Problem with numerator in a series expansion

In summary, the student is trying to solve a recurrence equation for a function with an unknown numerator. He is unsure of how to get a closed form for the numerator, and is seeking help from the tutor.
  • #1
Airsteve0
83
0

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


[itex]a_{n+2}=\frac{n(n+3)-\lambda}{R^{2}(n+2)(n+3)}[/itex] where [itex]a_{o}=1[/itex]

[itex]λ=\frac{2m^{2}}{\omega_{o}^{2}}[/itex] where m is the separation constant

The Attempt at a Solution


[itex]a_{2n}=\frac{something}{(R^{2})^{n}(2n+1)!}[/itex]
 

Attachments

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  • #2
Airsteve0 said:

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


[itex]a_{n+2}=\frac{n(n+3)-\lambda}{R^{2}(n+2)(n+3)}[/itex] where [itex]a_{o}=1[/itex]

[itex]λ=\frac{2m^{2}}{\omega_{o}^{2}}[/itex] where m is the separation constant

The Attempt at a Solution


[itex]a_{2n}=\frac{something}{(R^{2})^{n}(2n+1)!}[/itex]
Let's a least make the image more accessible.
attachment.php?attachmentid=43285&d=1327929002.gif
 
  • #3
Much better, so any ideas?
 
  • #4
Part (d) only asks for the recurrence relation. You don't need to get a closed form for an.
 
  • #5
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
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
is there a way to determine at what specific point it should vanish?
 
  • #8
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
Oh ok, I will work at this and see what I can do. Thank you for your help!
 

1. What is a numerator in a series expansion?

The numerator in a series expansion refers to the top number in a fraction that represents the number of equal parts being considered.

2. What is the problem with the numerator in a series expansion?

The problem with the numerator in a series expansion is that if it is too large, it can cause the series to diverge and become infinite, making it difficult to accurately calculate the sum of the series.

3. How can the problem with the numerator be solved?

The problem with the numerator can be solved by using methods such as truncating the series, using convergence tests, or adjusting the series to make it more manageable.

4. How does the numerator affect the convergence of a series?

The numerator plays a crucial role in determining the convergence of a series. A larger numerator can make the series diverge, while a smaller numerator can lead to faster convergence.

5. How can a scientist determine the appropriate numerator for a series expansion?

A scientist can determine the appropriate numerator for a series expansion by considering the desired accuracy and using convergence tests to determine the optimal value for the numerator.

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