MVT inequality
Let $f$ be continuous on $[a,b]$ and suppose $f$ is continuously differentiable on $(a,b)$.
There is a number $c$ in $(a,b)$ such that
$$f(b)-f(a) = f'(c)(b-a).$$
For all $a,b\in\mathbb{R}, \vert\sin b - \sin a \vert \leq |b-a|$.
The MVT implies that for any real numbers $a$ and $b$,
$$\sin b - \sin a = \cos (c)(b-a)$$ for some $c$ between $a$ and $b$.
Taking absolute values and recalling $|\cos(x)| \leq 1$ for any $x$ gives the desired result.
Fixed points
A fixed point of a function $g$ is a number $p$ in the domain of $g$ such that $g(p)=p$.
Think of a function as a rule producing a specified output for each given input. A point is fixed by $g$ when the output matches the input.
Graphically, fixed points correspond to intersections of the graph of $g$ with that of the line $y=x$.
Simple example
$\sin \color{var(--emphColor)}{0}=\color{var(--emphColor)}{0}$.
The sine function fixes the input zero.
Logistic growth
Population model given by
\begin{equation}
p_{n+1}= cp_n (1-p_n),
\end{equation}
where $c$ is a growth constant in $(0,4)$ and $p_n$ is the population at the $n$th time step, normalized relative to the environment's capacity.
Any fixed point of this model satisfies
\begin{equation}
p= cp (1-p).
\end{equation}
Solving this quadratic gives fixed points $p=0,\frac{c-1}{c}$.
Visualizing fixed points
$f(x)=e^{-x}$.
The fixed point equation $x=e^{-x}$ cannot be solved analytically, but the graph shows it has a solution. How can we compute it?
Tough question, partly because a function may or may not have fixed points,
and when it does, it may have many.
However, if the function satisfies some hypotheses, we're in good shape.
Fixed point theorem
If $g:[a,b]\to[a,b]$ is continuous, then $g$ has a fixed point $p\in [a,b]$. Further, if $g$ is differentiable on $(a,b)$ and $|g'(x)|\leq k< 1$ for $x\in(a,b)$ and some $k$, the fixed point is unique.
“ A smooth function that maps an interval to itself
must have a fixed point, and if the function does not grow
too quickly,the fixed point is unique. ”
If $g:[a,b]\to[a,b]$ is continuous, then $g$ has a fixed point $p\in [a,b]$. Further, if $g$ is differentiable on $(a,b)$ and $|g'(x)|\leq k< 1$ for $x\in(a,b)$ and some $k$, the fixed point is unique.
(Existence) Use auxiliary device.
Define $h(x) = g(x) - x$ and note $h$ is continuous. The fixed points of $g$ are precisely the roots of $h$. (Think of IVT here.)
Since $\min_{x \in [a,b] }g(x)\geq a$, $\max _{x \in [a,b]}g(x)\leq b $, $h(a)=g(a)-a\geq 0$, $h(b)=g(b)-b\leq0$.
If $h(a)=0$ or $h(b)=0$, we obtain a fixed point of $g$. If not, then $h(b) < 0 < h(a)$ and by the IVT there is $p$ between $a,b$ such that $h(p)=0$, giving a fixed point of $g$.
(Uniqueness) By contradiction. Suppose there are two fixed points. Then
\begin{align}
|p-q| & =|g(p)-g(q)|=|g'(c)||p-q| \\
& \leq k|p-q|<|p-q|
\end{align}
OK, so how can I compute it?
Our fixed point theorem
tells us when a function is guaranteed to have a fixed point.
Suppose we know $g$ has a fixed point in $[a, b]$. How can we approximate it?
Procedure
Given $g$ and a starting point $p_0$, compute
$$p_{n+1}=g(p_{n})$$
for each $n \geq 0$.
(“ Babylonian ” divide and average) Let $g(x) = \frac{1}{2}\big(x + {2 \over x}\big)$.
Note that
$$g(\color{var(--emphColor)}{\sqrt{2}}) = \frac{1}{2}\bigg(\sqrt{2} + {2 \over \sqrt{2}}\bigg) = \color{var(--emphColor)}{\sqrt{2}}$$
so $p = \sqrt{2}$ is a fixed point of $g$.
$p_{n+1} = g(p_n) = \frac{1}{2}\big(p_n + {2 \over p_n}\big)$