[Update] Comparision of 5 methods of interpolation to continuous tetration

[sheldonison]

....
Here the same for start at z0=1:

f(h) = 1.00000000000 + 1.30087612467*h + 0.613490605591*h^2 + 0.462613730384*h^3 + 0.258026678269*h^4 + 0.163196550177*h^5 + 0.0896232963846*h^6 + 0.0521088351004*h^7 + 0.0278728768121*h^8 + 0.0153988549222*h^9 + 0.00803787843644*h^10 + 0.00428567564173*h^11 + 0.00218991589639*h^12 + 0.00113674868744*h^13 + 0.000570258372500*h^14 + 0.000289775759114*h^15 + 0.000143052447222*h^16 + 0.0000714303563040*h^17 + 0.0000347673520279*h^18 + 0.0000171062518351*h^19 + 0.00000822180736115*h^20 + 0.00000399449536057*h^21 + 0.00000189817814974*h^22 + 0.000000912157817172*h^23 + 0.000000428985593627*h^24 + 0.000000204179146341*h^25 + 0.0000000951111473740*h^26 + 0.0000000448892717431*h^27 + 0.0000000207246449571*h^28 + 0.00000000970923382052*h^29 + 0.00000000444495348205*h^30 + 0.00000000206897297403*h^31 + 0.000000000939569121455*h^32 + 0.000000000434896014592*h^33 + 1.95950745547 E-10*h^34 + 9.02701746055 E-11*h^35 + 4.03579535562 E-11*h^36 + 1.85202423681 E-11*h^37 + 8.21532589062 E-12*h^38 + 3.75895249924 E-12*h^39 + 1.65398790921 E-12*h^40 + 7.55348988152 E-13*h^41 + 3.29536565950 E-13*h^42 + 1.50386648676 E-13*h^43 + 6.50048867954 E-14*h^44 + 2.96864069036 E-14*h^45 + 1.27005340533 E-14*h^46 + 5.81424087541 E-15*h^47 + 2.45836244673 E-15*h^48 + 1.13062762175 E-15*h^49 + 4.71495264346 E-16*h^50 + 2.18452094165 E-16*h^51 + 8.96002942713 E-17*h^52 + 4.19710010287 E-17*h^53 + 1.68674972326 E-17*h^54 + 8.02580312930 E-18*h^55 + 3.14409639911 E-18*h^56 + 1.52906227353 E-18*h^57 + 5.79803950799 E-19*h^58 + 2.90599482712 E-19*h^59 + 1.05635889369 E-19*h^60 + 5.51749837772 E-20*h^61 + 1.89731889651 E-20*h^62 + 1.04844959488 E-20*h^63 + 3.34780766727 E-21*h^64 + O(h^65)

No guarantee that this is correct; I just let Pari/GP expand the symbolic expression of the indicated computation with fixed x=<number> and h indeterminate.

Gottfried

The difference between your taylor series, and the Kneser solution taylor series is reasonably small. All the coefficients differ by less than 2.5^10-6. To me, it does indeed look like the Kneser solution, both in the complex plane and the real axis. Of course, there is the limit in accuracy to 10^-6, which makes it difficult to say if it really is the same function in the limit, but this is much more accurate than some of the other non-equivalent solutions, like the base change solution, or Tommy's 2sinh solution. I don't have the time right now, to dig into your algorithm to understand why it might be the same as the Kneser approach.
- Sheldon

Code:

differenece at sexp(0) where sexp(z)=1
{difference=
        0
+x^ 1*  0.00000182148778692911855
+x^ 2*  0.00000236395128074178906
+x^ 3*  0.0000000364681675375429581
+x^ 4* -0.000000488269140029040618
+x^ 5* -0.000000715322531855772069
+x^ 6* -0.000000894405252808757919
+x^ 7* -0.000000604527450774569225
+x^ 8* -0.000000566747124818283086
+x^ 9* -0.000000325020053083984524
+x^10* -0.000000262436257811905389
+x^11* -0.000000138691634039540176
+x^12* -0.000000101836267062431268
}
{kneser=
        1.00000000000000000
+x^ 1*  1.30087430318221307
+x^ 2*  0.613488241639719258
+x^ 3*  0.462613693915832462
+x^ 4*  0.258027166538140029
+x^ 5*  0.163197265499531856
+x^ 6*  0.0896241907898528088
+x^ 7*  0.0521094396278507746
+x^ 8*  0.0278734435592248183
+x^ 9*  0.0153991799422530840
+x^10*  0.00803814087269781191
+x^11*  0.00428581433336403954
+x^12*  0.00219001773265706243
}
{gottfried=
        1.00000000000000000
+x^ 1*  1.30087612467000000
+x^ 2*  0.613490605591000000
+x^ 3*  0.462613730384000000
+x^ 4*  0.258026678269000000
+x^ 5*  0.163196550177000000
+x^ 6*  0.0896232963846000000
+x^ 7*  0.0521088351004000000
+x^ 8*  0.0278728768121000000
+x^ 9*  0.0153988549222000000
+x^10*  0.00803787843644000000
+x^11*  0.00428567564173000000
+x^12*  0.00218991589639000000
}

I also looked at your Taylor series at 0, which is accurate to 10^-4, so it isn't nearly as accurate. To some extent, this might be because the inherent radius of converegence around sexp(-1)=0 is radius=1, so one must be more careful go get an accurate Taylor series. At sexp(0)=1, the radius of convergence is 2, which is sometimes easier to work with.
- Sheldon

Code:

differnce at sexp(-1) where sexp(z)=0
{differncem1=
        0
+x^ 1*  0.000000356817567666656170
+x^ 2* -0.00000779165247553834254
+x^ 3*  0.0000223778203088194483
+x^ 4* -0.0000392692063114664084
+x^ 5*  0.0000588782841305703862
+x^ 6* -0.0000798939836470416862
+x^ 7*  0.000101728675963082681
+x^ 8* -0.000123762693382960570
+x^ 9*  0.000145474277203847197
+x^10* -0.000166438511665979836
+x^11*  0.000186279179283386476
+x^12* -0.000204687037326049608
}