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Thread: A constant function!

  1. #1
    MHF Contributor

    May 2008

    A constant function!

    True or False:

    x \ + \ \sin x \ + \ \sin(x \ + \ \sin x) \ + \ \sin(x \ + \ \sin x \ + \ \sin(x \ + \ \sin x)) \ + \ \cdots=\pi for all 0 < x < 2\pi.
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  2. #2
    Mar 2009
    Let f(x)=x+\sin x for 0<x<2\pi. Then f'(x)=1+\cos x\geq0, so f(x) is one-one, mapping I=(0,2\pi) to itself.

    Define f_0(x)=f(x), f_n(x)=f(f_{n-1}(x)). The given function is G(x)=\lim_{n\to\infty}f_n(x), where the limit exists, and G:I\to I.

    It follows that f(G(x))=G(x), and so G(x)+\sin G(x)=G(x).

    Therefore \sin G(x)=0, which implies that G(x)=\pi on I.

    As for the limit, we know that 0<\frac{\sin x}x\leq 1 for -\pi<x<\pi, and so 0\leq 1-\frac{\sin(\pi-x)}{\pi-x}=1-\frac{\sin x}{\pi-x}<1 for x\in I.

    Given \epsilon>0 we may choose k<1 depending on \epsilon such that 1-\frac{\sin x}{\pi-x}<k for all x\in[\epsilon,2\pi-\epsilon].

    It follows that |\pi-x-\sin x|<k|\pi-x| for all x\in[\epsilon,2\pi-\epsilon].

    This means that |\pi-f(x)|<k|\pi-x| in this interval. Also if f maps [\epsilon,2\pi-\epsilon] to itself, since f(x) is closer to \pi than x is. Thus

    |\pi-f_n(x)|<k|\pi-f_{n-1}(x)|<k^2|\pi-f_{n-2}(x)|<\dots<k^n|\pi-f_0(x)|. It follows that G(x)=\lim_{n\to\infty}f_n(x)=\pi for all x in the interval [\epsilon,2\pi-\epsilon].

    Since \epsilon is arbitrary (get this!) we have G(x)=\pi for all x\in I.

    If this works I will definitely LOL.
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