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I'm trying to evaluate $\sum_{a=-\infty}^{\infty} e^{-2 \pi z a^2}$ using Euler-Maclaurin, but I get $\frac{1}{\sqrt{2z}}$. The only alternative I have is to calculate the remainder term directly for a small level of approximation, but I don't want to do this if there's a simple mistake I'm making and fixing it would let me avoid doing so, or if for some reason Euler-Maclaurin doesn't work at all.

SUMMARY:

By Euler-Maclaurin, we have:

$\sum_{a=-\infty}^{\infty}e^{-ca^2}-\int_{-\infty}^{\infty} e^{-ca^2} da = \sum_{k=2}^{\infty} \frac{b_k}{k!}(e^{-cx^2})^{(k-1)'}\vert_{-\infty}^{\infty}$, where $b_k=B_k(0)$ are the Bernoulli numbers. Implicitly, the remainder term (due to the large factorial) and the halved endpoints of the sum (due to $e^{-ca^2}$ vanishing in the infinite limits) vanish.

First, observe $\frac{d^n}{da^n}e^{-c a^2}$ is of the form $e^{-c a^2}p(a)$ for a polynomial $p$. (1)

Secondly, observe that $lim_{a \rightarrow \infty} \frac{p(a)}{e^{c a^2}} = 0$ for any polynomial $p$ and constant $c \gt 0$. (2)

Thirdly, observe that $e^{-ca^2}$ is even, and hence its odd derivatives (the derivatives taken for even $k$) are odd, and its even derivatives (odd $k$) are even.

The latter implies that for odd $k$, $(e^{-cx^2})^{(k-1)'}\vert_{-\infty}^{\infty}$ vanishes, and for even k, $(e^{-cx^2})^{(k-1)'}\vert_{-\infty}^{\infty} = 2 \lim_{a \rightarrow \infty} (e^{-cx^2})^{(k-1)'}(a) = 2 \lim_{a \rightarrow \infty} e^{-ca^2}p(a) = 0.$

Therefore all terms of the series vanish, implying $\sum_{n=-\infty}^{\infty}e^{-cx^2} = \int_{-\infty}^{\infty} e^{-cx^2} da = \frac{\sqrt{\pi}}{\sqrt{c}}$, and in this case $\frac{\sqrt{\pi}}{\sqrt{2 \pi z}} = \frac{1}{\sqrt{2z}}.$


DETAILS:

(1) follows by induction from $\frac{d}{da}e^{-ca^2}p(a) = e^{-ca^2}(\frac{d}{da}p(a)-2ca\ p(a)),$ which is of the form $e^{-ca^2}p(a)$.

Proof of (2): $\frac{\frac{d}{dx}a^n}{\frac{d}{dx}e^{c a^2}} = \frac{a^{n-1}}{2cae^{c a^2}} = \frac{a^{n-2}}{2ce^{c a^2}}, \implies \frac{\frac{d^m}{dx^m}a^n}{\frac{d^m}{dx^m}e^{c a^2}} = \frac{a^{n-2m}}{(2c)^m e^{ca}}, \implies lim_{a \rightarrow \infty} \frac{a^n}{e^{c a^2}} = 0 \implies lim_{a \rightarrow \infty} \frac{p(a)}{e^{c a^2}} = 0$

I have also derived an explicit formula for the derivatives from Faà di Bruno's formula together with $\frac{d^n}{dx^n}f(cx)=c^nf^{(n)}(cx)$, which yields a polynomial multiple of $e^{-ca^2}$, which by (2) vanishes in the limit, and is again even for even derivatives and odd for odd ones.

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    $\begingroup$ A relevant computation that it might be useful to consult can be found at this MSE link. $\endgroup$ Apr 15, 2014 at 23:15

2 Answers 2

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I just answered a very similar question here. Without writing everything out again, let me point out that you can use Euler-Maclaurin to deduce, for any $N>0$, that $$\sum_{a=-\infty}^{\infty} e^{-2 \pi z a^2} = \frac{1}{\sqrt{2z}} + O(z^N) \ \mbox{as}\ z \to 0.$$

See the fourth example in Chapter 9.6 of Concrete Mathematics for the details.

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  • $\begingroup$ So, what I said about the remainder term vanishing is false. To calculate the remainder term properly, I'm guessing I should take the $\int_n^m p_{2k}(x)\frac{B_{2k}(x-\lfloor x \rfloor)}{(2k)!} dx$ term to be the integral over $\mathbb{R}$ of the limit as $k$ goes to infinity of each taylor series, and if I have uniform convergence of that sequence of functions that would show I have the Taylor series for the function inside the integral. Or is there a better way to get more terms of the infinite sum, perhaps treated in that book? $\endgroup$
    – Loki Clock
    Mar 23, 2014 at 15:16
  • $\begingroup$ It seems that you can't take the limit of the integrals over $\mathbb{R}$ as the level of approximation goes to infinity, because otherwise we can replace the periodic Bernoulli functions' products with a sum over integrals from 0 to 1 of products of the Bernoulli polynomials with shifted versions of the factors, leaving inside each integral a ratio of two polynomials of the same degree, multiplied by the inverse-exponential and inverse-factorial factors. If I remember correctly, at that point the factorial still wins out. $\endgroup$
    – Loki Clock
    Mar 23, 2014 at 15:28
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I've found out Euler-Maclaurin only works for functions of some exponential type ($\lt \pi$?), and I assume that any Gaussian has no exponential type, and is bounded by functions of too large an exponential type to apply the formula to those, too.

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  • $\begingroup$ I never did say, but a function has exponential type a if it's bounded by $e^{ax}$ in the whole complex plane. So on the imaginary line, instead of a nice, constant-bounded Gaussian you get $e^{x^2}$. $\endgroup$
    – Loki Clock
    Mar 24, 2014 at 21:47

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