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It can be easily checked that both the complete elliptic integrals $K(k), K'(k)$ satisfy the same second order differential equation $$kk'^{2}\frac{d^{2}y}{dk^{2}} + (1 - 3k^{2})\frac{dy}{dk} - ky = 0$$ and hence from the theory of second order differential equations there is a relation of the form $$K'(k) = cK(k)\cdot\log k + f(k)$$ where $c$ is some constant and $f(k)$ is some analytic function of $k$. The exact relation between $K(k)$ and $K'(k)$ is given by $$K'(k) = \frac{2K(k)}{\pi}\log\left(\frac{4}{k}\right) - 2\left[\left(\frac{1}{2}\right)^{2}\left(\frac{1}{1\cdot 2}\right)k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\left(\frac{1}{1\cdot 2} + \frac{1}{3\cdot 4}\right)k^{4} + \cdots\right]$$ It can be verified with some patience that the RHS does satisfy the differential equation given above and thereby the relation between $K'(k)$ and $K(k)$ can be established.

However is there an alternative proof based on the definition of $K'(k)$ and $K(k)$ as complete elliptic integrals or using the hypergeometric relation $$\frac{2K(k)}{\pi} =\,_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right)$$ which can be presented to someone unaware of the theory of second order differential equations?

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  • $\begingroup$ Think $\frac{\mathrm{i}K'}{K}=\tau=\frac{\log q}{\pi\mathrm{i}}$ and consider that $k^2 = 16\,q + \mathrm{O}(q^2)$ which allows local inversion. Is that what you are looking for? $\endgroup$ – ccorn Jul 13 '13 at 11:50
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    $\begingroup$ What is $kk'^2$ in the diff eq, before $\frac{d^2y}{dk^2}$ ? I don't know elliptic integrals $\endgroup$ – vesszabo Jul 13 '13 at 12:59
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    $\begingroup$ @vesszabo: $k'^2=1-k^2$, and $K'(k)=K(k')$. $\endgroup$ – ccorn Jul 13 '13 at 13:51
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Let $\tau$ be a suitable period ratio with positive imaginary part (suitable meaning that $\tau$ is in a certain fundamental domain) and let $q=\mathrm{e}^{\pi\mathrm{i}\tau}$ be the nome. Then $$\frac{\mathrm{i}K'}{K} = \tau = \frac{1}{\pi\mathrm{i}}\log q$$ Now consider that $$k^2 = \frac{\vartheta_2^4(q)}{\vartheta_3^4(q)} = 16 q + \mathrm{O}(q^2)$$ where $\vartheta_2$, $\vartheta_3$ are known Jacobi thetanull functions. The relationship between $k^2$ and $q$ allows local inversion, so $q = \frac{k^2}{16}\left(1+\mathrm{O}(k^2)\right)$, hence $$\frac{\mathrm{i}K'}{K}=\frac{2}{\pi\mathrm{i}}\log\frac{k}{4}+\mathrm{O}(k^2)$$ which is the sought relation.

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  • $\begingroup$ ccorn, I would like to get a proof of the formula for $K'$ with full expansion in series (as given in my question) of powers of $k$ not just as $O(k^{2})$. I am wondering if there is a proof available without recourse to the theory of differential equations. $\endgroup$ – Paramanand Singh Jul 13 '13 at 19:08
  • $\begingroup$ Your formula can be seen as a special case of ${}_2F_1$ series around $z=1$, logarithmic case, but I see that you want a nice proof. $\endgroup$ – ccorn Jul 14 '13 at 8:11

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