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I want to be able to approximate the following in closed form $$L=\sqrt{\gamma ^2+4 \pi ^2 \psi ^2} \sum _{j=0}^{\infty } \frac{\left(\frac{(2 j-1)\text{!!} (2 \pi \psi )^j}{(2 j)\text{!!} \left(\gamma ^2+4 \pi ^2 \psi ^2\right)^{j/2}}\right)^2}{1-2 j}$$ but I would also like to accelerate the series since it takes about 50 terms to converge to the desired accuracy.

This derives from $$L=\frac{2 \sqrt{\gamma ^2+4 \pi ^2 \psi ^2} E\left(\frac{2 \psi \pi }{\sqrt{\gamma ^2+4 \pi ^2 \psi ^2}}\right)}{\pi }$$

So the solution to this problem is very similar to finding an approximation for the circumference of an ellipse.

If I can accelerate the summation before hand the approximation won't need to be so precise. Any hints are greatly appreciated!

J.D'Almbert
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  • This older answer of mine may contain material of interest – Semiclassical Jul 29 '14 at 22:45
  • Also, about how large is the argument of $E(x)$ for your parameters (closer to 0 or 1?) It would also help if you stated which convention you're using for $E(x)$. – Semiclassical Jul 29 '14 at 23:02
  • $x$ can range from 0 to 1 so how would I know how large my argument of $E(x)$ is? $4 E(x)=\int_0^{2 \pi } \sqrt{1-\frac{4 \pi ^2 \psi ^2 \left(\sin ^2(\theta)\right)}{\gamma ^2+4 \pi ^2 \psi ^2}} , d\theta =4 \int_0^{\frac{\pi }{2}} \sqrt{1-x^2 \sin ^2(x)} , d\theta$ – J.D'Almbert Jul 30 '14 at 12:28
  • Well, presumably you have a sense of how big $\gamma$ is compared to $2\pi \psi$ in your problem of interest.I stress that because you'll want to use different approximations if $x\approx 0$ vs $x \approx 1$. – Semiclassical Jul 30 '14 at 12:36
  • Sorry I misjudged the situation, $x$ ergo $E(x)$ must be closer to 1 since when $\psi$ and $\gamma$ are close together, for example when $\psi=1$ and $\gamma=2$ $x \approx 0.952890514$ and when $\psi$ and $\gamma$ are further apart, for example when $\psi=10000$ and $\gamma=2$ $x\approx 0.9999999995$ – J.D'Almbert Jul 30 '14 at 12:58
  • So, to be clear, you want $x$ in the nbhd of one? The infinite series expression in $x^2$ is then particularly ill-suited; an infinite series in $1-x^2$, on the other hand, is far more appropriate. (Another way to put it is that you really want to deal with the complementary elliptic integral $E'(x)=E(\sqrt{1-x^2})$.) It would also be helpful if you could put some of that info into your question specifically so that your audience is on the same page. – Semiclassical Jul 30 '14 at 13:17
  • My point was $x$ is always in the nbhd of 1, regardless of how close or far apart $\psi$ and $\gamma$ are. I want to be able to do what you did for your older answer for this problem. – J.D'Almbert Jul 30 '14 at 13:24
  • Not if $\gamma \gg \psi$. In that case $x\approx 0$ (and in which case your series converges a good deal more quickly.) – Semiclassical Jul 30 '14 at 13:26
  • Let us continue this discussion in chat. – J.D'Almbert Jul 30 '14 at 13:28

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