The balanced hyperop sequence
#6
The iterational formula for parabolic iteration (like for \( f(x)=x^x \)) is quite simple, the principal Abel function is:

\( \alpha(x)=\lim_{n\to\infty} \frac{f^{\circ n}(x)-f^{\circ n}(x_0)}{f^{\circ n+1}(x_0)-f^{\circ n}(x_0)} \) for an attracting fixed point at 0 and an arbitrary starting point \( x_0 \) in the attracting domain of the fixed point.
The formula remains still valid for an arbitrary attracting fixed point. The regular iteration is then

\( f^{\circ t}(x)=\alpha^{-1}(t+\alpha(x)) \).
I am not completely sure about the convergence but this should be equivalent (if we substitute \( \alpha \) with its approximations) to
\( f^{\circ t}(x)=\lim_{n\to\infty}f^{\circ -n}(t(f^{\circ n+1}(x_0)-f^{\circ n}(x_0))+f^{\circ n}(x)) \)

In our case however the fixed point at 1 is repelling, so we take advantage of \( f^{-1} \) having an attracting fixed point together with \( f^{\circ t}=(f^{-1})^{\circ -t} \):

\( f^{\circ t}(x)=\lim_{n\to\infty}f^{\circ n}(t(f^{\circ -n}(x_0)-f^{\circ -(n+1)}(x_0))+f^{\circ -n}(x)) \)

here \( f(x)=x^x \) and \( f^{-1}(x)=e^{W(\ln(x))} \).

However the convergence is so fucking slow, already for the Abel function of \( f^{-1} \) that I am not able to post a graph yet!

andydude Wrote:No you didn't. Perhaps it never got through.

Please resend it, have also a look at your private messages.
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Messages In This Thread
The balanced hyperop sequence - by bo198214 - 04/14/2008, 08:44 AM
RE: The balanced hyperop sequence - by andydude - 04/18/2008, 05:23 PM
RE: The balanced hyperop sequence - by bo198214 - 04/18/2008, 05:58 PM
RE: The balanced hyperop sequence - by bo198214 - 04/18/2008, 06:20 PM
RE: The balanced hyperop sequence - by andydude - 04/20/2008, 02:28 AM
RE: The balanced hyperop sequence - by bo198214 - 04/26/2008, 07:20 PM
RE: The balanced hyperop sequence - by bo198214 - 11/30/2009, 11:37 PM

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