent. – Mikhail Berlinkov Jul 2 '13 at 18:17
• It's helps to double check before posting an answer. In fact this is a general rule no matter what sort of science you are researching. – T.A.Tarbox May 2 '17 at 3:09
• @T.A.Tarbox, there was a wrong sign in the original posting, which the OP corrected after I posted my answer. If you think it is preferable to delete this answer I can certainly do that. – Mikhail Katz May 2 '17 at 17:05
• @Misha I see, in my comment I didn't intend or mean to suggest anything untoward. However this is an excellent example of how errors can compound and cascade and lead to further error. I endeavor to exercise "intuitive reasoning" (or at least I try) whenever I encounter a mathematical problem and this approach requires a certain degree of discipline. One needs to ask whether the problem statement makes sense logically in the context of precedent and past history of such problems. Errors in problem statements are usually unintended and accidental and hence correctable. – T.A.Tarbox May 5 '17 at 2:41
• @T.A.Tarbox, user "Misha" is different from user "Mikhail Katz". P.S. What is a "tinhorn"? – Mikhail Katz May 5 '17 at 6:48