Working with Series: Solve Bessel Function Diff. Eq.

[king vitamin]

Homework Statement

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

$$\jmath_{1}(x) = \sum \frac{(-1)^{n}x^{2n+1}}{n!(n+1)!2^{2n+1}}$$

Show that $$\jmath_{1}(x)$$ satisfies the differential equation

$$x^{2}\jmath_{1}''(x) + x\jmath_{1}'(x) + (x^{2} - 1)\jmath_{1}(x) = 0$$

Homework Equations

Simply differentiating $$\jmath_{1}(x)$$ twice:

$$\jmath_{1}'(x) = \sum \frac{(-1)^{n} (2n+1) x^{2n}}{n! (n+1)! 2^{2n+1}}$$

$$\jmath_{1}''(x) = \sum \frac{(-1)^{n} (4n^{2}+2n) x^{2n-1}}{n! (n+1)! 2^{2n+1}}$$

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.

$$\sum \frac{(x^{2}(-1)^{n}(4n^{2}+2n)x^{2n-1})+(x(-1)^{n}(2n+1)x^{2n})+((x^{2}-1)(-1)^{n}x^{2n+1})}{n!(n+1)!2^{2n+1}}$$

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

$$\sum \frac{(-1)^{n}x^{2n+1}(4n^{2}+4n+x^{2})}{n!(n+1)!2^{2n+1}}$$

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?

 

[Dick]

You don't have to show it converges to zero. You have to show it's EXACTLY zero, i.e. the coefficient of all powers of x are zero. Try this for a warmup. The beginning of your series is x/2-x^3/16+x^5/384-x^7/18432+... Put that into the ODE and show all terms up to x^5 cancel exactly. While you are doing that you'll want to notice that the terms you've denoted as x^(2n+1), x^(2n) and x^(2n-1) can cancel among themselves. For example for n=1, they are x^3, x^2 and x^1. For n=2, they are x^5, x^4 and x^3. The x^3 can cancel between the two different n. You'll need to reindex your sums so the power of x in each is equal.