# How to find the asymptotic behavior of this function

I have a function that I want to study it's asymptotic behavior.

The function is

$$f(k) = - \frac{k^2}{4} - \frac{\log\pi}{2} + \log\left( \frac12 \left| \mathrm{Erfi}(\frac{k}{2} - \pi i) - \mathrm{Erfi}(\frac{k}{2} + \pi i)\right|\right)$$

and k takes only positive integer.

The plot of this function looks like:

ListPlot[Table[f[k], {k, 0, 100}], PlotRange -> All]


It looks like that the function has a asymptotic behavior (looks like ~ Log[k]), could we find a way to extract the main term (the coefficient of the Log[k] if behave like log)?

From here and here, I find the mainTerm function, but it seems to have ignored the big Log term and not work in this case:

mainTerm[f[k], k]
(* -(k^2/4) *)


## migrated from mathematica.stackexchange.comMay 7 '14 at 23:00

This question came from our site for users of Wolfram Mathematica.

• Normal@Series[f[k], {k, \[Infinity], 3}] – Sektor May 7 '14 at 5:27
• possible duplicate of Growth of functions – Sektor May 7 '14 at 5:28
• @Sektor I guess I don't understand what's the point to expand it to the third order, are you saying that it's divergent? Also after reading the post you linked, I still can't figure out the answer to my question. I'm looking for an expression that can match the function f at large k. – xslittlegrass May 7 '14 at 6:40
• While not a complete Mathematica answer, it looks like -2 Log[k]. Sektor has given you the right start, but you will also need to help Mathematica a bit as well. – chuy May 7 '14 at 14:44
• @chuy I still don't quiet understand. Could you say more of what I need to help Mathematica to do? – xslittlegrass May 7 '14 at 15:07

This is more of a Mathematica answer than a Mathematics answer. As george indicated we can write

$$f(k) = \frac{1}{4}\log\left\{ \frac{\exp(-k^2)}{(2\pi)^2} \operatorname{Re}\left[\operatorname{erf}\left(\frac{ik}{2}-\pi\right)\right]^4\right\} + \frac{\log 2}{2}.$$

(Note the constant term $\log 2/2$ which was missing from george's answer.) The question really comes down to estimating $\operatorname{Re}\left[\operatorname{erf}\left(\frac{ik}{2}-\pi\right)\right]$ as $k \to \infty$.

In Mathematica we enter

Series[Erf[I k/2 - Pi], {k, \[Infinity], 2}]


which yields

$$i^{2 \text{Floor}\left[\frac{\text{Arg}\left[-i k \pi +\pi ^2\right]}{2 \pi }\right]}+e^{\frac{k^2}{4}+i k \pi } \left(\frac{2 i e^{-\pi ^2}}{\sqrt{\pi } k}+\frac{4 e^{-\pi ^2} \sqrt{\pi }}{k^2}+O\left[\frac{1}{k}\right]^3\right).$$

The reason we expand to second order is that we know $k$ will be an integer and hence $e^{ik\pi} = \pm 1$. So, when we take the real part, the first term in the parentheses, being purely imaginary, will disappear.

We continue with

Re[% // Normal] // ComplexExpand


which yields

$$\frac{4 e^{\frac{k^2}{4}-\pi ^2} \sqrt{\pi } \text{Cos}[k \pi ]}{k^2}+\text{Cos}\left[\pi \text{Floor}\left[\frac{\text{Arg}\left[-i k \pi +\pi ^2\right]}{2 \pi }\right]\right]-\frac{2 e^{\frac{k^2}{4}-\pi ^2} \text{Sin}[k \pi ]}{k \sqrt{\pi }}.$$

Now the second term is $O(1)$ which is much smaller than the first term and the third term is zero since $\sin(k\pi) = 0$, so to find the leading order we'll just take the first term.

The argument in the $\log$ term in $f$ is then, to first order,

E^(-k^2)/(2 Pi)^2 ((4 E^(k^2/4 - \[Pi]^2) Sqrt[\[Pi]] Cos[k \[Pi]])/k^2)^4


$$\frac{64 e^{-4 \pi ^2} \text{Cos}[k \pi ]^4}{k^8}.$$

Of course $\cos(k\pi)^4 = 1$ so we conclude that

$$f(k) \approx \frac{1}{4} \log\left(\frac{64 e^{-4 \pi ^2}}{k^8}\right) + \frac{\log 2}{2}.$$

To simplify this a little we could use

Simplify[Log[(64 E^(-4 \[Pi]^2))/k^8]/4 + Log[2]/2, Assumptions -> k > 0]


to get

$$f(k) \approx -2\log k - \pi^2 + \log 4.$$

Here's a plot of this approximation in red versus the numerical points in blue.

• Thanks a lot, but how do you determine that "the second term is $O(1)$"? – xslittlegrass May 14 '14 at 19:11
• You're welcome. It's $\cos$ of something real, and $-1 \leq \cos(x) \leq 1$. – Antonio Vargas May 14 '14 at 19:23
• Thanks :) Sorry for stupid question. – xslittlegrass May 14 '14 at 19:50

If it helps your expression simplifies to this:

 f[k_] = Log[Exp[-k^2]/(2 Pi)^2 (Erf[I k/2 - Pi] - Erf[I k/2 + Pi])^4]/4


or

 f[k_] = Log[Exp[-k^2]/(2 Pi)^2 (Re[Erf[I k/2 - Pi]])^4]/4


(making the assumption k is real )