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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!

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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
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4 Answers

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 $x$, $$ \frac1{2\sqrt{\pi x}}-1\leqslant S(x)\leqslant\frac1{2\sqrt{\pi x}}. $$

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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
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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 $x \rightarrow 0^+$.

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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
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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.

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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. $e^{-1/x}$.

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Great answer! Thanks. –  John Nov 14 '11 at 23:02
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