# What's the value of this Vieta-style product involving the golden ratio?

One way of looking at the Vieta product $${2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$ is as the infinite product of a series of successive 'approximations' to 2, defined by $a_0 = \sqrt{2}$, $a_{i+1} = \sqrt{2+a_i}$ (or more accurately, their ratio to their limit 2). This allows one to see that the product converges; if $|a_i-2|=\epsilon$, then $|a_{i+1}-2|\approx\epsilon/2$ and so the terms of the product go as roughly $(1+2^{-i})$.

Now, the sequence of infinite radicals $a_0=1$, $a_{i+1} = \sqrt{1+a_i}$ converges exponentially to the golden ratio $\phi$, and so the same sort of infinite product can be formed: $$\Phi = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$ and an equivalent proof of convergence goes through. The question is, what's the value of $\Phi$? The usual proof of Vieta's product by way of the double-angle formula for sin doesn't translate over, and from what I know of the logistic map it seems immensely unlikely that there's any function conjugate to the iteration map here in the same way that the trig functions are suitably conjugate to the version in the Vieta product. Is there any other approach that's likely to work, or is $\Phi$ just unlikely to have any formula more explicit than its infinite product?

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FWIW, attempts to find `0.509490972847535755...` in the Plouffe inverter yielded nothing. – J. M. Oct 7 '10 at 1:27
As to how I got that number in Mathematica: `SequenceLimit[FoldList[Times, 1, NestList[Sqrt[1 + #] &, N[1, 50], 50]/GoldenRatio]]`. – J. M. Oct 7 '10 at 1:28

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Yeah; I should've mentioned that I did a quick lookup on the value myself to little avail. My wild guess for the value would be along the lines of a hypergeometric function (or conceivably a theta function) evaluated at some Q[sqrt(5)] argument, possibly with some exponential factor, but that's purely speculation. I'm not even sure how to attack it. – Steven Stadnicki Oct 7 '10 at 3:28

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– J. M. Oct 7 '10 at 4:51

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I feel like the key to solving this problem will be similar to the ideas discussed in the article posted by J.M. If that is the case, then the function $f(z)=\frac{\sqrt{1+z}}{\phi} \cdot \frac{\sqrt{1+\sqrt{1+z}}}{\phi}\cdots$ should be of some interest. – Eric♦ Feb 24 '11 at 0:20
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## 1 Answer

What you're basically looking for is a function $f(x)$ such that $f(2x)=f^2(x)-1$ and $f(0)=\phi$, from there: \begin{align} 2f'(2x)&=2f(x)f'(x)\\ \frac{f'(2x)}{f'(x)}&=f(x)\\ \frac{f'(x)}{f'(x/2)}&=f(x/2)\\ \frac{f'(x)}{f'(x/2^n)}&=\prod_{k=1}^n f(x/2^k) \end{align} and, given a value $x_0$ such that $f(x_0)=1$, \begin{align} \Phi&=\prod_{k=1}^{\infty} \frac{f(x_0/2^k)}{\phi}\\ &=\lim_{n\rightarrow\infty}\phi^{-n} \prod_{k=1}^n f(x_0/2^k)\\ &=\lim_{n\rightarrow\infty}\phi^{-n} \frac{f'(x_0)}{f'(x_0/2^n)}\\ &=\lim_{h\rightarrow0}h^\alpha \frac{f'(x_0)}{f'(hx_0)} \end{align}

where $\alpha=\frac{ln(\phi)}{ln(2)}$. Unfortunately, I have no idea how to get $f(x)$, and the fact that $f(x)=1+O(x^{1+\alpha})$ does not make finding this function look easy.

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The paper I linked to in the comments constructs appropriate $f$'s for certain Vieta products, but I have not had the time to study this problem thoroughly. – J. M. Oct 31 '10 at 23:52
Unfortunately, I'm pretty sure that the functions in the paper all have Taylor series at all points on which they are defined, which means a function like the $f$ defined above in which $lim_{h\rightarrow0}f'(h)/h^\alpha=C>0$ would not be in that paper. – WAS Nov 1 '10 at 0:25

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This boils down to the comment I made towards the end of the post; finding such a $f$ would be tantamount to being able to find a closed-form for the Logistic map for a particular value of r (or equivalently the monic centered quadratic map), and AFAIK there are only a couple of very special values for which closed forms are known... – Steven Stadnicki Nov 1 '10 at 17:48
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