# Homework Help: Problem with numerator in a series expansion

1. Jan 30, 2012

### Airsteve0

1. The problem statement, all variables and given/known data
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!

2. Relevant equations
$a_{n+2}=\frac{n(n+3)-\lambda}{R^{2}(n+2)(n+3)}$ where $a_{o}=1$

$λ=\frac{2m^{2}}{\omega_{o}^{2}}$ where m is the separation constant

3. The attempt at a solution
$a_{2n}=\frac{something}{(R^{2})^{n}(2n+1)!}$

#### Attached Files:

• ###### series.gif
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2. Jan 30, 2012

### SammyS

Staff Emeritus
Let's a least make the image more accessible.

3. Jan 30, 2012

### Airsteve0

Much better, so any ideas?

4. Jan 30, 2012

### vela

Staff Emeritus
Part (d) only asks for the recurrence relation. You don't need to get a closed form for an.

5. Jan 30, 2012

### Airsteve0

ugh now I feel dumb. Guess I should have read it most closely. Thanks though. I don't suppose you have any ideas about part (e). I realize that if r=R then the series is simply an expansion of terms that go to infinity but I am unsure of how I would truncate the series.

6. Jan 30, 2012

### vela

Staff Emeritus
You need the numerator in the recurrence relation to vanish at some point so that all the subsequent coefficients will be 0. That requirement tells you the possible values of λ.

7. Jan 30, 2012

### Airsteve0

is there a way to determine at what specific point it should vanish?

8. Jan 30, 2012

### vela

Staff Emeritus
Nope, because you're obtaining a family of solutions. If the series terminates after one term, that's one mode of oscillation. If it terminates after two terms, that's another mode of oscillations, and so on.

9. Jan 30, 2012

### Airsteve0

Oh ok, I will work at this and see what I can do. Thank you for your help!