Modified Bessel Equation

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SUMMARY

The discussion focuses on the asymptotic behavior of modified Bessel equations, specifically for small values of x approaching 0 with complex order α. The user references the Wikipedia page on Bessel functions, which provides a limit for |z| << √(α + 1). The approximation for the modified Bessel function of the first kind is given as Iα(z) ≈ (1/Γ(α + 1))(z/2)α, treating α as an analytic function for small z. The user seeks clarification on the limits for Kα when α is complex, interpreting that α must not equal 0 for the function to remain analytic.

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thatboi
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Hey all,
I wanted to know if anyone knew somewhere I could find the asymptotic behavior for small x (i.e x approaching 0) limit of the modified Bessel equations with complex order. The wikipedia page for Bessel functions (https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_Iα,_Kα) provides a limit when ##|z| \ll \sqrt{\alpha+1}## but in the case of complex-valued ##\alpha## I am not sure how to interpret this bound.
Any assistance would be appreciated.
 
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They give

##I_\alpha(z) \approx \frac{1}{\Gamma(\alpha+1)}\left(\frac{z}{2}\right)^\alpha##

just treat this as an analytic function of ##\alpha## for small z.
 
Paul Colby said:
They give

##I_\alpha(z) \approx \frac{1}{\Gamma(\alpha+1)}\left(\frac{z}{2}\right)^\alpha##

just treat this as an analytic function of ##\alpha## for small z.
For the ##K_{\alpha}## limits, it is defined for ##\alpha=0## and ##\alpha>0##, so for the case of complex ##\alpha##, does this just mean ##\alpha\neq 0##?
 
That’s the way I read it. ##z=0## is a regular singular point and ##\alpha## is just a parameter. So it should be analytic.
 

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