From trigonometry to elliptic functions - UCI

From trigonometry to elliptic functions

Zhiqin Lu

The Math ClubUniversity of California, Irvine

March 31, 2010

Zhiqin Lu, The Math Club University of California, Irvine From trigonometry to elliptic functions 1/24

Question

What is the area of a triangle withside lengths a, b, and c?


Heron’s formula [A.D. 60]

A =√

p(p − a)(p − b)(p − c),

where p = (a + b + c)/2.


By the Pythagorean theorem, we have

b2 − d2 = a2 − (c − d)2.

Therefore, we have

d =1

2

(c +

b2 − a2

c

).

By the Pythagorean theorem again, wehave

h2 = b2 − 1

4

(c +

b2 − a2

c

)2

.

The formula follows from the fact that

A =1

2ch.

Figure: Triangle withaltitude h cutting base cinto d and (c - d).


Using Trigonometry, the formula is a lot easierto prove:

A =1

2bc sinα.

By the law of cosine, we have

sinα =√

1− cos2 α =

√1−

(b2 + c2 − a2

2bc

)2

.


1 What is sin 20◦?

2 We have to use the concept function.

3 A function is an assignment: f : A→ B.

4 How to define trigonometric functions?

















Definition:

sinA = a/h

cosA = b/h

tanA = a/b

· · · · · ·

Such a definition can hardly be used in Calculus!


Definition:

sinA = a/h

cosA = b/h

tanA = a/b

· · · · · ·

Such a definition can hardly be used in Calculus!


How to find the derivative of sin x?We have to use/assume the fact

limθ→0

sin θ

θ= 1,

which is the derivative of sin x at 0.


In Stewart’s or Minton/Smith’s Calculus book, there are two ways todefine the exp/log functions

1 Define ex first, and log x is the inverse function of ex

1 Define ex for x integers;2 Define ex for x rational numbers;3 Define ex for any real number x by continuity.

2 Define log x first and then define ex as the inverse of log x

log x =

∫1

xdx + C .


Can we do that same thing for trigonometric functions?

arcsin x =

∫1√

1− x2dx + C

Compare with the integral∫1√

1 + x2dx = log(x +

√1 + x2) + C

We get the following mysterious formula

log(x +√

1 + x2) = arcsin(√−1x)



arcsin x =

∫1√

1− x2dx + C


1 + x2dx = log(x +

√1 + x2) + C


log(x +√

1 + x2) = arcsin(√−1x)



arcsin x =

∫1√

1− x2dx + C


1 + x2dx = log(x +

√1 + x2) + C


log(x +√

1 + x2) = arcsin(√−1x)


A complex number z is a pair of two real numbers. But we are more usedto writing it as

z = x + iy = x +√−1y ,

where x , y are the two real numbers.

The production of two complex numbers is somewhat interesting

(x1 + iy1)(x2 + iy2) = x1x2y1y2 + i(x1y2 + x2y1).

The negative sign costs 99% of errors!



z = x + iy = x +√−1y ,

where x , y are the two real numbers.The production of two complex numbers is somewhat interesting

(x1 + iy1)(x2 + iy2) = x1x2 − y1y2 + i(x1y2 + x2y1).




z = x + iy = x +√−1y ,

where x , y are the two real numbers.The production of two complex numbers is somewhat interesting

(x1 + iy1)(x2 + iy2) = x1x2 − y1y2 + i(x1y2 + x2y1).



Euler’s formulae iz = cos z + i sin z

cos z =e iz + e−iz

2

sin z =e iz − e−iz

2i


Euler’s formulae iz = cos z + i sin z

cos z =e iz + e−iz

2

sin z =e iz − e−iz

2i


Theorem: cos(x + y) = cos x cos y − sin x sin y .

Proof.

cos x cos y − sin x sin y

=e ix + e−ix

2· e

iy + e−iy

2− e ix − e−ix

2i· e

iy − e−iy

2i

=1

4(e i(x+y) + e−i(x+y) + e i(x−y) + e i(y−x))

+1

4(e i(x+y) + e−i(x+y) − e i(x−y) − e i(y−x))

= cos(x + y).

Q.E.D.

(=quod erat demonstrandum=that which was to bedemonstrated)



Proof.


=e ix + e−ix

2· e

iy + e−iy


2i· e

iy − e−iy

2i

=1


+1


= cos(x + y).

Q.E.D. (=quod erat demonstrandum

=that which was to bedemonstrated)



Proof.


=e ix + e−ix

2· e

iy + e−iy


2i· e

iy − e−iy

2i

=1


+1


= cos(x + y).

Q.E.D. (=quod erat demonstrandum=that which was to bedemonstrated)


The above discussion hint us that we need to study∫1√

1− z2dz ,

where z is a complex number.

Or more precisely,

φ(z) =

∫ z

0

1√1− t2

dt.


The above discussion hint us that we need to study∫1√

1− z2dz ,

where z is a complex number.

Or more precisely,

φ(z) =

∫ z

0

1√1− t2

dt.


1 The function 1√1−z2 is multi-valued.

2 In order to make it single-valued, we need to construct a Riemannsurface.

3 We cut two copies of C along [−1, 1] and glue them. The space C iscalled a Riemann surface. 1√

1−z2 is a single-valued function on the

Riemann surface C .

4 φ(z) is multivalued even if on C , 1√1−z2 is single-valued,

5 because of the existence of Residue.

6 We have ∮|z|=R

dz√1− z2

= 2π






Riemann surface C .



6 We have ∮|z|=R

dz√1− z2

= 2π






Riemann surface C .



6 We have ∮|z|=R

dz√1− z2

= 2π






Riemann surface C .



6 We have ∮|z|=R

dz√1− z2

= 2π






Riemann surface C .



6 We have ∮|z|=R

dz√1− z2

= 2π






Riemann surface C .



6 We have ∮|z|=R

dz√1− z2

= 2π


1 Therefore, φ : C → C/2πZ

2 φ−1 : C/2πZ→ C exists.

3 C is the set of all points (t,±√

1− t2). Therefore, C can berepresented by the set

x2 + y2 = 1

in C2.

4 C is a group

(x1, y1)⊕ (x2, y2) = (x1y2 + x2y1, x1x2 − y1y2).

Check:(x1y2 + x2y1)2 + (x1x2 − y1y2)2 = 1.


1 Therefore, φ : C → C/2πZ2 φ−1 : C/2πZ→ C exists.



x2 + y2 = 1

in C2.

4 C is a group

(x1, y1)⊕ (x2, y2) = (x1y2 + x2y1, x1x2 − y1y2).

Check:(x1y2 + x2y1)2 + (x1x2 − y1y2)2 = 1.





x2 + y2 = 1

in C2.

4 C is a group

(x1, y1)⊕ (x2, y2) = (x1y2 + x2y1, x1x2 − y1y2).

Check:(x1y2 + x2y1)2 + (x1x2 − y1y2)2 = 1.





x2 + y2 = 1

in C2.

4 C is a group

(x1, y1)⊕ (x2, y2) = (x1y2 + x2y1, x1x2 − y1y2).

Check:(x1y2 + x2y1)2 + (x1x2 − y1y2)2 = 1.





x2 + y2 = 1

in C2.

4 C is a group

(x1, y1)⊕ (x2, y2) = (x1y2 + x2y1, x1x2 − y1y2).

Check:(x1y2 + x2y1)2 + (x1x2 − y1y2)2 = 1.


The map φ−1 : C/2πZ→ C is a group isomorphism:

(α, β) −−−−→ α + βy y((x1, y1), (x2, y2)) −−−−→ (x1x2 − y1y2, x1y2 + x2y1)

,

where x1 = cosα, y1 = sinα, etc.

Such an isomorphism is called Addition Theorem.




,






,




The following is another version of the addition theorem forthe sine function:∫ sin u

0

1√1− x2

dx +

∫ sin v

0

1√1− x2

dx =

∫ sin(u+v)

0

1√1− x2

dx

If we set sin u, z = sin v , then we have∫ y

0

1√1− x2

dx +

∫ z

0

1√1− x2

dx =

∫ T (y ,z)

0

1√1− x2

dx ,

whereT (y , z) = y

√1− z2 + z

√1− y2.

Any other functions satisfy the above addition theorem? EllipticFunctions!



0

1√1− x2

dx +

∫ sin v

0

1√1− x2

dx =

∫ sin(u+v)

0

1√1− x2

dx


0

1√1− x2

dx +

∫ z

0

1√1− x2

dx =

∫ T (y ,z)

0

1√1− x2

dx ,

whereT (y , z) = y

√1− z2 + z

√1− y2.




0

1√1− x2

dx +

∫ sin v

0

1√1− x2

dx =

∫ sin(u+v)

0

1√1− x2

dx


0

1√1− x2

dx +

∫ z

0

1√1− x2

dx =

∫ T (y ,z)

0

1√1− x2

dx ,

whereT (y , z) = y

√1− z2 + z

√1− y2.

Any other functions satisfy the above addition theorem?

EllipticFunctions!



0

1√1− x2

dx +

∫ sin v

0

1√1− x2

dx =

∫ sin(u+v)

0

1√1− x2

dx


0

1√1− x2

dx +

∫ z

0

1√1− x2

dx =

∫ T (y ,z)

0

1√1− x2

dx ,

whereT (y , z) = y

√1− z2 + z

√1− y2.



Consider the integral ∫dt√

t(t − 1)(t − λ),

where 0 < λ < 1.

Historic RemarksLegendre studied these kind of integral extensively and wrote 3 volumes(4 volumes?) of Traite des fonctions elliptiques. But he failed in findingthe most important properties of the integrals: they are the inversefunctions of some doubly periodic functions! This fact was found by Abeland Jocobi independently.


Consider the integral ∫dt√

t(t − 1)(t − λ),

where 0 < λ < 1.

Historic RemarksLegendre studied these kind of integral extensively and wrote 3 volumes(4 volumes?) of Traite des fonctions elliptiques. But he failed in findingthe most important properties of the integrals: they are the inversefunctions of some doubly periodic functions! This fact was found by Abeland Jocobi independently.


The Riemann surface of the function

1√z(z − 1)(z − λ)

y2 = z(z − 1)(z − λ)



1√z(z − 1)(z − λ)

y2 = z(z − 1)(z − λ)



1√z(z − 1)(z − λ)

y2 = z(z − 1)(z − λ)


Let

ω1 =

∮A

dz√z(z − 1)(z − λ)

ω2 =

∮B

dz√z(z − 1)(z − λ)

Then the inverse function of the elliptic integral∫ z dt√t(t − 1)(t − λ)

is a doubly period function of periods ω1, ω2.


Let

ω1 =

∮A

dz√z(z − 1)(z − λ)

ω2 =

∮B

dz√z(z − 1)(z − λ)

Then the inverse function of the elliptic integral∫ z dt√t(t − 1)(t − λ)

is a doubly period function of periods ω1, ω2.


Basic property of elliptic functions

Theorem: Entire (holomorphic) doubly periodic functions are constants.

Proof. Bounded holomorphic functions are constant by Liouivile Theorem.

Therefore, elliptic functions are meromorphic (singularities are poles).





Therefore, elliptic functions are meromorphic

(singularities are poles).





Therefore, elliptic functions are meromorphic (singularities are poles).


TheoremAny two tori T (ω1, ω2) and T (ω3, ω4) are biholomorphic if and only ifthere exists integers a, b, c , d such that ad − bc = 1, and

ω4

ω3=

aω1 + bω2

cω1 + dω2.


References:

1 http://www.rose-hulman.edu/mathjournal/archives/2009/vol10-n2/paper2/v10n2-2pd.pdf

2 http://websites.math.leidenuniv.nl/algebra/ellipticfunctions.pdf


From trigonometry to elliptic functions - UCI

Documents