f1=sinxf_1 = \sin x

f2=sin(sinx)f_2 = \sin(\sin x)

f3=sin(sin(sinx))f_3 = \sin(\sin(\sin x))

fn=sin(fn−1)f_n = \sin(f_{n-1})

Just from playing around on a graphing calculator, I was able to guess the limit of this would be the zero function as n→∞n \rightarrow \infty, but I’m not certain and I don’t know how I could prove this.

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You could try using |sin(x)|≤1|sin(x)|\leq 1 for all x, so sin(−1)≤sin(sin(x))≤sin(1)\sin(-1)\leq \sin(\sin(x))\leq \sin(1)

– Fede Poncio

Oct 21 at 3:31

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1 Answer

1

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We proceed via the following steps:

Show that the sequence fnf_n has a limit, for any choice of xx.

Show that the limit is zero, again for any choice of xx.

1:

Assume x∈[0,1]x\in[0,1]; the case x∈[−1,0]x\in[-1,0] can be dealt with separately, and we can reduce to one of these cases by starting after one iteration because sinx∈[−1,1]\sin x\in[-1,1]. We may also assume without loss of generality that x≠0x\neq 0, because for this choice of xx the iterates fn=sinn(x)f_n = \sin^n(x) are all 00.

Because x∈(0,1]x\in(0,1], it follows that sinx∈(0,1]\sin x\in(0,1] as well, and by induction we see that fn∈(0,1]f_n\in(0,1] for all n≥0n\geq 0. Moreover, from the mean-value theorem we know that for 0

f_{n+1} = \sin f_n = \sin f_n – 0 = \int_0^{f_n} \cos t~dt \leq \int_0^{f_n} 1~dt = f_n.

f_{n+1} = \sin f_n = \sin f_n – 0 = \int_0^{f_n} \cos t~dt \leq \int_0^{f_n} 1~dt = f_n.

For x<0x<0 the same idea works, but you need to take care of minus signs.