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Math Help - Another Uniform Continuity Problem

  1. #1
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    Another Uniform Continuity Problem

    Prove that f(x)=cos^3x is either uniform continuous or not.

    I claim that this function is not, since its derivative is not bounded. But how do I show that using the definition?

    Proof so far:

    Let  \epsilon = 1 , for every  \delta > 0 , I need to find x,y \in \mathbb {R} , |x-y| < \delta implies  |f(x)-f(y)| = |cos^3x - cos^3y| \geq 1

    What kind of x and y should I pick?

    Thanks.
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  2. #2
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    You made a mistake. The derivative is bounded.
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  3. #3
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    Theorem: a continuous periodic function is uniformly continuous.

    Indeed, it is uniformly continuous on two periods [0,2T] by Heine's Theorem (a continuous function defined on a compact is uniformly continuous), and by periodicity this implies the uniform continuity on \mathbb{R} (choose \delta<T so that, if |x-y|<\delta, then x and y lie in the same interval of length T and the situation matches a similar situation inside [0,2T]).

    And if you don't know this theorem by Heine, you can write |\cos^3 x -\cos^3 y|=|\cos x - \cos y||\cos^2 x + \cos x \cos y + \cos^2 x|\leq 3|\cos x-\cos y| =6|\sin\frac{x+y}{2}\sin\frac{x-y}{2}|\leq 6|\sin\frac{x-y}{2}|\leq 3|x-y|.
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  4. #4
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    I understand all but this last part, how is  6|sin \frac {x-y}{2}| \leq 3|x-y| ?
    Last edited by tttcomrader; October 12th 2008 at 12:18 PM.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    I understand all but this last part, how is  6|sin \frac {x-y}{2}| \leq 3|x-y| ?
    Because |\sin \alpha| \leq |\alpha|.
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  6. #6
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    Another question, how does 3| \cos x- \cos y| \leq 6| \sin \frac {x+y}{2} \sin \frac {x-y}{2}| ? Thanks.
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  7. #7
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    Quote Originally Posted by tttcomrader View Post
    Another question, how does 3| \cos x- \cos y| \leq 6| \sin \frac {x+y}{2} \sin \frac {x-y}{2}| ? Thanks.
    It is an identity,
     \cos A  - \cos B = 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2}
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    It is an identity,
     \cos A  - \cos B = 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2}
    Actually, it is  \cos A  - \cos B = - 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2} (which in absolute value is the same).
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