Elliptic functions revisited

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by  Jean-Christophe Feauveau

Elliptic functions revisited
Jean-Christophe Feauveau1
January 30th, 2017
Abstract.
7 Ellipticfunctionsarelargelystudiedandstandardizedmathematicalobjects.Thetwousualapproaches
1
are due to Jacobi and Weierstrass.
0
2 From a contour integral which allowed us to unify many summation formulae (Euler-MacLaurin, Pois-

n son, Voronoï or Circle formulae), we will find the entirety of the elliptic functions, proposed either in
a the shape of Jacobi or Weierstrass. But with one translation which appears in their natural form.
J
Whatcouldseemadefectwillleadustoarenormalisationoftheellipticfunctionsmakingitpossibleto
0
3 determine,inarathersimple way,aFourierseriesrepresentationandafactorizationofthesefunctions.

]
V
C
Key words. elliptic functions, contour integral, Fourier series representation, factorization.
.
h
t Classification A.M.S. 2010 : 33E05, 11G16,11J89.
a
m
[
Introduction

2
v Since Abel, Jacobi and Weiertsrass, the bases of the theory of the elliptic functions have been well
0 established. Abel proceeds by a length inversion of arc of ellipse (which justifies the denomination of
9 these functions), Jacobi introduces theta functions to define the elliptic functions, finally Weierstrass
8
defines those using series which deliver in an immediate way the double periodicity of those functions.
7
0 Following Weierstrass, the modern presentation of the elliptic functions rest on two pillars : the mero-
.
1 morphic character and the double periodicity. This last property naturally involves well-known repre-
0 sentations of the trigonometricaltype for Jacobi’s elliptic functions - Cf. [Halp], [Whit] et [Armi] - but
7
not for Weierstrass’s elliptic functions.
1
: We propose in this article a unified framework to obtain, by the same formalism, the elliptic functions
v
i of Weierstrassand Jacobi.According to the ideas developed in [Feau], a calculationof contour integral
X
will give a direct access to the Fourier series representations of these elliptic functions. We will also
r deduce integral representations as well as factorizations of these renormalized elliptic functions.
a
I – Study of a typical case
In order to justify the whole of the study, let us start with a typical case.
s
As we will establish it with part III, the function H(s) = decrease rather quickly ad
sin(s)sin(is)
infinitum when its poles are avoided.
Precisely, for sequences of contours γ , which end up including C, one will establish
n
1 1 1 1 s
∀z ∈]− , [+i]− , [, lim e2iszds=0.
2 2 2 2 n→+∞Zγn sin(s)sin(is)
1. Jean-Christophe Feauveau,
Professeurenclassespréparatoires aulycéeBellevue,
135,route deNarbonneBP.44370, 31031ToulouseCedex4,France,
email:[email protected]
1
The poles of H are localized at 0, πZ∗ and iπZ∗, and an easy calculation of residue gives
+∞ +∞
1 1 1 1 1 cos(2nπz) 1 ch(2nπz)
∀z ∈]− , [+i]− , [, + (−1)nn =− − (−1)nn .
2 2 2 2 4π sh(nπ) 4π sh(nπ)
n=1 n=1
X X
The1-periodicandi-periodiccharacterofthetwomembersofthisequalitymakespossibleaholomorphic
prolongationof the function
+∞
cos(2nπz)
℘˜(z)=π+4π2 (−1)nn
sh(nπ)
n=1
X
on |Re(z)| < 1, |Im(z)| < 1 then on C−(1+i +Z+iZ) and by double periodicity : ∀(n,m) ∈ Z2,
2 2 2
℘˜(z+n+im)=℘˜(z).
By integrating 3 times ℘˜, the singularity into 1+i becomes artificial : the order of poles are inferior or
2
equal to 3. It is thus about an elliptic function (Cf. II for a recall of the definition and the elementary
properties of these functions).
Usingperiodicity,onededucestheexistenceofasinglepoleontheelementarydomain[0,1]+i[0,1].The
parity of ℘˜makes it possible to conclude to double poles located at Λ= 1+i +Z+iZ. It is Weierstrass
2
type elliptic function. The parity and nullity at 0 of ℘˜ then make it possible to assert the existence of
α∈C (it will be shown that α=1) such as
1 1
℘˜(z)=α − .
(z−ω)2 ω2
ω∈Λ
X
Afirstnaturalresultisthusobtained:theFourierseriesrepresentationsofaWeierstrasstypefunction,
which is not current in the literature. The reason comes certainly from the usual choice to place the
pole at0.We willdiscuss thereafterthe utility of the translationof 1+i operatedhere,itis anessential
2
point.
II – Notations and generalities about the elliptic functions
Being given ω and ω , two non-zero complex numbers such as ω /ω 6∈R, and let us denote Λ =
1 2 1 2 ω1,ω2
ω Z+ω Z. The lattices Λ and Λ are identical, and we can impose Im(ω /ω ) > 0, which
1 2 ω1,ω2 ω1,−ω2 2 1
we will do systematically thereafter.
We pose ∆ = ω [−1,1[+ω [−1,1[, this is the basic elementary domain and any translation also
ω1,ω2 1 2 2 2 2 2
will be named elementary domain. It is noticed that (a+∆ ) is a partition of C.
ω1,ω2 a∈Λω1,ω2
ω +ω
Finally, we set ω = 1 2 and ∆˜ =∆ −{−ω }.
0 2 ω1,ω2 ω1,ω2 0
Being given ω and ω , two non-zero complex numbers such that ω /ω 6∈ R (and accordingly
1 2 1 2
Im(ω /ω ) > 0), we say as f : C → C ∪ {∞} is an elliptic function based on the lattice Λ
2 1 ω1,ω2
if
(i) f is a meromorphic function on C,
(ii) f is périodic in two directions, says ω and ω .
1 2
The well-known theory of the elliptic functions gives the following results :
(i) If f is elliptic and holomorphic on C, then f is a constant function.
(ii) Thenumberofpolesandzerosinanelementarydomainisfinite.Moreover,thesumoftheresidues
being in an elementary domain with no pole on its border is null : if (a ) are the poles of f at
i i∈I
a+∆ , beyond the border, then :
ω1,ω2
Res(f,a )=0.
i
i∈I
X
2
(iii) The sum of the multiplicities of the poles (counted negatively)andthe zeros (counted positively)
of f in an elementary domain is null.
To be brief, (i) comes from Liouville’s theorem, (ii) and (iii) are obtained by calculating the integrals
of f then f′/f on the border of an elementary domain without pole or zero on its border, those being
null by symmetry. One will be able, for example, to consult [Whit] or [Armi] on these points.
Finally,letusnoticethatz 7→f(z)isanellipticfunctiononthelatticeΛ ifandonlyifz 7→f(ω z)
ω1,ω2 1
ω
2
is an elliptic function on Λ , with τ = .
1,τ
ω
1
III - Weierstrass’s modified elliptic function on Λ
ω1,ω2
s
To generalize the results of part I, the function H(s)= is considered.
sin(ω s)sin(ω s)
1 2
For n ∈ N, we define a contour C connecting by segments, and in the trigonometrical direction, the
n
points (n+ 1)π(ω−1−ω−1), (n+ 1)π(−ω−1−ω−1), (n+ 1)π(−ω−1+ω−1) et (n+ 1)π(ω−1+ω−1).
2 1 2 2 1 2 2 1 2 2 1 2
Bythis choice,the contourpassesexactly“between” the polesofH onthe axesω R andω R.The four
1 2
segments are noted respectively Cnj, 16j 64 (on figure ω1 =1 and ω2 =i) :
✻
C niπ+iπ/2
n1 ✛
C
n4
✻
nπ+π/2
✲
O
❄
C
n2
✲
C
n3
Theoreme 1 : for z ∈ ]− 1,1[ω +]− 1,1[ω , we have lim H(s)e2iszds=0.
2 2 1 2 2 2 n→+∞
ZCn
Proof : Letus verifythe resultforthe segmentC ,the threeother casesresultingsomeby symmetry.
n,1
s
We have to show that lim e2iszds = 0, which will give
n→+∞Z[(n+12)π(−1−τ−1),(n+12)π(1−τ−1)] sin(s)sin(τs)
the result by change of variables τ =ω /ω , for z ∈ ]− 1,1[+]− 1,1[τ.
2 1 2 2 2 2
Let z =a+bτ, (a,b)∈ ]− 1,1[2. For s=u+(n+ 1)πτ−1, with −(n+ 1)π 6u6(n+ 1)π,
2 2 2 2 2
e2isz e2ias e2ibsτ e2ia(n+21)πτ−1 e2ibuτ
= = .
We denote τ−1 =(cid:12)(cid:12)(cid:12)(cid:12)siαn(+s)isβin((τlest)u(cid:12)(cid:12)(cid:12)(cid:12)s n(cid:12)(cid:12)(cid:12)(cid:12)ostien(tsh)a(cid:12)(cid:12)(cid:12)(cid:12)t(cid:12)(cid:12)(cid:12)(cid:12)sβin<(τ0s))(cid:12)(cid:12)(cid:12)(cid:12)and(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)sin(u+(n+ 12)πτ−1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)cos(τu)(cid:12)(cid:12)(cid:12)(cid:12)
e2ia(n+12)πτ−1 e−2a(n+12)πβ
=
(cid:12)sin(u+(n+ 1)πτ−1)(cid:12) (cid:12)sin((u+(n+ 1)πα)+i(n+ 1)πβ)(cid:12)
(cid:12) 2 (cid:12) (cid:12) 2 2 (cid:12)
(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)=(cid:12)(cid:12)(cid:12) e−(n+21)πβ+i(u+(n+21)πα) (cid:12)(cid:12)(cid:12)e(1−2a)(n+12)πβ.
(cid:12)sh((n+ 1)πβ−i(u+(n+ 1)πα))(cid:12)
(cid:12) 2 2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
3
ev e2ibuτ
The function 7→ is bounded on |Re(v)| > 1, whereas u 7→ is bounded on R, because
sh(v) cos(τu)
(cid:12) (cid:12) (cid:12) (cid:12)
|2b|<1.Now,we(cid:12)(cid:12)canfin(cid:12)(cid:12)dM >0dependingonlyonz andτ suchas(cid:12)(cid:12) e2i(cid:12)(cid:12)sz 6Me(1−2a)(n+12)πβ,
(cid:12) (cid:12) (cid:12) sin(s)si(cid:12)n(τs)
(cid:12) (cid:12)
and the result is thus deduced. (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
We have to integrate on segments of direction ω and ω passing exactly between the poles of H and
1 2
to use the method of residues.
All the pole of H are simple :
1
(i) s=0, and the residue is worth .
ω ω
1 2
(−1)nnπ
(ii) s=nπ/ω , n∈Z∗, and the residue is worth e2inπz/ω1.
1 ω2sin(nπω /ω )
1 2 1
(−1)nnπ
(iii) s=nπ/ω , n∈Z∗, and the residue is worth e2inπz/ω2.
2 ω2sin(nπω /ω )
2 1 2
The calculus of residues proposed to theorem 1 gives finally : ∀z ∈ ]− 1,1[ω +]− 1,1[ω ,
2 2 1 2 2 2
ω +∞ (−1)nn 1 ω +∞ (−1)nn
2 1
cos(2nzπ/ω )=− − cos(2nzπ/ω ).
1 2
ω sin(nπω /ω ) 2π ω sin(nπω /ω )
1 2 1 2 1 2
n=1 n=1
X X
Let Θ(z) be this common value, the periodic character of each member of the equality makes possible
a holomorphic prolongation of Θ on |Im(z/ω )| < 1Im(ω /ω ), |Im(z/ω )| < −1Im(ω /ω ) then
1 2 2 1 2 2 1 2
C−(ω +ω Z+ω Z) by double-periodicity : Θ(z+ω n+ω m)=Θ(z), (n,m)∈Z2.
0 1 2 1 2
After three integrations,it is noted that the poles ofΘ arealldoubles with nullresidues andplaced on
the relocated lattice Λ⋆ =ω +Λ , with ω = ω1+ω2.
ω1,ω2 0 ω1,ω2 0 2
According to Liouville’s theorem, except for a multiplicative constant and an additive constant, it is
about
1 1
℘˜ (z)= −
ω1,ω2 (z−a)2 a2
a∈Λ⋆
Xω1,ω2
1 1
= − .
(z+(n+ 1)ω +(m+ 1)ω )2 ((n+ 1)ω +(m+ 1)ω )2
(nX,m)∈Z 2 1 2 2 2 1 2 2
The meromorphic function ℘˜ presents double poles with null residues on Λ⋆ , as well as double
ω1,ω2 ω1,ω2
roots on Λ . This is the elliptic function which appears naturally here and which is studied now.
ω1,ω2
Fourier series representations of ℘˜
ω1,ω2
There thus exists α∈C∗ such as ℘˜ (z)=α(Θ(z)−Θ(0)).
ω1,ω2
1
We will find α while studying Θ(z)= +O(1) in the vicinity of ω .
α(z−ω )2 0
0
cos(2nπz/ω )
While developing, it is noted that (−1)n 1 +ie−2inπ(z−ω0)/ω1 enπIm(ω2/ω1) is bounded in-
sin(nπω /ω )
(cid:12) 2 1 (cid:12)
dependently of n ∈ N∗ and z ∈ [0,(cid:12)(cid:12)(cid:12)21[ω1+[0,21[ω2. If we introduce ψ(z) =(cid:12)(cid:12)(cid:12)−iωω2 +∞ne−2inπ(z−ω0)/ω1,
1
n=1
the function Θ−ψ is bounded on [0,1[ω +[0,1]ω . X
2 1 2 2
4
ω e−2iπ(z−ω0)/ω1 iω ω
A direct calculation shows that ψ(z)=−i 2 and lim (z−ω )2ψ(z)= 1 2.
ω1(1−e−2iπ(z−ω0)/ω1)2 z→ω0 0 4π2
4π2
We deduce ℘˜(z)=−i (Θ(z)−Θ(0)).
ω ω
1 2
Thus, the function ℘˜ is even, admits a double pole at ω , cancels itself twice at 0 and it comes :
ω1,ω2 0
i4π2 +∞ (−1)nn z +∞ (−1)nn
℘˜ (z)=− cos(2nπ )− (1)
ω1,ω2 ω2 sin(nπω /ω ) ω sin(nπω /ω )
1 n=1 2 1 1 n=1 2 1 !
X X
i4π2 +∞ (−1)nn z +∞ (−1)nn
= cos(2nπ )− (2)
ω2 sin(nπω /ω ) ω sin(nπω /ω )
2 n=1 1 2 2 n=1 1 2 !
X X
π2 1 1
= − . (3)
ω2 sin((z−ω −ω n)π/ω )2 sin((ω +ω n)π/ω )2
1 n∈Z(cid:18) 0 2 1 0 2 1 (cid:19)
X
π2 1 1
= − . (4)
ω2 sin((z−ω −ω n)π/ω )2 sin((ω +ω n)π/ω )2
2 n∈Z(cid:18) 0 1 2 0 1 2 (cid:19)
X
The first two equalities are valid for |Im(z/ω )| < 1Im(ω /ω ) and |Im(z/ω )| < −1Im(ω /ω )
1 2 2 1 2 2 1 2
1
respectively. The two last ones come from Euler’s equality sin(z)−2 = and are valid on
(z−nπ)2
n∈Z
C−Λ⋆ . X
ω1,ω2
It should be noted that in the typical case studied with part I, the symmetry of the function gives an
+∞ (−1)nn +∞ (−1)nn −i
explicit formula for the constant : =−i = .
sin(inπ) sh(nπ) 4π
n=1 n=1
X X
Link with Weierstrass’s elliptic function
One points out the definition of Weiertrass’s function :
1 1 1
℘ (z)= + − . (5)
ω1,ω2 z2 (z+nω +mω )2 (nω +mω )2
1 2 1 2
(n,mX)6=(0,0)
It comes immediately :
℘˜ (z)=℘ (z+ω )−℘ (ω ).
ω1,ω2 ω1,ω2 0 ω1,ω2 0
Let us note the differences :
(i) Poles of ℘ are localised on Λ whereas those of ℘˜ are on Λ⋆ .
ω1,ω2 ω1,ω2 ω1,ω2 ω1,ω2
(ii) Zeros of ℘˜ are localised on Λ , whereas those of ℘ are unknown.
ω1,ω2 ω1,ω2 ω1,ω2
(iii) Todefine℘ ,oneparticularizesthepoleat0whichdisymmetrizesthefunctionwhereas℘˜
ω1,ω2 ω1,ω2
is symmetrical by construction.
(iv) The choice of normalization of ℘ is closely related to the parameterization of a specifical
ω1,ω2
family of the cubic ones. We will see that the choice retained for ℘˜ allows to parameterize
ω1,ω2
another family of cubic curves, important also.
IV - Weiertrass functions on the reduced lattice Λ =Z+τZ
τ
As previously noted, the study of the elliptic functions on the lattice Λ is brought back under
ω1,ω2
investigationonthe lattice Λ =Λ =Z+τZ. This willbe largelyexploitedfor the study ofmodular
1,τ τ
forms.
Let us summarize knowledge concerning the functions ℘ =℘ and ℘˜ =℘˜ on lattice Λ .
τ 1,τ τ 1,τ τ
5
1 1 1
℘ (z)= + − , ∀z ∈C−Λ (6)
τ z2 (z+n+mτ)2 (n+mτ)2 τ
(n,mX)6=(0,0)
and
1 1 1+τ
℘˜ (z)= − , ∀z ∈C−( +Λ ). (7)
τ (z+(n+ 1)+(m+ 1)τ)2 ((n+ 1)+(m+ 1)τ)2 2 τ
(n,Xm)∈Z2 2 2 2 2
These functions are connected by ℘˜ (z)=℘ (z+ 1 + τ)−℘ (1 + τ).
τ τ 2 2 τ 2 2
• The ℘˜ function
- The integral of the functional equation : ∀z ∈ ]− 1,1[+]− 1,1[τ,
2 2 2 2
s
e2iszds=0.
sin(s)sin(τs)
ZC∞
- The calculus of residues : ∀z ∈ ]− 1,1[+]− 1,1[τ,
2 2 2 2
1 +∞ (−1)nn 1 +∞ (−1)nn
+τ cos(2nzπ)+ cos(2nzπ/τ)=0.
2π sin(nπτ) τ sin(nπ/τ)
n=1 n=1
X X
- The Fourier series representations : ∀z =α+βτ, (α,β)∈R2,
+∞ (−1)nn +∞ (−1)nn 1
℘˜ (z)=4iπ2 − cos(2nπz) when |β|< (8)
τ
sin(nπτ) sin(nπτ) 2
!
n=1 n=1
X X
4iπ2 +∞ (−1)nn +∞ (−1)nn 1
= − + cos(2nπz/τ) when |α|< . (9)
τ2 sin(nπ/τ) sin(nπ/τ) 2
!
n=1 n=1
X X
- The integral representations : ∀z =α+βτ, (α,β)∈R2,
s 1
℘˜ (z)=−8iπ2 (cos(2sz)−1)ds when |β|<
τ
sin(s)sin(τs) 2
ZC1
s 1
=8iπ2 (cos(2sz)−1)ds when |α|<
sin(s)sin(τs) 2
ZCτ
where C is a directed contour which includes the poles nπ, n ∈ N∗, and C a directed contour
1 τ
which includes the poles nπ, n∈N∗.
τ
• La fonction ℘
- The integral of the functional equation : ∀z ∈ ]0,1[+]0,1[τ,
s
eis(2z−1−τ)ds=0.
sin(s)sin(τs)
ZC∞
- The calculus of residues : ∀z ∈ ]0,1[+]0,1[τ,
+∞ +∞
1 n 1 n
+τ cos(nπ(2z−τ))+ cos(n(2z−1)π/τ)=0.
2π sin(nπτ) τ sin(nπ/τ)
n=1 n=1
X X
6
- The Fourier series representations : ∀z =α+βτ, (α,β)∈R2,
+∞
n
℘ (z)=c −4iπ2 cos(nπ(2z−τ)) when β ∈]0,1[
τ τ
sin(nπτ)
n=1
X
4iπ2 +∞ n
=d + cos(n(2z−1)π/τ) when α∈]0,1[.
τ τ2 sin(nπ/τ)
n=1
X
The terms c and d depending only on τ and satisfying c −d =−2.
τ τ τ τ τ
- The integral representations : ∀z =α+βτ, (α,β)∈R2,
s
℘ (z)=c −8iπ2 cos(s(2z−1−τ))ds when β ∈]0,1[
τ τ
sin(s)sin(τs)
ZC1
s
=d +8iπ2 cos(s(2z−1−τ))ds. when α∈]0,1[
τ
sin(s)sin(τs)
ZC1
where C is a directed contour which includes the poles nπ, n ∈ N∗, and C a directed contour
1 τ
which includes the poles nπ, n∈N∗.
τ
V - Jacobi’s elliptic functions
Historically preceding Weieirstrass’s theory, the Jacobi elliptic functions constitute the other standard
family of elliptic functions.
We return to [Armi] and [Whit] for the definitions of these functions. Let us note simply that the
periods are related to two real numbers K and K′ (defined themselves by a parameter k), that one
pose τ = iK′/K and q = eiπτ (caution : it is not the usual convention in modular forms theory). To
establishthe linkwithwhatfollows,onefinds the FourierseriesrepresentationsofSn,Cnanddn(with
the notation in q) in [Armi] and [Whit].
The approach developed in part III with contour calculus lim H(s)e2iszds = 0 remains valid
n→+∞
ZCn
to obtain the functions of Jacobi. The demonstrations of convergence are similar to those developed
partly III and will be omitted. In all the cases,after a transformationz 7→αz where α∈C, the double
periodicity is reduced to the periods (1,2τ) or (2,2τ) with Im(τ) > 0. Following the example of the
choiceoflatticeΛ intheprevioussection,thereducedformwillbeveryusefulforthestudyofmodular
τ
forms.
• Jacobi’s dn function
- The integral of the functional equation : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
1
eis(2z−1)ds=0.
sin(s)cos(τs)
ZC∞
- The calculus of residues : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
+∞ cos(2nπz) 2 +∞ (−1)ncos((n+ 1)π(2z−1)/τ)
1+2 − 2 =0.
cos(nπτ) τ sin((n+ 1)π/τ)
n=1 n=0 2
X X
- The Fourier series representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
7
2K +∞ qncos(2nπz) 1
DN(z,τ)= dn(2Kz)=1+4 when |β|< (10)
π 1+q2n 2
n=1
X
+∞
cos(2nπz) 1
=1+2 when |β|< (11)
cos(nπτ) 2
n=1
X
2 +∞ (−1)ncos((n+ 1)π(2z−1)/τ) 1
= 2 when |α|< . (12)
τ sin((n+ 1)π/τ) 2
n=0 2
X
- The integral representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
2K 1 1
DN(z,τ)= dn(2Kz)=1+4iπ cos(s(2z−1))ds when |β|<
π sin(s)cos(τs) 2
ZC1
1 1
=−4iπ cos(s(2z−1))ds when |α|< ,
sin(s)cos(τs) 2
ZCτ
where C includes the poles nπ, n∈N, and C the poles (n+ 1)π.
1 τ 2 τ
• Jacobi’s sn function
- The integral of the functional equation : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
1
eis(2z−1)ds=0.
cos(s)sin(τs)
ZC∞
- The calculus of residues : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
+∞ (−1)ncos(nπ(2z−1)/τ) +∞ sin((2n+1)πz)
1+2 −2τ =0.
cos(nπ/τ) sin((n+ 1)πτ)
n=1 n=0 2
X X
- The Fourier series representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
Kk +∞ qn+21sin((2n+1)πz) 1
SN(z,τ)= sn(2Kz)= when |β|< (13)
2π 1−q2n+1 2
n=0
X
+∞
i sin((2n+1)πz) 1
= when |β|< (14)
2 sin((n+ 1)πτ) 2
n=0 2
X
i i +∞ (−1)ncos(nπ(2z−1)/τ) 1
= + when |α|< . (15)
4τ 2τ cos(nπ/τ) 2
n=1
X
- The integral representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
Kk 1 1
SN(z,τ)= sn(2Kz)=π cos(s(2z−1))ds when |β|<
2π cos(s)sin(τs) 2
ZC1
i 1 1
= −π cos(s(2z−1))ds when |α|< ,
4τ cos(s)sin(τs) 2
ZCτ
where C includes the poles (n+ 1)π, n∈N, and C the poles nπ/τ, n∈N∗.
1 2 τ
8
• Jacobi’s cn function
- The integral of the functional equation : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
1
eis(2z−1)ds=0.
cos(s)cos(τs)
ZC∞
- The calculus of residues : ∀z ∈ 1+]− 1,1[+]− 1,1[τ,
2 2 2 2 2
+∞ (−1)nsin((n+ 1)π(2z−1)/τ) +∞ cos((2n+1)πz)
2 −τ =0.
cos((n+ 1)π/τ) cos((n+ 1)πτ)
n=1 2 n=0 2
X X
- The Fourier series representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
Kk +∞ qn+21cos((2n+1)πz)
CN(z,τ)= cn(2Kz)= (16)
2π 1+q2n+1
n=0
X
+∞
1 cos((2n+1)πz) 1
= when |β|< (17)
2 cos((n+ 1)πτ) 2
n=0 2
X
1 +∞ (−1)nsin((n+ 1)π(2z−1)/τ) 1
= 2 when |α|< . (18)
2τ cos((n+ 1)π/τ) 2
n=0 2
X
- The integral representations : ∀z = 1 +α+βτ, (α,β)∈R2,
2
Kk 1 1
CN(z,τ)= cn(2Kz)=iπ sin(s(2z−1))ds when |β|<
2π cos(s)cos(τs) 2
ZC1
1 1
=−iπ sin(s(2z−1))ds when |α|< .
cos(s)cos(τs) 2
ZCτ
where C includes the poles (n+ 1)π, n∈N, and C the poles (n+ 1)π/τ, n∈N∗.
1 2 τ 2
• Jacobi’s nd, cd and sd functions
Thereaderwillverifywithoutdifficulty thatoneobtainsinthesamewayFourierseriesrepresentations
and integralrepresentations for the other Jacobi’s elliptic functions. For an applicationto the modular
forms, the renormalized functions nd, cd and sd will be useful. One considers respectively :
1 1 1
e2iszds=0, e2iszds=0 et e2iszds=0,
sin(s)cos(τs) cos(s)sin(τs) cos(s)cos(τs)
ZC∞ ZC∞ ZC∞
i.e., by changing 2z − 1 by 2z in the previous integrals. As previously, the following Fourier series
representations are valid on z =α+βτ for |β|< 1 then for |α|< 1. It comes :
2 2
9
2k′K +∞ (−1)nqncos(2nπz) +∞ (−1)ncos(2nπz)
ND(z,τ)= nd(2Kz)=1+4 =1+2 (19)
π 1+q2n cos(nπτ)
n=1 n=1
X X
2 +∞ (−1)ncos((2n+1)πz/τ)
= (20)
τ sin((n+ 1)π/τ)
n=0 2
X
Kk +∞ (−1)nqn+21cos((2n+1)πz) i +∞ (−1)ncos((2n+1)πz)
CD(z,τ)= cd(2Kz)= = (21)
2π 1−q2n+1 2 sin((n+ 1)πτ)
n=0 n=0 2
X X
i i +∞ (−1)ncos(2nπz/τ)
= + (22)
4τ 2τ cos(nπ/τ)
n=1
X
Kkk′ +∞ (−1)nqn+21sin((2n+1)πz) 1 +∞ (−1)nsin((2n+1)πz)
SD(z,τ)= sd(2Kz)= = (23)
2π 1+q2n+1 2 cos((n+ 1)πτ)
n=0 n=0 2
X X
1 +∞ (−1)nsin((2n+1)πz/τ)
=− (24)
2τ cos((n+ 1)π/τ)
n=0 2
X
V - Factorization of elliptic functions
The zeros and the poles of the elliptic functions studied previously are quite localised which will make
it possible to factorize these functions. The results presented here will have a significant impact in
the studies of the modular forms which result from it. We start by factorizing ℘˜ then the Jacobi’s
τ
functions, these functions being renormalized from the point of view to come.
1
• Factorization of (cid:8)(z,τ)=− ℘˜ (z)
16π2 τ
In order to simplify the expressionsin the modular equalities, one standardizes℘˜ by also highlighting
τ
the variable τ :
1
(cid:8)(z,τ)=− ℘˜ (z).
16π2 τ
OnepointsoutthedefinitionofthereducedlatticeΛ =Z+τZandoneintroducesΛ⋆ = 1+τ +Z+τZ.
τ τ 2
The variable τ being fixed, in order to compensate zeros and poles of z 7→ (cid:8)(z,τ), the following
meromorphic function is introduced :
z+m+nτ 2
F(z,τ)= .
z+ω +m+nτ
n m (cid:18) 0 (cid:19)
YY
Whilegatheringaccordingtothevariablemandwhilesymmetrizinginordertohavetermsconverging,
N
z+n
the well known Euler’s relation sin(πz)=πz lim leads us to :
N→+∞ n
n=−N, n6=0(cid:18) (cid:19)
Y
N sin(π (z+nτ))2
F(z,τ) = lim
N→+∞ sin(π (z−1/2−1/2τ +nτ))sin(π (z+1/2+1/2τ +nτ))
n=−N
Y
+∞ sin(π (z+nτ))2
=
sin(π (z−1/2−1/2τ +nτ))sin(π (z+1/2+1/2τ +nτ))
n=−∞
Y
10