A series

1. May 21, 2008

Kb1jij

I just finished a final in my differential equations class. One of the problems had me solve a second order homogeneous differential equation using series. I boiled it down to this recursion relation:

$$a_{n+2}=\frac{(n+3)a_{n}}{2(n+2)(n+1)}$$

I found that the even coefficients work out nicely to the following sum:

$$y=a_{0}+\Sigma^{\infty}_{n=2}\frac{(n+1)a_{0}}{4*6^{n-2}}x^{n}$$

I couldn't get a nice result for the odd coefficients and still can't find one. It's kind of bothering me now. Is it even possible? I can boil it down to this series:

$$\frac{1}{3},\frac{1}{20},\frac{1}{210},\frac{1}{3024},\frac{1}{55440},...$$

2. May 21, 2008

Kb1jij

Wow, never mind that recursion relation is wrong. Silly me.

If anyone feels like it they are still welcome to find an expression for that sequence

3. May 21, 2008

Crosson

Putting this recursion relation in closed form requires the use of the double factorial function (or equivalently, a half integer gamma function). Here is my solution for the odd coefficients:

$$a_{n,odd} = a_1 \frac{2^{-\frac{n}{2}-\frac{3}{2}} \left(-1+(-1)^n\right) (n+1) }{n (n-2)!!}$$

You can see the definition of the double factorial here:

http://mathworld.wolfram.com/DoubleFactorial.html

It is just like a regular factorial, instead of steps of one it decrements in steps of 2.

The first time I solved a series problem with a double factorial it took me 4 hours of really difficult thinking. After a bit of practice you can get used to how the 2^n factors combine with the odd integer double factorial, but it's a mind bender at first.