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Thread: Proof of differentiability

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

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

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

    So I did this:

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

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

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

    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 $\displaystyle \theta \in (0,1) $ such that $\displaystyle 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:

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

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

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

    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 $\displaystyle f(x)=x.$

    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:

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

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

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

    However, this isn't a proof!

    Can someone help me finish this?
    Use Cauchy's mean value theorem with f and $\displaystyle sin(\pi /2) $
    Last edited by ramdayal9; May 22nd 2009 at 11:07 AM.
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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 $\displaystyle f(0)=0$ and $\displaystyle f(1)=1$.

    This makes the questions significantly easier!

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

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

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

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

    Rearranging gives:

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