Cryptography Reference
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are needed for convergence. With some care, one could justify rearranging the
sum, but it is easier to do the following. Differentiating
℘
(
z
) term by term
yields
℘
(
z
)=
−
2
ω
1
ω
)
3
.
(
z
−
∈
L
Note that
ω
= 0 is included in the sum. This sum converges absolutely (by
comparison with the case
k
= 3 in Lemma 9.4) when
z ∈ L
, and changing
z
to
z
+
ω
shifts the terms in the sum. Therefore,
℘
(
z
+
ω
)=
℘
(
z
)
.
This implies that there is a constant
c
ω
such that
℘
(
z
+
ω
)
− ℘
(
z
)=
c
ω
,
for all
z
∈
L
. Setting
z
=
ω/
2 yields
c
ω
=
℘
(
−
ω/
2)
−
℘
(
ω/
2) = 0
,
by (3). Therefore
℘
(
z
+
ω
)=
℘
(
z
). This proves (4).
Let
f
(
z
) be any doubly periodic function. Then
f
(
z
)=
f
(
z
)+
f
(
−
z
)
2
+
f
(
z
)
−
f
(
−
z
)
2
expresses
f
(
z
) as the sum of an even function and an odd function. Therefore,
it su
ces to prove (5) for even functions and for odd functions. Since
℘
(
−
z
)=
℘
(
z
), it follows that
℘
(
℘
(
z
), so
℘
(
z
) is an odd function. If
f
(
z
)
is odd, then
f
(
z
)
/℘
(
z
) is even. Therefore, it su
ces to show that an even
doubly periodic function is a rational function of
℘
(
z
).
Let
f
(
z
) be an even doubly periodic function. We may assume that
f
is
not identically zero; otherwise, we're done. By changing
f
, if necessary, to
−
z
)=
−
af
+
b
cf
+
d
for suitable
a, b, c, d
with
ad−bc
=0,wemayarrangethat
f
(
z
)doesnothave
a zero or a pole whenever 2
z ∈ L
(this means that we want
f
(0)
=0
, ∞
and
f
(
ω
i
/
2)
=0for
i
=1
,
2
,
3). If we prove (
af
+
b
)
/
(
cf
+
d
) is a rational function
of
℘
,thenwecansolvefor
f
and obtain the result for
f
.
Since
f
(
z
) is even and doubly periodic,
f
(
ω
3
−
z
)=
f
(
z
), so
ord
w
f
=ord
ω
3
−w
f.
We can therefore put the finitely many elements in
F
where
f
(
z
)=0or
where
f
(
z
) has a pole into pairs (
w, ω
3
− w
). Since we have arranged that
w
=
ω
3
/
2, the two elements of each pair are distinct.
There is a slight
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