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 ...
... you want to show, for $\displaystyle t<0$, $\displaystyle f(t)<t$, which is equivalent to $\displaystyle f(t)-t<0$. Let $\displaystyle g(t)=f(t)-t$. You have $\displaystyle g(0)=0$. If you can prove that $\displaystyle g$ is strictly increasing, this will imply what you need. Can you carry on?
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 $\displaystyle g$ is differentiable on $\displaystyle (-\infty,0]$ and $\displaystyle g'(x)<0$ for every $\displaystyle x<0$ then $\displaystyle g$ is strictly decreasing on $\displaystyle (-\infty,0]$". This would give you $\displaystyle g(x)>g(0)$ for $\displaystyle x<0$, which is what you want.
Now, using the mean value theorem... If you apply it to $\displaystyle f=\sin$ on $\displaystyle [x,0]$ where $\displaystyle x<0$, you get $\displaystyle \sin x-\sin 0=(\cos c)(x-0)$ for some $\displaystyle c\in(x,0)$, hence $\displaystyle \sin x= x \cos c\geq x$ (since $\displaystyle x<0$ and $\displaystyle \cos c\leq 1$). It doesn't give you the strict inequality though (you may a priori have $\displaystyle \cos c=1$). I don't know how you could get it easily from the mean value theorem.
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?
Define $\displaystyle f(t):=sin(t)-t$Show $\displaystyle sin(x)>x$ for $\displaystyle x<0$.
By algebra of continuity if $\displaystyle g(t)=sin(t)$ and $\displaystyle h(t)=-t$ then since $\displaystyle f(t)=g(t)+h(t)$ then $\displaystyle f(t)$ is continuous on $\displaystyle \mathbb{R}$.
Say $\displaystyle f(t)$ is continuous on $\displaystyle [a,b]$ where $\displaystyle a<b<0$ and $\displaystyle a,b \in \mathbb{R}$.
$\displaystyle f(t)$ is differentiable on $\displaystyle (a,b)$. In fact $\displaystyle f'(t)=cos(t)-1$
Hence, by the Mean Value Theorem $\displaystyle \exists \ t_0 \in (a,b) \ s.t \ f'(t_0)=\frac{f(b)-f(a)}{b-a}$.
$\displaystyle f'(t)=cos(t)-1<1-1=0$.
Hence $\displaystyle f'(t_0)=\frac{f(b)-f(a)}{b-a}<0 \Rightarrow \ f(b)<f(a)$ (ie. strictly decreasing)
$\displaystyle f(b)=sin(b)-b$ and $\displaystyle f(a)=sin(a)-a$.
Now we have $\displaystyle sin(b)-b<sin(a)-a \Rightarrow \ a-b<sin(a)-sin(b)$
$\displaystyle a-b<2cos(\frac{a+b}{2})sin(\frac{a-b}{2})<2sin(\frac{a-b}{2})$
Hence $\displaystyle 2sin(\frac{a-b}{2})>a-b \Rightarrow \ sin(\frac{a-b}{2})>\frac{a-b}{2} $.
Therefore define $\displaystyle x:=\frac{a-b}{2}$ so we get $\displaystyle sin(x)>x$ as required.