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Math Help - Prove that function is identically zero?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that function is identically zero?

    Let f(x) be a continuous function in [-1, 1] and satisfies 2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]. Prove that f(x) is identically zero \forall\ x\in [-1, 1].
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Let f(x) be a continuous function in [-1, 1] and satisfies 2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]. Prove that f(x) is identically zero \forall\ x\in [-1, 1].
    2f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1]

    2 cancels out both sides, why it is there?

    I think it may be f(2x^2 - 1) = 2x\cdot f(x)\ \forall\ x\in [-1, 1].
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  3. #3
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    Well, I'm having a go at f(2x^2-1)=2xf(x), which seems more sensible. To start with, putting -x for x shows that f(-x)=-f(x), so we can concentrate on the interval [0,1].

    So put x=\cos\theta for \theta\in[0,\pi/2] to derive f(\cos 2\theta)=2\cos\theta f(\cos\theta) (1).

    Putting x=\sin\theta gives f(-\cos 2\theta)=2\sin\theta f(\sin\theta), or f(\cos 2\theta)=-2\sin\theta f(\sin\theta) (2).

    It is clear from (2) that f(1)=0.

    Putting \theta=\pi/4 in (1) and (2) gives f(0)=\surd 2f(1/\surd2)=-\surd2f(1/\surd2), so now f(\sin 0)=f(\sin \frac\pi4)=f(\sin\frac\pi2)=0.

    Put \theta=\pi/8 in (1) and (2) now shows that f(\sin\frac\pi8)=f(\sin\frac{3\pi}8)=0.

    Now direct substitutions, the identity \sin(\pi/2-\theta)=\cos\theta and previous results show that f(\sin\frac\pi{16})=f(\sin\frac{3\pi}{16})=f(\sin\  frac{5\pi}{16})=f(\sin\frac{7\pi}{16})=0.

    So far we know that

    f(\sin 0)=f(\sin\frac\pi{16})=f(\sin\frac{2\pi}{16})=f(\s  in\frac{3\pi}{16})=f(\sin\frac{4\pi}{16})=f(\sin\f  rac{5\pi}{16})=f(\sin\frac{6\pi}{16}) =f(\sin\frac{7\pi}{16})=f(\sin\frac{8\pi}{16})=0.

    You can see how to proceed. We can prove by induction that for all positive integers n, f(\sin \theta)=0 whenever \theta\in S_n=\{\frac{k\pi}{2^{n+1}}:0\leq k\leq 2^n\}.

    The function g(\theta)=f(\sin\theta) is continuous on [0,\pi/2] and vanishes on S=\bigcup_{n=1}^\infty S_n. Moreover S is dense in [0,\pi/2] (anyone else want to prove this?)

    It follows that g(\theta)=0 on [0,\pi/2] and therefore f(x)=0 on [0,1] and therefore on [-1,1].
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