M HYPE SPLASH
// updates

What is the equation for a Bessel function of order zero?

By Emily Wilson
$\begingroup$

I am currently submitting a paper in a distantly related field (experimental psychology) in which we are using a von Mises distribution to model certain aspects of perceptually-driven behavior. One of our reviewers has requested that we write out the full equation to the Bessel function of zero order, which is a component of the former. I'm happy to oblige, but I've been unable to find a clear example of this equation. Would any of you folks be so kind as to point me in the right direction?

Thanks!

$\endgroup$ 5

2 Answers

$\begingroup$

Your question should be more specific. First, don't confuse "Bessel functions" and "Modified Bessel functions": they are different. In one of each of whose two sets of functions, they can be of the "first kind" or of the "second kind" : again different sub-sets of functions. And in each one of these sub-sets, they are different Bessel functions of various order.

For examples:

The modified Bessel function of the first kind and order $0$ is $I_0(x)$. One integral definition is : $$I_0(x)=\frac{1}{\pi}\int_0^\pi \exp\left(x\cos(t)\right)dt$$

The modified Bessel function of the second kind and order $0$ is $K_0(x)$. One integral definition is : $$K_0(x)=\int_0^\infty \cos\left(x \sinh(t) \right)dt$$

Series expressions can be found in :

and related differential equation :

$\endgroup$ $\begingroup$

Perhaps is is best to give both informations. The modified Bessel functions $I_n$ of the first kind of order $n$ are solutions of the modified Bessel differential equation: $$x^2y''(x) + xy'(x)-(x^2+n^2) y(x)=0$$ Their Taylor series are $$I_n(x) = (\tfrac{1}{2}x)^n \sum\limits_{k=0}^{\infty} \frac{(\tfrac{1}{4}x^2)^k}{k!(n+k)!}$$ Note that in not only $I_0$ appears but all functions $I_n$.

But this is only the formal part, I have no clue why the $I_n$ appear in the von Mises distribution.

$\endgroup$ 2

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

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