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

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?

#### Solutions Collecting From Web of "What's the value of this Viète-style product involving the golden ratio?"

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.