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Math Help - Chain of Sine and Cosine

  1. #1
    Super Member redsoxfan325's Avatar
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    Chain of Sine and Cosine

    As I was messing around in Maple (yes I'm a nerd), I noticed something.

    It appears as if the infinite chain \sin(\cos(\sin(\cos(\sin(\cos(.....(\sin(\cos(x)))  )))))) converges for all x.

    It appears to be converging to a number very close to \ln(2), but I don't think it actually does, unless it's really, really slowly.

    The number (calculated for 7 sine-cosine pairs) is approximately 0.69481969.

    Also, if you let the first term be cosine instead of sine, this chain gets close to 0.768169156736795977462.

    Does anyone know anything about this?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by redsoxfan325 View Post
    As I was messing around in Maple (yes I'm a nerd), I noticed something.

    It appears as if the infinite chain \sin(\cos(\sin(\cos(\sin(\cos(.....(\sin(\cos(x)))  )))))) converges for all x.

    It appears to be converging to a number very close to \ln(2), but I don't think it actually does, unless it's really, really slowly.

    The number (calculated for 7 sine-cosine pairs) is approximately 0.69481969.

    Also, if you let the first term be cosine instead of sine, this chain gets close to 0.768169156736795977462.

    Does anyone know anything about this?
    This kind of thing is called fixed point iteration. Starting with x_0 being whatever, you let x_1=\sin(\cos x_0) and more generally x_{n+1}=\sin(\cos x_n) for every n\geq 0.

    There are various assumptions that ensure that the sequence converges (for instance, here the function f(x)=\sin\cos x is contracting: |f(x)-f(y)|\leq k|x-y| where k\in(0,1), hence Banach's fixed point theorem applies). Then, if x_n\to\ell, necessarily we have \ell=\sin\cos\ell (taking the limit in x_{n+1}=\sin\cos x_n).

    So that your constants are just the fixed points of \sin\cos and \cos\sin: the numbers c_1,c_2 such that \sin\cos c_1=c_1 and \cos\sin c_2=c_2 (they are unique). I don't think there exits a simple explicit formula for them.

    You probably can find more about this if you look for "Picard iteration", or "Banach fixed point theorem" with Google.
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