4

Is it possible to find an upper bound (or even an exact value) for

$$\sum_{x\in\mathbb{Z}^n} e^{-x^Tx}\;?$$

In particular, is this sum less then $2^n$?

Approximate numerical answers:

  • For $n=1$ we get 1.7726
  • For $n=2$ we get 3.1422
  • For $n=3$ we get 5.5701
  • For $n=4$ we get 9.8739
  • For the case $n=1$: http://math.stackexchange.com/questions/174877/evaluation-of-sum-x-0-infty-e-x2 for the other cases you can reduce it to this case with simple algebraic manipulations – s.harp Mar 15 '16 at 18:52
  • Your approximate numerical answers can all be written as $1.7726^n$. – Semiclassical Mar 15 '16 at 20:29
  • @Semiclassical Which is approx $\sqrt{\pi}^n$.. but why? –  Mar 15 '16 at 20:30
  • There's no mystery regarding the exponent $n$: As both answers below indicate, the sums over $\mathbb{Z}n$ factorize as $(\sum_k e^{-k^2})^n$. So the only question is why $\sum{k=-\infty}^\infty e^{-k^2} \approx \sqrt{\pi}$ (the true result is greater by about $0.01%$). – Semiclassical Mar 15 '16 at 20:32
  • More generally, it looks like $$\sum_{k\in\mathbb{Z}}e^{-tk^2} = \vartheta_3(e^{-t})\sim \sqrt{\frac{\pi}{t}}$$ as $x\to 0^+$ where $\vartheta_3(x)$ is a Jacobi theta function. So evidently $t=1$ is small enough for that approximation. – Semiclassical Mar 15 '16 at 20:43
  • While playing around with the numerics, I realized that I'd covered the same ground in this earlier answer of mine: http://math.stackexchange.com/a/892009/137524. (That answer is a bit more general, but includes the case above when $x=0$.) – Semiclassical Mar 15 '16 at 21:03
  • @Semiclassical I don't understand your claim as we know that $\sum_{k\in\mathbb{Z}}e^{-tk^2} \geq 1$ by just considering the $k=0$ term. So the estimate must be wrong for $t=4$ for example. –  Mar 16 '16 at 20:00
  • @Lembik: It's an approximation as $t\to 0$ (there was a typo in my earlier comment.) Evidently $t=1$ is small enough to work but not $t=4.$ – Semiclassical Mar 16 '16 at 22:14

3 Answers3

3

$$\sum_{k \in \mathbb{Z}} e^{-k^2} < 1+2\sum_{k=1}^\infty e^{-k} -2e^{-2} \\ =1+\frac{\frac{2}{e}}{1-\frac{1}{e}}-2e^{-2}\\ =1+\frac{2}{e-1}-2e^{-2} < 1.893$$

Now use the fact that $\sum_{x \in \mathbb{Z}^n} e^{-\| x \|^2} = \left ( \sum_{k \in \mathbb{Z}} e^{-k^2} \right )^n$.

This is still somewhat of a "calculator answer", but you can get the $2^n$ bound only knowing, say, $e>2.7$.

Ian
  • 101,645
2

Hint: $$ \sum_{x \in \Bbb Z^n} e^{-x^Tx} = \sum_{x_n \in \Bbb Z} e^{-x_n^2} \sum_{y \in \Bbb Z^{n-1}} e^{-y^Ty} $$

Ben Grossmann
  • 225,327
2

Note that $\displaystyle \sum_{x \in [\![-N,N]\!]^n} e^{-||x||^2} = \left(\sum_{-N \le k \le N} e^{-k^2}\right)^n = \left(1+ 2\sum_{k = 1}^N e^{-k^2}\right)^n \le \left(1+2 \int_{0}^N e^{-t^2} dt \right)^n$. Therefore, we have the inequality $\displaystyle \sum_{x \in [\![-N,N]\!]^n} e^{-||x||^2} \le \bigg(1+\sqrt{\pi}\bigg)^n$ and, passing to the limit $$\displaystyle \sum_{x \in \mathbb Z^n} e^{-||x||^2} \le \bigg(1+\sqrt{\pi}\bigg)^n$$

user323127
  • 36
  • Thank you. How tight do you think this is? –  Mar 15 '16 at 19:54
  • 1
    I don't think it is that good (because of the +1), I think $\sum_{k=-\infty}^{\infty} e^{-k^2}$ should be closer to $\sqrt{\pi}$ because it should not be very far $\int_{\mathbb R} e^{-t^2}dt$ (Wolfram Alpha actually gives a value that is not $\sqrt{\pi}$ but coincides up to two decimal places). However, I do not see an obvious way of improving this. – user323127 Mar 15 '16 at 20:20
  • $\sum_{k=1}^N e^{-k^2}$ is a right hand sum for $\int_0^N e^{-x^2} dx$ or a left hand sum for $\int_1^{N+1} e^{-x^2} dx$. Because you have monotonicity, the right hand sums are smaller and the left hand sums are bigger. So you can get a lower bound by $\int_1^{N+1} e^{-x^2} dx$. Thus there is an upper bound of $(1+\sqrt{\pi})^n$ and a lower bound of $\left ( 1+\sqrt{\pi}-\int_0^1 e^{-x^2} dx \right )^n$. – Ian Mar 15 '16 at 20:26
  • @Ian Thank you. $(1+\sqrt{\pi})^n$ is too big an upper bound for me sadly. –  Mar 15 '16 at 20:29
  • @Arnold My lower bound, if correct, implies your desired result is false, in that $1+\sqrt{\pi}-\int_0^1 e^{-x^2} dx > 2.02$. – Ian Mar 15 '16 at 20:30
  • @Ian Oh!. But that seems to contradict the numerical results for $n=1\dots 4$ doesn't it? –  Mar 15 '16 at 20:31
  • Oh, I'm sorry, I dropped a factor of 2. My lower bound is actually $\left ( 1+\sqrt{\pi}-2\int_0^1 e^{-x^2} dx \right )^n \approx 1.28^n$. So these two estimates don't do very much for us together. – Ian Mar 15 '16 at 20:32