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Let $a,b$ be positive real numbers and we define two sequences $\{a_{n}\},\{b_{n}\}$ as follows: $$a_{0}=a,b_{0}=b,a_{n+1}=\frac{a_{n}+2b_{n}}{3},b_{n+1}=\sqrt[3]{\frac{(a_{n}^{2}+a_{n}b_{n}+b_{n}^{2})b_{n}}{3}}\tag{1}$$ Then it is not so difficult to show that both the sequences $\{a_{n}\},\{b_{n}\}$ tend to a common limit which may be denoted by $M_{3}(a,b)$, the cubic arithmetic geometric mean of $a$ and $b$. It also turns out that there is an integral representation of this cubic arithmetic geometric mean. Let us define $$I(a,b)=\int_{0}^{\infty}\frac{t\,dt}{\sqrt[3]{(t^{3}+a^{3})(t^{3}+b^{3})^{2}}}\tag{2}$$ Then we have the relation $$M_{3}(a, b) = \frac{I(1, 1)}{I(a, b)}\tag{3}$$ The above result $(3)$ can be proved easily if we can show that $$I(a_{n},b_{n})=I(a_{n+1},b_{n+1})\tag{4}$$ Given the definition $(2)$ of integral $I(a,b)$ how do I go about proving the identity $(4)$?

Paramanand Singh
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  • How did you figure out the relation in the first place? There is no idea for the proof? I also wouldn't call this function a 'mean' since it's obviously asymmetric. But great result nonetheless – Yuriy S Aug 18 '16 at 09:55
  • @YuriyS: This cubic analogue of arithmetic-geometric mean is the development of Borwein brothers. I got these results from http://www.cecm.sfu.ca/organics/vault/comp_aid_analy/paper/html/node5.html and I wanted to have an easy / elementary proof of $(4)$ which leads to a proof of $(3)$. From this we can see that $I(a,b) = I(b, a)$ and hence $M_{3}(a, b) = M_{3}(b, a)$ so that it is symmetric. Moreover their usefulness comes in establishing a theory of elliptic/theta functions to base $3$. The default theory is for base $2$ where usual arithmetic-geometric mean plays a role. – Paramanand Singh Aug 18 '16 at 11:09
  • @YuriyS: You can see the traditional theory in my blog posts (https://paramanands.blogspot.in/p/archives.html). Search for AGM, elliptic integrals, theta functions etc. – Paramanand Singh Aug 18 '16 at 11:12
  • Thank you for the links. When you state no references in your question the impression is that you obtained these results yourself. Which is why I was confused. On a related note, I would link this question, which is also a cubic generalization of AGM, but symmetric and for three numbers. So, this seems to be a particular case of that. See the answer there, it links to a paper – Yuriy S Aug 18 '16 at 11:36
  • @YuriyS: Sorry for the confusion. I am still a learner of mathematics and not a creator. And yes I should have added links for more context. Will take care in future. And thanks for that related question. I will have a look. – Paramanand Singh Aug 18 '16 at 12:59
  • See specifically this paper about the generalization http://www.sciencedirect.com/science/article/pii/S0022314X06002058 – Yuriy S Aug 21 '16 at 17:04

1 Answers1

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I worked out a solution using only substitutions.

First we transform $I(a,b)$ into an elliptic integral.
$$ I(a,b)=\int_0^{\infty}\frac{t\,\mathrm dt}{\sqrt[3]{(t^3+a^3)(t^3+b^3)^2}} =\frac{1}{a}\int_0^{\infty}\frac{u\,\mathrm du}{\sqrt{u^6+\left(\frac{4b^3}{a^3}-2\right)u^3+1}}\\ =\frac{1}{a}\int_0^1\frac{(v+1)\,\mathrm dv}{\sqrt{v^6+\left(\frac{4b^3}{a^3}-2\right)v^3+1}} =\frac{1}{a}\int_0^{\infty}\frac{\mathrm dz}{\sqrt{\frac{4b^3}{a^3}z^3+9z^2+6z+1}},\\ u=\frac{t}{a}\sqrt[3]{\frac{t^3+a^3}{t^3+b^3}},\ z=\frac{v}{(1-v)^2}. $$ Now to prove $\displaystyle I(a,b)=I\left(\frac{a+2b}{3},\sqrt[3]{\frac{a^2b+ab^2+b^3}{3}}\right)$ we only need to show that $$ \frac{1}{a}\int_0^{\infty}\frac{\mathrm dz}{\sqrt{\frac{4b^3}{a^3}z^3+9z^2+6z+1}} =\frac{3}{a+2b}\int_0^{\infty}\frac{\mathrm dx}{\sqrt{\frac{36(a^2b+ab^2+b^3)}{(a+2b)^3}x^3+9x^2+6x+1}} $$ and we can use the transformation $$ z=\frac{3 a x \left(3(a-b)^2x^2+(3(a + b)x+2 a+4 b)^2\right)}{4 (a+2 b) (3 b x+a+2 b)^2} $$ which is of order 3.

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