# Does This Sequence Converge to the Fixed Point?

• GregA
In summary, Homework Equations attempts to find a function which when graphed has a fixed point at the given point. It is easy to prove that the function has a fixed point if the equation for the given point is replaced by x_{n+1} and alpha.
GregA
[SOLVED] Proof of convergence

Let $\alpha[/tex] be a fixed point of $$x = g(x)$$ and let $$(x_n)$$ be the sequence generated by the fixed point iteration scheme. Using the first two terms of the Taylor series for $$g(x)$$ about [itex]\alpha[/tex] we can get an approximation for $$g(x_n)$$: [itex]g(x_n) = g(\alpha) + (x_n - \alpha)g'(\alpha)[/tex] (Assuming the terms in the sequence are close to α we have neglected non- linear terms.) Show that [itex]|x_n - \alpha| = |x_0 - \alpha||g'(\alpha)|^{n}[/tex] for all n $$\geq 1$$ hence show that the sequence $$(x_n)$$ converges to [itex]\alpha[/tex] for $$g'(x)< 1$$ ## Homework Equations ## The Attempt at a Solution I'm not so worried about the proving second part (looks obvious if the first part is true) but before trying to prove the first I want to try a couple of examples and see what's happening if I can. If I suppose that my function $$g(x) = \sqrt{x+1}$$ then the value of a fixed point [itex] \alpha = \frac{1+\sqrt{5}}{2}[/tex] Now if I let $$x_0 = 2$$ and set n = 1 then $$x_1 = \sqrt{2+1}$$ and I am under the impression that I can now show: [itex]|\sqrt{3} - \alpha| = |2 -\alpha||\frac{1}{2}(\frac{1}{\sqrt{\alpha+1}})|[/tex] $$\Rightarrow |\sqrt{3} - \frac{1+\sqrt{5}}{2}| = |2 - \frac{1+\sqrt{5}}{2}||\frac{1}{2}(\frac{1}{\sqrt{\frac{1+\sqrt{5}}{2}+1}})|$$ But this is false!...How am I misinterpreting the given statement or what have I done wrong? Please don't prove the problem for me. Last edited: You are right, it is not true- it is only approximately true. After all, you neglected powers higher than the first in the Taylor expansion. Proving [itex]|x_n- \alpha|=|x_0- \alpha||g'(\alpha)^n$, to first order, by induction, is pretty easy if you replace $g(x_n)$ and $g(\alpha)$ in $g(x_n) = g(\alpha) + (x_n - \alpha)g'(\alpha)$ by $x_{n+1}$ and $\alpha$.

Last edited by a moderator:
Cheers for that HallsofIvy, it was the presence of an equals sign, not approximately equals that was throwing me; I couldn't see why it should have been true

## What is "Proof of Convergence: Solved!"?

"Proof of Convergence: Solved!" is a scientific concept that refers to the ability to prove that a system or process will eventually reach a desired outcome or state.

## Why is "Proof of Convergence: Solved!" important?

It is important because it allows scientists to predict and understand the behavior of complex systems and processes, which can have real-world applications in fields such as physics, engineering, and computer science.

## What are the key components of "Proof of Convergence: Solved!"?

The key components of "Proof of Convergence: Solved!" are a defined starting point or initial conditions, a set of rules or equations that govern the system's behavior, and a proven method or algorithm for reaching the desired outcome.

## How is "Proof of Convergence: Solved!" different from "Proof of Convergence"?

The addition of "Solved!" indicates that a solution or method for proving convergence has been found, whereas "Proof of Convergence" simply refers to the concept itself.

## What are some real-world examples of "Proof of Convergence: Solved!"?

Some examples include the proof of convergence for the Newton-Raphson method in mathematics, the proof of convergence for the Jacobi method in linear algebra, and the proof of convergence for the gradient descent algorithm in machine learning.

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