Fixed point iteration convergence

[lemonthree]

Question: For the following functions, does the fixed point iteration for finding the fixed point in $\left [ 0,1 \right ]$ converge for all first points $ p_{0} \in \left [ 0,1 \right ]$?
Justify your answer.

a.$ g(x) = e^{\frac{-x}{2}}$
b.$ g(x) = 3x - 1$

Let me attempt for part a first. For fixed points, g(p) = p. I believe it is a yes, because it fulfils the conditions of having a convergence in a fix point iteration.
From the graph of $ e^{\frac{-x}{2}}$, I know that g(x) is continuous from $\left [ 0,1 \right ]$. So By Intermediate Value Theorem, I know that there exists a fixed point on $\left [ 0,1 \right ]$. Furthermore, $g'(x) = -0.5e^{\frac{-x}{2}}$, so$ \left | g'(x) \right | \leq 1$. Therefore, fixed point converges and there is a unique fixed point.

Am I right to be saying this? What else should I include to justify my answer, or is the above not helpful at all?

 

[jvicens]

your answer is correct and well-supported. To further justify your answer, you can also mention that the fixed point iteration method is based on the Banach fixed point theorem, which states that for a function g(x) that is continuous on a closed interval [a,b] and satisfies $ \left | g'(x) \right | < 1 $ for all x in [a,b], the iteration process will converge to the unique fixed point in [a,b] for any initial point p0 in [a,b]. This theorem reinforces your argument that since g(x) is continuous and satisfies the condition $ \left | g'(x) \right | \leq 1 $ for all x in [0,1], the fixed point iteration will converge for all initial points p0 in [0,1]. Additionally, you can also mention that the convergence rate of the fixed point iteration for this function is quadratic, making it a fast and efficient method for finding the fixed point.