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Elliptic Integral

by G01
Tags: elliptic, integral
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G01
#1
Feb25-11, 01:12 PM
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1. The problem statement, all variables and given/known data

Sub problem from a much larger HW problem:

From previous steps we arrive at a complete elliptic integral of the second kind:

[tex]E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x}[/tex]

In the next part of the problem, I need to expand this integral and approximate it by truncating at the first order term. (k is large)

2. Relevant equations

[tex]E(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n)!}{2^{2 n} (n!)^2}\right]^2 \frac{k^{2n}}{1-2 n}[/tex]

3. The attempt at a solution

I believe I should use the expansion quoted above.

Here is my question. Based of the previous steps I know that k^2 has to be large. Also, the sign of k^2 is opposite of what is is in the standard form of E(k).

So, 1. Does this expansion truncated at first order approximate the integral well if k^2 is large?

I think not. Is there another expansion, one for large k^2, that I can potentially use?

2. Can I just change the sign in the odd terms of the expansion to account for the sign change of k?

I think this should work.
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fzero
#2
Feb25-11, 01:17 PM
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The expansion above is not valid for large [tex]k[/tex]. Simply rewrite

[tex]
E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x} = \pm \int_0^{\pi/2} dx k \sin x\sqrt{1+\frac{1}{k^2\sin^2x}}
[/tex]

and expand this in [tex]1/k[/tex] yourself. Note that taking the square root introduces a sign ambiguity that should be chosen to give the correct sign to [tex]E(k)[/tex].
G01
#3
Feb25-11, 03:05 PM
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Quote Quote by fzero View Post
The expansion above is not valid for large [tex]k[/tex]. Simply rewrite

[tex]
E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x} = \pm \int_0^{\pi/2} dx k \sin x\sqrt{1+\frac{1}{k^2\sin^2x}}
[/tex]

and expand this in [tex]1/k[/tex] yourself. Note that taking the square root introduces a sign ambiguity that should be chosen to give the correct sign to [tex]E(k)[/tex].
Hmm. This makes sense, however, it is only a good approximation when sin(x) is larger than 1/m.

However, the contributions to the integral in [0,1/m] are small, so can I fix the approximation by changing the lower bound on the integral to 1/m?

fzero
#4
Feb25-11, 08:10 PM
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Elliptic Integral

Quote Quote by G01 View Post
Hmm. This makes sense, however, it is only a good approximation when sin(x) is larger than 1/m.

However, the contributions to the integral in [0,1/m] are small, so can I fix the approximation by changing the lower bound on the integral to 1/m?
I'd write the expansion as a power series, do the integrals and then check convergence.
G01
#5
Mar1-11, 01:25 PM
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Yeah you were right. The higher order terms don't necessarily converge with a lower bound of 1/m.

Checked with my prof. Looks like if you use 1/m as the lower bound and expand to first order, you get the correct result up to a multiplicative constant on the resulting ln(m) term.

I think can finish the rest of the problem now. Thanks fzero!


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