2

I'm aware of the approximation $K_1(x)\approx {\frac{1}{x}}$, for $x\ll1.$

Is there something similar that can be used for $K_o(x)$?

N A
  • 75
  • 1
    Maple gives $K_0(x) \approx \ln \frac{1}{x} + \ln 2 - \gamma$ for small $x$. You may get more terms in Maple. – River Li Jan 30 '21 at 04:33
  • @River Li, thanks for the help. Is $\gamma$ a variable here? I don't know much about Maple - could you point me in the direction of a resource that would help me learn how to get this info from it? – N A Jan 30 '21 at 06:54
  • 1
    $\gamma$ is the Euler–Mascheroni constant. Perhaps you can find the result in literature. You can do it in https://www.wolframalpha.com/. Input: asymptotic expansion Besselk[0,1/x] – River Li Jan 30 '21 at 08:06
  • This works well, thanks. It would be perfect for what I need, except that for x>1, ln(1/x) becomes negative instead approaching zero like Ko(x). I wonder if there's a straight forward way to help with that. The only thing I can think of is a heaviside function but that would just make things more complicated again. – N A Feb 12 '21 at 00:39
  • It depends on the region of $x$: small $x$, large $x$, $x$ near 1, etc. – River Li Feb 12 '21 at 01:25
  • It works well for x<=1. For x>1 the approximation becomes negative, which causes problems for my equation. If there was something that made $ln \frac{1}{x} = 0$ for $x>1$ I might be able to use that. – N A Feb 12 '21 at 05:05
  • 1
    It only works for small $x$. If $x$ is large, we have $K_0(x) \approx \frac{1}{2}\sqrt{\frac{2\pi}{x}}\mathrm{e}^{-x}$. You need to figure out the region of $x$. – River Li Feb 12 '21 at 06:37
  • @RiverLi Thanks. That one works well for x>1. My answer is most sensitive in the approximate range $1e-2$ $$ $1e2$, so the approximation for the larger x doesn't quite do the trick. If that first approximation trended towards 0 instead of become negative for x>1 it'd be perfect. Reminds me of goldilocks and the three bears...Both approximations you provide are good ones though. Do you want to post them as answers to the question? – N A Feb 13 '21 at 09:26
  • It is fine as a comment. – River Li Feb 13 '21 at 13:12

0 Answers0