Also, Leo
If you are interested in hard numerical values; I'd turn to fractional calculus. If \( f(\xi) = -\xi(1-\xi) \), then in the attracting petal \( \mathcal{P} \) about zero, \( f^{\circ n}(\xi) \to 0 \). Then on this basin,
\(
f^{\circ 1/2}(\xi) = \frac{d^{-1/2}}{dw^{-1/2}}|_{w=0} \sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{w^n}{n!}\\
= \frac{1}{\sqrt{\pi}}\int_0^\infty (\sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{(-x)^n}{n!})x^{-1/2}\,dx\\
\)
Which is a consequence of Ramanujan's theorem; which is just a representation theorem.
If \( f = \sin(\xi) \) the above method is the equivalent of doing the Taylor expansion about zero; to make the fractional iterates about 0 (on the real line).
I'll explain this better, if you need me to; but most of this comes from my handwritten papers and its difficult to summarize pages. I haven't culminated this research perfectly yet.
If you are interested in hard numerical values; I'd turn to fractional calculus. If \( f(\xi) = -\xi(1-\xi) \), then in the attracting petal \( \mathcal{P} \) about zero, \( f^{\circ n}(\xi) \to 0 \). Then on this basin,
\(
f^{\circ 1/2}(\xi) = \frac{d^{-1/2}}{dw^{-1/2}}|_{w=0} \sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{w^n}{n!}\\
= \frac{1}{\sqrt{\pi}}\int_0^\infty (\sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{(-x)^n}{n!})x^{-1/2}\,dx\\
\)
Which is a consequence of Ramanujan's theorem; which is just a representation theorem.
If \( f = \sin(\xi) \) the above method is the equivalent of doing the Taylor expansion about zero; to make the fractional iterates about 0 (on the real line).
I'll explain this better, if you need me to; but most of this comes from my handwritten papers and its difficult to summarize pages. I haven't culminated this research perfectly yet.