06/05/2014, 08:21 AM
A good id(z) has been found !
id(z)/z = 1 sometimes but that is no big problem.
id(z) = z * fake(1) on the real line
id(z) = z * fake(exp(exp(i z))^O(z)) on the imag line.
id(z) is of the type z * f(z i) where f(z i) is similar too
\( f(z i)=\int_0^\infty \frac{e^{zt i}}{t^t}\mathrm{d}t \)
Yes the same function that occured in the fake exp^[1/2] thread.
It seems " fake function theory " is developping fast.
regards
tommy1729
id(z)/z = 1 sometimes but that is no big problem.
id(z) = z * fake(1) on the real line
id(z) = z * fake(exp(exp(i z))^O(z)) on the imag line.
id(z) is of the type z * f(z i) where f(z i) is similar too
\( f(z i)=\int_0^\infty \frac{e^{zt i}}{t^t}\mathrm{d}t \)
Yes the same function that occured in the fake exp^[1/2] thread.
It seems " fake function theory " is developping fast.
regards
tommy1729