08/13/2007, 10:47 PM
\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \) is the general solution where
\( f^0(z) = z \)
\( f^1(z) = e^z-1 \) and
\( f^m(f^n(z)) = f^{m+n}(z) \).
\( f^0(z) = z \)
\( f^1(z) = e^z-1 \) and
\( f^m(f^n(z)) = f^{m+n}(z) \).
Daniel
