According to wiki: https://en.wikipedia.org/wiki/Dedekind_eta_function, Dedekind eta function is defined in many equivalent forms. But none of them is an explicit description (say in algorithmic format) on how to computing it. Where to find such one? Thanks!

1There are some efficient Mathematica programs described in a post on Mathematica.SE – მამუკა ჯიბლაძე May 5 at 4:28

Thanks! But it seems there is no an explicit description on how to compute Dedekind eta function on that page. – Licheng Wang May 5 at 6:18

2is this helpful for you? (also includes Pari code): Evaluation of the Dedekind eta function – Carlo Beenakker May 5 at 7:59

Yes, it is very useful! Thanks a lot! – Licheng Wang May 5 at 9:06

1@LichengWang Hi. Did you download the PARI/GP tarball which has the complete source code? – Somos May 5 at 18:40
My maple code of GatteschiSokal algorithm for computing $R(t,x)=\prod_{n=1}^{\infty}(1tx^n)$:
GS:=proc(t,x,prec)
local R0, a0, b0, Rn, an, bn, d, c, i, N, r, Rd;
N := 100;
d := 1/2; if d = t*x then d := (1/2)*d end if;
r := evalf$[prec]$(1+d/(1x));
a0 := 1; b0 := evalf$[prec]$(d/(dt*x));
R0 := evalf$[prec]$(r*a0+(1r)*b0);
i := 0;
while i < N do
c := evalf$[prec]$(a0*(d*a0+(1.0d)*b0));
an := evalf$[prec]$(c/b0);
bn := evalf$[prec]$(c/(x*a0+(1.0x)*b0));
Rn := evalf$[prec]$(r*an+(1.0r)*bn);
Rd := evalf$[prec]$(abs(RnR0));
if Rd < 10^(prec) then i := N else
a0 := an; b0 := bn; R0 := Rn; i := i+1 end if;
end do;
return Rn;
end proc;
eta:=(t,prec)>evalf$[prec]$(GS(1,exp($2*Pi*I*t$),prec));
Dedekind_eta := (t, prec)> evalf$[prec]$(exp($1/12*Pi*I*t$)*eta(t, prec));
Test:
t:=0.3*I:
eta(t,40); Dedekind_eta(t,40);
0.8251926470787677741036466781518992636742
0.7628619270903183863013294748250092216042
Almost same with PARI/GP output:
eta(0.3*I,0)=0.82519264707876777410364667815189926367
eta(0.3*I,1)=0.76286192709031838630132947482500922160
Euler's formula
$$ \sum\limits_{n \in \mathbb{Z}} {(  1)^n q^{\frac{{(3n^2  n)}} {2}} } = \prod\limits_{n = 1}^\infty {(1  q^n ),} $$
(which can be proven from Jacobi’s triple product identity by using the fact that $\prod\limits_{n = 1}^\infty {(1  q^{3n} )(1  q^{3n  2} )} (1  q^{3n  1} ) = \prod\limits_{n = 1}^\infty {(1  q^n )} $) provides a good way of numerically computing
$$ \eta (\tau )=e^{\frac {\pi {\rm {{i}\tau }}}{12}}\prod _{n=1}^{\infty }(1e^{2n\pi {\rm {{i}\tau }}})=q^{\frac {1}{24}}\prod _{n=1}^{\infty }(1q^{n}). $$
I hope this answers your question.