Fixed point iteration, locally convergent

AI Thread Summary
The discussion revolves around determining the local convergence of fixed point iteration for the function g(x) = 1/x^2 - 1. The key point is that the solution xbar is identified as 1, but it is not locally convergent since |g'(1)| = -2, indicating divergence. Participants express confusion about the definition of a fixed point and the conditions for convergence, particularly the requirement that |g'(xbar)| < 1. The conversation also touches on the need to graph the function and numerically find the fixed point, with some suggesting that the original g(x) may not lead to convergence. Ultimately, the goal is to clarify the conditions for convergence and accurately identify xbar for the given function.
John Harris
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Homework Statement


For which of them will the corresponding fixed point iteration xk+1 = g(xk) be locally convergent to the solution xbar in [0, 1]? (The condition to check is whether |g'(xbar)| < 1.)
A) 1/x2 -1
B)...
C)...

compute xbar to within absolute error 10-4.

Homework Equations


3. The Attempt at a Solution
I[/B] don't think I understand the question in general terms, but this is what I think I have to do.
The solution xbar is 1 because g(1)=0. It's not locally convergent because |g'(1)|=-2

Since xbar isn't locally convergent I can't computer xbar. But calculating it would look like:
k | xk
1 1
2 -2
3 .25
4 128

Can you please help me understand what this problem is asking for, or similar examples of this problem, or what I'm doing wrong. Thank you
 
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John Harris said:
The solution xbar is 1 because g(1)=0.
That is not a fixed point as your point changes (from 1 to 0) and the next iteration becomes impossible because you would have to divide by zero.
What does "fixed point" mean? What is fixed?
It's not locally convergent because |g'(1)|=-2
You'll have to check that again once you find the right xbar.
Since xbar isn't locally convergent I can't computer xbar.
Oh sure you can, just not directly with the given iteration.
 
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mfb said:
That is not a fixed point as your point changes (from 1 to 0) and the next iteration becomes impossible because you would have to divide by zero.
What does "fixed point" mean? What is fixed?
You'll have to check that again once you find the right xbar.
Oh sure you can, just not directly with the given iteration.

I'm guessing a fixed point is a point that can't change. My teacher said xbar is the solution, but I'm not sure of what and how to get it.
 
John Harris said:
I'm guessing a fixed point is a point that can't change.
It does not change if you apply g to it, so the next iteration starts with the same value again. What does that mean in terms of equations?
 
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mfb said:
It does not change if you apply g to it, so the next iteration starts with the same value again. What does that mean in terms of equations?
Is it 0 because g(0)=0
 
##g(0)=\frac{1}{0^2}-1## is undefined.
 
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mfb said:
##g(0)=\frac{1}{0^2}-1## is undefined.
Sorry I was thinking 2x for no reason.

I'm looking at the graph and all the values of x seem to have different y's. I'm not sure what I could put into g(x) that wouldn't change.

Where does the 1+51/2/2 come from in the attached picture.
 

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I did fixed point iteration, but it's definitely not converging. I wish I knew what xbar is and how to get it.
 
John Harris said:
Sorry I was thinking 2x for no reason.

I'm looking at the graph and all the values of x seem to have different y's. I'm not sure what I could put into g(x) that wouldn't change.

Where does the 1+51/2/2 come from in the attached picture.
Here's that image:
capture-png.79090.png

I would have never guessed this from the OP.

Apparently, ##\displaystyle\ \frac{1+\sqrt{5}}{2} \ ## is xbar and is a solution to the equation, g(x) = x .
 
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  • #10
SammyS said:
Here's that image:
capture-png.79090.png

I would have never guessed this from the OP.

Apparently, ##\displaystyle\ \frac{1+\sqrt{5}}{2} \ ## is xbar and is a solution to the equation, g(x) = x .

That's the root of g'(x), but my problem doesn't have any roots, and my problem has an interval. So I have no idea.
 
  • #11
John Harris said:
I did fixed point iteration, but it's definitely not converging. I wish I knew what xbar is and how to get it.
Did you draw a sketch? You can get a rough idea where the fixed point is, and solve the equation numerically.
 
  • #12
mfb said:
Did you draw a sketch? You can get a rough idea where the fixed point is, and solve the equation numerically.
Yes I graphed it. I don't know what the I 'm looking for though. I tried the roots of the of f(x) and g'(x). And points close to the intersection of the graphs, but nothing is converging when I do it numerically.

edit:
I just tried it with another g(x) and it's converging to the root of f(x). So I'm pretty sure the root of f(x) is what I'm looking for and the original g(x) is not convergent.
 
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  • #13
What is f(x)?

It is not convergent, right, but it still has a fixed point. Can you write down the equation for the fixed point? What has to be true at this point? We said it in words, but you'll need the equation.
 
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  • #14
John Harris said:
That's the root of g'(x), but my problem doesn't have any roots, and my problem has an interval. So I have no idea.
I take it that the image you posted, (and what I responded to), was for an example with the function ## \displaystyle \ g(x)=1+\frac{1}{x} \ ##. For this example problem, ## \displaystyle \ x_\text{bar}=\frac{1+\sqrt{5}}{2} \ ## because for that value of xbar, g(xbar) = xbar . In other words, xbar is a root of the equation, g(x) = x. (It's one of two roots.)

xbar is not a root (zero) of g'(x). Note that ## \displaystyle \ g'(x)=-\frac{1}{x^2} \ ## which has no root. However, as stated in the example, ## \displaystyle \ \left|g'(x_\text{bar})\right|<1 \, ,\ ## so the fixed point iteration gives a sequence converging to xbar for the example problem.

It's not entirely clear to me what the problem is that you're asked to solve. I guess that you're to do similarly for ## \displaystyle \ \frac{1}{x^2}-1 \ ## (or is it ## \displaystyle \ \frac{1}{x^2-1}\, ? \ ##) . It's likely the former since that gives xbar in [0, 1] .

Are you calling the function f(x) or calling it g(x) for your problem?
 
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  • #15
I figured it out. Thank for the help
 
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