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Elliptic Integral

  1. Feb 25, 2011 #1

    G01

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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.
     
  2. jcsd
  3. Feb 25, 2011 #2

    fzero

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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].
     
  4. Feb 25, 2011 #3

    G01

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    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?
     
    Last edited: Feb 25, 2011
  5. Feb 25, 2011 #4

    fzero

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    I'd write the expansion as a power series, do the integrals and then check convergence.
     
  6. Mar 1, 2011 #5

    G01

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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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