PDE Series Problem: Solving for Numerator

[Airsteve0]

Homework Statement

The problem I am having has to do with part (d) in the picture which I have attached. I have managed to get as far as to determine that the coefficients in the series expansion have the recurrence relation shown below in part (2). From this I think that I have been able to determine that the general form of the coefficients must what is shown in part (3) below. The issue is I am unsure of how to get the proper form of the numerator. Any assistance would be greatly appreciated, thanks!

Homework Equations

$a_{n+2}$=($\frac{n(n+3)-\lambda}{R^{2}(n+2)(n+3)}$)$a_{n}$ -> assume $a_{o}$=1

$\lambda$=$\frac{2m^{2}}{\omega_{o}^{2}}$ -> m is the separation constant

The Attempt at a Solution

$a_{2n}$=$\frac{something}{(R^{2})^{n}(2n+1)!}$

 

[mighty2000]

Thank you for sharing your progress and the attached picture. It looks like you are on the right track in determining the general form of the coefficients for part (d). To get the proper form of the numerator, you can use the recurrence relation you have determined and the given value for lambda.

Substituting the value for lambda into the recurrence relation, we get:

a_{n+2}=(\frac{n(n+3)-\frac{2m^{2}}{\omega_{o}^{2}}}{R^{2}(n+2)(n+3)})a_{n}

Expanding the numerator, we get:

a_{n+2}=(\frac{n^{2}+3n-\frac{2m^{2}}{\omega_{o}^{2}}}{R^{2}(n+2)(n+3)})a_{n}

Now, we can see that the numerator has a quadratic term in n, a linear term in n, and a constant term. We can simplify this further by factoring out the coefficient of the quadratic term, which is 1:

a_{n+2}=(\frac{n(n+3)-\frac{2m^{2}}{\omega_{o}^{2}}}{R^{2}(n+2)(n+3)})a_{n}=(\frac{1}{R^{2}})(\frac{n^{2}+3n-2m^{2}/\omega_{o}^{2}}{(n+2)(n+3)})a_{n}

Now, we can see that the numerator is in the form of a polynomial, and we can use the binomial theorem to expand it. The binomial theorem states that for any real number x and any positive integer n, (1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+... The numerator in your problem is in the form of (n+2)(n+3)-2m^{2}/\omega_{o}^{2}, which can be rewritten as (1+(n+2))(1+(n+3))-2m^{2}/\omega_{o}^{2}=(1+(n+2)+(n+3)+2)^{2}-2m^{2}/\omega