Ansus Wrote:\( \Delta[f]=\exp f - f \)
Since \( \Delta[f] = -\frac{1}{2\pi i}
\int_{-1-i\infty}^{-1+i\infty} \frac{f(z+x)}{z(z-1)}\, dz \), we derive f(x).
Though I dont know where from you got this formula, if I assume that the formula is correct and slightly reformulate it:
\( \exp(f(z_0)) - f(z_0) = -\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{f(it+z_0-1)}{(it-1)(it-2)} dt \)
for \( z_0 \) on the imaginary axis \( z_0=is \):
\( f(i s)-\exp(f(i s))= - \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{f(it+is-1)}{(it-1)(it-2)} dt \)
then it can also be used to iteratively compute the superexponential (base \( e \)) on the imaginary axis:
\( \fbox{f(is)=\exp(f(i s)) + \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\log(f(i(t+s)))}{(it-1)(it-2)} dt} \)
Any volunteer to implement this formula?
PS: This formula needs no assumption about the value of convergence of \( f \) for \( z\to i\infty \), the only arbitrarity is the choosen branch of logarithm.
