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§4. The Differential Equation for ℘(z) 175

1 2k−1
= − G k (L)z ,
z
2≤k
−2k
where G k (L) = ω converges for k ≤ 2 by (3.1). Observe that the odd
ω∈L−0
powers for k ≥ 2 sum to zero
−2k+1
ω = 0.
ω∈L−0
Since ℘(z) =−ζ (z) we derive the Laurent series expansions for ℘(z) and ℘ (z) by

differentiation.
(4.1) Laurent Series Expansions.
1 2k−1
ζ(z, L) = − G k (L)z ,
z
2≤k
1 2k−2
℘(z, L) = + G k (L)(2k − 1)z ,
z 2
2≤k

−2
2k−3
℘ (z, L) = + G k (L)(2k − 1)(2k − 2)z ,
z 3
2≤k
−2k
where G k (L) = ω .
ω∈L−0
In order to derive the differential equation for ℘(z), we write out the first few
terms of the expansions at 0 for the elliptic functions ℘(z), ℘ (z), and various com-

binations of these functions:
1 2 4
℘(z) = + 3G 2 z + 5G 3 z +· · · ,
z 2
2
3
℘ (z) =− 3 + 6G 2 z + 20G 3 z +· · · ,
z
4 24G 2
2
℘ (z) = 6 − 2 − 80G 3 +· · · ,
z z
1
3
2
4℘(z) = 4℘(z) + 6G 2 + 10G 3 z +· · ·
z 4
4 36G 2
= + + 60G 3 +· · · ,
z 6 z 2
60G 2 2 2
60G 2 ℘(z) = + 180 (G 2 ) z +· · · ,
z 2
Hence the following equation
2 3
℘ (z) = 4℘(z) − 60G 2 ℘(z) − 140G 3