Proof of differentiability

• May 16th 2009, 10:39 AM
Showcase_22
Proof of differentiability
Quote:

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?
• May 16th 2009, 12:50 PM
heathrowjohnny
You would probably use the Mean Value Theorem. You know that there is a $\displaystyle \theta \in (0,1)$ such that $\displaystyle f'(\theta) = f(b)-f(a)$.
• May 16th 2009, 12:54 PM
NonCommAlg
Quote:

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:

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

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

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.
• May 22nd 2009, 09:18 AM
ramdayal9
Quote:

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:

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

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

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 $\displaystyle sin(\pi /2)$
• May 22nd 2009, 10:06 AM
Showcase_22

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:

$\displaystyle \frac{f'(\theta)}{\cos \left( \frac{\pi}{2} \theta \right)}=\frac{2}{\pi}$
• May 22nd 2009, 11:06 AM
ramdayal9
no problem Showcase 22