(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

First of all, I'm encountering this problem in a Calc 2/Sequences and Series class, so I havn't taken Diff. Eq's yet. Also, I've never used latex before so i'm iffy on it, but all the sigmas are from n = 0 to infinity. Here's the problem statement:

The Bessel function of order 1 is defined by

[tex]\jmath_{1}(x) = \sum \frac{(-1)^{n}x^{2n+1}}{n!(n+1)!2^{2n+1}}[/tex]

Show that [tex]\jmath_{1}(x)[/tex] satisfies the differential equation

[tex]x^{2}\jmath_{1}''(x) + x\jmath_{1}'(x) + (x^{2} - 1)\jmath_{1}(x) = 0[/tex]

2. Relevant equations

Simply differentiating [tex]\jmath_{1}(x)[/tex] twice:

[tex]\jmath_{1}'(x) = \sum \frac{(-1)^{n} (2n+1) x^{2n}}{n! (n+1)! 2^{2n+1}}[/tex]

[tex]\jmath_{1}''(x) = \sum \frac{(-1)^{n} (4n^{2}+2n) x^{2n-1}}{n! (n+1)! 2^{2n+1}}[/tex]

3. The attempt at a solution

Being very unsure of where to start, I simply plugged in the series into the equation - since they all have a common denominator I figured the answer would appear.

[tex] \sum \frac{(x^{2}(-1)^{n}(4n^{2}+2n)x^{2n-1})+(x(-1)^{n}(2n+1)x^{2n})+((x^{2}-1)(-1)^{n}x^{2n+1})}{n!(n+1)!2^{2n+1}} [/tex]

Now gathering and setting aside terms of (-1)^n and x^(2n+1)

[tex] \sum \frac{(-1)^{n}x^{2n+1}(4n^{2}+4n+x^{2})}{n!(n+1)!2^{2n+1}} [/tex]

Is this the correct way to approach this problem? Am I supposed to prove that this converges to 0 for all x (and how would I do that?)? Am I way off-base?

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# Homework Help: Working with series

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