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Thread: Equation

  1. #1
    Junior Member
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    Equation

    HI everybody,

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

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

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

    CAN YOU HELP ME PLEASE?

    AND THANKS ANYWAY.
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  2. #2
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    For $\displaystyle f(0)\times f(\pi)$ consider

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

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

    and

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

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

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