Prove Newtons method converges quadratically f(x)= cosh(x) +cos(x) -3 
Let x* be the none negative root of f. Prove that Newton's Method applied to f converges quadratically to x*.
Really confused where to start for a proof. I understand that if the second derivative of f(x*) doesn't equal 0 it converges precisely quadratic. 
But don't no how to start or what to include. Looked through a few books and it gets very confusing!
 A: I will explain. Take as true the following statement:
If the second derivative of $f(x^*)$ is nonzero, then newton's theorem converges quadratically to $x^*$. 
Now, to show that newton's method converges quadratically we only have to show that $f''(x^*)\neq 0$. The problem is we don't know what $x^*$ is so we have to consider all the possible values it could take on. The good news is that $f''(x) \neq 0$ for all $x\neq 0$. Thus, as long as $x^*\neq 0$, $f''(x^*) \neq 0$. How can we conclude $x^*\neq 0$? We can substitute 0 into the function and see it is not 0, $f(0)=1+1-3=-1\neq 0$. Thus, whatever $x^*$ is, it is nonzero, and thus we can conclude that $f''(x^*)\neq 0$ and thus newton's method converges quadratically to $x^*$.
A: And,
of course,
the reason for the
quadratic convergence
is that
$f(x+h)
=f(x)+hf'(x)+O(h^2)
$
so,
if
$h = -f(x)/f'(x)
$,
$\begin{array}\\
f(x-f(x)/f'(x))
&=f(x)+f'(x)(-f(x)/f'(x))+O((f(x)/f'(x))^2)\\
&=O((f(x)/f'(x))^2)\\
&=O((f(x))^2)\\
\end{array}
$
if $f'(x)$
is bounded away from zero.
