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Is there a general closed form or the integral representation for the limit of the sequence: $$a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}=\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1}=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}$$ in terms of $a_0,b_0,c_0$?

$$L(a_0,b_0,c_0)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n$$

For the most simple case $a_0=b_0$ we have some interesting closed forms in terms of inverse hyperbolic or trigomonetric functions:

$$L(1,1,\sqrt{2})=\frac{1}{\ln(1+\sqrt{2})}$$

$$L(1,1,1/\sqrt{2})=\frac{2 \sqrt{2}}{\pi}$$

$$L(1,1,2)=\frac{\sqrt{3}}{\ln(2+\sqrt{3})}$$

$$L(1,1,1/2)=\frac{3 \sqrt{3}}{2 \pi }$$

$$L(1,1,3)=\frac{\sqrt{2}}{\ln(1+\sqrt{2})}$$

$$L(1,1,1/3)=\frac{2 \sqrt{2}}{3 \arcsin \left( \frac{2 \sqrt{2}}{3} \right)}$$

$$L(1,1,\sin 1)=\frac{2 \cos 1}{\pi -2}$$

$$L(1,1,\sin 1/2)=\frac{2 \cos 1/2}{\pi -1}$$

$$L(1,1,\cosh 1)=\sinh 1$$

$$L(1,1,\cosh 2)=\frac{\sinh 2}{2}$$

These values are obtained by Wolfram Alpha and Inverse Symbolic Calculator and checked with Mathematica.

This result seems familiar to me, but I'm pretty sure I don't know how to obtain a general closed form or even integral representation.

This question is closely related, but the sequence is very different.

Update:

It seems likely that the closed form for the special case $a_0=b_0=1$ is:

$$L(1,1,x)=\frac{\sinh \left(\cosh ^{-1}\left(x\right)\right)}{\cosh ^{-1}\left(x\right)}$$

However, the proof would be nice as well as the more general case.

Thanks to Sangchul Lee I now see that this sequence is exactly Carlson's transformation. Change:

$$a^2=A,\quad b^2=B,\quad c^2=C$$

$$A_{n+1}=\frac{1}{4} (A_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$

$$B_{n+1}=\frac{1}{4} (B_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$

$$C_{n+1}=\frac{1}{4} (C_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$

Accoding to Wikipedia and references wherein, we have:

$$R_F(A_{n+1},B_{n+1},C_{n+1})=R_F(A_n,B_n,C_n)$$

$$R_F(A,B,C)=\frac{1}{2} \int_0^\infty \frac{dt}{\sqrt{(t+A)(t+B)(t+C)}}$$

Which is exactly the 'closed form' I wanted.

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Yuriy S
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  • squaring and adding the constituents gives after a parametrization in polar coordinates an interesting constraint namely $1=\sin(\phi_n)\sin(\theta_n)\cos(\phi_n)+\sin(\phi_n)\cos(\theta_n)\cos(\phi_n)+\sin(\phi_n)\sin(\theta_n)\sin(\phi_n)\cos(\theta_n)$ – tired Jan 23 '17 at 21:34
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    @tired, So, this would allow us to get rid of two variables and leave only one? – Yuriy S Jan 23 '17 at 23:32
  • yes indeed!$\quad$ – tired Jan 24 '17 at 00:34
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    Denote the symmetric/antisymmetric combinations by

    $$ \delta^{\pm}{ij,n+1}=x{i,n}\pm x_{j,n} $$

    where $x_i\in {a,b,c}$ and $i\neq j$ then

    $$ \delta^{+}{ij,n+1}\delta^{-}{ij,n+1}=(\delta^+{ij,n})^2((\delta{ik,n}^+)^2-(\delta_{jk,n}^+)^2)=(\delta^+{ij,n})^2(\delta^-{ij,n})^2 $$

    which means that the problem can be reduced to

    $$ 4(a_{n+1}^2-b^2_{n+1})=a_{n}^2-b^2_{n}\ 4(a_{n+1}^2-c^2_{n+1})=a_{n}^2-c^2_{n}\ 4(c_{n+1}^2-b^2_{n+1})=c_{n}^2-b^2_{n} $$

     – tired Jan 24 '17 at 01:07
  • assume furthermore $i>j$ because there is no distiniction between $a-b$ and $b-a$ etc. The problem seems to be quite manageable now – tired Jan 24 '17 at 01:36
  • @tired, I wouldn't say it's manageable. It appears there is no closed form for the general case and the proof of the closed form for the particular case $a_0=b_0$ is not obvious to be either – Yuriy S Jan 24 '17 at 07:18
  • Ok this surprises me. The recursions look so simple now...i will see what i can do, at least for the case $a=b$ should be manageble – tired Jan 24 '17 at 10:04

2 Answers2

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Here is an observation: As in your link, if $b_0 = c_0$ then we can prove recursively that $b_n = c_n$ for all $n \geq 0$. Then plugging this to OP's recurrence relation, we find that the sequence $(a_n, b_n)$ satisfies the Schwab-Borchardt recurrence relation

$$ a_{n+1} = \frac{a_n + b_n}{2}, \qquad b_{n+1} = \sqrt{a_{n+1}b_n}. $$

So if we write $(a_0, b_0) = (a, b)$, then the limit is given by the Schwab-Borchardt mean

$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = SB(a_0, b_0) = \begin{cases} \dfrac{\sqrt{a^2 - b^2}}{\operatorname{arccosh}(a/b)}, & a > b \\ \dfrac{\sqrt{b^2 - a^2}}{\operatorname{arccos}(a/b)}, & a < b \\ a, & a = b \end{cases} \tag{*} $$

Anyway, let me write down my failed attempt to mimic Carlson's proof of $\text{(*)}$ so that future me do not repeat this mistake.

Define $I(a,b,c)$ by

$$ I(a,b,c) := \lim_{R\to\infty} \int_{-R}^{R} \frac{dx}{(x+a^2)^{1/3}(x+b^2)^{1/3}(x+c^2)^{1/3}}, $$

where $z^{1/3} = \exp(\frac{1}{3}\log z)$ with the branch cut $-\pi < \arg(z) \leq \pi$. Then we can check that $I(a, b, c)$ converges. Now using the substitution $x \mapsto 4x + ab + bc + ca$, we find that

$$ I(a, b, c) = I\left( \frac{\sqrt{(a+b)(a+c)}}{2}, \frac{\sqrt{(b+c)(b+a)}}{2}, \frac{\sqrt{(c+a)(c+b)}}{2} \right). $$

This tells us that $I(a_n,b_n,c_n) = \cdots = I(a_0,b_0,c_0)$. If we recall how the AGM is computed from Landen's transformation, this may possibly allow us to analyze $L$ using $I$.

Well, it turns out that the function $I$ has a serious issue, which is that $I$ is identically zero! This can be checked either by using the fact $I(a,a,a) = 0$ or by interpreting $I(a,b,c)$ as a value of the Schwarz–Christoffel mapping.

Sangchul Lee
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  • Very nice! I didn't know this particular class of means... – tired Jan 24 '17 at 01:40
  • @tired, I was also surprised that this has closed form, considering that AGM involves elliptic functions. – Sangchul Lee Jan 24 '17 at 01:58
  • @SangchulLee, but the sequence you gave is the same as in this answer, even though the sequences in the OP are different. Ah, I see, they are actually the same... – Yuriy S Jan 24 '17 at 12:50
  • @YuriyS, Assuming $b_0 = c_0$, you can prove recursively that $b_n = c_n$ for all $n \geq 0$ and that your recurrence relation reduces to what I wrote down. For instance, $$ a_{n+1} = \frac{\sqrt{(a_n + b_n)(a_n + b_n)}}{2} = \frac{a_n + b_n}{2} $$ and similarly $$ b_{n+1} = \sqrt{\frac{(b_n + a_n)(b_n + b_n)}{4}} = \sqrt{\frac{a_n+b_n}{2} \cdot b_n} = \sqrt{a_{n+1}b_n}. $$ – Sangchul Lee Jan 24 '17 at 12:54
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    @SangchulLee, I've read you edit and it's great - you regonized the Carlson's transformation. I'll make an addition to the OP. – Yuriy S Jan 25 '17 at 08:49
  • @YuriyS, Thank you. Thinking that the Carlson's transformation works in a very subtle way, however, my naive transformation using linear substitutions such as my one $x \mapsto 4x + ab+bc+ca$ may lead nowhere. – Sangchul Lee Jan 25 '17 at 08:56
  • @SangchulLee, well, I've read Carlson's papers and his proof was not very simple. I'll check them again, because I admire his work very much – Yuriy S Jan 25 '17 at 08:57
  • @YuriyS, Oh, I was not aware of the Carlson duplication theorem, but you are right. This seems the right ingredient for us. Thank you! I also learned from it. – Sangchul Lee Jan 25 '17 at 09:03
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I can show that the sum of the squares of the terms decreases at each step.

I think this implies convergence, but I am not completely sure.

$a_{n+1} =\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}= \frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1} =\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2} $

I'm going to play around and see if anything interesting happens, with a goal of proving convergence, maybe.

Multiplying, we get

$a_{n+1}b_{n+1}c_{n+1} =(a_n+b_n)(a_n+c_n)(b_n+c_n)/8 =(a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2)/8 $ (omitting the "_n") so that

$\begin{array}\\ \dfrac{a_{n+1}b_{n+1}c_{n+1}}{abc} &=(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + 2 + \frac{c}{b} + \frac{b}{a} + \frac{c}{a})/8\\ &=(\frac{a}{c} + \frac{c}{a} + \frac{a}{b} + \frac{b}{a} + \frac{b}{c}+ \frac{c}{b} + 2 )/8\\ &=(\frac{a}{c} -2+ \frac{c}{a} + \frac{a}{b}-2 + \frac{b}{a} + \frac{b}{c}-2+ \frac{c}{b} + 8 )/8\\ &=(g(\dfrac{a}{c}) + g(\dfrac{a}{b}) + g(\dfrac{b}{c}))/8+1 \qquad\text{where } g(x) = (\sqrt{x}-\dfrac1{\sqrt{x}})^2=x+\dfrac1{x}-2\\ \end{array} $

Note that $g'(x) =1-\dfrac1{x^2} $.

Don't know what to do with this, so I'll try something else.

$\begin{array}\\ a_{n+1}-a &=\frac{\sqrt{(a+b)(a+c)}}{2}-a\\ &=a(\frac{\sqrt{(1+b/a)(1+c/a)}}{2}-1)\\ \end{array} $

Again nothing.

Let's try this.

Since $(a+b+c)^2 \le 3(a^2+b^2+c^2) $ with equality iff $a=b=c$,

$\begin{array}\\ a_{n+1}^2+b_{n+1}^2+c_{n+1}^2 &=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\ &=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\ &=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\ &=(\frac32(a^2+b^2+c^2+2ab+2ac+2bc)-\frac12(a^2+b^2+c^2))/4\\ &=(\frac32(a+b+c)^2-\frac12(a^2+b^2+c^2))/4\\ &\le(\frac32 3(a^2+b^2+c^2)-\frac12(a^2+b^2+c^2))/4\\ &=a^2+b^2+c^2\\ \end{array} $

Therefore, if at least two of $a, b, c$ are distinct, the sum of their squares decreases.

This is not a proof of convergence, but a start.

Let's try a variation on this last result.

$\begin{array}\\ a_{n+1}^2+b_{n+1}^2+c_{n+1}^2 &=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\ &=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\ &=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\ \end{array} $

so

$\begin{array}\\ a_{n+1}^2+b_{n+1}^2+c_{n+1}^2 -(a^2+b^2+c^2) &=(-3a^2-3b^2-3c^2+3ab+3ac+3bc)/4\\ &=-3(a^2+b^2+c^2-ab-ac-bc)/4\\ &=-3(2a^2+2b^2+2c^2-2ab-2ac-2bc)/8\\ &=-3((a-b)^2+(a-c)^2+(b-c)^2)/8\\ \end{array} $

This shows precisely how the sum of squares decreases at each step.

I think that this implies convergence.

marty cohen
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