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The legendre's relation can be stated as follows

$$ K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) = \frac{\pi}{2} $$ where $k^* = \sqrt{1 - k^2}$ is the complimentary modulus, and $E$ and $K$ are respectively complete elliptic integral's of the first and second kind

$$ K(k) = \int_0^{\pi/2} \frac{\mathrm{d}\theta}{\sqrt{1-k^2 \sin^2\theta}}\ ,\quad E(k) = \int_0^{\pi/2}\sqrt {1-k^2 \sin^2\theta}\ \mathrm{d}\theta\,. $$

Now sorry for not attempting to solve this problem myself, but I have tried both myself and finding sources online. Alas it seems this relation has been somewhat forgotten. I did however find an article claiming to show the relation, but I do not have access to check it's validity.

Can someone provide sources for a proof of this relation, or outline a proof? Hopefully not using hypergeometric functions, but a proof more in the spirit of Legendre. Any help would be greatly appreciated.

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3 Answers 3

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Here is an outline of a proof:

First use the integral definitions to show that $$\frac{\mathrm d E(k)}{\mathrm dk} = \frac{E(k)-K(k)}{k} \tag{1}$$

and $$ \frac{\mathrm d K(k)}{\mathrm d k} = \frac{E(k) - (k^{*})^{2} K(k)}{k(k^{*})^{2}} \tag{2}.$$

Next use those identities to show that $$ \frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \Big) =0 .$$

This will show that $K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) $ is a constant.

Finally take the limit as $k$ goes to $0$ from the right to show that the constant is $ \frac{\pi}{2}$.

EDIT:

The second identity is not obvious. See the question here.

SECOND EDIT:

I'm going to fill in the gaps a bit.

If we differentiate $E(k^{*})$ or $K(k^{*})$ with respect to $k^{*}$, we'll get similar formulas to the ones above since all we're really doing is relabeling.

Using the chain rule, we get $$ \begin{align} \frac{\mathrm d E(k^{*})}{ \mathrm dk} &= \frac{\mathrm d E(k^{*})}{\mathrm dk^{*}} \frac{\mathrm d k^{*}}{\mathrm dk} = \frac{E(k^{*})-K(k^{*})}{k^{*}} \frac{-k}{\sqrt{1-k^{2}}} \\ &= \frac{E(k^{*})-K(k^{*})}{k^{*}} \frac{-k}{k^{*}} = -\frac{k}{(k^{*})^{2}} \Big( E(k^{*})-K(k^{*}) \Big) , \end{align}$$

and

$$ \begin{align} \frac{\mathrm d K(k^{*})}{\mathrm dk} &= \frac{\mathrm d K(k^{*})}{\mathrm dk^{*}} \frac{\mathrm d k^{*}}{\mathrm dk} = \frac{E(k^{*}) - k^{2} K(k^{*})}{k^{*} k^{2}} \frac{-k}{k^{*}} \\ &= - \frac{1}{k(k^{*})^{2}} \left( E(k^{*}) - k^{2} K(k^{*}) \right). \end{align}$$

Therefore,

$$ \begin{align} &\frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \Big) \\ &= \frac{E(k) - (k^{*})^{2}K(k)}{k (k^{*})^{2}}E(k^{*}) - K(k) \frac{k}{(k^{*})^{2}} \Big( E(k^{*})-K(k^{*}) \Big) \\ &+ \frac{E(k)-K(k)}{k} K(k^{*}) - E(k) \frac{1}{k(k^{*})^{2}} \left( E(k^{*}) - k^{2} K(k^{*}) \right) \\ &- \frac{E(k) - (k^{*})^{2}K(k)}{k (k^{*})^{2}} K(k^{*}) + K(k) \frac{1}{k(k^{*})^{2}} \left( E(k^{*})-k^{2}K(k^{*}) \right) \\ &= E(k) E(k^{*}) \left( \frac{1}{k (k^{*})^{2}} - \frac{1}{k (k^{*})^{2}} \right) + K(k) E(k^{*}) \left(- \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}}\right) \\ &+ K(k) K(k^{*}) \left( \frac{k}{(k^{*})^{2}} - \frac{1}{k} + \frac{1}{k} - \frac{k}{(k^{*})^{2}}\right) + E(k) K(k^{*}) \left( \frac{1}{k} + \frac{k}{(k^{*})^{2}} - \frac{1}{k (k^{*})^{2}} \right) \\ &= K(k) E(k^{*}) \left(- \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}}\right) - E(k) K(k^{*}) \left( -\frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k (k^{*})^{2}} \right) . \end{align}$$

But since

$$ - \frac{1}{k} - \frac{k}{(k^{*})^{2}} + \frac{1}{k(k^{*})^{2}} = \frac{-(k^{*})^{2} - k^{2} +1}{k(k^{*})^{2}} = \frac{-(1-k^{2})-k^{2}+1}{k(k^{*})^{2}} = 0,$$

we have

$$ \frac{\mathrm d}{\mathrm dk} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \Big) =0. $$

And

$$ \begin{align} \lim_{k \to 0^{+}} \Big( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \Big) &= \frac{\pi}{2} (1) + \lim_{k \to 0^{+}} \Big( E(k) -K(k) \Big) K(k^{*}) \\ & \overset{(3)}{=} \frac{\pi}{2} + \lim_{k \to 0^{+}} \mathcal{O}(k^{2}) \left(-\ln (k) + \mathcal{O}(1) \right) \\ &= \frac{\pi}{2} + 0 \\ &= \frac{\pi}{2}. \end{align}$$


$(3)$ The answer here shows that $K(k) = -\log(k^{*})+ \mathcal{O}(1) $ as $k \to 1^{-}$. Replace $k$ with $k^{*}$.

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  • $\begingroup$ Nice RV.I was looking for a proof :) $\endgroup$ May 27, 2014 at 0:49
  • $\begingroup$ @ZaidAlyafeai Thanks. I just started reading about the Jacobi elliptic functions (the inverse functions of elliptic integrals). Previously I was only familiar with the Weierstrass elliptic functions. $\endgroup$ May 27, 2014 at 1:08
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    $\begingroup$ Woow,Good luck with that. I was thinking of reading about Jacobi theta functions. Things are getting very advanced easily. $\endgroup$ May 27, 2014 at 1:24
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Although the proof presented by User Random Variable is enough, I would like to provide a separate proof.

This proof does not expect you to already know Legendre's Relation.

(As I read in a paper, Legendre himself accidentally stumbled upon this relation by deriving it for $k=1/\sqrt{2}$ and then proved it by differentiating both sides and showing it is a constant to derive the relation.)

When studying functions it is common to study their Differential Equations as well, so we start with the D.E. for $K$

$$\frac{d}{dk}\left[k(1-k^2)\frac{d}{dk}f(k)\right]-kf(k)=0$$

then one can verify that $f(k)=K'$ is also a solution to this.

First replace $f(x)$ by $K$ and multiply the whole equation by $K'$ and repeat the same other way, then subtract the two resulting equations which gets us,

$$K'\frac{d}{dk}\left[k(1-k^2)\frac{d}{dk}K\right]-K\frac{d}{dk}\left[k(1-k^2)\frac{d}{dk}K'\right]=0$$

Our goal here is to write the Left Hand Side as a single differential.

Now it can be shown that,

$$[hf']'g-[hg']'f=[f'gh-g'fh]'$$

Applying this results in,

$$\frac{d}{dk}\left[k(1-k^2)K'\frac{d}{dk}K-k(1-k^2)K\frac{d}{dk}K'\right]=0$$

One can now use Derivatives of $K$ and $K'$ to write it as:

$$\frac{d}{dk}(EK'+E'K-KK')=0$$

We now quite naturally have arrived at Legendre's Relation.

One can find the constant as based on the answer by Random Variable.

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We set shortened notations $D^nf(k)$ for $\text{d}^nf(k)/\text{d}k^n$, and consider the Wronskian: $$ W=\text{det}\begin{pmatrix} D^0K & D^0K^\prime\\ D^1 K & D^1K^\prime \end{pmatrix}. $$ And $$ D^1W=\text{det}\begin{pmatrix} D^0K & D^0K^\prime\\ D^ 2K & D^2K^\prime \end{pmatrix}. $$ Since $K,K^\prime$ satisfies the same second-order differential equation, namely $\mathcal{L}K=\mathcal{L}K^\prime=0$, where $\mathcal{L}=k(k^2-1)D^2+(3k^2-1)D^1+kD^0$, after breaking the determinant and applying the linear relation, one gets $$ D^1W=\frac{3k^2-1}{k(1-k^2)} W. $$ From this we predict $W=\frac{C}{k(1-k^2)}$ for a certain constant $C$. Expanding $W$ using normal derivatives of $K$, it turns out that $$ KE^\prime+K^\prime E-KK^\prime=-C. $$ This gives an explanation on how @Miracle Invoker's method did work and at the meantime, could be generalized to $n$-orders by considering $n\times n$ Wronskians.

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