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tommy's simple solution ln^[n](2sinh^[n+x](z))
#1
Consider f(z,x) = Lim(n --> oo)  ln^[n] ( 2sinh^[n+x] (z) ).

This simple Function satisfies exp(f(z,x)) = f(z,x+1).

So we have a simple superfunction that requires only the real iterations of 2sinh(z).

Notice lim ( n --> oo) 2sinh^[n]( 2^(z-n)) is a superf for 2sinh.

f(z,x) could be analytic for re(z) > 1.

Also , is it really new ?

Or is it the ( analytic continuation ? ) of the 2sinh method ?
It sure is very similar.

----

Mick wondered if F^[n] ( g^[n] ) is analytic for f = sqrt and g = x^2 +1.

---

Regards

Tommy1729
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#2
(01/16/2017, 01:29 PM)tommy1729 Wrote: Consider f(z,x) = Lim(n --> oo)  ln^[n] ( 2sinh^[n+x] (z) ).

This simple Function satisfies exp(f(z,x)) = f(z,x+1).
...
Or is it the ( analytic continuation ? ) of the 2sinh method ?
consider ignoring z, and using the formal inverse Schroeder series (below) for putting 2sinh^[ox] into correspondence with 2^x. 
Then your sexp2sinh/TommySexp function is exactly:

k=0.0678383660707522254065
This also happens to be the definition I personally used for your TommySexp function, but numerically, they are all exactly the same; infinitely differentiable but conjectured nowhere analytic.  
TommySexp(-0.5)=0.498743364531671
Kneser's sexp(-0.5)= 0.498563287941114

Since f(x) is only defined at the real axis, the term analytic continuation has no meaning.

"Mick wondered if F^[n] ( g^[n] ) is analytic for f = sqrt and g = x^2 +1", 
yes it is.  So long as you restrict yourself to the region where |g^[n]|>>1 then it will converge.  Have Mick ask on Mathstack if he wants more details.

Code:
2sinh^[0] = formal2sinh_ischroeder(1) = 1.05804904330694441126
{formal2sinh_ischroeder=
 x +
+x^ 3*  1/18
+x^ 5*  13/5400
+x^ 7*  1193/14288400
+x^ 9*  219983/87445008000
+x^11*  225002297/3280062250080000
+x^13*  3624242332901/2095369366596105600000
+x^15*  294797208996087793/7208971629918239589408000000
+x^17*  532541776280711150089/581464560943620715682280960000000
+x^19*  4423796286922654904342141267/225896613039975363731770463347368960000000 ...}
- Sheldon
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