On trigonometric expansions of the Weierstrass function $\wp(z)$ and applications

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On trigonometric expansions of the Weierstrass function

℘(z) and applications

Abd Raouf Chouikha

To cite this version:

Abd Raouf Chouikha. On trigonometric expansions of the Weierstrass function ℘(z) and applications.

2021. �hal-03226340�

1

ON TRIGONOMETRIC EXPANSIONS OF THE WEIERSTRASS FUNCTION ℘(z) AND APPLICATIONS

ABD RAOUF CHOUIKHA

Abstract. Thanks to representation of the Jacobi theta functions

θ

(v, τ ) = f

(v, τ)f

(v + 1, τ ) we deduce a trigonometric expression of the Weierstrass function ℘(z, τ) with primitive periods (2, 2τ).

It allows us to find again more directly many known identities. In particular we derive the following identity

4℘(2z, 2τ) + 4η(2τ) = ℘(z, τ) + ℘(z + 1, τ) + 2η(τ) where

η(τ) = π

2 [ 1

6 + X

n≥1

1 (sinnπτ )

].

1. Introduction

In a preceding paper [2] we state the following representation for the Jacobi theta function

θ j (v, τ) = θ j (0, τ )f j (v, τ)f j (v + 1, τ ), j = 3, 4;

θ 1 (v, τ) = (π sin(πv) θ 1

(0, τ )f 1 (v, τ)f 1 (v + 1, τ );

θ 2 (v, τ) = cos(πv) θ 2 (0, τ) f 2 (v, τ ) f 2 (v + 1, τ ).

We derive their expansions as infinite products, their Fourier series expansions, as well as for log(f j (v, τ )) and for ^f _f

j

(v, τ).

Let q = e ^iπτ , | q |< 1, we prove in particular ([2,Cor 2-5]) the functions θ 1 , θ 4

may be expressed as infinite products

θ ¹ (v, τ) = (π sin πv) θ 1

(0, τ ) Y

k≥ 1

"

1 −

sin πv sin kπτ

² #

θ ⁴ (v, τ ) = θ ⁴ (0, τ ) Y

k≥ 0

"

1 −

sin πv sin(k + ¹ ₂ )πτ

² #

here θ ⁴ is defined in the band | Imv |< ¹ 2 Imτ, while θ ¹ is defined for

| Imv |< Imτ.

Moreover, we have carried out θ 4 (v, τ )

θ ⁴ (0, τ ) = f ⁴ (v, τ ) f ⁴ (−v, τ ) = f ⁴ (v, τ ) f ⁴ (v + 1, τ ) where

f ⁴ (v, τ) = Y

k≥ 0

1 −

sin πv sin(k + ¹ 2 )πτ

.

1

chouikha@math.univ-paris13.fr. 4, Cour des Quesblais 35430 Saint-Pere, France Key words and phrases. theta functions, elliptic functions, trigonometric expansions.

1

In this paper we are interested in various aspects of trigonometric expressions of the Weierstrass function ℘(z) with primitive periods (2, 2τ). We produce another one derived from the infinite product of the theta function θ ⁴ (v, τ )

℘(z + τ) = −η(τ) + π ² X

k≥ 0

1 − cos πz cos(2k + 1)πτ [cos(2k + 1)πτ − cos πz] ² . As a corollary one deduces some identities as

℘(z + τ + [

1 2, τ + 1) + η(τ + 1) = ℘(z + τ, τ) + η(τ).

2. Trigonometric expansions of ℘(z)

In this section we propose to achieve trigonometric expansions for the elliptic Weierstrass function ℘(z) analogous to that we did before for the Jacobi θ function.

Recall that ℘(z) which has primitive periods 2 and 2τ relative to g 2 and g 3 , may be written as ([1,4,6]

℘(z) = ℘(z; g 2 , g 3 ) = ℘(z, τ ) = 1 z ² + X

m,n

1

(z − 2m − 2nτ) ² − 1 (2m + 2nτ) ²

.

A direct consequence of the preceding definition is the fact that the Weierstrass elliptic function is an even function ℘(−z, τ) = ℘(z, τ ).

The original constructions of elliptic functions are due to Weierstrass and Jacobi [1]. Excellent approaches on the subject of elliptic functions are the classic book by Watson and Whittaker [7]. Useful reference handbooks with many details on tran- scendental functions including those used in this paper are provided by Bateman and Erdelyi, [4], which is freely available online.

We define the values of the Weierstrass elliptic function at the half-periods 1, 1 + τ, τ

e ¹ (τ) = ℘(1, τ ), e ² (τ) = ℘(1 + τ, τ), e ³ (τ) = ℘(τ, τ).

These e i obey the relations

e 1 + e 2 + e 3 = 0, e 1 e 2 + e 3 e 1 + e 2 e 3 = −g ²

4 , e 1 e 2 e 3 = g ³ 4 . The Weierstrass elliptic function verifies the homogeneity relation

℘(z; g ² , g ³ ) = µ ² ℘(µz; g ² µ ⁴ , g ³

µ ⁶ ).

Finally, when two of the roots e ¹ , e ² and e ³ coincide, the Weierstrass elliptic func- tion degenerates to a simply periodic function.

2

On the other hand, the Weierstrass function ℘(z, τ) is related to the theta func- tions θ i (v) where v = ^z 2 :

℘(z) = ( 1

2 ) ² [−4η − d ² logθ ¹ (v) dv ² ] η = η(τ) = − 1

12 θ

1 (0)

θ 1

(0) = π ² 2 [ 1

6 + X

n≥ 1

1 (sin nπτ) ² ].

In the same way we have

℘(z + τ) = ( 1

2 ) ² [−4η − d ² logθ ⁴ (v) dv ² ].

= ( 1

2 ) ² [−4η − d ² logf ⁴ (v)

dv ² − d ² logf ⁴ (−v) dv ² ].

Therefore, we may deduce

e 3 (τ) = ℘(τ) = −η(τ) + π ² X

k≥ 0

1

1 − cos(2k + 1)πτ , e ³ (τ) = ( π

2 ) ²

θ 1

(0)

3θ 1

(0) − θ 4

(0) θ ⁴ (0)

. By the same way

e ² (τ) = −η(τ + 1) + π ² X

k≥ 0

1

1 + cos(2k + 1)πτ = ( π 2 ) ²

θ 1

(0)

3θ 1

(0) − θ

3 (0) θ ³ (0)

.

e ¹ (τ) = η(τ) + η(τ + 1) − 2π ² X

k≥ 0

1

(sin(2k + 1)πτ) ² = ( π 2 ) ²

θ

1 (0)

3θ

1 (0) − θ 2

(0) θ ² (0)

.

Notice the following theta function identity [7]

θ 1

(0)

θ

1 (0) = θ 2

(0)

θ 2 (0) + θ 3

(0)

θ 3 (0) + θ

4 (0) θ 4 (0) .

On the other hand the Weierstrass function ℘(z, τ) may be expressed in different ways (see [5], [7]). In particular, we obtain the Fourier expansions

℘(z + 1, τ ) = −η + (π) ²



 1

4(cos πz) ² − 2 X

k≥ 1

k(−1) ^k q ^{2 k}

1 − q ^2k cos kπz





℘(z + 1 + τ, τ) = −η − 2(π) ² X

k≥ 1

k(−1) ^k q ^{2 k}

1 − q ^{2 k} cos kπz

℘(z + τ, τ) = −η − 2(π) ² X

k≥ 1

kq ^2k

1 − q ^{2 k} cos kπz.

Other expansions will be obtained differently (see [7, p.183])

℘(z, τ ) = −η + (π) ² 4

X

k∈IN

1 [sin( ^πz 2 − kπτ)] ²

3

One trigonometric expansion called by [5] a first q-expansion where q = e ^iπτ : 1

4π ² ℘(z, τ ) = − 1

12 − 1

sin ² πz − X

m,n≥ 1

nq

(cos nπz − 1).

This series is defined in the band | q |< e ^iπz <

¹ , Imτ > 0.

[5] also defined a second q-expansion : 1

4π ² ℘(z, τ ) = − 1

12 + 2 X

n≥ 1

q ^{2 n}

(1 − q ²ⁿ ) ² − 2 X

n≥ 1

(−1) ⁿ nq ⁿ 1 − q ²ⁿ cos nπz .

Notice that the last expression yields an isomorphism between the multiplicative group of complex numbers and the complex points of the Tate curve parametrized by (℘(z, τ), ℘

(z, τ )).

Another alternative trigonometric expansion of ℘(z) is given by

Theorem 2-1 The Weierstrass elliptic function ℘(z) = ℘(z, τ ) with primi- tive periods 2 and 2τ and Imz < ¹ ₂ Imτ may be expressed under the form

℘(z + τ ) = ℘ ¹ (z) + ℘ ¹ (−z) = −η(τ ) + π ² X

k≥ 0

1 − cos πz cos(2k + 1)πτ [cos(2k + 1)πτ − cos πz] ²

℘(z) = −η(τ) + π ² X

k≥ 0

1 − cos π(z − τ) cos(2k + 1)πτ [cos(2k + 1)πτ − cos π(z − τ)] ² where

℘ 1 (z) = −η(τ) 2 + π ²

4 X

k≥ 0

1 − sin( ^πz ₂ ) sin(k + ¹ ₂ )πτ (sin( ^πz ₂ ) − sin(k + ¹ ₂ )πτ) ² .

Proof Start from the following where 2v = z and ω = 1

℘(z + τ) = ( 1

2ω ) ² [−4ηω − d ² logθ ⁴ (v)

dv ² ] = ℘ ¹ (z) + ℘ ² (z)

= ( 1

2ω ) ² [−4ηω − d ² log f 4 (v)

dv ² + d ² log f 4 (−v) dv ² ].

By Corollary 3-5 of [2], deriving 1

θ ⁴

∂θ 4

∂v (v, τ ) = 1 f ⁴

∂f 4

∂v (v, τ ) + 1 f ⁴

∂f 4

∂v (−v, τ)

= −π X

k≥ 0

cos πv

sin(k + ¹ 2 )πτ − sin πv + π X

k≥ 0

cos πv sin(k + ¹ 2 )πτ + sin πv

= −π X

k≥ 0

sin 2πv sin(k + ¹ 2 )πτ ²

− (sin πv) ² . We then obtain (using Maple for example)

d ² logf 4 (v)

dv ² = − π ² sin (k + ¹ ₂ )πτ

sin (π v) − 1

−2 + cos (k + ¹ 2 )πτ ²

+ 2 sin (k + ¹ 2 )πτ

sin (π v) + (cos (π v)) ²

4

= π ² X

k≥ 0

1 − sin(πv) sin(k + ¹ 2 )πτ (sin(πv) − sin(k + ¹ 2 )πτ) ² d ² logf ⁴ (−v)

dv ² = π ² sin (k + ¹ 2 )πτ

sin (π v) + 1

−2 sin (k + ¹ 2 )πτ

sin (π v) − 2 + cos (k + ¹ 2 )πτ ²

+ (cos (π v)) ²

= π ² X

k≥ 0

1 + sin(πv) sin(k + ¹ 2 )πτ (sin(πv) + sin(k + ¹ 2 )πτ) ² .

d ² logθ ⁴ (v) dv ² =

4π ²

− (cos (π v)) ² − cos (k + ¹ 2 )πτ ²

+ 2 cos (k + ¹ 2 )πτ ²

(cos (π v)) ² cos (k + ¹ 2 )πτ ⁴

− 2 cos (k + ¹ 2 )πτ ²

(cos (π v)) ² + (cos (π v)) ⁴ . Since

−2 (cos (π v)) ² − 2

cos

(k + 1 2 )πτ

² + 4

cos

(k + 1

2 )πτ ²

(cos (π v)) ²

= −1 + cos (2 π v) cos ((2 k + 1) π τ ) therefore

d ² logθ ⁴ (v)

dv ² = X

k≥ 0

4 π ² (cos (2 π v) cos ((2k + 1)π τ) − 1)

(cos ((2k + 1)π τ )) ² − 2 cos (2 π v) cos ((2k + 1)π τ ) + (cos (2 π v)) ²

= 4π ² X

k≥ 0

(cos (2 π v) cos ((2k + 1)τ) − 1) [cos(2k + 1)πτ − cos 2πv] ² Then

℘(z+τ)+η = −π ² X

k≥ 0

(cos (2 π v) cos ((2k + 1)πτ) − 1)

[cos(2k + 1)πτ − cos 2πv] ² = π ² X

k≥ 0

1 − (cos πz cos ((2k + 1)πτ)) [cos(2k + 1)πτ − cos πz] ² . On the other hand one has ℘ ¹ (−z) = ℘ ² (z) and the following equality hold

4 X

k≥ 0

1 − (cos (π z) cos ((2k + 1)πτ )) [cos(2k + 1)πτ − cos πz] ² = X

k≥ 0

1 − sin(π ^z ₂ ) sin(k + ¹ ₂ )πτ (sin(π ^z ₂ ) − sin(k + ¹ ₂ )πτ) ² + X

k≥ 0

1 + sin(π ^z ₂ ) sin(k + ¹ ₂ )πτ (sin(π ^z ₂ ) + sin(k + ¹ ₂ )πτ) ² .

(1) Or equivalently replacing z by z + 1 and τ by τ + 1

4 X

k≥ 0

1 − (cos (π z ) cos ((2k + 1)πτ)) [cos(2k + 1)πτ − cos πz] ² = X

k≥ 0

1 − (−1) ^k cos(π ^z 2 ) cos(k + ¹ 2 )πτ (cos(π ^z 2 ) − (−1) ^k cos(k + ¹ 2 )πτ) ²

+ X

k≥ 0

1 + (−1) ^k cos(π ^z ₂ ) cos(k + ¹ ₂ )πτ (cos(π ^z ₂ ) + (−1) ^k cos(k + ¹ ₂ )πτ) ²

= X

k≥ 0

1 − cos(π ^z ₂ ) cos(k + ¹ ₂ )πτ (cos(π ^z ₂ ) − cos(k + ¹ ₂ )πτ) ² + X

k≥ 0

1 + cos(π ^z ₂ ) cos(k + ¹ ₂ )πτ (cos(π ^z ₂ ) + cos(k + ¹ ₂ )πτ) ² .

5

Replacing τ by τ + ¹ 2 one also obtains the identity 4 X

k≥ 1

1 − (cos (π z) cos (2kπτ)) [cos 2kπτ − cos πz] ² = X

k≥ 1

1 − cos(π ^z 2 ) cos kπτ (cos(π ^z ₂ ) − cos kπτ) ² + X

k≥ 1

1 + cos(π ^z 2 ) cos kπτ (cos(π ^z ₂ ) + cos kπτ ) ² .

(2) Corollary 2-2 Consider the function

φ(z, τ) = X

k≥ 0

1 − (cos (π z) cos ((2k + 1)πτ)) [cos(2k + 1)πτ − cos πz] ² = − 1

4π ²

d ² logθ ⁴ (v, τ ) dv ²

defined for Imτ > 0, and Imz < ^Imτ 2 . Then φ verifies the following functional equation :

4φ(2z, 2τ) = φ(z, τ ) + φ(z + 1, τ ) = φ(z, τ ) + φ(z, τ + 1).

and θ verifies the following identities (i) 4 d ² logθ ⁴ (2v, 2τ)

dv ² = d ² logθ ⁴ (v, τ )

dv ² + d ² logθ ⁴ (v + ¹ 2 , τ ) dv ²

= d ² logθ 4 (v, τ)

dv ² + d ² logθ 4 (v, τ + 1) dv ² (ii) d ² logθ ⁴ (v, τ )

dv ² = d ² logθ ⁴ (v + ¹ 2 , τ + 1)

dv ² .

Indeed, notice that by ([2,Cor 2-5]) θ ⁴ and θ ³ may be expressed as infinite products

θ ⁴ (v, τ ) = θ ⁴ (0, τ ) Y

k≥ 0

"

1 −

sin πv sin(k + ¹ 2 )πτ

² #

θ ³ (v, τ ) = θ ³ (0, τ) Y

k≥ 0

"

1 −

sin πv cos(k + ¹ 2 )πτ

² # .

We then deduce θ 4 (v, τ ) θ ⁴ (0, τ )

θ 3 (v, τ) θ ³ (0, τ) = Y

k≥ 0

"

1 −

sin πv sin(k + ¹ ₂ )πτ

² # "

1 −

sin πv cos(k + ¹ ₂ )πτ

² # .

However,

"

1 −

sin πv sin(k + ¹ 2 )πτ

² # "

1 −

sin πv cos(k + ¹ 2 )πτ

² #

= 1−

sin πv sin(k + ¹ 2 )πτ

²

−

sin πv cos(k + ¹ 2 )πτ

²

+

sin πv sin(k + ¹ 2 )πτ

²

sin πv cos(k + ¹ 2 )πτ

²

= 1 −

sin πv sin(k + ¹ ₂ )πτ

²

−

sin πv cos(k + ¹ ₂ )πτ

² + 4

(sin πv) ² sin(2k + 1)πτ

²

= 1 −

(sin πv)

(sin(k + ¹ 2 )πτ)(cos(k + ¹ 2 )πτ ) ²

+ 4

(sin πv) ² sin(2k + 1)πτ

²

6

= 1 −

2(sin πv) sin(2k + 1)πτ

² +

2(sin πv) ² sin(2k + 1)πτ

²

=

"

1 −

sin 2πv sin(2k + 1)πτ

² #

Thus we derive the well known identity ([4,6,7]) θ ⁴ (2v, 2τ )

θ 4 (0, 2τ) = θ ³ (v, τ ) θ 3 (0, τ )

θ ⁴ (v, τ) θ 4 (0, τ) = Y

k≥ 0

"

1 −

sin 2πv sin(2k + 1)πτ

² # .

Corollary 2-2 may be deduced from the Landen transformation by logarithmic differentiation of this identity.

Or equivalently,

θ ⁴ (2v, 2τ) = θ 3 (v, τ)θ 4 (v, τ)

θ ⁴ (0, 2τ) = θ 4 (v + 1/2, τ )θ 4 (v, τ) θ ⁴ (0, 2τ) since (θ 4 (0, 2τ)) ² = θ 3 (0, τ )θ 4 (0, τ ).

By the same way we may prove

"

1 −

sin πv sin kπτ

² # "

1 −

sin πv cos kπτ

² #

=

"

1 −

sin 2πv sin 2kπτ

² # .

Since

θ 4 (0, 2τ) = p

θ 3 (0, τ )θ 2 (0, τ) = θ

1 (0, τ )θ 2 (0, τ ) 2θ

1 (0, 2τ) we then deduce

θ ¹ (2v, 2τ)

θ ⁴ (0, 2τ) = θ ¹ (v, τ ) θ ³ (0, τ )

θ ² (v, τ)

θ ⁴ (0, τ ) = (π sin πv) Y

k≥ 0

"

1 −

sin 2πv sin 2kπτ

² # .

Corollary 2-3 The Weierstrass function ℘(z) = ℘(z, τ) with primitive periods 2 and 2τ satisfies the following identities

(i) ℘(z + τ + 1, τ + 1) + η(τ + 1) = ℘(z + τ, τ) + η(τ), (ii) 4℘(z + τ, τ) + 4η(τ) = ℘( z + τ

2 , τ

2 ) + ℘( z + τ + 2 2 , τ

2 ) + 2η( τ 2 )

= ℘( z + τ 2 , τ

2 ) + ℘( z + τ 2 , τ

2 + 1) + η( τ 2 ) + η( τ

2 + 1).

Replacing z + τ by z one gets

℘(z + 1/2, τ + 1) + η(τ + 1) = ℘(z, τ) + η(τ), 4℘(z, τ ) + 4η(τ) = ℘( z

2 , τ

2 ) + ℘( z + 2 2 , τ

2 ) + 2η( τ 2 )

= ℘( z 2 , τ

2 ) + ℘( z 2 , τ

2 + 1) + η( τ 2 ) + η( τ

2 + 1).

7

Corollary 2-4 The Weierstrass function ℘(z) = ℘(z, τ) with primitive periods 2 and 2τ verifies the following identity

4℘(2z, 2τ) + 4η(2τ) = ℘(z, 2τ ) − e ¹ (τ) + 2η(τ).

Indeed, notice that since [4, p.372]

℘(z + τ, τ) = e ³ + (e ³ − e ¹ )(e ³ − e ² )

℘(z) − e 3

and ℘(z, τ

2 ) = ℘(z, τ) + (e ³ − e ¹ )(e ³ − e ² )

℘(z) − e 3

. Therefore

℘(z, τ

2 ) = ℘(z, τ ) + ℘(z + τ, τ) − e ³ (τ).

By the same way we have

℘(z, τ + 1

2 ) = ℘(z, τ ) + ℘(z + τ + 1, τ) − e ² (τ),

℘(z, 2τ) = ℘(z, τ ) + ℘(z + 1, τ) − e ¹ (τ).

Replacing the last equality in (ii) of Corollary 2-3 we obtain Corollary 2-4. Recall the duplication formula

℘(2z, 2τ) + ℘(z, 2τ ) =

℘

(z, 2τ) 2℘(z, 2τ)

² .

8

References

[1] P. Appell, E. Lacour, Fonctions elliptiques et applications Gauthiers- Villard ed., Paris (1922).

[2] A.R. Chouikha Functions related to Jacobi Theta Functions and applica- tions https://hal.archives-ouvertes.fr/hal-03170818. (2021)

[3] A.R. Chouikha Expansions of Theta Functions and Applications ArXiv, math/0112137, http://front.math.ucdavis.edu/0112.5137, (2011).

[4] A.Erdelyi, W.Magnus, F.Oberhettinger, F.Tricomi, Higher transcendental functions Vol II. Based on notes left by H. Bateman. Robert E. Krieger Publish.

Co., Inc., Melbourne, Fla., (1981).

[5] S. Lang, Elliptic functions Addison-Wesley, Springer, 1970.

[6] D.F. Lawden, Elliptic functions and applications, Springer-Verlag, vol 80, 1980.

[7] E.T. Whittaker,G.N. Watson A course of Modern Analysis Cambridge (1963).

9