Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I mean that

$$ K_{a} (x)= CJ_{a}(ix).$$

Here $C$ is a complex number, and $a$ is real.

So is the Macdonald function a Bessel function in disguise (or proportional) of complex argument??

share|improve this question

1 Answer 1

up vote 2 down vote accepted

In fact $$K_a(x) = \frac\pi2 i^{a+1} H^{(1)}_a(ix) = \frac\pi2 i^{a+1} [J_a(ix) + i Y_n(i x)]$$ so it is closer related to the Hankel $H^{(1)}$ than to the Bessel function (of course the Hankel function is just a linear combination of the two Bessel functions $J$ and $Y$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.