Aitken's and Steffenson's Process

[rootX]

Homework Statement

For
cos(x)-1 = 0, the Newton raphson function is g(x) = x-tan(x/2)
Use Steffenson's algorithm with g(x) and start with p0, and find
p1, p2, p3, then find p4, p5, p6

Homework Equations

Using Aitken's process with fixed point iteration

The Attempt at a Solution

Find p1,p2,p3 using fixed point:
So p1 = g(p0); p2 = g(p1);...

plug these values into aitken's formula:
q0 = p0 - (p1-p0)^2/...

And here I get q0
What should I do next?

Thanks a lot.
If you can also provide an example that demonstrates the use of Aitken's process with Newton's Method and second for Steffenson? I am using Matthew's Numerical Methods and it really does not provide any information about how to use them.

 

[TheoMcCloskey]

I guess I'm a little fuzzy on how to answer based on the wording of the problem. Normally, you would procede as follows

As you said, you have $p_0$. Then, let

$$p_1 = g(p_0), p_2 = g(p_1), p_3 = g(p_2), \ldots, p_n = g(p_{n-1})$$

Now apply the Steffensen Algo (Aitken's process)

$$\^p_0 = p_0 - \frac{(p_1 - p_0)^2}{(p_2-2p_1-p_0)}$$

$$\^p_1 = p_1 - \frac{(p_2 - p_1)^2}{(p_3-2p_2-p_1)}$$

$$\ldots$$

$$\^p_n = p_n - \frac{(p_{n+1} - p_{n})^2}{(p_{n+2} - 2p_{n+1} - p_{n})}$$

These p-hats are the revised estimates to the original p's. More simply, you could have used the subscripted variable "q".

The wording of the problem statement makes me wonder. Is this the exact wording? It almost reads as if they want you to project the next three iterates (?). Are they asking for p4, p5, p6 to be the Aitken estimates using p0, p1, p2, p3 (not eneough) ? or whatever, .. its not clear.

Note, in all cases, you'll need n+2 iterates to generate n Aitken's iterates

If I was doing this problem, I'd generate a table of $p_i$ iterates for i = 0 to 8. I would then apply the Steffensen Algo to generate $q_i=\^p_i$ for i = 0 to 6.

But, that's just me - doesn't mean its what the problem statement is looking for.

 

[TheoMcCloskey]

RootX - please disregard what I said in previous post - it's clear to me now (way to earlier for me this Sunday)

Sorry for confusion - please DO NOt do the procedure I suggested. I'll provide a much better example in a few minutes of how we do this in practice.

Again sorry for the confusion.

 

[TheoMcCloskey]

What this problem is asking for is the application of Steffensen Algorithm to speed up the convergence of the fixed point iteration.

What follows is my understanding of the problem statement.

What they want you to do is use p0 (given), compute p1=g(p0) and p2=g(p1), and then to construct the first Aitken's iterate p3.

Then, repeat this process replacing p0 by p3. This will generate three more iterates, p4, p5, p6. The method to compute p4 is the same as p1, ie p4=g(p3), and the method to compute p5 is the same as the method for p2, namely, p5=g(p4). The value for p6 is the second Aitken's iterate.

The process can be repated of course, but your only asked to generate up to p6.

This can all be diagram in a nice short table

i = 0 => p0, p1, p2
i = 1 => p3, p4, p5
i = 2 => p6

Note that

$$p_3 = p_0 - \frac{(p_1 - p_0)^2}{(p_2-2p_1-p_0)}$$

and

$$p_6 = p_3 - \frac{(p_4 - p_3)^2}{(p_5-2p_4-p_3)}$$

Give it a try, I think you'll find it works. An interseting comparision would be to compare this process with the original Newton-Raphson method and see how the convergence compares between the two methods.

I hope this clears things up a bit.

 

[rootX]

Yes, it does.
Thanks a lot!