156
12 Weierstraß Functions
e 3 = −
π 2
12ω 2
1
2ϑ
4
2 (0, q) + ϑ
4
4 (0, q)
.
(12.6c)
With ˜
z =
πz
2ω 1
,
(12.6d)
℘ (z) − e 1 =
πϑ 3 (0, q)ϑ 4 (0, q)ϑ 2 (˜ z, q)
2ω 1 ϑ 1 (˜ z, q)
2
,
(12.6e)
℘ (z) − e 2 =
πϑ 2 (0, q)ϑ 4 (0, q)ϑ 3 (˜ z, q)
2ω 1 ϑ 1 (˜ z, q)
2
,
(12.6f)
℘ (z) − e 3 =
πϑ 2 (0, q)ϑ 3 (0, q)ϑ 4 (˜ z, q)
2ω 1 ϑ 1 (˜ z, q)
2
.
(12.6g)
To evaluate the ζ(z) and σ (z) functions, we need, in addition, Jacobi’s identity
d
dz
ϑ 1 (0, q) = ϑ 2 (0, q) ϑ 3 (0, q) ϑ 4 (0, q) , and
(12.6h)
η j = −ω j e j + 2RG(0, e j − e k , e j − e l ),
(12.6i)
with (j, k, l) cyclic permutations.
The Weierstraß function ℘ (z) could be evaluated via Eqs. (12.6a) and (12.6e);
σ (z) is given by
σ (z) = 2ω 1 exp
η 1 z 2
2ω 1
ϑ 1 (˜ z, q)
π
d
dz ϑ 1 (0, q)
,
(12.7)
and ζ(z) within the half-period parallelogram is given by
ζ(z) = −z℘ (z) + 2RG(℘ (z) − e 1 , ℘ (z) − e 2 , ℘ (z) − e 3 )
(12.8a)
and in general by
ζ(z) =
η 1
ω 1
z +
π
2ω 1
d
dz
log ϑ(˜ z, τ ).
(12.8b)
We will use the first equation only if (τ ) < 0.02 and z sufficiently small. The
logarithmic derivative could be evaluated by
d
dz
log ϑ(z, τ ) = cot(z) + 4
∞
n=1
q 2n
1 − q 2n sin(2nz).
(12.8c)