Given the sequence $\{x_n\}$ generated from the iterative function $\Phi(x)$, $\{x_n\}$ converges with order $p$ to the fixed point $\alpha$ if: $$ \exists \lim_{n \to \infty} \frac{|x_{n+1}-\alpha|}{|(x_n-\alpha)|^{p}} = c $$
The theorem states:
Let $\Phi(x) \in C^p([a,b])$ with $\alpha$ fixed point of $\Phi$ then:
$$ \underbrace{\phi \text{ converges to } \alpha \text{ with order } p}_{a} \iff \underbrace{\phi'(\alpha)=0, \quad ..., \quad \phi^{p-1}(\alpha)=0 \land \phi^p(\alpha)\neq0}_{b} $$ I was able to prove $b\implies a $ using $\Phi$ taylor's expansion and a minimal amount of algebra, but i can't prove $a \implies b$. I made some attemps from my intuition but I'm not able to write formally the proof' steps (sorry I'm a CS guy). Anyone can give me some tips to solve the problem? Thanks.
Hints: you can show that $\phi'(\alpha)=0$ as follows: note that $ \frac {\phi (x_n)-\alpha} {|x_n-\alpha|^{p}}=\frac { x_{n+1}-\alpha} {|x_n -\alpha|^{p}} \to c$. Hence $ \frac {\phi (x_n)-\alpha} {x_n-\alpha}=\frac { \phi (x_n)-\alpha} {|x_n -\alpha|^{p}} \frac {|x_n -\alpha|^{p}} {x_n -\alpha}\to c \times 0=0$ proving that $\phi '(\alpha) =0$. Repeat this argument for higher derivatives.