TPID 4

[sheldonison]

What ? I never saw any exceptions to the chain rule for analytic functions ?

sexp(w-1) chain law works. sexp(w+1) chain law works.
In fact the idea that the derivative of sexp relates to a product is the result of the chain law.

So you say :
1) sexp(w+k) is analytic in w,k,w+k and sexp(w) is analytic in w.
2) sexp(w+k) = exp^[k](sexp(w))
3) exp^[k] is also analytic.
4) Yet the derivative of exp^[k](sexp(w)) IS NOT exp^[k] ' (sexp(w)) * sexp ' (w) despite that all functions involved are analytic and the conditions for the chain law are fullfilled ?

The flaw in your proof is that exp^[k] may not be analytic, if k is not an integer, even if sexp(z) is analytic. If k is an integer, exp^[k] is well defined. But you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". A simple counter example to your proof is f(z), which has f'(n)=0 and f''(n)=0 for all integers>-2.

\( f(z) = \text{sexp}(z - \frac{sin(2\pi z)}{2\pi})\;\;\; f(z+1)=\exp(f(z)) \)

The reason why is because \( f^{-1}(z) \) has a cube root branch singularity for n>=0, at \( z=\exp^{[n]}(0) \). This is relevant since the exp^[k](z) function used implicitly assumes \( f^{-1}(z) \) is analytic.

\( \exp^{[k]}(z) = f(f^{-1}(z)+k) \)

Also, the f(z) function is the "seed" value used to generate tet_alt(z) from the secondary fixed point, and is a rough approximation for tet_alt, and has the same tet'(n)=0 and tet''(n)=0 for integers>-2.