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Searching for an asymptotic to exp[0.5]
(08/03/2014, 04:54 AM)sheldonison Wrote:
(08/02/2014, 11:48 PM)tommy1729 Wrote: ....
f(x) is the half-iterate from the sinh method.

I assume since sinh is an odd function, that f(x), the asymptotic half iterate, would also be an odd function, though this is not required. As I remember, the closest singularity to the origin for sinh^{0.5} is on the imaginary axis.

For f(x) asymptotic to exp^{0.5}, the branch cut is on the negative real axis. For the asymptotic to sinh^{0.5}, one possible branch cut is on the imaginary axis, in both directions, which would lead to an odd function, with even Taylor series coefficients=0. Is this what you had in mind?

Just as you use kneser in post 9 , I use the 2sinh here.
So just for numerical reasons.
Im still on the real line.

It is convenient since it satisfies ln(f(exp(x))) = f(x) what simplifies the equations.



Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/03/2014, 08:46 AM

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