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I took Viète's infinite product for $\frac{2}{\pi}$:

$\displaystyle \dfrac{2}{\pi} = \dfrac{\sqrt2}{2} . \dfrac{\sqrt{2+\sqrt2}}{2} . \dfrac{\sqrt{2+\sqrt{2+ \sqrt2}}}{2} \dots$

and made it generic:

$\displaystyle v = |\dfrac{\sqrt{z}}{z} . \dfrac{\sqrt{z+\sqrt{z}}}{z} . \dfrac{\sqrt{z+\sqrt{z+ \sqrt{z}}}}{z} . \dots |$

Checked whether other converging values for $v$ exist and found three more (by trial & error):

$z=-1$ yielding $1.29425...$

$z=i-1$ yielding $0.92741...$

$\displaystyle z=i+ \frac13 (1+ \sqrt[3]{(28-3 \sqrt{87})}+\sqrt[3]{(28+3 \sqrt{87})})$ yielding $0.64801...$

Tested the necessary (but not sufficient) convergence of the 'final' factor to 1 by using:

$\displaystyle a = \sqrt{z+\sqrt{z+ \sqrt{z} \dots}}$

then solving $a$ via:

$a = \sqrt{z + a}$

and indeed $\displaystyle |\frac{a}{z}|=1$ for all these three values.

How could this be proven? Are there more converging values $v$?

Are these values for $v$ connected in any way e.g. on a circle in the complex plane or to other mathematical constants (did check Plouffe's, Wolfram Math and Sloane's but without any result)?

Thanks!

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1 Answer 1

up vote 6 down vote accepted

If $a$ satisfies $a=\sqrt{z+a}$ then $a/z = 1/(a-1)$ which usually does not have modulus $1$. So for $1/(a-1) = e^{i\theta}$ we get $z = e^{-2i\theta}+e^{-i\theta}$.

picture

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1  
Very nice, Gerald. Not really a circle after all, but quite an intriguing shape. If this is the solution space for values of $z$ that make $v$ converge, I now wonder what the output shape for $v$ looks like. Many thanks! –  Agno Dec 3 '11 at 18:59

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