3
$\begingroup$

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

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$ Apr 8, 2011 at 6:18
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.