Fixed point iteration, locally convergent

[John Harris]

Homework Statement

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

compute x_bar 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 x_bar is 1 because g(1)=0. It's not locally convergent because |g^'(1)|=-2

Since x_bar isn't locally convergent I can't computer x_bar. But calculating it would look like:
k | x_k
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

 

[mfb]

The solution x_bar 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 x_bar.

Since x_bar isn't locally convergent I can't computer x_bar.

Oh sure you can, just not directly with the given iteration.

 

[John Harris]

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 x_bar.
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 x_bar is the solution, but I'm not sure of what and how to get it.

 

[mfb]

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?

 

[John Harris]

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

 

[mfb]

$g(0)=\frac{1}{0^2}-1$ is undefined.

 

[John Harris]

$g(0)=\frac{1}{0^2}-1$ is undefined.

Sorry I was thinking 2^x 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+5^1/2/2 come from in the attached picture.

 

[John Harris]

I did fixed point iteration, but it's definitely not converging. I wish I knew what x_bar is and how to get it.

 

[SammyS]

Sorry I was thinking 2^x 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+5^1/2/2 come from in the attached picture.

Here's that image:

I would have never guessed this from the OP.

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

 

[John Harris]

Here's that image:

I would have never guessed this from the OP.

Apparently, $\displaystyle\ \frac{1+\sqrt{5}}{2} \$ is x_bar 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.

 

[mfb]

I did fixed point iteration, but it's definitely not converging. I wish I knew what x_bar 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.

 

[John Harris]

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.

 

[mfb]

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.

 

[SammyS]

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 x_bar, g(x_bar) = x_bar . In other words, x_bar is a root of the equation, g(x) = x. (It's one of two roots.)

x_bar 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 x_bar 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 x_bar in [0, 1] .

Are you calling the function f(x) or calling it g(x) for your problem?

 

[John Harris]

I figured it out. Thank for the help