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Numerically, I found that if one builds a sequence of $sin(sin(sin(....(sin(x)...))$ with $n$ being the number of times sin operation is performed, then with n going to infinity the product of this operation multiplied by the square root of n approaches the value, which is remarkably close to the square root of 3 for any $0<x<\pi$. The same if at every step instead of using sin function I use Taylor series. However, starting from scratch I could not find analytical approximation because for every power of x in Taylor series I stop at, the result diverges; it goes to negative infinity if $sin(x)$ is approximated as $x-x^3/3!$ and to positive infinity if $sin(x)$ is approximated as $x-x^3/3!+x^5/5!$ and so forth.

Please, advise if anyone observed this yet and has analytical proof/disproof.

Thanks,

Anthony

user36225
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  • it seems very interesting.. note that the same thing does not happen with cos, (and of course it does not with tan, sinh and cosh) but it does happen with tanh; I've found (empirically) that $$\lim_{n \to \infty} \sqrt{n} \cdot \tanh(\tanh(... \tanh(x) ... ))_{n \ times} = \frac{\sqrt{6}}{2}$$ – Ant Nov 21 '13 at 19:33

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You are looking for $\lim_{n \to \infty}\sqrt n \sin^n(x)$, where the power on $\sin$ is iteration and claiming that it approaches $\sqrt 3$. This cannot be true for all $x$, as with $x=0$ we will always have a value of $0$. Also for $-\frac \pi 2 \lt x \lt 0$ the value will always be negative. You can rescue that by claiming that the limit will then be $-\sqrt 3$

Then we can give an argument that supports this. Suppose $n \sin^{n^2}x=L$ and $n$ is large enough that $L$ is small. Then $\sin \frac Ln \approx Ln-\frac {L^3}{6n^3}$, but that is not much below $L$. Each application of $\sin$ will decrease the value by (through third order) the same amount. Then $(n+1) \sin^{(n+1)^2}x\approx (n+1)(\frac Ln-\frac {L^3}{6n^3}(2n+1))$ We want to find the stable value of $L$, so $$L=(n+1)(\frac Ln-\frac {L^3}{6n^3}(2n+1))\\ L=L+\frac Ln-\frac {L^3}{6n^3}(2n+1)(n+1)$$ which to leading order in $n$ gives $L=\sqrt 3$

To really prove this, you would have to show that if you have some error from $L$ at $n$, it is reduced at $n+1$ in a way that takes it to zero as $n \to \infty$

Ross Millikan
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