08/12/2022, 05:59 AM
(08/12/2022, 01:51 AM)tommy1729 Wrote: ...
tommy1729
Thank u tommy
About Julia equation, we know \(\lambda(z)=\frac{1}{\alpha'(z)}\)
and we know that for any 1-periodic function \(\theta(z)\) we know \(\alpha(z)+\theta(\alpha(z))\) is also an abel function.
So we combine these 2 and then have \(\lambda(z)\frac{1}{1+\theta'(\alpha(z))}\) is also a julia function.
and wlog \(\theta'(z)\) is also arbitrarily 1-periodic, so \(\lambda(z)\frac{1}{1+\theta(\alpha(z))}\) is also a julia function and so is \(\lambda(z)\theta(\alpha(z))\).
It seems now that we can only discern 2 different julia function by a multiplication by \(\theta(\alpha(z))\), still lack of some Fourier technique.
Regards, Leo

