Cyclic complex functions and uniqueness
#3
jaydfox Wrote:To illustrate, consider a sexp solution \( S(z) \) and its inverse \( S^{\small -1}(z) \). Let's say that, in the vicinity of the upper fixed point c_0, \( S^{\small -1}(z) \approx \log_{c_0}\left(z-c_0\right) \).

Now let's say we have an alternate solution \( R(z) \) and its inverse \( R^{\small -1}(z) \). Let's say that \( R(z)=S\left(z+\frac{\sin(2\pi z)}{200\pi}\right) \).

I forgot to get back to my point with the logarithm.

To simplify my calculations, I'm going to reverse the relationship of R and S, though I suspect the difference is nearly trivial:

\(
\begin{eqnarray}
S^{\small -1}(z) & \approx & \log_{c_0}\left(z-c_0\right) & \text{for } \left\|z-c_0\right\| < \delta \\
\\[5pt]

\\
S(z) & = & R\left(z+\frac{\sin(2\pi z)}{200\pi}\right) \\
\\[5pt]

\\
R^{\small -1}\left(S(z)\right) & = & z+\frac{\sin(2\pi z)}{200\pi}
\end{eqnarray}
\)

Bear in mind, the logarithmic approximation was in the immediate vicinity of the upper primary fixed point. Here, I've specified that z is within some arbitrarily small disk centered at the fixed point.

Then:
\(
\begin{eqnarray}
R^{\small -1}(z)
& = & R^{\small -1}\left(S\left(S^{\small -1}(z)\right)\right) \\
& = & R^{\small -1}\left(S(w)\right) \text{ with } w = S^{\small -1}(z) \\
& = & w+\frac{\sin(2\pi w)}{200\pi} \\
& = & S^{\small -1}(z)+\frac{\sin\left(2\pi S^{\small -1}(z)\right)}{200\pi} \\
\\[5pt]

\\
& \approx & \log_{c_0}\left(z-c_0\right) + \frac{\sin\left(2\pi \log_{c_0}\left(z-c_0\right)\right)}{200\pi}
\end{eqnarray}
\)

Given that \( \log_{c_0}\left(z-c_0\right) \) will have a large imaginary part near the fixed point, the sine of the logarithm will have the large oscillations I previously described.
~ Jay Daniel Fox
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Messages In This Thread
RE: Cyclic complex functions and uniqueness - by jaydfox - 10/29/2007, 05:31 AM

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