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exp^[3/2](x) > sinh^[1/2](exp(x)) ?
#5
(04/23/2015, 04:38 PM)tommy1729 Wrote: ....
Therefore 2sinh^[1/2] < exp^[1/2].
I use Koenig's for 2sinh^[1/2] and Kneser for exp^[1/2] where 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 where is the Abel function for 2sinh. converges very quickly (super-exponentially) to a 1-cyclic function as z increases, rather than converging to a constant. Wherever the 2sinh^[1/2](sexp(z)) function is larger than exp^[1/2](sexp(z)). Here is a graph of The zeros correspond exactly to the zeros above.
   
- Sheldon
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RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by sheldonison - 04/24/2015, 12:40 AM

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