# Evaluating a series to order “three halves”

In doing some calculations related to one-dimensional Brownian Motion confined to a finite interval, I have come across functions such as

$$f(t) = \sum_{n=1}^\infty\frac{\exp(-n^2t)}{n^4}.$$ I am interested in the behavior of this function for small $t$. By expanding the exponential it is easy to get the terms of order 0 and 1 in $t$. However, the term of order $t^2$ diverges. I take this to mean that $f(t)$ does not have a second derivative at $t = 0$. I think (although I'm not sure) that in some sense there is a "3/2" order term. How does one define and calculate this term? Perhaps the limit $$\lim_{t \to 0}\frac{f(t) - f(0) - t f'(0)}{t^{3/2}}$$ is finite and can be calculated? Is there an unambiguous way of expanding such functions in power series with fractional powers?

Background: Consider a particle undergoing one-dimensional Brownian Motion in the interval $\left[0,L\right]$ with reflecting boundary conditions. The particle has diffusion coefficient $D$ and its position is $x(t)$. Its initial position is distributed uniformly in the interval. Then one can show that the mean squared displacement at time $t$ is $$\text{MSD}(t) \equiv \langle \left[x(t) - x(0)\right]^2\rangle = \frac{L^2}{6} + \frac{8 L^2}{\pi^4}\sum_{n=1}^\infty\frac{-1 + (-1)^n}{n^4} \cdot \exp(-t/\tau_n),$$ where $$\tau_n = \frac{L^2}{n^2 \pi^2 D}$$ As I mentioned, one can calculate this up to first order in $t$. The constant term is of course zero (the particle can't travel any distance in zero time) and the first order term turns out to be $2 D t$, which makes sense since this is the MSD of a free particle (i.e. before it encounters any walls).

Further Notes Here is a plot of the function $f(t)$ given by the series, as well as the approximations from Marko Riedel's answer:

• I think a naive Taylor expansion is perhaps not the best way to go here. Take a look at this page for a related question on how to construct an approximation for theta functions (note that $f''(t) = \sum e^{-n^2t} \equiv \frac{\vartheta(0;\frac{it}{\pi})-1}{2}$ is a theta function) close to $t=0$. – Winther Sep 16 '14 at 4:12
• @Daniel Fischer: Can you tell us what's wrong with my answer? I think it is a legitimate question! – Mhenni Benghorbal Dec 26 '14 at 3:24

As noted in other answers here, $f''(t)=\sum e^{-n^2t}$, which is the Riemann sum for $\int_0^{\infty}e^{-x^2t}dx=\sqrt{\pi/4t}$. So $f(t)$ will be the integral of the integral of that, or $(2/3)\sqrt{\pi}t^{3/2}$, plus At+B.

• Thanks Michael. Could you elaborate on what you mean by "the" Riemann sum? It's been a long time since I was in analysis class. I thought that there was not a single Riemann sum for an integral, but a family of such sums (with finer and finer partitions). In what sense is $f''(t)$ "the" Riemann sum? – Greg P Sep 16 '14 at 22:37
• Whoops, I should have said 'a Riemann sum', as you say. A partition of width 1, with rectangles whose x-coordinates are each centred at an integer. – Empy2 Sep 17 '14 at 7:29
• It is interesting that although this is just a particular Riemann sum (as you say, partition of width 1), it still gives the exact coefficient of $t^{3/2}$ (compare with Marko Riedel's answer). In fact, this makes a lot of sense. If you plot the function $\exp(-x^{2}t)$ for smaller and smaller values of $t$, you see that its spatial variation happens on increasingly large length scales (in particular, larger than 1). Thus that particular approximation gets better and better, giving the exact behavior at small $t$. Very nice! – Greg P Sep 17 '14 at 16:44

If you start with Taylor expansion $$e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$$ replace $x$ by $-n^2t$ and then $$e^{-n^2t}=1-n^2 t+\frac{n^4 t^2}{2}-\frac{n^6 t^3}{6}+\frac{n^8 t^4}{24}-\frac{n^{10} t^5}{120}+O\left(t^6\right)$$ $$\frac{\exp(-n^2t)}{n^4}=\frac{1}{n^4}-\frac{t}{n^2}+\frac{t^2}{2}-\frac{n^2 t^3}{6}+\frac{n^4 t^4}{24}-\frac{n^6 t^5}{120}+O\left(t^6\right)$$ Summing over $n$ from $1$ to $\infty$ makes a problem with the third and higher terms.

So, in my opinion, if we need more than the first order, we should use $$\sum_{n=1}^\infty\frac{\exp(-n^2t)}{n^4}\sim \frac {\pi^4}{90} -\frac {\pi^2}{6}t+ \alpha t^{\beta}$$ with ($1<\beta<2$) and parameters $\alpha$ and $\beta$ would be adjusted by nonlinear regression for a specified range of $t$.

Based on eleven equally spaced data points ($0\leq t\leq 1$), I obtained $\alpha=0.934254$, $\beta=1.39772$ leading to an $R^2=0.999987$. So the idea, as proposed by Antonio Vargas, of using $\beta=\frac{3}{2}$ seems to be very interesting. Using this last value, I found $\alpha=0.954250$ for $R^2=0.999302$. By the end, why not to simply use $$\sum_{n=1}^\infty\frac{\exp(-n^2t)}{n^4}\sim \frac {\pi^4}{90} -\frac {\pi^2}{6}t+ t^{\frac{3}{2}}$$ which is basically Antonio Vargas's idea and to whom the credit should be given.

• The power of $3/2$ is also suggested by the integral analogue for the sum, which according to Mathematica has an expansion like $$\int_1^\infty e^{-x^2 t} x^{-4}\,dx = \tfrac{1}{3} - t + \tfrac{2}{3}\sqrt{\pi} t^{3/2} - \tfrac{1}{2}t^2 + \tfrac{1}{18}t^3 - \tfrac{1}{120}t^4 + \cdots$$ as $t \to 0$. To get the numerical value of the coefficient I used the commands pts=Table[{t,NSum[E^(-n^2 t)/n^4,{n,1,\[Infinity]},NSumTerms->50000,WorkingPrecision->20]-Zeta[4]+t Zeta[2]},{t,1/10000000,1/1000,1/10000}]; and a/.FindFit[pts,a t^(3/2),a,t,WorkingPrecision->9], giving 1.17479810. – Antonio Vargas Sep 16 '14 at 7:24
• In particular I sampled very small values of $t$ -- from the interval $[1/10000000,1/1000)$. – Antonio Vargas Sep 16 '14 at 7:27
• @AntonioVargas. This is very interesting ! Thanks – Claude Leibovici Sep 16 '14 at 8:05

It seems to have escaped attention that this sum may be evaluated using harmonic summation techniques which can be an instructive exercise.

Introduce $S(x)%$ given by $$S(x) = \sum_{n\ge 1} \frac{1}{n^4}\exp(-(nx)^2).$$

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^4}, \quad \mu_k = k \quad \text{and} \quad g(x) = \exp(-x^2).$$ We need the Mellin transform $g^*(s)$ of $g(x)$, which is

$$\int_0^\infty e^{-x^2} x^{s-1} dx = \int_0^\infty e^{-t} t^{s/2-1/2} \frac{1}{2} t^{-1/2} dt \\ = \frac{1}{2} \int_0^\infty e^{-t} t^{s/2-1} dt = \frac{1}{2} \Gamma(s/2).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \frac{1}{2} \Gamma(s/2) \zeta(s+4) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^4} \frac{1}{k^s} = \zeta(s+4)$$ for $\Re(s) > -3.$

The fundamental strip of the transform of the base function is $\langle 0, \infty \rangle$ and intersecting this with $\langle -3, \infty\rangle$ we obtain that the Mellin inversion integral here is

$$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

Fortunately the poles of the gamma function term at even integers $\le -6$ are cancelled by the trivial zeros of the zeta function, so that just four poles remain: $s=0, s=-2$ from the gamma function, $s=-3$ from the zeta function and again $s=-4$ from the gamma function.

We have $$\mathrm{Res}\left(Q(s)/x^s; s=0\right) = \frac{1}{2} \times 2 \times \zeta(4) = \frac{\pi^4}{90},$$ and $$\mathrm{Res}\left(Q(s)/x^s; s=-2\right) = \frac{1}{2} \times (-2) \times \zeta(2) \times x^2 = - \frac{\pi^2}{6} x^2,$$ and $$\mathrm{Res}\left(Q(s)/x^s; s=-3\right) = \frac{1}{2} \times \frac{4}{3}\sqrt{\pi} \times x^3 = \frac{2}{3}\sqrt{\pi} x^3$$ and finally $$\mathrm{Res}\left(Q(s)/x^s; s=-4\right) = \frac{1}{2} \times 1 \times \zeta(0) \times x^4 = - \frac{1}{4} x^4.$$

This gives the following expansion in a neighborhood of zero: $$S(x) \sim \frac{\pi^4}{90} - \frac{\pi^2}{6} x^2 + \frac{2}{3}\sqrt{\pi} x^3 - \frac{1}{4} x^4.$$

Since the sum being asked for is actually $S(\sqrt{t})$ we obtain $$S(\sqrt{t}) \sim \frac{\pi^4}{90} - \frac{\pi^2}{6} t + \frac{2}{3}\sqrt{\pi} t\sqrt{t} - \frac{1}{4} t^2.$$

This answer matches the results from the earlier posts, which deserve the credit, with this contribution being enrichment only.

• But enrichment it is indeed! I will need to read up on these things. Thanks! – Greg P Sep 16 '14 at 22:39
• Is the last relation exact or just an asymptotic? – Antonio Vargas Sep 17 '14 at 13:35
• It's an asymptotic. – Marko Riedel Sep 17 '14 at 16:59
• As I'm sure is often the case, I wish I could check (accept) both this answer and Michael's. This one seems to provide a general method to approach many problems of this kind, while Michael's answer gives a straightforward way to get the $t^{3/2}$ coefficient. In the end I checked Michael's answer once I realized that the Riemann sum approximation becomes exact in the limit $t \to 0$, so that it gives the correct coefficient. But again, Marko's answer is very helpful. – Greg P Sep 17 '14 at 17:11
• The remainder is in the left side of the rectangular contour. (The top and bottom contributions vanish as the height goes to $\pm\infty.$) For more information consult "Mellin Transform Asymptotics" by Sedgewick and Flajolet, INRIA Rapport de Recherche 2956, which explains it better than I ever could. – Marko Riedel Sep 18 '14 at 17:36