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Math Help - Proof of differentiability

  1. #1
    Super Member Showcase_22's Avatar
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    Proof of differentiability

    Let f:[0,1] \rightarrow \mathbb{R} be a continuous function.

    If also f is differentiable on (0,1) prove that \exists \ \theta \in (0,1) such that \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi}.
    I wasn't sure how to start this but I thought it had something to do with a Taylor series.

    So I did this:

    \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi} \Rightarrow f'(\theta)=\frac{2}{\pi} \cos \frac{\pi}{2} \theta

    f'(\theta)=\frac{2}{\pi} \left( 1- \frac{ \left( \frac{\pi}{2}  \theta \right)^2}{2!} + \frac{\left( \frac{\pi}{2} \theta \right)^4}{4!}-....\right)

    and I know that this is only valid for \theta \in (0,1).

    However, this isn't a proof!

    Can someone help me finish this?
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  2. #2
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    You would probably use the Mean Value Theorem. You know that there is a  \theta \in (0,1) such that  f'(\theta) = f(b)-f(a) .
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    I wasn't sure how to start this but I thought it had something to do with a Taylor series.

    So I did this:

    \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi} \Rightarrow f'(\theta)=\frac{2}{\pi} \cos \frac{\pi}{2} \theta

    f'(\theta)=\frac{2}{\pi} \left( 1- \frac{ \left( \frac{\pi}{2} \theta \right)^2}{2!} + \frac{\left( \frac{\pi}{2} \theta \right)^4}{4!}-....\right)

    and I know that this is only valid for \theta \in (0,1).

    However, this isn't a proof!

    Can someone help me finish this?
    there's something missing here because the claim is not true in general. for example it fails for f(x)=x.

    by the way don't put your question in "Quote" because, as you see, when we quote you, your question won't appear.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    I wasn't sure how to start this but I thought it had something to do with a Taylor series.

    So I did this:

    \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi} \Rightarrow f'(\theta)=\frac{2}{\pi} \cos \frac{\pi}{2} \theta

    f'(\theta)=\frac{2}{\pi} \left( 1- \frac{ \left( \frac{\pi}{2}  \theta \right)^2}{2!} + \frac{\left( \frac{\pi}{2} \theta \right)^4}{4!}-....\right)

    and I know that this is only valid for \theta \in (0,1).

    However, this isn't a proof!

    Can someone help me finish this?
    Use Cauchy's mean value theorem with f and  sin(\pi /2)
    Last edited by ramdayal9; May 22nd 2009 at 12:07 PM.
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  5. #5
    Super Member Showcase_22's Avatar
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    Ah, i'm so sorry about this but I haven't told you the entire question!

    I thought that they had redefined the function f when they haven't.

    The question also states that f(0)=0 and f(1)=1.

    This makes the questions significantly easier!

    Let g(x)=\sin \left( \frac{\pi}{2} x \right). It is apparent that g is continuous over [0,1] and differentiable over (0,1).

    Therefore by Cauchy's mean value theorem \exists \ \theta \in (0,1) \ s.t \ \frac{f'(\theta)}{g'(\theta)}=\frac{f(1)-f(0)}{g(1)-g(0)}

    We know that g'(\theta)=\frac{\pi}{2} \cos \left( \frac{\pi}{2} \theta \right) and g(1)-g(0)=1.

    This gives: \frac{f'(\theta)}{g'(\theta)}=\frac{f'(\theta)}{\f  rac{\pi}{2} \cos \left( \frac{\pi}{2}x \right)}=f(1)-f(0)=1

    Rearranging gives:

    \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi}
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  6. #6
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    no problem Showcase 22
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