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Given two positive real numbers $a$ and $b$, one can define their arithmetic-geometric mean as $M(a,b):=\ell$, where $$ (\ell,\ell)=\lim_{k\to\infty}(a_k,b_k)\quad\text{and}\quad\begin{cases}a_{k+1}=\frac{a_k+b_k}{2} \\ b_{k+1}=\sqrt{a_kb_k} \\ a_0=a,\ b_0=b.\end{cases} $$ On the other hand, one can define the elliptic integral $$ I(a,b):=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}. $$ According to this interesting blog post by Paramanand, the famous formula $$\boxed{M(a,b)I(a,b)=\frac{\pi}{2}}$$ was discovered by Gauss, thanks to some (actually very hidden) numerical evidence. Using a clever change of variables, it can be shown that $I(a,b)=I(\frac{a+b}{2},\sqrt{ab})$, from which the identity follows. Actually, Gauss' first proof was a rather tedious computation with power series.

Is there a simple, natural proof of the identity, explaining how one could guess it (without numerical evidence) in the first place?

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    $\begingroup$ the blog post mentions such a simplified derivation ("Later authors simplified the substitution [to prove $I(a,b)=I(\frac{a+b}{2},\sqrt{ab})$] into two parts...") --- is it simple enough? $\endgroup$ – Carlo Beenakker Jul 6 '17 at 10:31
  • $\begingroup$ I am interested in a proof which leads naturally to the identity $M(a,b)I(a,b)=\frac{\pi}{2}$. The proofs arguing that $I(a,b)$ has the same invariance properties as $M(a,b)$ resemble too much an a posteriori justification to me. $\endgroup$ – Mizar Jul 6 '17 at 10:50
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    $\begingroup$ @Mizar I disagree. Invariance is essentially a definition of $M(a,b)$. $\endgroup$ – Fedor Petrov Jul 6 '17 at 11:05
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    $\begingroup$ You are absolutely right, I phrased my last comment in an ambiguous way. What I dislike in those proofs is that one already knows the correct guess (i.e. the identity) and merely checks that $\frac{\pi}{2I(a,b)}$ satisfies the same "axioms" defining $M(a,b)$. I would like a simple proof which explains how the elliptic integral pops out. $\endgroup$ – Mizar Jul 6 '17 at 11:56
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Is there a simple, natural proof of the identity, explaining how one could guess it (without numerical evidence) in the first place?

IF you are asking for the intuition behind the formally proven identity, then consider the following:

  • We are all familiar with the formulae expressing the circle's area and circumference in terms of its one and only radius.

  • An ellipse is a “circle” with two (unequal) “radii”.

  • This being said, let us try and “guess”, as you've put it, the formulae for the elliptic area and “circumference”.

  • It stands to reason, then, that $A=\pi~r^2=\pi~r~r$ should generalize into $A=\pi~r_1~r_2,$ since products of lengths yield areas.

  • But what about $L=2\pi~r$ ? Should the $d=2r$ be interpreted as $d=r+r,$ yielding a possible generalization of $L=\pi~(r_1+r_2)$ ? Or, on the contrary, should the $2$ be coupled with $\pi,$ and the stand-alone r be regarded as a “mean” elliptical radius, ultimately related to an integral, which concept is intimately connected to that of area, which would then, in turn, lend credence to using $r=\sqrt{r_1~r_2}~$ here as well, as opposed to $r=\dfrac{r_1+r_2}2$ ? As it happens, ellipses themselves, and not just the people studying them, seem unable to make up their little elliptic minds on this issue, and, in their confusion (or perhaps out of a politically correct desire to reconcile both parts and offend none), they have opted neither for the former $r=AM(r_1,r_2),$ nor for the latter $r=GM(r_1,r_2),$ but for something in between, namely $r=AGM(r_1,r_2).$

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  • $\begingroup$ Isn't the circumference of an ellipse the elliptic integral of the 2nd kind, but $AGM$ is related to the elliptic integral of the 1st kind? $\endgroup$ – user49822 Mar 27 at 10:37
  • $\begingroup$ @user49822: The two types of elliptic integrals are also connected with one another, which is why we use similar terminology and mathematical expressions to describe and define them. The main reason for divergence (from the much simpler case of the circle) lies in the fact that $d(r^2)=2r~dr=(r+r)~dr,~$ whereas $d(r_1~r_2)=r_1~dr_2+r_2~dr_1\neq (r_1+r_2)~dr.$ $\endgroup$ – Lucian Apr 23 at 12:01

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