# To expand Li2 in Taylor series about the point -1/2.
Li2(z) = sum[k positive integer]z^k/k^2.
We differentiate Li2 n times term wise:
d^n/(dz)^n = sum[k >= n](k!/(k-n)!)z^(k-n)/k^2
dli2(n, z) below computes (the nth derivitive of li2 at z)/n!.
#
MODE C = LONG COMPLEX;
MODE NC = LONG COMPLEX;
PROC sq = (REAL z)REAL: z*z;
PROC m2 = (C z)REAL: sq(SHORTEN RE z) + sq(SHORTEN RE z);
PROC li2 = (C z) C: IF m2(z) < 1 THEN C s := 0, f := z; FOR n WHILE REAL fn = n;
   C t = f/(fn*fn); ABS t > 1.e-16 DO s +:= t; f *:= z OD; s
   ELSE 0 FI;
PROC dli2 = (INT n, NC z) NC:
  (NC s := 0, f := 1; FOR k FROM n WHILE NC t = f/(k*k); ABS t > 1.e-18 DO
      s +:= t; f *:= z*((k+1)/(k-n+1)) OD; s);
NC nc = -0.5 I 0.01; INT ts = 70;
[0:ts] NC u;
u[0] :=  li2(nc);
FOR n TO ts DO u[n] := dli2(n,  nc); print((n, u[n], newline)) OD;
(PROC li2h = (C zo)C: (C s := 0, f := 1; C z = zo - nc;
  FOR n FROM 0 WHILE C t = f*u[n]; ABS t > 1.e-16 DO s +:= t; f *:= z OD; s);
 PROC s = (STRING m, C z) VOID: print((z, " ", m, newline));
 PROC t = (C z)VOID: (s("z", z); s("li2h", li2h(z)); s("li2", li2(z)); print(newline));
 t(0); t(-0.5); t(0.1); t(-0.99); t((-0.52) I .452); t(-1);
 t(long cos(1) I long sin(1)))