$f(x)$ is a function such that for every $x$, $f'(x)$ is well defined and $0<m<f'(x)<M$. We are using a Newton-like method to find its root. consider sequence $x_{n+1} = x_n - \lambda f(x_n)$ where $0<\lambda<2/M$. We want to proof that these sequence converge to a root of $f(x)$. i.e. $\lim_{x \rightarrow \infty} f(x_n) = 0$.
This question is for a quiz of an introductory Numerical Method course and I know only simple undergraduate Calculus concepts like Bolzano Theorem and Taylor Series.
I searched for the question in Internet and came up with some articles about Newton-Kantorovich methods. It seems they are related to this problem but they use a lot of advanced topics I am unaware of.
Another thing I found was something that used a method called Newton's Inexact Method with formula: $$x_{n+1} = x_n − F′(x_n)^{-1}F(x_n) + y_n$$ and substituted $$y_n = (1 − λ)F′(x_n)^{−1}F(x_n)$$
but Newton's Inexact Method isn't covered in the course and it seems it uses more advanced tools than just simple Calculus theorems.
Is there any simple and elementary solution to solve the problem? I know some simple methods like Newton-Raphson and fixed-point method.
Let $g(x) = x - \lambda f(x)$.
For any $u < v$, apply MVT to $g(x)$ over $[u,v]$, you can find a $w \ in (u,v)$ such that $$g(u) - g(v) = (u-v)(1 - \lambda f'(w))$$
Notice $$-1 = 1 - \frac{2}{M}M < 1 - \lambda M < 1 - \lambda f'(w) < 1 - \lambda m < 1$$ We have $$| 1 - \lambda f'(w)| \le K \quad\text{ where }\quad K = \max(|1-\lambda m|,|1 - \lambda M|) < 1$$ This leads to $|g(u) - g(v)| \le K |u-v|$ and hence $g(x)$ is a contraction.
By contractive mapping theorem, the function $g(x)$ has a unique fixed point $x^*$and every sequence defined recursively by iteration of $g$:
$$x_{n} = \begin{cases} g(x_{n-1}), & n > 0\\ a, & n = 0\end{cases}$$
converges to this fixed point $x^*$ independent of the seed value $a$.
Since $x^* - \lambda f(x^*) = g(x^*) = x^*$, we have $\lambda f(x^*) = 0 \implies f(x^*) = 0$, this unique fixed point $x^*$ is a root of $f(x)$.