# Why do these two sums define functions that are asymptotically so close in value?

Let $$f_n(\epsilon)=(2n+1)e^{-n(n+1)\epsilon},\qquad Z_\text{odd}(\epsilon)=\sum_{p\ge0}f_{2p+1}(\epsilon),\qquad Z_\text{even}(\epsilon)=\sum_{p\ge0}f_{2p}(\epsilon).$$ (These functions are relevant to describing rotation of diatomic molecules in statistical physics.)

When $$\epsilon\searrow0$$, these functions are extremly close in value; numerically: $$Z_\text{even}(0.1)-Z_\text{odd}(0.1)\simeq 3\,10^{-9},\qquad Z_\text{even}(0.05)-Z_\text{odd}(0.05)\simeq 2\,10^{-19}.$$ Is there a reason for this? What is an asymptotic equivalent of the difference?

I tried to obtain a small $$\epsilon$$ expansion using the Euler-MacLaurin expansion and Mathematica, and I found that they both have the same expansion. Notice that Euler-MacLaurin involves derivatives of $$f$$ at 0 for $$Z_\text{odd}$$ and derivatives of $$f$$ at 1 for $$Z_\text{even}$$, so it is not obvious how this should lead to the same expansion.

[Paragraph added in an edit] In fact, to compare the expansions, it is easier to write $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)=2Z_\text{even}(\epsilon)-Z_\text{total}(\epsilon)$$, where $$Z_\text{total}(\epsilon)=\sum_{n\ge0} f_n(\epsilon)$$. Then, applying Euler-MacLaurin to order $$m$$ to $$Z_\text{even}$$ and $$Z_\text{total}$$, one obtains $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)=\frac{f_0(\epsilon)}2+\sum_{p=0}^m \frac{b_{2p}(2^{2p}-1)}{(2p)!}f^{(2p-1)}(\epsilon)+R_m(\epsilon).$$ The first terms in this summation are $$\frac12$$, $$\frac14(-2+\epsilon)$$, $$\frac1{48}(-12\epsilon+12\epsilon^2-\epsilon^3)$$, $$\frac1{480}(-120\epsilon^2+180\epsilon^3-30\epsilon^4+\epsilon^5)$$, $$\frac{17}{80\,640}(-1680 \epsilon ^3+3360 \epsilon ^4-840 \epsilon ^5+56 \epsilon ^6-\epsilon ^7)$$. Notice how each new term cancels the lowest power of $$\epsilon$$ remaining. Finally, the result is the remainder term $$R_m(\epsilon)$$ plus a polynomial of $$\epsilon$$ with powers going from $$\epsilon^{m+1}$$ to $$\epsilon^{2m+1}$$. I find it quite fascinating, but maybe the Euler-MacLaurin approach is not the good one here. [End of paragraph added in the edit]

A last remark: this result is not universal. If I pick the not-so-different function $$\tilde f_n(\epsilon)=2n e^{-n^2\epsilon}$$, then the corresponding $$\tilde Z_\text{odd}$$ and $$\tilde Z_\text{even}$$ differ by a quantity of order $$\epsilon$$.

So, what is so special about my choice of $$f_n(\epsilon)$$ ? Why are my two $$Z$$ functions so close? How can I compute the asymptotic equivalent of $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)$$ as $$\epsilon\searrow0$$ ?

We have (for $$\epsilon>0$$) $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)=\sum_{n\geqslant 0}(-1)^n f_n(\epsilon)=\frac12\sum_{n\in\mathbb{Z}}(-1)^n f_n(\epsilon)$$ since clearly $$f_{-n-1}(\epsilon)=-f_n(\epsilon)$$. Now apply the Poisson summation formula $$\sum_{n\in\mathbb{Z}} g(n)=\sum_{n\in\mathbb{Z}}\hat{g}(n),\qquad\hat{g}(y)=\int_{-\infty}^\infty g(x)e^{-2i\pi xy}\,dx$$ to $$g(x)=(2x+1)e^{-x(x+1)\epsilon-i\pi x}$$. To compute $$\hat{g}(y)$$, use the integral $$\int_{-\infty}^\infty te^{-at^2+2bt}\,dt=\frac{b\sqrt\pi}{a^{3/2}}e^{b^2/a}\qquad(a\in\mathbb{R}_{>0},b\in\mathbb{C})$$ (obtained via $$t=z+b/a$$ and rectangular contours). This way we get $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)=\left(\frac\pi\epsilon\right)^{3/2}e^{\epsilon/4}\sum_{n\geqslant 0}(-1)^n(2n+1)e^{-(2n+1)^2\pi^2/(4\epsilon)}.$$ This is very small $$\asymp(\pi/\epsilon)^{3/2}e^{-\pi^2/(4\epsilon)}$$ when $$\epsilon\to 0^+$$.
• This is very nice, thanks. A comment on the last expression: if I expand the $(2n+1)^2$, I get that $$Z_\text{even}(\epsilon)-Z_\text{odd}(\epsilon)=\Big(\frac\pi\epsilon\Big)^{3/2}e^{\epsilon/4-\pi^2/(4\epsilon^2)}\Big[Z_\text{even}(\pi^2/\epsilon)-Z_\text{odd}(\pi^2/\epsilon)\Big]$$ Mar 3, 2022 at 13:27