I'm trying to show that the Bessel function of the first kind satisfies the Bessel differential equation for m greater of equal to 1.(adsbygoogle = window.adsbygoogle || []).push({});

The Bessel function of the first kind of order m is defined by

[tex]J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n} = x^m \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}(x^2)^n[/tex]

and suposedly it satisfies the Bessel differential equation, which would then write

[tex]x^2\frac{d^2 J}{dx^2} + x\frac{dJ}{dx} + (x^2-m^2)J = 0[/tex]

I'm trying to help myself in doing that by following the steps laid down in the mathworld page on this subject. There is one detail however, it is that on the site, they proceed "backward". I.e. they start by saying that the Bessel function is defined by the solution to the Bessel D.E. and then proceed to find its form.

My main trouble with the steps they take is that in equation (30), they state that a_1 = 0. But a_n, as they will find at the end, is

[tex]a_n = \frac{(-1)^n}{2^{m+2n}n!(n+m)!}[/tex]

and of course, a_1 is not 0 ! So what the greasy poop's going on here? :grumpy:

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# Bessel function and Bessel D.E.

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