One way of looking at the Viète (Viete?) product $${2\over\pi} = {\sqrt{2}\over 2}\times{\sqrt{2+\sqrt{2}}\over 2}\times{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\times\dots$$ 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: $$\Phi = {\sqrt{1}\over\phi}\times{\sqrt{1+\sqrt{1}}\over\phi}\times{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\times\dots$$ and an equivalent proof of convergence goes through. The question is, what's the value of $\Phi$? The usual proof of Viète'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 Viète 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?

Tito Piezas III
  • 47,981
  • 5
  • 96
  • 237
Steven Stadnicki
  • 50,003
  • 9
  • 78
  • 143
  • FWIW, attempts to find `0.509490972847535755...` in the Plouffe inverter yielded nothing. – J. M. ain't a mathematician Oct 07 '10 at 01:27
  • As to how I got that number in *Mathematica*: `SequenceLimit[FoldList[Times, 1, NestList[Sqrt[1 + #] &, N[1, 50], 50]/GoldenRatio]]`. – J. M. ain't a mathematician Oct 07 '10 at 01:28
  • @J.M. I guessed it would be $\frac{1}{\sqrt{5}}$ and was guessing that approximating the numerator terms as the ratio of consecutive Fibonacci numbers might work. Apparently not. – Aryabhata Oct 07 '10 at 02:45
  • 2
    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 07 '10 at 03:28
  • 5
    After a cursory look at [this article](http://dx.doi.org/10.1007/s11139-005-4852-z), I suspect that this may come out in terms of an (Jacobi or Weierstrass) elliptic function, but I'm still trying to digest the formulae given. Maybe somebody with better analytical powers than me may be able to quickly figure out a closed form from the stuff in that paper. – J. M. ain't a mathematician Oct 07 '10 at 04:51
  • if this isn't answered by tomorrow or so I'll stab at it – Quadrescence Oct 21 '10 at 07:25
  • 3
    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 Naslund Feb 24 '11 at 00:20

1 Answers1


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.

  • 9,613
  • 7
  • 37
  • 57
  • 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. ain't a mathematician 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. – W. Cadegan-Schlieper Nov 01 '10 at 00:25
  • 3
    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 01 '10 at 17:48