I am having a hard time understanding this proof (leading up to Newtons method) about why a sequence must converge. English is not my native language so I will do my best to use the correct mathematical terms.
Theorem: Suppose that a Function $f(x)$ has the property $f(x) = x$ Suppose that this function also has the property of being differentiable in the interval $I$ and that there is a root $a$ to $f(x) = x$ in $I$. Further suppose that:
(1) There is a number $c < 1 $ & $|f'(x)| < c$ when $x \in I$
(2) $f$ "mirrors" the interval $I$ in itself.
then $a$ is the only root to $I$ and $x_n = f(x_{n-1})$ converge towards $a$ for every $x_o$ in $I$.
Now what is should be proved is:
When only (1) is true on $f'$ the sequence will still converge if you start close enough to the root we are searching for. More precisely:
(1) is true and the root $a$ is localized to a subinterval $I_1 = ]a-\delta, a+\delta[$ of $I$ which is small enough that the interval $I_2 = ]a-3\delta, a+3\delta[$ is also contained within $I$.
Then the recursion sequence $x_n = f(x_{n-1})$ converge for every start value $x_o$ in $I_1$
In my book they prove this by saying:
Let $I' = [a-2\delta, a+2\delta]$ then $I_1 \subseteq I' \subseteq I_2$. $f$ will then "mirror" the interval $I'$ in itself. If $x \in I'$ then:
$|f(x) - a| = |f(x) - f(a)| = f'(\xi)(x-a) \le c|x-a|$
This makes that $f(x) \in I'$. According to the theorem this will make the $x_n = f(x_{n-1})$ converge for each $x_o$ in $I'$ and especially for each $x_o \in I_1$.
If someone could explain the proof to me I would be very happy! Thank you for your time!