1. ## Mean Value Theorem

Hi,

I would like to know how to proof the following:

Show that sin x > x if x < 0.

I think I have to start by
let x<0.
let f(t) = Sin t
and f(t) is continuous on [- infinity,0) and differentiable on (-infinity,0)
and ...

2. Originally Posted by zxcv
Hi,

I would like to know how to proof the following:

Show that sin x > x if x < 0.

I think I have to start by
let x<0.
let f(t) = Sin t
and f(t) is continuous on (- infinity,0] and differentiable on (-infinity,0] (note I changed the intervals)
and ...
... you want to show, for $t<0$, $f(t), which is equivalent to $f(t)-t<0$. Let $g(t)=f(t)-t$. You have $g(0)=0$. If you can prove that $g$ is strictly increasing, this will imply what you need. Can you carry on?

3. So what I want is f(t)>t and f(t)-t>0.
Let g(t) = f(t)-t. Let g(0) =0.

By the Mean Value Theorem, since f(t) is cont. and differentiable, then there is a c in (- infinity, 0) such that

f(0) - f(-infinity) = f '(c) (0 - -infinity)

And for any x1 and x2 in [- infinity, 0] with x1<x2, there exists a c in (x1,x2) such that g(x2)-g(x1) = g '(x)(x2-x1).
g ' (x) = cos x-1 <0 for x in (- infinity, 0).
if g '(x) <0 for all x in (- infinity, 0), then f is strictly increasing.

Can you guide me more, I find myself a bit confusing at times ...

4. Originally Posted by zxcv
So what I want is f(t)>t and f(t)-t>0.
Let g(t) = f(t)-t. Let g(0) =0.

By the Mean Value Theorem, since f(t) is cont. and differentiable, then there is a c in (- infinity, 0) such that

f(0) - f(-infinity) = f '(c) (0 - -infinity)

And for any x1 and x2 in [- infinity, 0] with x1<x2, there exists a c in (x1,x2) such that g(x2)-g(x1) = g '(x)(x2-x1).
g ' (x) = cos x-1 <0 for x in (- infinity, 0).
if g '(x) <0 for all x in (- infinity, 0), then f is strictly increasing.

Can you guide me more, I find myself a bit confusing at times ...
First you shouldn't write "f(-infinity)" or take x1=-infinity... this makes no sense indeed (especially since f has no limit at -infinity).

I didn't take the title "Mean value theorem" into account when I wrote my previous post; I thought you could use the result "If $g$ is differentiable on $(-\infty,0]$ and $g'(x)<0$ for every $x<0$ then $g$ is strictly decreasing on $(-\infty,0]$". This would give you $g(x)>g(0)$ for $x<0$, which is what you want.

Now, using the mean value theorem... If you apply it to $f=\sin$ on $[x,0]$ where $x<0$, you get $\sin x-\sin 0=(\cos c)(x-0)$ for some $c\in(x,0)$, hence $\sin x= x \cos c\geq x$ (since $x<0$ and $\cos c\leq 1$). It doesn't give you the strict inequality though (you may a priori have $\cos c=1$). I don't know how you could get it easily from the mean value theorem.

5. If using your method, when you mention using "If g is differentiable on (-infinity,0] and g '(x)<0 for every x<0 then g is strictly decreasing on (-infinity,0]".

I will then start by saying that let g be differentiable on (-infinity,0], that is, g ' (x) = cos x-1 <0 for x in (- infinity, 0).
Then do you know how I should continue on so to really show that g '(x)<0 from your way?

6. Show $sin(x)>x$ for $x<0$.
Define $f(t):=sin(t)-t$

By algebra of continuity if $g(t)=sin(t)$ and $h(t)=-t$ then since $f(t)=g(t)+h(t)$ then $f(t)$ is continuous on $\mathbb{R}$.

Say $f(t)$ is continuous on $[a,b]$ where $a and $a,b \in \mathbb{R}$.

$f(t)$ is differentiable on $(a,b)$. In fact $f'(t)=cos(t)-1$

Hence, by the Mean Value Theorem $\exists \ t_0 \in (a,b) \ s.t \ f'(t_0)=\frac{f(b)-f(a)}{b-a}$.

$f'(t)=cos(t)-1<1-1=0$.

Hence $f'(t_0)=\frac{f(b)-f(a)}{b-a}<0 \Rightarrow \ f(b) (ie. strictly decreasing)

$f(b)=sin(b)-b$ and $f(a)=sin(a)-a$.

Now we have $sin(b)-b

$a-b<2cos(\frac{a+b}{2})sin(\frac{a-b}{2})<2sin(\frac{a-b}{2})$

Hence $2sin(\frac{a-b}{2})>a-b \Rightarrow \ sin(\frac{a-b}{2})>\frac{a-b}{2}$.

Therefore define $x:=\frac{a-b}{2}$ so we get $sin(x)>x$ as required.