Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$\displaystyle |sin(b)-sin(a)|$ $\displaystyle \leq$ $\displaystyle | b-a |$

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- Jan 26th 2009, 03:39 PMLarriotoMean Value Theorem Proof
Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :

$\displaystyle |sin(b)-sin(a)|$ $\displaystyle \leq$ $\displaystyle | b-a |$ - Jan 26th 2009, 04:03 PMNonCommAlg
well, if you let $\displaystyle f(x)=\sin x,$ then by mean value theorem there exists $\displaystyle c$ between $\displaystyle a,b$ such that $\displaystyle f(b)-f(a)=(b-a)f'(c).$ that means $\displaystyle \sin b - \sin a = (b-a)\cos c.$

so $\displaystyle |\sin b - \sin a|=|b-a||\cos c| \leq |b-a|,$ because $\displaystyle |\cos c| \leq 1.$ it wan't very difficult, was it? (Shake) - Jan 26th 2009, 04:15 PMLarrioto
There is a second part to this problem :

Assume that f : $\displaystyle \Re \mapsto \Re $ is a differentiable function and that there exists a real number M > 0 such that |f '(x)|$\displaystyle \leq M$ for all x in the reel then show that :

$\displaystyle |f(a) - f(b)|\leq M|b-a|$for all x in the reel - Jan 26th 2009, 04:25 PMMathstud28
Choose any value $\displaystyle c$, we know there exists a corresponding interval $\displaystyle [a,b]$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $\displaystyle c$ was arbitrary this must work for every interval $\displaystyle [a,b]$.

Another almost identical method would be to assume that there exists an interval $\displaystyle [a,b]$ such that $\displaystyle \left|\frac{f(b)-f(a)}{b-a}\right|>M$. But since $\displaystyle f$ is differentiable we can find a $\displaystyle c\in(a,b)$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.

This is called Lipschitz's continuity by the way.

If you have any questions at all feel free to ask :) - Jan 26th 2009, 04:38 PMLarrioto
- Jan 26th 2009, 04:59 PMMathstud28
- Jan 26th 2009, 05:09 PMLarrioto
ok , i get it , it's just that f(c) can be replaced by f(x), that was confusing me ...

- Jan 26th 2009, 05:46 PMNonCommAlg
this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is

reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible! - Jan 26th 2009, 06:10 PMMathstud28
- Jan 26th 2009, 07:15 PMNonCommAlg
here's a simple counter-example: let $\displaystyle f(x)=x^3$ and choose $\displaystyle c=0.$ geometrically it's clear that there are no $\displaystyle a \neq b$ such that $\displaystyle f'(0)=\frac{f(b)-f(a)}{b-a}.$ you can also prove it very easily:

if there exist such $\displaystyle a,b,$ then: $\displaystyle 0=\frac{b^3-a^3}{b-a}=a^2+ab+b^2,$ which is possible iff $\displaystyle a=b=0,$ because: $\displaystyle 0=2a^2 + 2ab+2b^2=(a+b)^2+a^2+b^2.$

regarding**Larrioto**'s question:

by mean value theorem for any real numbers $\displaystyle a,b,$ there exists $\displaystyle c$ (between $\displaystyle a,b$) such that $\displaystyle f(b)-f(a)=(b-a)f'c).$ thus: $\displaystyle |f(b)-f(a)|=|b-a||f'(c)| \leq M|b-a|.$