02/24/2013, 08:36 PM
Perhaps formal derivation:
Attractive fixpoint \( z_0 \) means
\( \left|\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\right|=c<1 \).
By continuity of the left side in \( z \) there is a disk \( D_\eps(z_0) \) with center \( z_0 \) such that \( \left|\lim_{z\to z_0}\frac{f(z)-z_0}{z-z_0}\right|=d<1 \) for all \( z\in D_\eps(z_0) \). (\( c<d<1 \))
So for any \( z\in D_\eps(z_0) \) we have
\( |f(z)-z_0|<d|z-z_0| \), particularly \( f(z) \) is again in \( D_\eps(z_0) \).
\( |f(f(z))-z_0|<d|f(z)-z_0|<d^2|z-z_0| \)
and so on
\( |f^{[n]}(z)-z_0|<d^n|z-z_0| \).
The right sight converges to 0 for \( n\to\infty \), hence by definition of convergence: \( f^{[n]}(z)\to z_0 \) for \( n\to\infty \).
Attractive fixpoint \( z_0 \) means
\( \left|\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\right|=c<1 \).
By continuity of the left side in \( z \) there is a disk \( D_\eps(z_0) \) with center \( z_0 \) such that \( \left|\lim_{z\to z_0}\frac{f(z)-z_0}{z-z_0}\right|=d<1 \) for all \( z\in D_\eps(z_0) \). (\( c<d<1 \))
So for any \( z\in D_\eps(z_0) \) we have
\( |f(z)-z_0|<d|z-z_0| \), particularly \( f(z) \) is again in \( D_\eps(z_0) \).
\( |f(f(z))-z_0|<d|f(z)-z_0|<d^2|z-z_0| \)
and so on
\( |f^{[n]}(z)-z_0|<d^n|z-z_0| \).
The right sight converges to 0 for \( n\to\infty \), hence by definition of convergence: \( f^{[n]}(z)\to z_0 \) for \( n\to\infty \).