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Dear Reader:

Let $K(k)$ and $E(k)$ be elliptic integrals of respectively the first and second kind, where $k$ is the elliptic modulus and $k'=\sqrt{1-k^2}$ is the complementary elliptic modulus.

I happened to encounter the following numerical "fact" (when solving an engineering problem regarding energy conversion):

When I chose a $k$ such that $K(k) \[ K(k')-E(k') \] = \pi/2$, then seemingly $K(k)/K(k')$ is quite close to $\pi/4$, if not exactly. I wonder whether there is an expansion like $K(k)/K(k')=\pi/4+(\text{small terms})$ for this particular $k$. I am just curious. Does someone know?

Thank you!

Best regards,

Hiroshi Okamoto

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2 Answers 2

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Given the Legendre relation, your question is equally about K - E. This is a difference of hypergeometric function values (see http://en.wikipedia.org/wiki/Elliptic_integral for all of this). You seem to be setting a condition on k that also is simpler when read out of the Legendre relation, on E and K'. I would think the truth would come out of the power series in k, though I haven't looked at details.

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  • $\begingroup$ Dear Dr. Matthews, Thank you for your helpful comment. Yes, the condition on $k$ could be simplified to $E(k)K(k')=\pi$. On the other hand, the first few terms in the power series of $E(k)$ and $K(k)$ results in $E(k)K(k)=\pi^2/4+O(k^4)$. Dividing the latter expression by the former, I obtained $K(k)/K(k') \sim \pi/4$, as desired. Would you mind if I mention your name in a paper I am writing up? Anyways, thanks again. Hiroshi Okamoto $\endgroup$
    – a guy on the street
    Commented Apr 8, 2011 at 6:18
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Note also an inequality from NIST book, page 494: $$ \ln\frac{(1+\sqrt{k'})^2}{k} < \frac{\pi K(k')}{2K(k)} < \ln\frac{2(1+k')}{k}. $$

On this page there are some references to more exact results.

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