exp^[3/2](x) > sinh^[1/2](exp(x)) ?

[sheldonison]

....
Therefore 2sinh^[1/2] < exp^[1/2].

I use Koenig's for 2sinh^[1/2] and Kneser for exp^[1/2] \( \;\;\alpha^{-1}(\alpha(z)+0.5)\;\; \) where \( \alpha(z) \) is the Abel or slog function
s2inh^[1/2](6) = 20.0717649
exp^[1/2](6) = 20.0860615
at x=6, exp^[1/2](x) is bigger

s2inh^[1/2](20) ~= 399.098221
exp^[1/2](20) ~= 398.247512
at x=20, 2sinh^[1/2](x) is bigger.

The first crossing where the two functions are equal to each other for positive reals occurs at x~=9.49535513; the second crossing where the two functions are equal occurs at x~=54.3741864, which is ~=exp^[1/2](9.49). The third crossing where the two functions are equal occurs at ~=13297.7591 which is near exp(9.49); the fourth crossing occurs near 4.11537228*10^23, which is almost exactly exp(54.3741864). This pattern of the two functions intersecting each other repeats infinitely....

To explain the pattern, you might look at \( \theta(z)=\alpha(\text{sexp}(z))-z\;\; \) where\( \alpha(z) \) is the Abel function for 2sinh. \( \theta(z) \) converges very quickly (super-exponentially) to a 1-cyclic function as z increases, rather than converging to a constant. Wherever \( \theta(z+0.5)-\theta(z)<0 \) the 2sinh^[1/2](sexp(z)) function is larger than exp^[1/2](sexp(z)). Here is a graph of \( \theta(z+0.5)-\theta(z)\; \) The zeros correspond exactly to the zeros above.