Newton's Method and One definition?

Question

We are some third year students on CS. We have encountered one definition that was not well formed for us. Could anyone describe it more simpler for us? (i.e: how this derivations is adopted?)

Function $f(x)$ in $x=\alpha$ has one repeated or double roots and $f"(x)$ has defined in neighborhood of $\alpha$. if the sequence {${x_n}$} is calculated for solving $f(x)=0$ with newton's method the following is correct ! ($a$ and $b$ is defined between $\alpha$ and $x_n$)

$\dfrac {x_{n+1}-\alpha}{x_n-\alpha} = \dfrac{f"(a)}{2f"(b)} $

Answer 1

Statement: if $f(x)$ is twice differentiable on $[0,t]$, $f(0)=f'(0)=0$ then $$\exists a \in (0,t) : tf'(t)-f(t) = {t^2 \over 2}f''(a)$$ Proof: let $g(x) = xf'(x)-f(x)$, then $g'(x) = xf''(x)$, $g(0)=0$. For $h(x) = x^2$, by Cauchy's mean value theorem: $$\exists a \in (0,t) : (g(t)-g(0))h'(a) = (h(t)-h(0))g'(a) \\ g(t) * 2a = t^2af''(a) \\ tf'(t)-f(t) = {t^2 \over 2}f''(a) \, |\blacksquare. $$

Now let's suppose $\alpha = 0$ (this is a simple coordinate shift). By mean value theorem, $\exists b \in (0,x_n) : f'(x_n) = x_nf''(b)$. Also, $x_n \neq 0$, $f'(x_n) \neq 0$ (otherwise Newton's method doesn't work).

So we have $${x_{n+1} \over x_n} = {x_n - {f(x_n) \over f'(x_n)} \over x_n} = {x_nf'(x_n)-f(x_n) \over x_nf'(x_n)}$$ Per two previous statements, $$\exists a,b \in (0,x_n) : {x_{n+1} \over x_n} = {{x_n^2 \over 2}f''(a) \over x_n^2f''(b)} = {f''(a) \over 2 f''(b)} \, |\blacksquare.$$

Answer 2

If $x_0$ is sufficiently near $\alpha$, but $\ne\alpha$, then there is a $b$ between $\alpha$ and $x_0$ such that $$f'(x_0)=f'(x_0)-f'(\alpha)=(x_0-\alpha )f''(b)\ne0\ .\tag{1}$$ Computing $f(\alpha)=0$ by means of a Taylor expansion of $f$ at $x_0$ we get $$0=f(x_0)+(\alpha-x_0)f'(x_0)+(\alpha-x_0)^2{f''(a)\over 2}\tag{2}$$ for a suitable $a$ between $x_0$ and $\alpha$. If we now divide $(2)$ by $f'(x_0)\ne0$ then the definition of $x_1$ together with $(1)$ leads to $$0=(x_0-x_1)+(\alpha-x_0)+(x_0-\alpha)^2{f''(a)\over2 f'(x_0)}=\alpha-x_1+(x_0-\alpha){f''(a)\over2 f''(b)}\ ,$$ which is equivalent to $${x_1-\alpha\over x_0-\alpha}={f''(a)\over2 f''(b)}\ .$$