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Thread: Number of roots

  1. #1
    Senior Member pankaj's Avatar
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    Number of roots

    Let $\displaystyle f(x)$ be a continuous function on the interval $\displaystyle [0,\pi].$If $\displaystyle \int_{0}^{\pi}f(x)\sin x dx=0$ and $\displaystyle \int_{0}^{\pi}f(x)\cos x dx=0$,then prove that $\displaystyle f(x)=0$ must have atleast two real roots on the interval $\displaystyle (0,\pi).$
    Last edited by pankaj; Aug 7th 2009 at 09:44 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pankaj View Post
    If $\displaystyle \int_{0}^{\pi}f(x)\sin x dx=0$ and $\displaystyle \int_{0}^{\pi}f(x)\cos x dx=0$,then prove that $\displaystyle f(x)=0$ must have atleast two real roots on the interval $\displaystyle (0,\pi).$
    Without any extra condition/s it need not have any roots.

    CB
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  3. #3
    Senior Member pankaj's Avatar
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    I guess one condition that must be there is that $\displaystyle f(x) $must be continuous on the interval $\displaystyle [0,\pi]$.I have edited my post.
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  4. #4
    Senior Member pankaj's Avatar
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    $\displaystyle \int_{0}^{\pi}f(x)\sin x dx=0$ implies that $\displaystyle f(x)=0$ must have atleast one root lying on the interval $\displaystyle (0,\pi)$(since $\displaystyle \sin x>0$ onthe interval $\displaystyle (0,\pi)$).

    Not able to proceed from this stage.What to do with the condition $\displaystyle \int_{0}^{\pi}f(x)\cos x dx=0$.

    This appears to be some challenge.
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  5. #5
    Senior Member pankaj's Avatar
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    Sorry,I could not update earlier.The solution is a joy to share.I found it on another website though but it doesn't really matter.One can't claim intellectual propriety all the time.So here it is.


    Assume $\displaystyle f$ is identically not zero,else the result is trivial.

    Since $\displaystyle \sin x>0$ on $\displaystyle (0,\pi)$ and $\displaystyle \int_{0}^{\pi}f(x)\sin x=0$,
    this leads to the conclusion that $\displaystyle f$ takes on positive and negative values in this interval and therefore due to Intermediate Value theorem,$\displaystyle f$ has atleast one root in the open interval $\displaystyle (0,\pi).$

    Suppose now $\displaystyle f(x)=0 $has exactly one root in this interval say $\displaystyle x=x_{0},i.e. f(x_{0})=0$.

    Without loss of generality,we can say that $\displaystyle f<0$ for $\displaystyle x<x_{0}$ and $\displaystyle f>0$ for $\displaystyle x>x_{0}.$

    Now consider $\displaystyle \int_{0}^{\pi}f(x)\sin(x-x_{0})dx$

    Clearly, $\displaystyle \int_{0}^{\pi}f(x)\sin(x-x_{0})dx=0$.

    For $\displaystyle x<x_{0},\sin(x-x_{0})<0$ and for $\displaystyle x>x_{0},\sin(x-x_{0})>0$.

    For $\displaystyle x\in (0,\pi),f(x)\sin(x-x_{0})\geq 0$

    and thus $\displaystyle \int_{0}^{\pi}f(x)\sin(x-x_{0})dx$ is strictly positive

    which is a contradiction since $\displaystyle \int_{0}^{\pi}f(x)\sin(x-x_{0})dx=0$


    Thus our assumption that $\displaystyle f(x)=0$ has exactly one root on $\displaystyle (0,\pi)$ is false.

    $\displaystyle \therefore f(x)=0$ has atleast two roots on $\displaystyle (0,\pi).$
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