Nesting the Sine Function


Let x(0,π). Consider the sequence of real numbers given by s1=sinx and sn+1=sinsn for n{2,3,}. Thus we consider iterations of sine nesting:


(a) Prove that {sn} is a convergent sequence.

(b) Determine its leading-order asymptotic behavior for large n.


(a) The restriction of x to (0,π) makes it so that s1=sinx>0. We would like to show that {sn} is decreasing and bounded below by zero. We may do it by induction: suppose sk>0 so that, because 0<sinu<u for u>0, we have sk>sk+1=sinsk>0. This is exactly the desired statement. Therefore the sequence is convergent by the monotone convergence theorem, and we may define s=limnsn.

Now to get the value of the limit s, we write 0=limn(sn+1sinsn)=ssin(limnsn)=ssins.

This equation is solved by s=0, and therefore s=limnsn=0,

independent of the value of x.

(b) For large n, sn is small and positive. We postulate an ansatz of algebraic decay, sn=αnr.

Now, the recursive definition of sn gives sn+1=sinsn=sn16sn3+O(sn5),

and, plugging in the ansatz, we find αnr(1+1n)r=αnr(116α2n2r+O(n5r)).

The left hand side is expanded according to (1+1n)r=1rn+O(1n2).

Forcing agreement to first order, we have rn=α26n2r. -\frac{r}{n}=-\frac{\alpha^2}{6n^{2r}}.

In order to match powers of n, we need r=12 and subsequently α=3. Therefore, the leading order asymptotic behavior is sn3n,

or equivalently limnnsin(n)(x)=3,x(0,π)

where the notation sin(n) refers to iterating rather than differentiating or exponentiating.


If we make another ansatz, sn=3n+βns,

we quickly find that it gives us no information. This failure indicates that the subsequent correction to the asymptotic form is not algebraic. In this case, the next term in the expansion turns out to be logarithmic, specifically sn3n+O(lnnn3/2).