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Math Help - Prove result (continuous derivative)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove result (continuous derivative)?

    Suppose that 'f' has a continuous derivative on \mathbb{R} and that f(x) + f(y) = f\left(\frac{x + y}{1 - xy}\right) for each x and y such that xy < 1. Prove that, for some constant C, f(x) = C\tan^{-1} x.
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  2. #2
    Senior Member pankaj's Avatar
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    Put x=y=0 to obtain f(0)+f(0)=f(0) and thus f(0)=0

    Put y=-x to obtain f(x)+f(-x)=f(0)=0 and thus f(-x)=-f(x)

    f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

    f'(x)=\lim_{h\to 0}\frac{f(x+h)+f(-x)}{h}

    f'(x)=\lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{h}

    f'(x)=\lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{\frac{h}{1+hx+x^2}}. \frac{1}{1+hx+x^2}

     <br />
f'(x)=\frac{f'(0)}{1+x^2}<br />

    since \lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{\frac{h}{1+hx+x^2}} =\frac{f(0+\frac{h}{1-hx+x^2})-f(0)}{\frac{h}{1+hx+x^2}}=f'(0)

    f(x)=\int\frac{f'(0)}{1+x^2}dx =f'(0)tan^{-1}x+k

    k=0 since f(0)=0

    f(x)=f'(0)tan^{-1}x

     <br />
C=f'(0)<br />

    f(x)=Ctan^{-1}x
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