Prove quadratic convergence and limit

[radutanasa]

Hello,

I have this problem that I simply cannot nail down. Please help.
xi - fixed point of g, g is twice continuously differentiable in a vecinity of xi.

z(n+1) = g(g(z(n))) - [g(g(z(n))) - g(z(n))]^2 / [g(g(z(n))) - 2*g(z(n)) + z(n)]

Using taylor series expansion of g(z(n)) and g(g(z(n))) in a vecinity of xi I have to prove that xi is z's limit and that the convergence is quadratic.

Thank you for your help!

Radu

 

[Crosson]

Interesting problem. I don't know a smart way to do it, so I just dive right in and calculate the series up to two terms around x_i (I won't bother to indicate the higher order terms):

<br /> g(z) = \frac{1}{2} g&#039;&#039;\left(x_i\right) \left(z-x_i\right){}^2+g&#039;\left(x_i\right) \left(z-x_i\right)+g\left(x_i\right)<br />

<br /> g(g(z)) = \frac{1}{2} \left(g&#039;&#039;\left(g\left(x_i\right)\right) g&#039;\left(x_i\right){}^2+g&#039;\left(g\left(x_i\right)\right) g&#039;&#039;\left(x_i\right)\right)<br /> \left(z-x_i\right){}^2+g&#039;\left(g\left(x_i\right)\right) g&#039;\left(x_i\right) \left(z-x_i\right)+g\left(g\left(x_i\right)\right)<br /> <br />

Now if we go ahead and replace these into your expression and use the fact that g(g(x_i) = g(x_i) = x_i we get the following for z_{n + 1}:

\frac{g&#039;&#039;\left(x_i\right){}^2 \left(x_i-z\right){}^3+4 x_i \left(g&#039;\left(x_i\right)-1\right){}^2+2 \left(z-x_i\right) \left(g&#039;\left(x_i\right)-1\right) \left(2 x_i+z<br /> g&#039;\left(x_i\right)\right) g&#039;&#039;\left(x_i\right)}{4 \left(g&#039;\left(x_i\right)-1\right){}^2+2 \left(z-x_i\right)<br /> \left(g&#039;\left(x_i\right){}^2+g&#039;\left(x_i\right)-2\right) g&#039;&#039;\left(x_i\right)}<br />

Note: I dropped the subscript z_n because of formatting problems with LaTeX in these forums.

I do not know how to simplify the expressions involving the derivatives, but maybe this is already to the point where simplifying into a particular form might make it easy to see the convergence.