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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]

enter image description here

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) *)
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    $\begingroup$ Normal@Series[f[k], {k, \[Infinity], 3}] $\endgroup$
    – Sektor
    May 7, 2014 at 5:27
  • $\begingroup$ possible duplicate of Growth of functions $\endgroup$
    – Sektor
    May 7, 2014 at 5:28
  • $\begingroup$ @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. $\endgroup$ May 7, 2014 at 6:40
  • $\begingroup$ 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. $\endgroup$
    – chuy
    May 7, 2014 at 14:44
  • $\begingroup$ @chuy I still don't quiet understand. Could you say more of what I need to help Mathematica to do? $\endgroup$ May 7, 2014 at 15:07

2 Answers 2

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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.

enter image description here

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  • $\begingroup$ Thanks a lot, but how do you determine that "the second term is $O(1)$"? $\endgroup$ May 14, 2014 at 19:11
  • $\begingroup$ You're welcome. It's $\cos$ of something real, and $-1 \leq \cos(x) \leq 1$. $\endgroup$ May 14, 2014 at 19:23
  • $\begingroup$ Thanks :) Sorry for stupid question. $\endgroup$ May 14, 2014 at 19:50
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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 )

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