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Thread: Proof

  1. #1
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    Proof

    I think this goes in this section.

    Let x be a real number. Prove that

    abs(sin nx) <= n*abs(sinx) for all positive integers n.
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  2. #2
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    Quote Originally Posted by Nichelle14
    I think this goes in this section.

    Let x be a real number. Prove that

    abs(sin nx) <= n*abs(sinx) for all positive integers n.
    I did not try to do the problem yet but you can use the fact that,
    $\displaystyle \sqrt{x^2}=|x|$
    Thus, you need to prove that,
    $\displaystyle \sqrt{\sin ^2 nx}\leq n\sqrt{\sin^2 x}$
    If and only if,
    $\displaystyle \sin^2 nx\leq n^2\sin ^2 x$
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  3. #3
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    I still don't understand what to do.
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  4. #4
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    Mathematical induction is the tool here.
    It is true for $\displaystyle n=1$ thus, there is a $\displaystyle k$ such as, $\displaystyle |\sin kx|\leq k|\sin x|$
    Thus,
    $\displaystyle 0\leq |\sin kx|\leq k|\sin x|$
    But,
    $\displaystyle 0\leq |\cos x|\leq 1$
    Thus, (mutiply inequalities notice they are non-negative),
    $\displaystyle |\sin kx||\cos x|\leq k|\sin x|$......(1)
    Also,
    $\displaystyle 0\leq |\sin x|\leq |\sin x|$
    And,
    $\displaystyle 0\leq |\cos kx|\leq 1$
    Thus, (mutiply inequalites notice they are non-negative),
    $\displaystyle |\cos kx||\sin x|\leq |\sin x|$......(2)
    Using the property $\displaystyle |x||y|=|xy|$ on (1) and (2) we have,
    $\displaystyle |\sin kx \cos x|\leq k|\sin x|$
    $\displaystyle |\cos kx \sin x|\leq |\sin x|$
    Now, add these inequalities,
    $\displaystyle |\sin kx \cos x|+|\cos kx \sin x|\leq k|\sin x|+|\sin x|$
    Now, by triangular inequality,
    $\displaystyle |\sin kx \cos x|+|\cos kx \sin x|\geq |\sin kx \cos x+\cos kx \sin x|$
    By transitivity ($\displaystyle a<b \mbox{ and }b<c \rightarrow a<c$),
    $\displaystyle |\sin kx \cos x+\cos kx \sin x|\leq k|\sin x|+|\sin x|$
    Thus, recognizing the sum for sine and simply the right side,
    $\displaystyle |\sin (k+1) x|\leq (k+1)|\sin x|$
    Proof is complete.
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  5. #5
    MHF Contributor Quick's Avatar
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    Is it possible to say $\displaystyle 5\leq5$
    (because it seems somewhat ridiculous)
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  6. #6
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    Quote Originally Posted by Quick
    Is it possible to say $\displaystyle 5\leq5$
    (because it seems somewhat ridiculous)
    Of course.

    $\displaystyle a\leq b$ is defined to be true whenever,
    $\displaystyle a<b$ OR $\displaystyle a=b$.
    Furthermore, in mathematics this expression is used a lot. There is one powerful theorem in set theory (Zorn's Lemma) which is based on the fact that $\displaystyle a\leq a $.
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