# Newton's Method convergence for all approximations in $[a,b]$

Given a function $$F(x)$$ defined on $$[a,b]$$ such that:

• $$f$$ $$\in$$ $$C^2([a,b])$$
• $$F(a)F(b)$$<0
• $$F'(x) \neq 0$$, $$\forall$$ $$x$$ $$\in$$ $$[a,b]$$
• $$F''(x)$$ $$\geq$$ $$0$$ or $$F''(x)$$ $$\leq$$ $$0$$, $$\forall$$ $$x$$ $$\in$$ $$[a,b]$$
• $$(*)\frac{|F(c)|}{|F'(c)|}$$ $$<$$ $$b-a$$, where $$c$$ is the endpoint of $$[a,b]$$ where $$|F'(x)|$$ reaches the minimum value among them $$\\$$ Then $$f$$ converges to the only zero in $$[a,b]$$ for any initial approximation $$\mathrm{p_{0}}$$ $$\in$$ $$[a,b]$$.

I'm having trouble in showing that statement $$(*)$$ guarantees that any initial approximation $$\in$$ $$[a,b]$$ will converge to the only root of $$F$$ in $$[a,b]$$. I could write $$G(x)=x-\frac{F(x)}{F'(x)}$$ and if I could guarantee that it's a contraction, i.e., $$G([a,b])=[a,b]$$ and that $$\exists$$ $$\$$ $$0<\lambda < 1$$: $$\forall x,y \in[a,b]$$, $$|G(x)-G(y)|$$ $$\leq$$ $$\lambda|x-y|$$ , then by the Fixed-Point Theorem, there is an unique fixed point $$p$$ in $$[a,b]$$ and $$G(p)=p \iff p- \frac{F(p)}{F'(p)}=p \iff \frac{F(p)}{F'(p)}=0 \iff F(p)=0$$. I know that $$G'(x)=\frac{F(x)F''(x)}{(F'(x))^2}$$ and it could also be shown that $$\exists$$ $$\$$ $$0<\lambda < 1$$: |$$G'(x)$$| $$\leq$$ $$\lambda$$, $$\forall$$ $$x$$ $$\in$$ $$[a,b]$$ leading to the same conclusion above, but I've not found a way to show it. Any suggestions?

• You can prove it by using the convergence of a monotone bounded sequence. Oct 25, 2023 at 21:11
• So the last statement assures that every element of the sequence will belong to the interval $[a,b]$?
– J P
Oct 25, 2023 at 21:14
• Yes. That's the purpose of it. Oct 25, 2023 at 21:20
• Have a look at Darboux theorem numdam.org/article/NAM_1869_2_8__17_0.pdf Oct 26, 2023 at 9:07