# Showing the bessel function is entire

• Sep 23rd 2013, 01:26 AM
99.95
Showing the bessel function is entire

I guess just by looking at it, you can tell it's always defined and thus always differentiable.

However, I'm not precisely sure how to show this? I was thinking perhaps applying the Louiville's theorem.

Do I have to evaluate the sum, and show that it converges? I really could use some direction, as I am unsure which theorems to apply,
or further yet, how to apply them to a sum.

As for the second question, I tried applying the quotient rule, but things got really messy and I couldn't get the answer. Was this the incorrect method, or just me making a silly mistake somewhere in my working? Surely there is a way to simplify the expression.

Any help is very much appreciated!

Thanks
• Sep 23rd 2013, 03:18 AM
chiro
Re: Showing the bessel function is entire
Hey 99.95.

You could show that the series converges by say a ratio test. If you show convergence for all z then you have shown its entire.

As for showing that it satisfies the DE, do a term by term differentiation twice and then show that all terms of the infinite series cancel out when you substitute the derivatives in with the DE equation. (Remember d/dz z^n = nz^(n-1) even when z is complex).
• Sep 26th 2013, 10:30 AM
99.95
Re: Showing the bessel function is entire
Moment of clarification,
in the given sum/function, are both j and x treated as variables, i.e will I need to do partial differentiation, or, are j and n parameters and x is the only variable?
• Sep 26th 2013, 04:32 PM
chiro
Re: Showing the bessel function is entire
Only z is a true variable: you must fix the order value n to be some constant.