User talk:MathFacts/Tetration Summary

Dubious

$\log _{a}{\frac {f'_{a}(x)}{f'_{a}(0)(\ln a)^{x}}}=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}(B_{n}(x)-B_{n})$

Could someone please provide a proof or references for this? AJRobbins (talk) 19:46, 3 September 2009 (UTC)

This is based on Faulhaber's formula.--MathFacts (talk) 22:46, 5 September 2009 (UTC)

formula 2 (lagrange)

There occurs an undefined letter "m" in the iteration-parameter of exp, Gottfried —Preceding unsigned comment added by Druseltal2005 (talk • contribs) 05:42, 14 June 2010 (UTC)

Thank you. Corrected.--MathFacts (talk) 00:33, 18 September 2010 (UTC)

couldn't compute the regular sexp-version

I tried your formula for regular iteration method in the following implementation without success. Maybe I've some parsing error...

Taken from:

4. Regular iteration method. It is derived using the technique of regular iteration at the fixed point of the exponential function.

$\operatorname {sexp} _{a}(x)=\lim _{n\to \infty }\log _{a}^{[n]}\left(\left(1-\left(\ln \left({\frac {W(-\ln a)}{-\ln a}}\right)\right)^{x}\right){\frac {W(-\ln a)}{-\ln a}}+\ln \left({\frac {W(-\ln a)}{-\ln a}}\right)\exp _{a}^{[n]}(1)\right)$

The expression ${\frac {W(-\ln a)}{-\ln a}}$ is the principal fixed point of the function $a^{x}$.

For brevity of the formula I assign the fixpoint-expression to the symbol t, and assign the symbol u for u=ln(t)

Then I parse the above sexp-equation in the following way:

$\operatorname {sexp} _{a}(x)=\lim _{n\to \infty }\log _{a}^{[n]}\left(\left(1-u^{x}\right)*t+u*\exp _{a}^{[n]}(1)\right)$

Implementation in Pari/GP is the following:

[t=2, u=log(t), lna = u/t, a=exp(lna)]   \\ set parameters , base a = sqrt(2)

expa_n(n=1,x) = for(k=1,n, x = exp(lna * x) );return(x);
loga_n(n=1,x) = for(k=1,n, x = log(x) / lna );return(x);

sexp_lim(x,n=10)= loga_n( n, (1 - u^x)*t + u* expa_n(n,1))

Then a sequence of approximations for sexp(1), n=1..12^2 is

Mat(vectorv(12,n,sexp_lim(1,n^2)))

[1.34523769828]
[1.39339857929]
[1.41100467230]
[1.41396843923]
[1.41420451351]
[1.41421340180]
[1.41421356100]
[1.41421356237]
[1.41421356237]
[1.41421356237]
[1.41421356237]
[1.41421356237]

For all other x it seems to approximate always sexp(x)→4.0; for instance x=1.1

Mat(vectorv(12,r,sexp_lim(1.1,r^2)))

[1.43416331892]
[1.70442895608]
[2.61205265747]
[3.74394549698]
[3.98512736599]
[3.99958842831]
[3.99999410618]
[3.99999995609]
[3.99999999983]
[4.00000000000]
[4.00000000000]
[4.00000000000]

What am I missing?

Gottfried Helms

--Gotti 16:40, 1 October 2010 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs)

This is a screenshot of Mathematica notebook which demonstrates the method: http://static.itmages.ru/i/10/1002/h_1286026077_9fee2a059e.png --MathFacts (talk) 13:28, 2 October 2010 (UTC)

It would make sense to discuss Kneser's Riemann mapping method

Kneser's Riemann transformation method is a theoretically proven method. The method starts with a regular superfunction solution developed from the fixed point for bases > exp(1/e), and transforms this entire solution to the real valued sexp at the real axis. The downside, is the difficulty in computing a Riemann mapping. Previously, Jay Daniels Fox had done some medium level precision calculations of a Riemann mapping. I have published a pari-gp program on math.eretrandre.org, entitled kneser.gp which calculates iteratively calculates the Riemann mapping over a wide range of bases, by reinterpreting the Riemann mapping as a mathematically equivalent theta(z) mapping. There are many comments on the website, but I have not yet published a paper. Sheldonison (talk) 11:39, 31 May 2011 (UTC) Sheldonison