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Let $f_0(x) = x, f_{n+1} = \sin f_n(x) (0 < x < \pi)$.
Then for $n \ge 3$, $$f_n(x) \gt \frac{f_2(x)}{n-1} ? $$

Here is what I have shown: For $x > 0$, $\sin x > x - x^3 / 6$. So to show the proposition, it suffices to show that, by induction, for $n$, $n f_n(x)^2 < 6$. And so it suffices to show that $n f_n(1)^2 < 6$. And using a calculator, this inequality seems to be true.

k.j.
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  • But $nf(1)^2<6$ for all $n$ would imply $f(1) = 0$, which is not the case. – Jakobian Jul 07 '19 at 14:20
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    This is related https://math.stackexchange.com/q/3215/476484 which means that for any $0<x<\pi$, $f_n(x)\sim \frac{\sqrt{3}}{\sqrt{n}}$, so the series diverges. – Jakobian Jul 07 '19 at 14:36
  • @Jakobian Sorry, I misunderstood. $f_n(x)$ goes to $0$. So $nf_n(1)^2 < 6$ makes no contradictions, I think. (Not $nf(1)^2 < 6$, but $nf_n(1)^2 < 6$. This is my typo. Sorry.) – k.j. Jul 07 '19 at 14:36

2 Answers2

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Analogously to $\sin(x)<x-x^3/6$, we have $\sin(x)>x-x^3/6+x^5/120$ (one additional term in the Taylor Series). Additionally, $\sin([0, 1])\subseteq [0,1]$ so that $0\leq f_n(1)\leq 1$ for all $n$.

We prove by induction that $nf_n(1)^2<6$. Base case: $f_1(1)^2=\sin(1)<6$. Now let $n\geq 2$ and assume the statement holds for $n-1$. Since $\sin(x)$ is increasing on $[0,1]$, we have $$nf_n(1)^2=n\sin^2(f_{n-1}(1))<n\sin^2\left(\sqrt{\frac{6}{n-1}}\right)<n\left(\sqrt{\frac{6}{n-1}}-\frac{1}{n-1}\sqrt{\frac{6}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{6}{n-1}}\right)^2=6\left(\sqrt{\frac{n}{n-1}}-\frac{1}{n-1}\sqrt{\frac{n}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{n}{n-1}}\right)=6g(n),$$ where $$g(x)=\sqrt{\frac{x}{x-1}}-\frac{1}{x-1}\sqrt{\frac{x}{x-1}}+\frac{3}{10(x-1)^2}\sqrt{\frac{x}{x-1}}.$$ We have to prove $g(n)\leq 1$ for all $n\geq 2$.

But the only local extremum of $g(x)$ in $\mathbb{R}^+$ is between $x=1$ and $x=2$ and there $g(x)<1$. Furthermore, we easily see $\lim\limits_{x\rightarrow\infty} g(x)=1,$ so that $g(x)\leq 1$ for all $x\geq 2$. Thus, the claim is proved.

mxian
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  • Thank you very much! I found some mistakes in my argument of this question, but your answer is still valid and I could solve it. – k.j. Jul 08 '19 at 12:30
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HINT.-One has $$(n-1){\underbrace{\sin(\sin(\cdots (\sin}_{\text{n times }}}(x))\cdots)\lt (n) {\underbrace{\sin(\sin(\cdots (\sin}_{\text{n+1 times }}}(x))\cdots)$$ for $x\in[0,\pi]$ which becomes in a clear way the inequality easy to prove $$(n-1)Y\lt n\sin(Y)\tag{*}$$ for $Y\in[0,\sin(\sin(\frac{\pi}{2})]=[0,0.84171]$ and $n\ge3$.

In short you do have $$2f_3(x)\lt3f_4(x)\lt4f_4(x)\lt\cdots\lt(n-1)f_n(x)$$ Consequently in order to prove $$(n-1)f_n(x)\gt f_2(x)$$ it is sufficient to prove $$2f_3(x)\gt f_2(x)$$ or to prove $$2\sin(X)\gt X$$ for $X\in[0,1]$ which does not present difficulty as a particular case of the inequality $(*)$ above (obviously with the inequality sign renversed).

Piquito
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