I know that Newton's method is $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ for n $\geq$ 0. Now, if I consider an interval [a, b] and I suppose that there exists a root $\alpha \in (a, b)(f{(\alpha)} = 0)$. I also assume that both f' and f'' are continuous functions and f'($\alpha$) $\neq$ 0. Using Taylor's Theorem, we have that for some $c_n$ between $\alpha$ and $x_n$ that:
f($\alpha$) = f($x_n$) + ($\alpha-x_n$)f'($x_n$) + $\frac{(\alpha - x_n)^2}{2}f''(c_n)$.
0 = f($x_n$) + ($\alpha-x_n$)f'($x_n$)$ + \frac{(\alpha - x_n)^2}{2}f''(c_n)$.
If we divide both sides of the equation by f'($x_n$), we get that:
0 = $\frac{f(x_n)}{f'(x_n)}$+ ($\alpha-x_n$) + $\frac{(\alpha - x_n)^2}{2}\frac{f''(c_n)}{f'(x_n)}$.
Since $x_{n+1}$ = $x_n-\frac{f(x_n)}{f'(x_n)}$, then, by rearranging these terms, we get $\frac{f(x_n)}{f'(x_n)}$ = $x_n - x_{n+1}$. We can substitute this in the equation above and solve for $\alpha - x_{n+1}$. We get: 0 = ($x_n- x_{n+1}$) + ($\alpha-x_n$) + $\frac{(\alpha - x_n)^2}{2}\frac{f''(c_n)}{f'(x_n)}$.
So, $\alpha-x_{n+1}$ = -$\frac{(\alpha - x_n)^2}{2}\frac{f''(c_n)}{f'(x_n)}$.
So I have a function f(x) = $x^2$ - a. I know that the first derivative of this fuction is 2x. If I substitute this into the above equation, I'd get this, $\alpha-x_{n+1}$ = -$\frac{(\alpha - x_n)^2}{2}\frac{f''(c_n)}{2x}$. Now I get this:
$\alpha-x_{n+1}$ = -$\frac{(\alpha - x_n)^2}{2x}f''(c_n)$
How do I go from this to this final equation:
$\sqrt{a}-x_{n+1} = \frac{-1}{2x_n}(\sqrt{a}-x_{n})^2$
