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Elliptic Integral 
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#1
Feb2511, 01:12 PM

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P: 2,688

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}}{12 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. 


#2
Feb2511, 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]. 


#3
Feb2511, 03:05 PM

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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? 


#4
Feb2511, 08:10 PM

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Elliptic Integral



#5
Mar111, 01:25 PM

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P: 2,688

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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