1828 CHAPTER 58. ELLIPTIC FUNCTIONS
This function is odd. Here is why.
ζ (−z) =1−z
+ ∑w ̸=0
1−z−w
− zw2 +
1w
while
−ζ (z) =1−z
+ ∑w̸=0
−1z−w
− zw2 −
1w
=1−z
+ ∑w̸=0
−1z+w
− zw2 +
1w.
Now consider 58.1.7. It will be used to find the Laurent expansion about the origin for ζ
which will then be differentiated to obtain the Laurent expansion for ℘ at the origin. Sincew ̸= 0 and the interest is for z near 0 so |z|< |w| ,
1z−w
+z
w2 +1w
=z
w2 +1w− 1
w1
1− zw
=z
w2 +1w− 1
w
∞
∑k=0
( zw
)k
= − 1w
∞
∑k=2
( zw
)k
From 58.1.7
ζ (z) =1z+ ∑
w̸=0
(−
∞
∑k=2
zk
wk+1
)
=1z−
∞
∑k=2
∑w̸=0
zk
wk+1 =1z−
∞
∑k=2
∑w ̸=0
z2k−1
w2k
because the sum over odd powers must be zero because for each w ̸= 0, there exists−w ̸= 0such that the two terms z2k
w2k+1 and z2k
(−w)2k+1 cancel each other. Hence
ζ (z) =1z−
∞
∑k=2
Gkz2k−1
where Gk = ∑w̸=01
w2k . Now with this,
−ζ′ (z) = ℘(z) =
1z2 +
∞
∑k=2
Gk (2k−1)z2k−2
=1z2 +3G2z2 +5G3z4 + · · ·
Therefore,
℘′ (z) =
−2z3 +6G2z+20G3z3 + · · ·