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exp^[3/2](x) > sinh^[1/2](exp(x)) ?
#7
(04/24/2015, 02:07 PM)tommy1729 Wrote: Sheldon, that is exactly as expected.
The reason you get this result is because you used kneser instead of my 2sinh method.

I assumed Kneser's exp^[1/2](z), since it is analytic in the complex plane Smile
- Sheldon
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RE: exp^[3/2](x) > sinh^[1/2](exp(x)) ? - by sheldonison - 04/24/2015, 02:59 PM

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