Numerical algorithm for Fourier continuum sum tetration theory

[nuninho1980]

your code:

Code:

- vt = ConvToFourier(vector(3^4, i, 1));
- for(i=1,5,vt = NormTetIter(32*I, exp(1), vt));
- TetApprox(32*I, vt, exp(1), Pi)

exp(1)^^Pi -> 45349385167682.05641577567842 - 3.105150749 E-12*I - it may not be good.

code by sheldonison:

Code:

- init(exp(1.))
base         2.71828182845905
fixed point  0.318131505204764 + 1.33723570143069*I
iterations   126 used for superf/isuperf

- loop(7)
2 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
60 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498455596179694
8.16051791246692 Riemann/sexp binary precision bits I=0.0157073173118207*I
-0.00163897418270385 recenter/renorm 0.00120350507213262
14 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
64 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498564041905756
15.5432727261501 Riemann/sexp binary precision bits I=0.0157073173118207*I
0.0000252463454086706 recenter/renorm -0.00000302943456839241
50 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
115 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498563283958465
22.9408135385566 Riemann/sexp binary precision bits I=0.0157073173118207*I
-0.000000152690627849965 recenter/renorm 0.00000000801647392410488
26 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
164 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498563287960307
30.5562661278055 Riemann/sexp binary precision bits I=0.0157073173118207*I
0.000000000772404436364769 recenter/renorm -2.53031553890253 E-11
31 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
213 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498563287941024
38.2678540771119 Riemann/sexp binary precision bits I=0.0157073173118207*I
-3.66939771466542 E-12 recenter/renorm 9.18690053231789 E-14
42 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
264 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498563287941115
46.0107842465053 Riemann/sexp binary precision bits I=0.0157073173118207*I
1.70948369274334 E-14 recenter/renorm -3.82049499757912 E-16
47 ctrm(s) out of 50 sexp(z) generates 800 Riemann samples
315 rtrm(s) out of 400 riemaprx(z) generates 100 sexp samples
sexp(-0.5)= 0.498563287941114
53.8101076400847 Riemann/sexp binary precision bits I=0.0157073173118207*I
-7.64624004450182 E-17 recenter/renorm 7.97835112074203 E-19

- sexp(Pi)

exp(1)^^Pi ~= 37149801960.5570 - it's good! because you remember Andrew' slog - this result is almost equal.

your code may be bad!?