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I Series for Elliptic Integral of the First Kind

  1. Sep 3, 2016 #1
    I'm not sure if this should go in the homework forum or not, but here we go.

    Hello all, I've been trying to find a series representation for the elliptic integral of the first kind. From some "research", the power series for the complete form (## \varphi=\frac{\pi}{2} ## or ## x=1 ##) seems to be (from the wiki page):

    ## \frac{\pi}{2}*\sum_{n=0}^\infty [\frac{(2n)!}{2^{2n}*(n!)^2}]^2 ## (eq 1) (I'll just say it now: I'm not that good with LaTeX, please excuse my mistakes).

    This is what I have for the Jacobi form (I used the variable s instead of t):

    ## \int_{0}^1 \frac{ds}{[(1-k^2*s^2)(1-s^2)]^{1/2}} = \sum_{0}^\infty \frac{g^{(n)}(0)}{(2n+1)(n!)} ## (eq 2) where

    ## g^{(n)}(0)=1/4*(2n-1)(1+k^2)^2-1/2(2n-3)(n-1)(k^2)(1+k^2) ##. (eq 3)

    I computed this by, first, letting ## s^2=t ## (don't worry; I took care of the differential; see (eq 4) below) in the Jacobi form. Then, I found the nth derivative of the integrand of the Jacobi form, except that I had the variable of integration to be linear instead of quadratic. I evaluated the nth derivative at 0 so that I could develop a Maclaurin series (I theoretically could have used some constant c, but for simplicity, I used c=0). So, we have

    Jacobi form: ## \int_{0}^1 \frac{ds}{[(1-k^2*s^2)(1-s^2)]^{1/2}} ## which is equivalent to, for ##t=s^2 ## (see footnote 1)

    ## 1/2 \int_0^1 \frac{1}{t^{1/2}} \frac{dt}{(1-k^2t)(1-t)^{1/2})} ## (eq 4) sorry, don't know why there's a black box around the expressionMod note (Mark44): I fixed the LaTeX

    Letting ## g(t)=[(1-k^2t)(1-t)]^{-1/2} ## (eq 5), we can show that

    ## g^{(n)}(0)=1/4*(2n-1)(1+k^2)^2-1/2(2n-3)(n-1)(k^2)(1+k^2)## (same as eq 3)

    So, (eq 4) is the same as

    ## 1/2*\int_{0}^1 [\frac{1}{t^{1/2}}][\sum_{0}^\infty \frac{g^{(n)}(0)*t^n}{n!}] ## (eq 5).

    Moving ## t^{(-1/2)} ## inside, multiplying by 1/2, and integrating, this gives

    ## \sum_{0}^\infty \frac{g^{(n)}(0)*t^{(n+1/2)}}{(2n+1)n!} + C ## (eq 6)

    If we back-substitute ## s^2=t ##, now we can say

    ##\sum_{0}^\infty \frac{g^{(n)}(0)*s^{(2n+1)}}{(2n+1)n!} + C ##. (eq 7)

    Here is where some confusion sets in. How do I find what C is if I'm just interested in the antiderivative? If I let s=0, I get C=s. Note that I didn't determine interval of convergence since this is probably wrong, so I probably shouldn't spend so much time applying the ratio test with such an ugly series.

    Of course, since we are dealing with the definite integral, we don't need to worry about C (but I am interested in what C would be and how to find it). So, if our integration limits for the Jacobi form are x=0 and x=1, we have

    ## \left. \sum_{0}^\infty \frac{g^{(n)}(0)*s^{(2n+1)}}{(2n+1)n!} \right|_0^1 = \sum_{0}^\infty \frac{g^{(n)}(0)}{(2n+1)(n!)} ##. (eq 8)

    What did I do wrong? Is there some manipulation that I'm not aware of that makes (eq 8) equivalent to (eq 1)? Your help is appreciated. Thank you so much!

    1: When doing this change of variables, when I had to take the square root, I just used the positive version. I'm not sure if that affects anything, but here is what it may have looked like:

    Let ##t=s^2##, hence ##dt=2sds## which implies ##\frac{dt}{2s}=ds## which is like saying ##\frac{dt}{2\sqrt(t)}=ds##, if we use the positive square root.

    2: Computation of ##g^{(n)}(0) ##:

    ## g(t)=[(1-k^2t)(1-t)]^{-1/2} ##

    ## g'(t)= -1/2*[k^2t^2-(1+k^2)t+1]^{-3/2}[2k^2t-(1+k^2)] ##

    Let ##a_t=[k^2t^2-(1+k^2)t+1]^{-3/2} ## and let ## b_t=[2k^2t-(1+k^2)] ##. It is important to note that ##(b_t)''=0 ##. Also note that ##(a_t)^{(n)}=(2n+1)g'(t) ##. This tells us that ##(a_t)^{(n-1)}=(2n-1)g'(t)## and ##(a_t)^{(n-2)}=(2n-3)g'(t)##. These substitutions made computing the nth derivative much, much, much, much less tedious. That is,

    ##g^{(n)}(t)=1/2*[(a_t)^{(n-1)}b_t+(n-1)(a_t)^{(n-2)}(b_t)'] ##. Back substituting for t and letting t=0 gives the result in (eq 3).
     
    Last edited: Sep 3, 2016
  2. jcsd
  3. Sep 4, 2016 #2
  4. Sep 6, 2016 #3
    I think you should expand

    $$\frac {1}{\sqrt {1-k^2x^2}}= \sum_{n=0}^\infty \frac {(2n)!}{2^{2n}(n!)^2} (xk)^{2n}$$

    $$K (k)= \sum_{n=0}^\infty \frac {(2n)!}{2^{2n}(n!)^2}k^{2n} \int^1_0 \frac {x^{2n}}{\sqrt {1-x^2}} dx = \sum_{n=0}^\infty \frac {(2n)!}{2^{2n}(n!)^2} \frac {\sqrt {\pi}\,\Gamma (n+1/2)}{2\,n!}k^{2n}$$

    Duplication Formula:

    $$\frac {\sqrt{\pi}(2n)!}{2^{2n}(n)!}=\Gamma\left(n+\frac{1}{2}\right).$$

    Which implies

    $$K (k)= \frac {\pi}{2}\sum_{n=0}^\infty \left[\frac {(2n)!}{2^{2n}(n!)^2} \right ]^2k^{2n}$$

    Also you can use the hypergeometric function

    https://zaidalyafeai.files.wordpress.com/2015/09/advanced-integration-techniques.pdf [Broken]

    See page 80.
     
    Last edited by a moderator: May 8, 2017
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