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Math Help - Equation

  1. #1
    Junior Member
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    Equation

    HI everybody,

    I must show that the following equation: cosx+cos2x+...+cosnx=0 (n\in \mathbb{N}*) accepts at least one solution in [0,\pi].

    1)-We know that f(x)=cosx+cos2x+...+cosnx is continious in [0,\pi].

    2)-So i must show that f(0)*f(\pi)<0, BUT HOW?

    CAN YOU HELP ME PLEASE?

    AND THANKS ANYWAY.
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  2. #2
    Master Of Puppets
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    For f(0)\times f(\pi) consider

     f(\pi)= \cos(\pi)+\cos(2\pi)+\cos(3\pi)+\cos(4\pi)+\dots

     f(\pi)= -1+1-1+1+\dots

    and

    f(0) = \cos(0)+\cos(2\times 0 )+\cos(3\times 0)+\cos(4\times 0)+\dots

    f(0) = 0+0+0+0+\dots

    Can you see a pattern? What are the implications for n being either even or odd for  f(\pi) ?
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