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Math Help - Number of roots

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

    Let f(x) be a continuous function on the interval [0,\pi].If \int_{0}^{\pi}f(x)\sin x dx=0 and \int_{0}^{\pi}f(x)\cos x dx=0,then prove that f(x)=0 must have atleast two real roots on the interval (0,\pi).
    Last edited by pankaj; August 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 \int_{0}^{\pi}f(x)\sin x dx=0 and \int_{0}^{\pi}f(x)\cos x dx=0,then prove that f(x)=0 must have atleast two real roots on the interval (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 f(x) must be continuous on the interval [0,\pi].I have edited my post.
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  4. #4
    Senior Member pankaj's Avatar
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    \int_{0}^{\pi}f(x)\sin x dx=0 implies that f(x)=0 must have atleast one root lying on the interval (0,\pi)(since \sin x>0 onthe interval (0,\pi)).

    Not able to proceed from this stage.What to do with the condition \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 f is identically not zero,else the result is trivial.

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

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

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

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

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

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

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

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

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


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

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