01/17/2017, 07:21 AM
(This post was last modified: 01/17/2017, 04:19 PM by sheldonison.)
(01/16/2017, 01:29 PM)tommy1729 Wrote: Consider f(z,x) = Lim(n --> oo) ln^[n] ( 2sinh^[n+x] (z) ).consider ignoring z, and using the formal inverse Schroeder series (below) for putting 2sinh^[ox] into correspondence with 2^x.
This simple Function satisfies exp(f(z,x)) = f(z,x+1).
...
Or is it the ( analytic continuation ? ) of the 2sinh method ?
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