Daniel Wrote:\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \) is the general solution
Which formula for the coefficients is this exactly?
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computing the iterated exp(x)-1
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