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Math Help - An Inequality?

  1. #1
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    An Inequality? (A Challenging One: Prove or disprove)

    Prove or disproe:
    For any positive integer n and x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0
    Last edited by elim; March 4th 2010 at 07:50 PM. Reason: Make the title more specific
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  2. #2
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    Quote Originally Posted by elim View Post
    Prove or disproe:
    For any positive integer n and x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0
    Dear elim,

    You could show that this statement is correct using mathamatical induction. Can you give it a try? If you need more help please don't hesitate to reply back.
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  3. #3
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    Some how I just could not see this clearly before

    Say n (>1) is the smallest integer such that for some minimal point x_0 \in (0,\pi),
    \sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0,
    then \sin nx_0 < 0

    On the other hand the critical points for the function f_n(x) = \sum_{k=1}^n \frac{1}{k}\sin kx are
    x = \frac{(2m+1)\pi}{n+1}, \quad m=0,\cdots,\left[\frac{n-1}{2}\right]
    They are the points in (0,\pi) such that
    \sum_{k=1}^n \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}} = 0
    (note that \sin \frac{nx_0}{2} \neq 0 by our finding above)
    But \sin \frac{n(2m+1)\pi}{n+1} = \sin \left( 2m\pi+\frac{n-2m}{n+1}\right) = \sin \frac{n-2m}{n+1} > 0

    Thus we get a contradiction and we actually proved that
    \forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left( \sum_{k=1}^n \frac{1}{k}\sin kx > 0\right)

    This is a bit stronger statement.

    Thanks Sudharaka for the direction. I thought this problem needs more advanced tech to solve.
    Last edited by elim; March 5th 2010 at 08:23 AM.
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    Quote Originally Posted by elim View Post
    It seems that I still not be able to figure out the way of using induction.
    say n (>1) is the smallest integer such that for some x_0 \in (0,\pi),
    we have \sum_{k=1}^n \frac{1}{k} \sin kx \leq 0,
    then what?

    I did get the critical points for the function f_n(x) = \sum_{k=1}^n \frac{1}{k}\sin kx.
    They are x = \frac{(2m+1)\pi}{n+1}, \quad m=0,\cdots,\left[\frac{n-1}{2}\right]
    They are the points in (0,\pi) such that
    \sum_{k=1}^n \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}} = 0
    Dear elim,

    We have to show that for any positive integer n and <br />
x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0

    When n=1,

    \sum_{k=1}^1 \frac{1}{k} \sin kx =sinx\geq{0}~Since,~x \in [0,\pi]

    Therefore the expression is true for n=1

    Suppose the expression is true for n=p where p\in{Z^{+}}

    Then, \sum_{k=1}^p \frac{1}{k} \sin kx\geq0

    Now, \frac{1}{(p+1)} \sin (p+1)x\geq{0}~Since,~p+1>0~and~x\in[0,\pi]

    Therefore, \sum_{k=1}^p \frac{1}{k} \sin kx+\frac{1}{(p+1)} \sin (p+1)x\geq{0}

    Hence, \sum_{k=1}^{p+1} \frac{1}{k} \sin kx\geq0

    Therefore our the expression is true for n=p+1~where~p\in{Z^+}

    Hence by mathamatical induction the expression is true for all n\in{Z^+}

    Hope this will help you.
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Now, \frac{1}{(p+1)} \sin (p+1)x\geq{0}~Since,~p+1>0~and~x\in[0,\pi]
    This is not true. And is the hard part.

    On the other hand, actually I did used math induction to prove the inequality is true for all positive integer n and all x \in [0,\pi] by showing that no such an integer n>1 and x_0 \in (0,\pi) to let
    \sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0 happen assuming

    (**) \sum_{k=1}^m \frac{1}{k} \sin kx_0 > 0 for 1 \leq m < n, \quad x \in (0,\pi) which is true for m=1
    Last edited by elim; March 6th 2010 at 09:33 AM.
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