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Let $f_2(x) = \sin(\sin x),$ and define $f_n(x) = \sin(f_{n-1}(x))$ for $n \geq 3.$ (So $f_n$ has sine nested $n$ times.) I am asked to prove that $$f_n(x) > \frac{f_2(x)}{n-1}$$ for $n \geq 3$ and $0 < x < \pi,$ but I'm having a hard time doing so.

At first, I considered the function $F(x) = (n-1)f_n(x) - f_2(x).$ Now sine is symmetric in the interval $[0,\pi],$ so it suffices to show that $F(x) > 0$ for $x \in (0, \frac{\pi}{2}).$ Furthermore, $F(0) = 0,$ so we will be done if we show that $F'(x) > 0$ in the interval $(0, \frac{\pi}{2}).$ Now $$F'(x) = (n-1)\cos(f_{n-1}(x))f'_{n-1}(x) - f'_2(x)$$ which is equal to $$F'(x) = (n-1)\cos(f_{n-1}(x))\cos(f_{n-2}(x))\cdots \cos(f_2(x))f'_2(x) - f'_2(x).$$ However, I'm not sure how I can show that this is positive in the interval given above.

I also tried using induction. If we can show that $$f_n(x) > \frac{n-2}{n-1}f_{n-1}(x),$$ then we will be done. I tried using the same method as above, but again, the derivative looks too complicated for me to decide whether it is positive or negative.

The previous part of the question also asks us to prove that $\sin x > x - \frac{x^3}{6},$ and I have tried to make use of this fact as well, but to no avail.

Any help will be much appreciated!

quasicoherent_drunk
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1 Answers1

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This is not an answer.

If they did not ask you to prove that $\sin (x) > x - \frac{x^3}{6}$, I should have used the expansion $$f_n(x)=x-\frac n 6 x^3-\big(\frac n{30}-\frac{n^2}{24}\big) x^5+O(x^7)$$ which makes $$(n-1)f_n(x)-f_2(x)=(n-2)x-\frac {(n-2)(n+1)} 6x^3+\frac{(n-2) \left(5 n^2+n+6\right)}{120} x^5+O\left(x^{7}\right)$$ $$(n-1)f_n(x)-(n-2)f_{n-1}(x)=x-\frac{(n-1)}{3} x^3+\frac{(n-1) (5 n-6) }{40} x^5+O\left(x^7\right)$$ whci does not show any real root for $n\geq 2$ making the last expression always positive.

Claude Leibovici
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  • Do you mind explaining why it does not have any real roots for $n \geq 2$? – quasicoherent_drunk Aug 13 '18 at 06:18
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    @dhk628. Divide the last ecpression by $x$; it reduces to a quadratic equation in $X=x^2$ for which $\Delta=-\frac 1 {90} (35 n^2-79 n+44)$ that is to say that $\Delta \geq 0$ if $1 \leq n \leq \frac{44}{35}$. Outside this range $\Delta \lt 0$, no real root for $X$ and no real root for $x$. Is this OK ? Cheers. – Claude Leibovici Aug 13 '18 at 06:26