08/17/2010, 05:39 AM
From the double-angle formulas for the Jacobi elliptic functions:
\(
cn(2x) = {\frac{cn(x)^2 - sn(x)^2 dn(x)^2} {1 - k^2 sn(x)^4}},
sn(2x) = {{2 sn(x)cn(x)dn(x)}\over{1 - k^2 sn(x)^4}},
dn(2x) = {{dn(x)^2 - k^2 (sn(x)^2 cn(x)^2)}\over{1 - k^2 sn(x)^4}}
\), we can get superfunctions to other rational functions.
For example, \( cn(2^n) \), if my computations are correct, is the superfunction for:
\(
f(z) = {{k - 1 + (2 - 2k) z^2 + k z^4}\over{1 - k^2 + 2 k^2 z^2 - k^2 z^4}}
\), or
\(
f(z) = \frac{\(z^2 - \frac{k+1}{k} + \frac{sqrt(1-k)}{k} \)\(z^2 - \frac{k+1}{k} - \frac{sqrt(1-k)}{k} \)}{-k\(z^2 - \frac{k+1}{k} \)\(z^2 - \frac{k-1}{k} \)}
\).
So for example, picking the modulus to be k=-3, we would get: \( f(z) = \frac{z^2}{3\(z^2 - \frac{2}{3} \)} \). This means that \( nc(2^n, k=-3) \) would be the superfunction for \( f(z) = \(3 - 2 z^2 \) \), which is well outside the Mandelbrot set.
\(
cn(2x) = {\frac{cn(x)^2 - sn(x)^2 dn(x)^2} {1 - k^2 sn(x)^4}},
sn(2x) = {{2 sn(x)cn(x)dn(x)}\over{1 - k^2 sn(x)^4}},
dn(2x) = {{dn(x)^2 - k^2 (sn(x)^2 cn(x)^2)}\over{1 - k^2 sn(x)^4}}
\), we can get superfunctions to other rational functions.
For example, \( cn(2^n) \), if my computations are correct, is the superfunction for:
\(
f(z) = {{k - 1 + (2 - 2k) z^2 + k z^4}\over{1 - k^2 + 2 k^2 z^2 - k^2 z^4}}
\), or
\(
f(z) = \frac{\(z^2 - \frac{k+1}{k} + \frac{sqrt(1-k)}{k} \)\(z^2 - \frac{k+1}{k} - \frac{sqrt(1-k)}{k} \)}{-k\(z^2 - \frac{k+1}{k} \)\(z^2 - \frac{k-1}{k} \)}
\).
So for example, picking the modulus to be k=-3, we would get: \( f(z) = \frac{z^2}{3\(z^2 - \frac{2}{3} \)} \). This means that \( nc(2^n, k=-3) \) would be the superfunction for \( f(z) = \(3 - 2 z^2 \) \), which is well outside the Mandelbrot set.