Res. Lett. Inf. Math. Sci., (2000) 1, 65{78
Available online at http://www.massey.ac.nz/?wwims/rlims/
The development of elliptic functions according
to Ramanujan and Venkatachaliengar
Shaun Cooper
I.I.M.S., Massey University Albany Campus, Auckland, N.Z.
s.cooper@massey.ac.nz
Abstract
These notes are based on the monograph Development of Elliptic Functions according to Ra-
manujan by K. Venkatachaliengar [2]. The goal of the notes is to show how some of the main
properties of Jacobian and Weierstrass elliptic functions can be developed in an elementary way
from the 1?1 function.
All of the ideas presented in these notes can be found in Venkatachaliengar?s book. The only
thing I have done is to rearrange the order in which the material is presented. I am entirely re-
sponsible for any errors in these notes, and would be very grateful to be informed about them,
whether they be large or small.
1 Introduction
Throughout these notes, let ? be a flxed complex number which satisfles Im ? > 0 and let q = ei?? ,
so that jqj < 1. We will make use of the following notation for products. Let
(a; q)n =
n?1
Y
j=0
(1? aqj);
(a; q)1 =
1
Y
j=0
(1? aqj);
(a1; a2; : : : ; an; q)1 = (a1; q)1(a2; q)1 : : : (an; q)1:
We will use the symbol
P
n to denote summation over all integer values of n from ?1 to 1, and
P0
n will be used to denote summation over all integer values of n from ?1 to1, excluding n = 0.
The 1?1 function is deflned to be
1?1(a; b; q; x) =
X
n
(a; q)n
(b; q)n
xn:
Ramanujan?s 1?1 summation formula is
X
n
(a; q)n
(b; q)n
xn =
(ax; q=ax; q; b=a; q)1
(x; b=ax; b; q=a; q)1
: (1.1)
For a proof of this result and for additional information about the 1?1 function, please see [1,
equation (3.15)].
The Jordan-Kronecker function, which is introduced by Venkatachaliengar on p.37, is a special
66 R.L.I.M.S. Vol. 1, Sept. 2000
case of the 1?1 function, and is deflned as follows.
Deflnition Let
f(a; t) =
1
X
n=?1
tn
1? aq2n : (1.2)
This series converges provided jq2j < jtj < 1, and so long as a 6= q2k, k = 0;?1;?2; : : :. Using the
1?1 summation formula (1.1) we obtain
f(a; t) =
(at; q2=at; q2; q2; q2)1
(t; q2=t; a; q2=a; q2)1
: (1.3)
This extends the deflnition of f to all values of a and t except for a; t = q2k, k = 0;?1;?2; : : :,
where there are simple poles. The following results are immediate from (1.3):
f(a; t) = f(t; a); (1.4)
f(a; t) = ?f(1=a; 1=t); (1.5)
f(a; t) = tf(aq2; t) = af(a; tq2): (1.6)
The twelve Jacobian elliptic functions correspond to the twelve functions f(A;Bei?), whereA = ?1; q
or ?q, and B = 1;?1; q or ?q. The precise identiflcations will be given at the end of these notes.
The Weierstrass ? and } functions are also related to the Jordan-Kronecker function f , and some
of these connections will be given in sections 3 and 4.
At the beginning of chapter 3 of [2], Venkatacheliangar derives a fundamental multiplicative iden-
tity for the function f . He uses this identity to develop the theory of the Weierstrass and Jacobian
elliptic functions. These notes describe how to obtain results such as the difierential equations and
addition formulas for the Jacobian elliptic functions, the connection between the Weierstrass }
function and the Jacobian elliptic functions, and the difierential equation for the } function, from
Venkatachaliengar?s fundamental multiplicative identity.
These notes deal with only a small part of the theory of elliptic functions. Topics such as hyperge-
ometric functions, modular transformations and the problem of inversion are not even mentioned
here. These topics, however, are taken up and developed in Venkatachaliengar?s book.
2 The fundamental multiplicative identity and the Weierstrass } func-
tion
Venkatachaliengar?s development of elliptic functions is based on the following result (see [2, p. 37]).
Theorem (Fundamental multiplicative identity)
f(a; t)f(b; t) = t
@
@t
f(ab; t) + f(ab; t)(?1(a) + ?1(b)); (2.1)
where the function ?1 is deflned by
?1(z) =
1
2
+
X
n
0 zn
1? q2n : (2.2)
Remark
The series (2.2) deflning ?1 converges in the annulus jq2j < jzj < 1. Shortly we will obtain the
S. Cooper, The development of elliptic functions 67
analytic continuation of ?1, so the identity (2.1) will be valid for all values of a; b and t.
Proof
For jq2j < jaj; jbj < 1, we have
f(a; t)f(b; t) =
1
X
m=?1
1
X
n=?1
ambn
(1? tq2m)(1? tq2n)
=
1
X
m=?1
ambm
(1? tq2m)2 +
X
m6=n
ambn
(1? tq2m)(1? tq2n) : (2.3)
The flrst sum is
1
X
m=?1
(ab)m
(1? tq2m)2 =
1
X
m=?1
@
@t
(ab=q2)m
(1? tq2m)
=
@
@t
f(ab=q2; t)
=
@
@t
[tf(ab; t)]
= t
@
@t
f(ab; t) + f(ab; t) (2.4)
The penultimate step above follows from (1.6). The interchange of difierentiation and summation
is valid as all series converge absolutely and uniformly (in t) on compact sets which aviod the poles
t = q2k, k = 0;?1;?2; : : : ; provided jq4j < jabj < jq2j. By analytic continuation, equation (2.4)
continues to remain valid for jq4j < jabj < 1.
Using partial fractions, the second sum on the right hand side of (2.3) becomes
X
m6=n
ambn
(1? tq2m)(1? tq2n)
=
1
X
m=?1
X
k
0 ambm+k
(1? tq2m)(1? tq2m+2k)
=
1
X
m=?1
X
k
0 ambm+k
(1? tq2m)(1? q2k) +
1
X
m=?1
X
k
0 ambm+k
(1? tq2m+2k)(1? q?2k)
=
1
X
m=?1
ambm
(1? tq2m)
X
k
0 bk
(1? q2k) +
X
k
0 a?k
(1? q?2k)
1
X
m=?1
am+kbm+k
(1? tq2m+2k)
= f(ab; t)
"
X
k
0 ak
1? q2k +
X
k
0 bk
1? q2k
#
= f(ab; t) [?1(a) + ?1(b)? 1] : (2.5)
All of the series in the derivation of (2.5) converge at least for jqj < jaj; jbj < 1 and t 6= q2k,
k = 0;?1;?2; : : :, and so the series rearrangements above are valid. Now combine (2.3), (2.4)
and (2.5). This gives (2.1) and proves the theorem.
The analytic continuation of ?1 can be obtained as follows.
?1(z) =
1
2
+
X
n
0 zn
1? q2n
68 R.L.I.M.S. Vol. 1, Sept. 2000
=
1
2
+
1
X
n=1
zn
1? q2n +
1
X
n=1
z?n
1? q?2n
=
1
2
+
1
X
n=1
zn(1? q2n + q2n)
1? q2n ?
1
X
n=1
z?nq2n
1? q2n
=
1
2
+
z
1? z +
1
X
n=1
znq2n
1? q2n ?
1
X
n=1
z?nq2n
1? q2n
=
1 + z
2(1? z) +
1
X
n=1
1
X
m=1
(znq2mn ? z?nq2mn)
=
1 + z
2(1? z) +
1
X
m=1
?
zq2m
1? zq2m ?
z?1q2m
1? z?1q2m
?
: (2.6)
This last series converges for all values of z except z = q2k, k = 0;?1;?2; : : :, where there are
poles of order 1. Thus (2.6) gives the analytic continuation of the function ?1, and so now the
fundamental multiplicative identity (2.1) is valid for all values of a; b and t.
The function ?1 is related to the Weierstrass } function in the following way. The Weierstrass
} function with periods 2? and 2?? is deflned by
}(?) =
1
?2
+
X
m;n
0
?
1
(? ? 2?n? 2??m)2 ?
1
(2?n+ 2??m)2
?
: (2.7)
The symbol
P
m;n
0 denotes a double sum over all integer values of m and n from ?1 to 1,
excluding (m;n) = (0; 0). Using the results
1
X
n=?1
1
(? ? 2?n)2 =
1
4 sin2 ?2
and
1
X
n=1
1
n2
=
?2
6
;
we have that
}(?) =
1
?2
+
X
n
0
?
1
(? ? 2?n)2 ?
1
(2?n)2
?
+
X
m
0 1X
n=?1
?
1
(? ? 2?n? 2??m)2 ?
1
(2?n+ 2??m)2
?
=
X
n
1
(? ? 2?n)2 ?
2
4?2
1
X
n=1
1
n2
+
X
m
0 1X
n=?1
?
1
(? ? 2?n? 2??m)2 ?
1
(2?n+ 2??m)2
?
=
1
4 sin2 ?2
? 1
12
+
X
m
0
"
1
4 sin2( ?2 ? ??m)
? 1
4 sin2 ??m
#
= ? 1
12
? 1
2
1
X
m=1
1
sin2 ??m
+
1
4
1
X
m=?1
1
sin2( ?2 + ??m)
:
S. Cooper, The development of elliptic functions 69
Recall that q = ei?? . Then
}(?) = ? 1
12
+ 2
1
X
m=1
1
(qm ? q?m)2 ?
1
X
m=?1
1
(ei?=2qm ? e?i?=2q?m)2
= ? 1
12
+ 2
1
X
m=1
q2m
(1? q2m)2 ?
1
X
m=?1
ei?q2m
(1? ei?q2m)2 : (2.8)
Continuing, we have
}(?) = ? 1
12
+ 2
1
X
m=1
q2m
(1? q2m)2
? e
i?
(1? ei?)2 ?
1
X
m=1
?
ei?q2m
(1? ei?q2m)2 +
e?i?q2m
(1? e?i?q2m)2
?
= ? 1
12
+ 2
1
X
m=1
q2m
(1? q2m)2 + i
d
d?
?1(ei?):
Formula (2.6) was used to obtain the last line. Thus if we let
P = 1? 24
1
X
m=1
q2m
(1? q2m)2 ; (2.9)
then we have
}(?) = i
d
d?
?1(ei?)?
P
12
: (2.10)
3 Jacobian elliptic functions
We will now look at three special cases of the function f(a; t), which we shall call f1; f2 and f3.
These functions will turn out to be the Jacobian elliptic functions cs; ns and ds, respectively, up
to rescaling. The precise identiflcations will be given at the end of the notes.
A number of properties of f1; f2 and f3 (Fourier series, inflnite product formulas, double pe-
riodicity, location of zeros and poles) will follow immediately from the deflnition of f(a; t) and
Ramanujan?s 1?1 summation formula. We will then use the fundamental multiplicative identity
(2.1) to obtain some of the other properties of these functions, namely the connection with the }
function, elliptic analogues of the formula sin2 ? + cos2 ? = 1, derivatives and addition formulas.
Deflnition Let
f1(?) =
1
i
f(ei?; ei?); (3.1)
f2(?) =
ei?=2
i
f(ei?? ; ei?); (3.2)
f3(?) =
ei?=2
i
f(ei?+i?? ; ei?): (3.3)
The factors 1=i and ei?=2=i are included so that f1; f2 and f3 will be real valued when ? is real.
The Fourier expansions follow directly from (1.2), the deflnition of f . For example
f1(?) =
1
i
1
X
n=?1
ein?
1 + q2n
70 R.L.I.M.S. Vol. 1, Sept. 2000
=
1
i
"
1
2
+
1
X
n=1
ein?
1 + q2n
+
1
X
n=1
e?in?
1 + q?2n
#
=
1
i
"
1
2
+
1
X
n=1
ein? ?
1
X
n=1
q2nein?
1 + q2n
+
1
X
n=1
q2ne?in?
1 + q2n
#
=
1
i
"
1
2
+
ei?
1? ei? ?
1
X
n=1
q2n
1 + q2n
(ein? ? e?in?)
#
=
1
2
cot
?
2
? 2
1
X
n=1
q2n
1 + q2n
sinn?: (3.4)
Similarly,
f2(?) =
ei?=2
i
1
X
n=?1
ein?
1? q2n+1
=
1
2
csc
?
2
+ 2
1
X
n=0
q2n+1
1? q2n+1 sin(n+
1
2
)?; (3.5)
f3(?) =
ei?=2
i
1
X
n=?1
ein?
1 + q2n+1
=
1
2
csc
?
2
? 2
1
X
n=0
q2n+1
1 + q2n+1
sin(n+
1
2
)?: (3.6)
Inflnite product formulas follow from (1.3). We flnd that
f1(?) =
1
i
(?ei?;?q2e?i?; q2; q2; q2)1
(ei?; q2e?i?;?1;?q2; q2)1
(3.7)
=
1
2
(q2; q2)21
(?q2; q2)21
cot
?
2
1
Y
n=1
(1 + 2q2n cos ? + q4n)
(1? 2q2n cos ? + q4n) ; (3.8)
f2(?) =
ei?=2
i
(qei?; qe?i?; q2; q2; q2)1
(ei?; q2e?i?; q; q; q2)1
(3.9)
=
1
2
(q2; q2)21
(q; q2)21
csc
?
2
1
Y
n=1
(1? 2q2n?1 cos ? + q4n?2)
(1? 2q2n cos ? + q4n) ; (3.10)
f3(?) =
ei?=2
i
(?qei?;?qe?i?; q2; q2; q2)1
(ei?; q2e?i?;?q;?q; q2)1
(3.11)
=
1
2
(q2; q2)21
(?q; q2)21
csc
?
2
1
Y
n=1
(1 + 2q2n?1 cos ? + q4n?2)
(1? 2q2n cos ? + q4n) : (3.12)
Either from the Fourier series (3.4) or from the inflnite product formula (3.7), we see that f1(? + 2?) = f1(?).
From the inflnite product (3.7), we obtain
f1(? + 2??)
f1(?)
=
(?q2ei?;?e?i?; q2)1
(q2ei?; e?i?; q2)1
? (?e
i?;?q2e?i?; q2)1
(ei?; q2e?i?; q2)1
=
(1 + e?i?)
(1 + ei?)
(1? ei?)
(1? e?i?) = ?1;
and therefore f1(? + 2??) = ?f1(?). Similar calculations can be done for f2 and f3. The results
are summarized below.
f1(? + 2?m+ 2??n) = (?1)nf1(?); (3.13)
S. Cooper, The development of elliptic functions 71
f2(? + 2?m+ 2??n) = (?1)mf2(?); (3.14)
f3(? + 2?m+ 2??n) = (?1)m+nf3(?): (3.15)
Here m and n are integers. Thus f1 is doubly periodic with periods 2? and 4?? , f2 is doubly
periodic with periods 4? and 2?? , while f3 is doubly periodic with periods 4? and 2? + 2?? .
From the inflnite product expansion (3.7) we see that f1 has zeros when 1 + q2nei? = 0, where n is
any integer. Remembering that q = ei?? , this implies that f1(?) = 0 when ? = (2m+ 1)? + 2n?? ,
for any integer values of m and n. The zeros of f2 and f3 are at ? = 2m? + (2n+ 1)?? and
? = (2m+ 1)? + (2n+ 1)?? , respectively. The poles of f1; f2 and f3 all occur when 1? q2nei? = 0,
that is, when ? = 2m? + 2n?? .
Before describing the connection of f1; f2 and f3 with the Weierstrass } function, we deflne
the Weierstrass invariants e1; e2 and e3.
Deflnition Let
e1 = }(?); (3.16)
e2 = }(??); (3.17)
e3 = }(? + ??): (3.18)
Explicit formulas for e1; e2 and e3 follow at once from equation (2.8). Speciflcally,
e1 =
1
6
+ 2
1
X
m=1
q2m
(1? q2m)2 + 2
1
X
m=1
q2m
(1 + q2m)2
; (3.19)
e2 = ?
1
12
+ 2
1
X
m=1
q2m
(1? q2m)2 ? 2
1
X
m=1
q2m?1
(1? q2m?1)2 ; (3.20)
e3 = ?
1
12
+ 2
1
X
m=1
q2m
(1? q2m)2 + 2
1
X
m=1
q2m?1
(1 + q2m?1)2
: (3.21)
Relation of f1, f2 and f3 to the Weierstrass } function
Let b! 1=a in the fundamental identity (2.1):
lim
b!1=a
f(a; t)f(b; t) = lim
b!1=a
t
@
@t
f(ab; t) + lim
b!1=a
f(ab; t)(?1(a) + ?1(b)): (3.22)
The left hand side is just f(a; t)f(1=a; t).
The flrst limit on the right hand side is
lim
b!1=a
t
@
@t
1
X
n=?1
tn
1? abq2n = limb!1=a
X
n
0 ntn
1? abq2n
=
X
n
0 ntn
1? q2n = t
d
dt
?1(t):
From equation (2.6) it follows that ?1(b) = ??1(1=b). Using this and the inflnite product
formula (1.3) for the function f , the remaining limit on the right hand side of equation (3.22)
becomes
lim
b!1=a
f(ab; t)(?1(a) + ?1(b))
72 R.L.I.M.S. Vol. 1, Sept. 2000
= lim
b!1=a
(1? ab)f(ab; t) lim
b!1=a
?1(a) + ?1(b)
1? ab
= lim
b!1=a
(1? ab) (abt; q
2=abt; q2; q2; q2)1
(t; q2=t; ab; q2=ab; q2)1
lim
b!1=a
?1(a)? ?1(1=b)
a? 1=b
?
?1
b
?
= (1)?01(a)(?a):
Thus
f(a; t)f(1=a; t) = t
d
dt
?1(t)? a
d
da
?1(a):
On letting a = eifi, t = ei? and using equation (2.10), this becomes
f(eifi; ei?)f(e?ifi; ei?) = }(fi)? }(?): (3.23)
Remark
This formula can also be obtained by combining the two terms on the right hand side of (3.23)
into a single series using (2.8), and then applying the 6?6 summation formula.
Letting fi = ?, fi = ?? and fi = ? + ?? in (3.23), respectively, and simplifying, gives
f21 (?) = }(?)? e1; (3.24)
f22 (?) = }(?)? e2; (3.25)
f23 (?) = }(?)? e3: (3.26)
Successively letting ? = ?? in (3.24), ? = ? + ?? in (3.25) and ? = ? in (3.26), and using the inflnite
products for f1, f2 and f3, gives
e1 ? e2 =
1
4
(?q; q2)41(q2; q2)41
(q; q2)41(?q2; q2)41
(3.27)
e3 ? e2 = 4q
(?q2; q2)41(q2; q2)41
(?q; q2)41(q; q2)41
(3.28)
e1 ? e3 =
1
4
(q; q2)41(q
2; q2)41
(?q2; q2)41(?q; q2)41
(3.29)
Note that since Im ? > 0 this implies that e1 6= e2 6= e3 6= e1. Further, if ? is purely imaginary,
then q is real, and so in this case we also have e1 > e3 > e2.
If we let
x =
e3 ? e2
e1 ? e2
= 16q
(?q2; q2)81
(?q; q2)81
; (3.30)
x0 =
e1 ? e3
e1 ? e2
=
(q; q2)81
(?q; q2)81
; (3.31)
then clearly x+ x0 = 1, and hence we obtain Jacobi?s formula
(q; q2)81 + 16q(?q2; q2)81 = (?q; q2)81:
If the equations (3.24), (3.25) and (3.26) are combined two at a time to eliminate the }(?) term,
we obtain
f22 (?)? f21 (?) = e1 ? e2; (3.32)
f22 (?)? f23 (?) = e3 ? e2; (3.33)
f23 (?)? f21 (?) = e1 ? e3: (3.34)
S. Cooper, The development of elliptic functions 73
These are the elliptic function analogues of the trigonometric identity sin2 ? + cos2 ? = 1. In fact,
from (3.4){(3.6) and (3.19){(3.21), we have
lim
q!0
f1(?) =
1
2
cot
?
2
; lim
q!0
f2(?) = lim
q!0
f3(?) =
1
2
csc
?
2
;
lim
q!0
e1 = 1=6; lim
q!0
e2 = lim
q!0
e3 = ?1=12:
Therefore when q = 0, (3.32) and (3.34) reduce to
1
4
csc2
?
2
? 1
4
cot2
?
2
=
1
4
;
while (3.33) reduces to a tautology.
Derivatives
In the fundamental multiplicative identity (2.1), let t = ei? to get
f(a; ei?)f(b; ei?) =
1
i
@
@?
f(ab; ei?) + f(ab; ei?)(?1(a) + ?1(b)): (3.35)
Now let a = ei? and b = ei?? . From (2.6) we have ?1(ei?) = 0, ?1(ei?? ) = 12 , hence
f(?1; ei?)f(q; ei?) = 1
i
@
@?
f(?q; ei?) + 1
2
f(?q; ei?): (3.36)
The left hand side of this is
f(?1; ei?)f(q; ei?) = if1(?)ie?i?=2f2(?) = ?e?i?=2f1(?)f2(?):
The right hand side of (3.36) is
1
i
@
@?
?
ie?i?=2f3(?)
?
+
i
2
e?i?=2f3(?)
= e?i?=2f 03(?)?
i
2
e?i?=2f3(?) +
i
2
e?i?=2f3(?)
= e?i?=2f 03(?):
Combining gives
f 03(?) = ?f1(?)f2(?): (3.37)
Similarly, letting a = ei?? , b = ei?+i?? and a = ei?+i?? , b = ei? in (3.35) leads, respectively, to
f 01(?) = ?f2(?)f3(?); (3.38)
f 02(?) = ?f3(?)f1(?): (3.39)
Venkatachaliengar shows how to obtain the difierential equation for the } function from the funda-
mental multiplicative identity. We will instead obtain it by putting together the previous results.
From (3.24) we have that
}(?) = e1 + f21 (?):
Difierentiate both sides and use (3.38) to simplify the result.
}0(?) = 2f1(?)f 01(?) = ?2f1(?)f2(?)f3(?):
Therefore, by (3.24), (3.25) and (3.26), we have
(}0(?))2 = 4f21 (?)f
2
2 (?)f
2
3 (?) = 4(}(?)? e1)(}(?)? e2)(}(?)? e3): (3.40)
74 R.L.I.M.S. Vol. 1, Sept. 2000
Addition formulas
The fundamental multiplicative identity (2.1) can be written in the form
f(eifi; ei?)f(eifl ; ei?) =
1
i
f(ei(fi+fl); ei?) + f(ei(fi+fl); ei?)(?1(eifi) + ?1(eifl)):
Apply @=@fi? @=@fl to both sides. The result is
@
@fi
f(eifi; ei?)f(eifl ; ei?)? @
@fl
f(eifi; ei?)f(eifl ; ei?)
= f(ei(fi+fl); ei?)
?
d
dfi
?1(eifi)?
d
dfl
?1(eifl)
?
:
Rearranging this and using (2.10) gives
f(ei(fi+fl); ei?) =
i
?
@
@fi
f(eifi; ei?)f(eifl ; ei?)? @
@fl
f(eifi; ei?)f(eifl ; ei?)
?
}(fi)? }(fl) : (3.41)
Let ? = ? in this to get
if1(fi+ fl) =
i [if 01(fi)if1(fl)? if1(fi)if 01(fl)]
}(fi)? }(fl) :
Simplify this using (3.24) and (3.38). The result is
f1(fi+ fl) =
f1(fi)f2(fl)f3(fl)? f1(fl)f2(fi)f3(fi)
f21 (fl)? f21 (fi)
: (3.42)
Similarly, letting ? = ?? and ? = ? + ?? in (3.41) leads to
f2(fi+ fl) =
f2(fi)f3(fl)f1(fl)? f2(fl)f3(fi)f1(fi)
f22 (fl)? f22 (fi)
; (3.43)
f3(fi+ fl) =
f3(fi)f1(fl)f2(fl)? f3(fl)f1(fi)f2(fi)
f23 (fl)? f23 (fi)
: (3.44)
The fundamental multiplicative identity (2.1) can also be used to derive addition formulas for
the Weierstrass } function. Venkatachaliengar?s derivation of the symmetric form of the addition
formula for the } function goes as follows.
Let t = ev and write the Jordan{Kronecker function as
f(a; t) = f(a; ev) =
1
X
n=?1
env
1? aq2n
=
1
1? a +
1
X
n=1
env(1? aq2n + aq2n)
1? aq2n +
1
X
n=1
e?nv
1? aq?2n
=
1
1? a ?
1
ev ? 1 ? 1 +
1
X
n=1
?
envaq2n
1? aq2n ?
e?nva?1q2n
1? a?1q2n
?
: (3.45)
The series (3.45) converges for jRe vj < Im 2?? . Hence in the annulus 0 < jvj < minf2?; Im 2??g,
the function f(a; ev) can be expanded further as a Laurent series in powers of v. Since
v
ev ? 1 =
1
X
k=0
Bk
k!
vk;
S. Cooper, The development of elliptic functions 75
where Bk are the Bernoulli numbers, we have
f(a; ev)
=
1
1? a ? 1?
1
v
1
X
k=0
Bk
k!
vk +
1
X
k=0
vk
k!
1
X
n=1
?
nkaq2n
1? aq2n ?
(?1)knka?1q2n
1? a?1q2n
?
= ?1
v
+
?
?1
2
+
1
1? a +
1
X
n=1
aq2n
1? aq2n ?
a?1q2n
1? a?1q2n
!
+
1
X
k=1
vk
k!
?
?Bk+1
k + 1
+
1
X
n=1
nkaq2n
1? aq2n ?
(?1)knka?1q2n
1? a?1q2n
!
: (3.46)
The term independent of v in this expansion is
?1
2
+
1
1? a +
1
X
n=1
aq2n
1? aq2n ?
a?1q2n
1? a?1q2n
which is precisely ?1(a), the same function as in equation (2.6). The reason why ?1 occurs both
here and in the fundamental multiplicative identity (2.1) will become clear below. For v ? 2, let
us deflne
?k(a) = ?
Bk
k
+
1
X
n=1
nk?1
?
aq2n
1? aq2n ?
(?1)k?1a?1q2n
1? a?1q2n
?
: (3.47)
Then equation (3.46) becomes
f(a; ev) = ?1
v
+
1
X
k=0
?k+1(a)vk
k!
: (3.48)
With t = ev, the fundamental multiplicative identity can be written as
f(a; ev)f(b; ev) =
@
@v
f(ab; ev) + f(ab; ev)(?1(a) + ?1(b)):
Substitute the expansion (3.48) into this to obtain
"
?1
v
+
1
X
k=0
?k+1(a)vk
k!
#"
?1
v
+
1
X
k=0
?k+1(b)vk
k!
#
=
1
v2
+
1
X
k=1
?k+1(ab)vk?1
(k ? 1)! +
"
?1
v
+
1
X
k=0
?k+1(ab)vk
k!
#
(?1(a) + ?1(b)):
(3.49)
Let us compare coe?cients of vk on both sides. Clearly the coe?cients of v?2 are equal. The
coe?cients of v?1 are both equal to ?(?1(a) + ?1(b)). Thus if ?1 is deflned to be the term
independent of v in the expansion (3.48), then this explains why the term ?1(a) + ?1(b) occurs on
the right hand side of the fundamental multiplicative identity (2.1). Equating coe?cients of v0 in
(3.49) gives
??2(a)? ?2(b) + ?1(a)?1(b) = ?2(ab) + ?1(ab)(?1(a) + ?1(b)): (3.50)
From (2.6) and (3.47) we have ?1(1=c) = ??1(c) and ?2(1=c) = ?2(c), respectively. Let c = 1=ab.
Then (3.50) becomes
?1(a)?1(b) + ?1(b)?1(c) + ?1(c)?1(a) = ?2(a) + ?2(b) + ?2(c): (3.51)
76 R.L.I.M.S. Vol. 1, Sept. 2000
Venkatachliengar gives a direct proof of this identity at the beginning of his book, and uses it to
derive a number of results about the Weierstrass } function. Applying a@=@a? b@=@b to both
sides gives
a?01(a)(?1(b) + ?1(c))? b?01(b)(?1(a) + ?1(c)) = a?02(a)? b?02(b):
Now apply ab@2=@a@b to this to obtain
a2b?001(a)?
0
1(b)? ab2?01(a)?001(b)
+ b2c?001(b)?
0
1(c)? bc2?01(b)?001(c)
+ c2a?001(c)?
0
1(a)? ca2?01(c)?001(a) = 0:
This can also be written in the form
det
2
4
1 1 1
a?01(a) b?
0
1(b) c?
0
1(c)
a2?001(a) b
2?001(b) c
2?001(c)
3
5 = 0:
Now let a = eifi; b = eifl and c = ei? , so that fi+fl+? = 0. Using (2.10) and elementary properties
of determinants, this equivalent to
det
2
4
1 1 1
}(fi) }(fl) }(?)
}0(fi) }0(fl) }0(?)
3
5 = 0;
provided fi+ fl + ? = 0. This is the symmetric form of the addition formula for the Weierstrass }
function. Venkatachaliengar also uses equation (3.51) to derive the addition formula in the form
}(fi+ fl) =
1
4
?
}0(fi)? }0(fl)
}(fi)? }(fl)
?2
? }(fi)? }(fl):
Please see [2, p. 9] for the details. This formula can also be obtained by taking the logarithm of
(3.23) and then applying
@
@fi
+
@
@fl
twice to both sides.
4 Identiflcation with the notation and formulas in Whittaker and Wat-
son
1. The modulus and complementary modulus.
The quantities x and x0 deflned in equations (3.30) and (3.31) correspond to the squares of
the modulus k and complementary modulus k0, respectively. We have
x = k2 = 16q
(?q2; q2)81
(?q; q2)81
;
x0 = k02 =
(q; q2)81
(?q; q2)81
:
See [2, p. 86, eqn. (5.47)] and [3, p. 479 or p. 488, ex. 9,10].
2. Ramanujan?s z and the complete elliptic integral K.
Although it was not introduced in these notes, Venkatachaliengar makes extensive use of
the quantity z that was introduced by Ramanujan. We mention z now, to aid with the
identiflcation of f1; f2 and f3 with Jacobian elliptic functions.
z =
? 1
X
n=?1
qn
2
!2
= (?q; q2)41(q2; q2)21;
S. Cooper, The development of elliptic functions 77
K =
?
2
z:
See [2, p. 86, eqn. (5.49)] and [3, p. 479].
3. f1; f2; f3 and Jacobian elliptic functions.
On comparing the Fourier series (3.4){(3.6) with those in [3, p. 511{512], we flnd
f1(?) = (K=?) cs(K?=?; k); cs(u; k) = (2=z)f1(2u=z);
f2(?) = (K=?) ns(K?=?; k); ns(u; k) = (2=z)f2(2u=z);
f3(?) = (K=?) ds(K?=?; k); ds(u; k) = (2=z)f3(2u=z):
4. The functions f(a; t); ?1(z) and the Weierstrass ?; ? and } functions.
Earlier, we showed that the Weierstrass } function with periods 2? and 2?? is related to the
function ?1 by equation (2.10). Here is the formula again.
}(?) = i
d
d?
?1(ei?)?
P
12
:
The corresponding Weierstrassian ? function is related to the function ?1 as follows.
?(?) = ?i?1(ei?) +
P?
12
:
See [3, pp. 445{447]. Recall that P is given by (2.9). In particular, this together with
equation (2.6) gives
?1 = ?(?) =
P?
12
;
?2 = ?(??) = ?
i
2
+
P??
12
;
from which Legendre?s identity (see [3, p. 446, sect. 20.411]) follows trivially:
?1?? ? ?2? =
1
2
?i:
The corresponding Weierstrass ? function is as follows.
?(?) = ie?i?=2eP?
2=24 (e
i?; q2e?i?; q2)1
(q2; q2)21
:
See [3, pp. 447{448]. The Jordan{Kronecker function f(a; t), (equation 1.2), is related to
the Weierstrass ? function by
f(eifi; ei?) = ie?Pfi?=12
?(fi+ ?)
?(fi)?(?)
:
Location of the main formulas in Venkatachaliengar?s book
1. Ramanujan?s 1?1 summation formula (1.1) is proved in [2, pp. 24{30].
2. The Jordan{Kronecker function f(a; t), (equation 1.2), is deflned in [2, p. 37]. The inflnite
product for f(a; t), (equation 1.3), is [2, p. 40, eq. 3.32]. See also [3, p. 460, ex. 34], which is
basically equation (1.3) in disguise. The fundamental multiplicative identity (2.1) is proved
in [2, p. 41].
3. The function ?1, (equation 2.2), is deflned in [2, p. 5] and some of its properties, including
the analytic continuation, are given there.
78 R.L.I.M.S. Vol. 1, Sept. 2000
4. The connection between the Weierstrass } function and the fuction ?1 given by equation
(2.10) is given in [2, p. 8, eq. 1.21]. Essentially the same formula is in [3, p. 460, ex. 35].
5. The functions f1; f2 and f3, (equations 3.1{3.3), are deflned in [2, p. 111]. The Fourier
series for essentially the same functions are given in [3, pp. 511{512].
6. The Weierstrass invariants e1; e2 and e3, (equations(3.16{3.18), are deflned in [2, p. 59].
7. Equation (3.23) is derived in [2, p. 112, eqn. 6.50]. It is equivalent to example 1 on p. 451
of [3].
8. Formulas (3.24){(3.26) are in [2, p. 112]. Compare these with [3, p. 451 example 4] and [3,
p. 505, section 22.351].
9. The inflnite products (3.27){(3.29) are given in [2, p. 66].
10. Formulas (3.30) and (3.31) for x and x0 are in [2, p. 86, eqn. 5.47]. Analogous formulas for
k and k0 are in [3, p. 479 and p. 488, ex. 9, 10].
11. The derivatives of f1; f2 and f3 are computed in [2, p. 111].
12. The difierential equation for the Weierstrass } function is in [2, p. 13, eqn. 1.49].
13. The addition formulas for f1; f2 and f3 are given in [2, pp. 112{113].
14. Formula (3.48) is given in [2, p. 43].
15. Formula (3.51) as derived here in the notes is given in [2, p. 43]. Also see [2, pp. 3{4].
16. Addition formulas for the } function are derived in [2, pp. 8{9].
References
[1] R. Askey. Ramanujan?s Extensions of the Gamma and Beta Functions. American Mathemat-
ical Monthly, pp.346{359, vol. 87, No.5, 1980.
[2] K. Venkatachaliengar. Development of Elliptic Functions according to Ramanujan. Depart-
ment of Mathematics, Madurai Kamaraj University, Technical Report 2, 1988.
[3] E. Whittaker and G. Watson A Course of Modern Analysis Cambridge University Press, 4th
edition, 1927.