I'm somewhat new to Mathematica, and I don't understand why I'm getting inconsistent series expansions for the modified Bessel Function of first kind near $x=\infty$.

First problem: I get different expansions if I multiply the modified Bessel function by any constant:


$$ e^{-x} \left(O\left(\left(\frac{1}{x}\right)^0\right)+e^{2 x} \left(\frac{\sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^1\right)\right)\right) $$

Series[2 BesselI[0,x],{x,\[Infinity],0}]

$$ e^{-x} \left(e^{2 x} \left(\sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{x}}+O\left(\left(\frac{1}{x}\right)^{3/2}\right)\right)+\left(i \sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{x}}+O\left(\left(\frac{1}{x}\right)^{3/2}\right)\right)\right) $$

Second problem: Even for $x \in \Re$, the expansion always gives me an imaginary component, for any order of my expansion. This is clearly wrong. For example:

Series[BesselI[0, x], {x, \[Infinity], 3}]

$$ e^{-x} \left(e^{2 x} \left(\frac{\sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}+\frac{\left(\frac{1}{x}\right)^{3/2}}{8 \sqrt{2 \pi }}+\frac{9 \left(\frac{1}{x}\right)^{5/2}}{128 \sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^{7/2}\right)\right)+\left(\frac{i \sqrt{\frac{1}{x}}}{\sqrt{2 \pi }}-\frac{i \left(\frac{1}{x}\right)^{3/2}}{8 \sqrt{2 \pi }}+\frac{9 i \left(\frac{1}{x}\right)^{5/2}}{128 \sqrt{2 \pi }}+O\left(\left(\frac{1}{x}\right)^{7/2}\right)\right)\right) $$

Is there a problem with Mathematica, or I am misunderstanding how Series works? For the reference, I'm using Mathematica


  • $\begingroup$ "for $x\in\mathbb R$" - you are assuming that, but Mathematica assumes all variables are complex unless told otherwise (e.g. via Assuming[]). $\endgroup$ – J. M. will be back soon Aug 5 '16 at 19:11
  • $\begingroup$ what I mean is that, if you evaluate that expression with $x \in \Re$, you get a complex number, which is obviously wrong. And if I explicitly include "Assumptions->x>0", I still get the same expansion. $\endgroup$ – jornada Aug 5 '16 at 19:16
  • $\begingroup$ I see what you mean now after doing my own tests. That is bizarre indeed. $\endgroup$ – J. M. will be back soon Aug 6 '16 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.