Any analytic techniques I can use to solve this problem?

I'm basically looking for some scale for which this function decays:

$$ f(x)=x^k e^{-\frac{x^2}{4\lambda^2}} $$ I'm currently trying the following method: The maximum of the function (for positive domain) is at $x_0=\sqrt{2k}\lambda$. Normalizing the function and setting it to the folding distance (such that the value of the function decays to $\frac{1}{e}$ times the maximum): $$ \frac{x^k e^{-\frac{x^2}{4\lambda^2}}}{(\sqrt{2k}\lambda)^k e^{-k/2}}=\frac{1}{e} \\ x^k e^{-\frac{x^2}{4\lambda^2}}=(\sqrt{2k}\lambda)^k e^{-\frac{k+2}{2}} \\ k\ln{x}-\frac{x^2}{4\lambda^2}=k\ln{(\sqrt{2k}\lambda)}-\frac{k}{2}-1 $$ Which I believe isn't solvable analytically... Any approximations or other techniques that could be used here?


Looking at the last equation you wrote, it seems that you want to solve for $x$ an equation looking like $$\log(x)-a x^2=b$$ The only analytical solution is given in terms of Lambert function.

Assuming that $x >0$, let $y=x^2$ making the equation $$\frac 12\log(y)-a y=b$$ and the solution is given by $$y=-\frac{1}{2 a}W\left(-2 a e^{2 b}\right)$$ Sooner or later, you will learn that any equation which can write as $A+B x+C \log(D+Ex)=0$ has solution(s) in terms of Lambert function.

Otherwise, only numerical methods would do the job.

  • $\begingroup$ Thanks! But which numerical methods could I use here? $\endgroup$ – user167289 Dec 12 '17 at 10:22
  • $\begingroup$ @user167289. Newton method would be the simplest (from far away) ... provided a reasonable guess of the root. If you want, give me some values of your parameters. Cheers. $\endgroup$ – Claude Leibovici Dec 12 '17 at 10:31

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