Proof of differentiability

**Showcase_22** · May 16th 2009, 10:39 AM

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?

**heathrowjohnny** · May 16th 2009, 12:50 PM

You would probably use the Mean Value Theorem. You know that there is a $\theta \in (0,1)$ such that $f'(\theta) = f(b)-f(a)$ .

**NonCommAlg** · May 16th 2009, 12:54 PM

Originally Posted by Showcase_22

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.

**ramdayal9** · May 22nd 2009, 09:18 AM

Originally Posted by Showcase_22

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 $sin(\pi /2)$

**Showcase_22** · May 22nd 2009, 10:06 AM

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:

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

**ramdayal9** · May 22nd 2009, 11:06 AM

no problem Showcase 22

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