# Sum of Gaussian pdfs

I learned from a colleague that if one sums translates of the Gaussian density $f(x)=(2\pi)^{-1/2}e^{-x^2/2}$ translated by the integers (i.e. one considers $F(x)=\sum_{n\in\mathbb Z}f(x+n)$), the resulting function is remarkably constant: that is, the function differs from its average value by less than one part in $10^8$. The significance of translating by multiples of 1 is that the inflection point of the Gaussian occurs at $\pm1$.

I attempted to repeat this for $g(x)=e^{-3x^4/4}$ which also has an inflection point at $\pm 1$, and found that the resulting sum is constant to one part in 200.

Can anyone offer any conceptual explanation for the remarkable degree of constancy of $F(x)$, or is this just a fluke?
• Are you sure that it is only remarkably close or may it be exact? (My crude numerical experiment gave exactly one, but then, it may be too crude…). – Dirk Oct 26 '17 at 6:48
• A equivalent question to ask is: Why does $\sum_{n\in\mathbf{Z}} e^{-xn-n^2/2}$ is almost (or exactly) equal to $e^{x^2/2}\sqrt{2\pi}$? – Dirk Oct 26 '17 at 6:50
• I'm pretty sure it's not exact. If you compute $F(0)-F(\frac 12)$, the result (using only $n$'s in the range $-10$ to 10) is $1.07012\times10^{-8}$. The error from truncating at the 10th term is something like $e^{-50}$, which is much smaller than this. – Anthony Quas Oct 26 '17 at 6:51
• I'm just a novice in the field, but I believe this is closely related to the smoothing parameter of a lattice in lattice-based cryptography. – Mark Oct 27 '17 at 2:18
• This phenomenon is the basis for "approximate approximations", a semi-analytical numerical method for solving differential equations, developed by V. Mazya and G. Schmidt. See Section 1.1.1 in their book: amazon.com/… – Hans Lundmark Oct 27 '17 at 9:58

First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$\sum_{n=-\infty}^\infty e^{-\alpha (x-n)^2}=2{\sqrt{\frac{\pi}{\alpha}}}\left(1+2\sum_{n=1}^\infty e^{-\frac{\pi^2}{\alpha}n^2}\cos2\pi n x\right),\quad \text{Re}~\alpha>0.\tag{1}$$ From this formula one can see that when $\alpha>0$ is small, the Fourier coefficients $a_n=2{\sqrt{\frac{\pi}{\alpha}}}e^{-\frac{\pi^2}{\alpha}n^2}$ of the resulting function decreases rapidly with increasing $n$. When $\alpha=1/2$ one has $$\frac{a_1}{a_0}=2e^{-2\pi^2}\approx 5.4\times 10^{-9}.$$

I don't know any conceptual explanation for this in mathematics, but there is such an explanation that comes from physics. Consider a quantum particle with mass $m$ on a ring of radius $a$. We assume the ring is pierced with magnetic flux $\phi$. We want to calculate partition function of this system at temperature $T>0$ in two different ways.

On the one hand, it is known that the energy spectrum of the particle is given by $$E_n=\frac{\hbar^2}{2m^2}\left(n+\frac{\phi}{\phi_0}\right)^2,$$ where $\phi_0$ is the so called magnetic flux quantum. Then partition function is given as the Gibbs sum $$Z=\sum_{n=-\infty}^\infty e^{-E_n/T}=\sum_{n=-\infty}^\infty e^{-\frac{\hbar^2}{2m^2T}\left(n+{\phi}/{\phi_0}\right)^2}.\tag{2}$$ Here one immediately recognizes the sum analogous to the LHS of $(1)$.

On the other hand, partition function is related to the trace of the density matrix $\rho_{\theta_1,\theta_2}$, i.e. to $\int\rho_{\theta,\theta}d\theta$. It is possible to calculate the density matrix of this system in imaginary time representation by solving a certain differential equation. Details of this calculation can be found for example in this book, chapter 4.3. Consider first the more simple case of unbounded line $-\infty<\theta<+\infty$; then the answer is $$\rho_{\theta_1,\theta_2}=Ce^{-\frac{mTa^2(\theta_1-\theta_2)^2}{2\hbar^2}},$$ where $C$ is some normalization constant. Magnetic flux does not enter this expression because there is not any nontrivial loop in the system. On a ring there is a nontrivial loop, and let $n$ be the winding number. It is known that the partition function is a sum over all homotopy classes multiplied by corresponding phase factor. In this case the phase factor comes from the magnetic flux (Aharonov-Bohm phase) and equals $e^{2\pi in\phi/\phi_0}$. As a result we have \begin{align} Z&=\sum_{n=-\infty}^\infty e^{2\pi in\phi/\phi_0}\int\rho_{\theta,\theta+2\pi n}d\theta\\ &=2\pi C \sum_{n=-\infty}^\infty e^{-\frac{2\pi^2mTa^2}{\hbar^2}n^2+2\pi in\phi/\phi_0}.\tag{3} \end{align} Combining $(2)$ and $(3)$ one obtains the transformation in $(1)$.

One can see from this physical interpretation that when the temperature $T$ increases, then the higher Fourier harmonics decrease. This corresponds to the physical intuition that the higher the temperature the more chaotic the system becomes and the effect of the Aharonov-Bohm phase averages out due to thermal fluctuations.

• This is a remarkable connection of a seemingly abstract mathematical fact to a physics problem related to my research topic, thank you very much – Yuriy S Oct 26 '17 at 21:20

Interesting! Put another way, the fractional part of the standard Gaussian variable closely approximates the standard uniform variable.

You can also closely approximate standard Gaussian from standard uniform: Add 12 independent uniform variables before subtracting 6. This approximation is also shockingly good; see John D. Cook's comment here. (12 isn't "large," but rather uses var(U)=1/12 to get unit variance.)

Notice that the fractional part of this approximate Gaussian is exactly uniform. You might think of this as a dual explanation for the phenomenon you observed.

• This is a nice observation. I followed up by calculating the total variation distance between $N(0,1)$ and the translated sum of 12 $U[0,1]$'s: it's 0.006, so this is close, but not ridiculously close. On the other hand, the total variation distance between $\langle N(0,1)\rangle$ and $U[0,1]$ is of the order $10^{-8}$, which is ridiculously close. – Anthony Quas Oct 26 '17 at 22:52
• If you add U[-1/2,1/2] to N(0,11/12), the fractional part is still exactly uniform, and it approximates N(0,1) about 10 times better, but it's still not "ridiculously" close. – Dustin G. Mixon Oct 26 '17 at 23:49

Too long for a comment...

I'm not sure why this is so surprising. If one sums translates by $\varepsilon \mathbb{Z}$ of the standard Gaussian then for $\varepsilon$ sufficiently small the result ought to be roughly constant. More generally, one can take any sufficiently nice function in lieu of the Gaussian, albeit with the awareness that this will affect $\varepsilon$.

The fact that $\varepsilon = 1$ is "small" for the standard Gaussian is the only thing that should be surprising. So let's set $z := -i\varepsilon^2/4\pi$ and $\tau := i\varepsilon^2/2\pi$ and compare

$\sum_n \exp(-[\varepsilon n]^2/2) = \vartheta(0;\tau) = \vartheta_{01}(z;\tau)$

and

$\sum_n \exp(-[\varepsilon(n-1/2)]^2/2) = \exp(-\varepsilon^2/8) \cdot \vartheta(z;\tau) = \exp(-\pi i[\tau/4+z]) \cdot \vartheta(z;\tau)$.

Now the question becomes when these two functions are approximately equal.

• So two comments: First when I was asking, I had in mind that this was specifically related to the fact that 1 was the inflection point; Nemo's answer disabuses me of that idea; Secondly: if 1 is replaced by 2, $10^{-8}$ becomes replaced by $10^{-2}$, which shows this is much more than just the kind of polynomial convergence you would get from the simplest considerations. – Anthony Quas Oct 26 '17 at 17:01