12/14/2008, 05:51 PM
(This post was last modified: 12/14/2008, 06:00 PM by Kouznetsov.)
bo198214 Wrote:\( \Psi(z)=\frac{1}{c}\log(z-c)+\sum_{0\le m\le n } c_{m,n} (z-c)^{m+\frac{2\pi i n}{c}} \)I do not understand why \( m\le n \). The sum with first 5 terms with \( n=0 \) gives the approximation of slog with function \( f \)
\( f=f(z)=\frac{1}{c}\log(z-c)+\sum_{m=0}^4 c_{m,0} (z-c)^{m} \)
shown in the figure below.
Levels \( \Re(f)=0,\pm 1, \pm2 \) are shown with thick black lines.
Levels \( \Re(f)=-1.9, .. -0.1 \) are shown with thin red lines.
Levels \( \Re(f)=0.1, .. 1.9 \) are shown with thin blue lines.
Levels \( \Im(f)=0.1, .. 1.9 \) are shown with thin dark green lines.
Levels \( \Im(f)=0 \) is shown with thick green line.
(Deviation of this line from the real axis indicates the error of the approximation.)
Level \( \Im(f)=-1 \) is shown with thick red line
Levels \( \Im(f)=1,2,3 \) are shown with thin dark blue lines lines.
For comparison, dashed lines show the precise evaluation for some of the levels above for the robust implementation of the slog function as inverse of tetration. While \( |z-L|<1 \), the deviation of these dashed lines from the levels for function \( f \) is not seen even at the strong zooming-in of the central part of the figure.
Here is the code I used for the evaluation of the coefficients:
fog := z -> log((exp(z)-1)/z)
M := 6
Digits := 41
slo := z-> sum(v[n]*z^n, n = 1 .. M)
F := series(slo(z)-slo((exp(z)-1)*L)-fog(z), z, M)
Le := solve(log(x) = x, x)
L := conjugate(evalf(Le, 40))
for m to M-1 do v[m] := solve(coeff(F, z^m) = 0, v[m]) end do
Can anybody suggest a similar code for evaluation of the "fractal-power" coefficients, for example, \( c_{0,1} \), \( c_{1,1} \) ?

