f( f(x) ) = exp(x) solved ! ! !

[tommy1729]

note that this can also be used to compute

f(f(f(x))) = exp(x).

But slowly, slowly. Before we continue we first have to see whether your idea works. Thats not really clear. The thing is that the regular iteration at a fixed point depends strongly on the derivatives at the fixed point. However your function \( \exp(-x^2) \) messes up the derivations of \( \exp \) at 0. From that point of view its doubtful whether it results in something usable.

well , i dont take the derivates of exp(-x^2)

to construct the taylor series of exp(-x^2), i take the expansion of exp(y) and substitute y = - x^2.

simple and efficient.

regards

tommy1729