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Math Help - problem of equation

  1. #1
    Senior Member Sambit's Avatar
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    Question problem of equation

    let \alpha,\beta,\gamma be 3 positive roots of the equation e^x\sin x -\frac{1}{2}=0 where \alpha<\beta<\gamma.
    show that there is a root of the equation f(x)=e^x\sin x +\frac{1}{2} in (\alpha,\gamma)

    help please....
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  2. #2
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    It depends on what you can use (any methods of calculus?) and the level of strictness. Basically, if you look at the graph of e^x\sin x, this is obvious, but the question is, how much handwaving you are allowed to use in referring to the graph.

    The function e^x\sin x behaves like \sin x in that it is positive on (2k\pi,(2k+1)\pi) and negative on ((2k+1)\pi,(2k+2)\pi) for k\in\mathbb{Z} and it is zero in k\pi, k\in\mathbb{Z}. However, the amplitude increases exponentially. You can see the graph in WolframAlpha. So if \alpha,\beta,\gamma are three positive roots of e^x\sin x -\frac{1}{2}=0, then there is an interval ((2k+1)\pi,(2k+2)\pi) either inside (\alpha, \beta) or inside (\beta,\gamma) (or both, if these are not consecutive roots), where e^x\sin x is negative. Also, one can show that even in the first negative dip between \pi and 2\pi, the function's minimum is far less than -1/2.
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by emakarov View Post
    It depends on what you can use (any methods of calculus?) and the level of strictness. Basically, if you look at the graph of e^x\sin x, this is obvious, but the question is, how much handwaving you are allowed to use in referring to the graph.

    The function e^x\sin x behaves like \sin x in that it is positive on (2k\pi,(2k+1)\pi) and negative on ((2k+1)\pi,(2k+2)\pi) for k\in\mathbb{Z} and it is zero in k\pi, k\in\mathbb{Z}. However, the amplitude increases exponentially. You can see the graph in WolframAlpha. So if \alpha,\beta,\gamma are three positive roots of e^x\sin x -\frac{1}{2}=0, then there is an interval ((2k+1)\pi,(2k+2)\pi) either inside (\alpha, \beta) or inside (\beta,\gamma) (or both, if these are not consecutive roots), where e^x\sin x is negative. Also, one can show that even in the first negative dip between \pi and 2\pi, the function's minimum is far less than -1/2.
    better to do by calculus in exam. what will be the approach then?
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