Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm wondering how to examine the asymptotic behaviour as $x$ goes to $0^{+}$ for the following series (which is the continuous function in the right neighbourhood of $0$):
$$\sum_{n=1}^{\infty}\sin^{2}\left(\frac{n\pi}{2}\right)\exp\left(-\frac{n^2\pi^2 x}{4}\right)$$ i.e. how to find a simple continuous function $g$ such that the above series is equivalent to $g(x)$ as $x$ goes to $0^{+}$. Thank you for any replies!

share|cite|improve this question
What do you think of $g(x)=\frac{1}{e^{\frac{\pi^2x}{4}}-1}=\sum_{n=1}^{\infty} \exp\left(\frac{- n \pi^2 x}{4}\right)?$ – Gardel Nov 12 '11 at 18:49
Nice function, but in a view of other answers it looks like it's not correct. – John Nov 13 '11 at 17:38

For every positive $u$, consider the two series $$ s(u)=\sum\limits_{n=1}^{+\infty}\mathrm e^{-u^2n^2},\qquad r(u)=\sum\limits_{n=0}^{+\infty}\mathrm e^{-u^2(2n+1)^2}. $$ Since the function $t\mapsto e^{-t^2}$ is decreasing on $t\geqslant0$, for every $n\geqslant1$, $$ \int_{n}^{n+1}\mathrm e^{-u^2t^2}\mathrm dt\leqslant\mathrm e^{-u^2n^2}\leqslant\int_{n-1}^n\mathrm e^{-u^2t^2}\mathrm dt. $$ Summing these, one gets $$ \int_{1}^{+\infty}\mathrm e^{-u^2t^2}\mathrm dt\leqslant s(u) \leqslant \int_{0}^{+\infty}\mathrm e^{-u^2t^2}\mathrm dt = \frac1u\int_{0}^{+\infty}\mathrm e^{-t^2}\mathrm dt=\frac{\sqrt{\pi}}{2u}. $$ The LHS is greater than the RHS minus $1$ hence, for every positive $u$, $$ \frac{\sqrt{\pi}}{2u}-1\leqslant s(u)\leqslant \frac{\sqrt{\pi}}{2u}, $$ which is more than enough to see that $us(u)\to \frac12\sqrt{\pi}$ when $u\to0$.

The series $r(u)$ is liable to the same treatment, which yields $r(u)\sim \dfrac{\sqrt{\pi}}{4u}$. Using $u^2=\pi^2x/4$, this yields $$ S(x)=\sum_{n=1}^{\infty}\sin^{2}\left(\frac{n\pi}{2}\right)\exp\left(-\frac{n^2\pi^2 x}{4}\right)=r(u)\sim\frac1{2\sqrt{\pi x}}. $$ One sees that this method yields the stronger result that, for every positive $x$, $$ \frac1{2\sqrt{\pi x}}-1\leqslant S(x)\leqslant\frac1{2\sqrt{\pi x}}. $$

share|cite|improve this answer
Thanks for that very clear answer. It uses the fact that $\sin^{2}(\frac{n\pi}{2})$ is always either $0$ or $1$. But what if the $S(x)$ was the following: $\sum\limits_{n=1}^{\infty}\sin^{2}(\frac{n\pi}{\sqrt{2}})\exp(-\frac{n^2 \pi^2 x}{4})$? If your solution can be modified also to that latter series? I can't see it. – John Nov 12 '11 at 22:33

Not sure how to actually derive it, but I can get the answer with Mathematica at least. Note $f(x) = g(e^{-\pi^2 x})$ where $g(z) = \sum_{k=0}^\infty z ^ {(k+1/2)^2}$. The latter is a special case of an Elliptic Theta function. Then since $f(x) \rightarrow \infty$ slower than $1/x$, I made an educated guess on the leading order, so evaluating with Mathematica, it turns out that $\lim_{x\rightarrow 0^+} f(x)x^{1/2}=\frac{1}{2 \sqrt{\pi}}$. (Links go to Wolfram Alpha). So $f(x) \sim \frac{1}{2 \sqrt{\pi x}}$ as $x \rightarrow 0^+$.

share|cite|improve this answer
Thanks for your answer. The only thing I don't understand is $f(x)=g(e^{-pi^2 x})$. If in the definition of $g$ the imaginary unit is ommited? Could you explain it in a little more detailed form? – John Nov 12 '11 at 21:22
The factor $\sin(n \pi/2)^2$ is 0 if n is even and 1 if n is odd. So replace $n^2/4$ with $(2k+1)^2/4$. – p.s. Nov 12 '11 at 22:28
OK, I see it. Thanks. – John Nov 12 '11 at 22:34

The sine factor only means that you only take odd sum indices.

The sum should be split into a finite initial part that is interpreted as a Riemann sum and scales to an approximation of the integral $\int_{0}^{big}c(\exp(-a t^2))dt$.

It will obviously take some work to choose the cut-off point correctly, so that both the rest of the sum and the rest of the integral to infinity are small.

Then you can use the known result for $\int_0^{\infty}\exp(-t^2)dt$ to conclude.

share|cite|improve this answer

Your series can be expressed by means of the theta function

$$\theta(x)\ :=\ \sum_{n=-\infty}^\infty e^{-n^2\pi x}\ .$$

Indeed, one has

$$\eqalign{f(x):=\sum_{n\geq1,{\rm odd}}\ \exp\Bigl({-n^2\pi^2 x\over 4}\Bigr)&= \sum_{n\geq1}\ \exp\Bigl({-n^2\pi^2 x\over 4}\Bigr) - \sum_{m\geq1}\ \exp\Bigl({-4m^2\pi^2 x\over 4}\Bigr) \cr &={1\over2}\Bigl(\theta\bigl({\pi x\over 4}\bigr) -\theta(\pi x)\Bigr)\ .\cr}$$

Now this theta function satisfies a famous functional equation (the "Jacobi transformation"), proven in Fourier analysis:

$$\theta(x)={1\over\sqrt{x}}\ \theta\Bigl({1\over x}\Bigr)\qquad(x>0)\ .$$

Therefore we get

$$f(x)={1\over 2\sqrt{\pi x}}\Biggl(2\theta\Bigl({4\over\pi x}\bigr)-\theta\Bigl({1\over \pi x}\Bigr)\Biggr)\ .$$

Taking $k=0$ in the theta series this not only confirms the result $g(x)={1\over 2\sqrt{\pi x}}$ of other answers. In addition, our formula shows that the approximation is unbelievably good when $x\to 0$: The terms corresponding to $k=\pm1$ in the two theta series are of order $e^{-4/x}$ resp. $e^{-1/x}$.

share|cite|improve this answer
Great answer! Thanks. – John Nov 14 '11 at 23:02

Your Answer


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.