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OutlineA Brief History of the Quintic

Transformations of the Quintic EquationDefinition and Properties of Theta Functions

The General SchemeThe Modular Equation

Solving the quintic

Theta Functions and the Quintic

Ali Uncu & Frank Patane

www.math.ufl.edu/∼akuncu www.math.ufl.edu/∼frankpatane

October 18, 2011

Ali Uncu & Frank Patane Theta Functions and the Quintic

Solving the quintic

1 A Brief History of the Quintic

2 Transformations of the Quintic EquationOur approachTschirnhaus TransformationInverse Transformation

3 Definition and Properties of Theta Functions

4 The General Scheme

5 The Modular Equation

6 Solving the quintic

Solving the quintic

A Quick Background on the Quintic

Let the general monic quintic be denoted byx5 + a4x

4 + a3x3 + a2x

2 + a1x + a0, ai ∈ Q.

Mathematicians such as Leibniz, Tschirnhausen, Euler,Vandermonde, Lagrange, and Ruffini, all tried to solve the generalquintic in radicals, but were succesful only in special cases.

In 1824 Abel delivered the first correct proof of what is now knownas the Abel-Ruffini Theorem or Abel’s Impossibility Theorem.

Solving the quintic

4 + a3x3 + a2x

2 + a1x + a0, ai ∈ Q.

Solving the quintic

4 + a3x3 + a2x

2 + a1x + a0, ai ∈ Q.

Solving the quintic

Abel’s proof is one of the first achievements of group theory, andlater evolved into what is now known as Galois Theory.

Indeed, we may use the notion of a solvable group to determinewhether a polynomial is solvable in radicals.

Any quadratic, cubic, or quartic is solvable in radicals because theGalois group of such a polynomial must be a subgroup of S2,S3, orS4 which are all solvable groups.

Solving the quintic

Any quadratic, cubic, or quartic is solvable in radicals because theGalois group of such a polynomial must be a subgroup of S2,S3, orS4 which are all solvable groups.

Solving the quintic

Any quadratic, cubic, or quartic is solvable in radicals because theGalois group of such a polynomial must be a subgroup of S2, S3, orS4 which are all solvable groups.

Solving the quintic

The general quintic is not solvable since it may have Galois groupisomorphic to A5 or S5.

Thus one can ask if a given quintic is solvable.

As we will soon see, we are able to transform the general quinticinto the Bring-Jerrard quintic: x5 + 5x − a

Solving the quintic

Theorem(Spearman Williams 1994) An irreducible quintic P(x) inBring-Jerrard form is solvable by radicals if and only if there existrationals r , s such that P(x) is of the form:

x5 +5r4(4s + 3)

s2 + 1− 4r4(2s − 11)

s2 + 1= 0.

Let us now see that we can transform the general quintic toBring-Jerrard form.

Solving the quintic

x5 +5r4(4s + 3)

s2 + 1− 4r4(2s − 11)

s2 + 1= 0.

Solving the quintic

x5 +5r4(4s + 3)

s2 + 1− 4r4(2s − 11)

s2 + 1= 0.

Solving the quintic

Our approachTschirnhaus TransformationInverse Transformation

General Quintic Equation:

x5 + a4x4 + a3x

3 + a2x2 + a1x + a0 = 0

Bring-Jerrard Form:

x5 + c1x + c0 = 0

x5 + 5x + a = 0Ali Uncu & Frank Patane Theta Functions and the Quintic

Solving the quintic

f (x) = x5 + a4x4 + a3x

3 + a2x2 + a1x + a0

Assume that f (x) has distinct roots x1, . . . , x5.

Suppose that

y = φ(x) = p4x4 + p3x

3 + p2x2 + p1x + p0

is the Tschirnhaus transformation where yi = y(xi ) beingdistinct roots of another quintic equation. When one is solving forf (x) = 0;

y = p4x4 + p3x

3 + p2x2 + p1x + p0

xy = p4x5 + p3x

4 + p2x3 + p1x

2 + p0xxy = p4(−a4x

4−a3x3−a2x

2−a1x−a0)+p3x4 +p2x

3 +p1x2 +p0x

xy = p′4x4 + p′3x

3 + p′2x2 + p′1x + p′0

Solving the quintic

f (x) = x5 + a4x4 + a3x

3 + a2x2 + a1x + a0

Suppose that

y = φ(x) = p4x4 + p3x

3 + p2x2 + p1x + p0

y = p4x4 + p3x

3 + p2x2 + p1x + p0

xy = p4x5 + p3x

4 + p2x3 + p1x

2 + p0xxy = p4(−a4x

4−a3x3−a2x

2−a1x−a0)+p3x4 +p2x

3 +p1x2 +p0x

xy = p′4x4 + p′3x

3 + p′2x2 + p′1x + p′0

Solving the quintic

f (x) = x5 + a4x4 + a3x

3 + a2x2 + a1x + a0

Suppose that

y = φ(x) = p4x4 + p3x

3 + p2x2 + p1x + p0

is the Tschirnhaus transformation where yi = y(xi ) beingdistinct roots of another quintic equation.

When one is solving forf (x) = 0;

y = p4x4 + p3x

3 + p2x2 + p1x + p0

xy = p4x5 + p3x

4 + p2x3 + p1x

2 + p0xxy = p4(−a4x

4−a3x3−a2x

2−a1x−a0)+p3x4 +p2x

3 +p1x2 +p0x

xy = p′4x4 + p′3x

3 + p′2x2 + p′1x + p′0

Solving the quintic

f (x) = x5 + a4x4 + a3x

3 + a2x2 + a1x + a0

Suppose that

y = φ(x) = p4x4 + p3x

3 + p2x2 + p1x + p0

y = p4x4 + p3x

3 + p2x2 + p1x + p0

xy = p4x5 + p3x

4 + p2x3 + p1x

2 + p0xxy = p4(−a4x

4−a3x3−a2x

2−a1x−a0)+p3x4 +p2x

3 +p1x2 +p0x

xy = p′4x4 + p′3x

3 + p′2x2 + p′1x + p′0

Solving the quintic

Similarly one can write

x2y = p′′4x4 + p′′3x

3 + p′′2x2 + p′′1x + p′′0

x3y = p′′′4 x4 + p′′′3 x3 + p′′′2 x2 + p′′′1 x + p′′′0

x4y = p′′′′4 x4 + p′′′′3 x3 + p′′′′2 x2 + p′′′′1 x + p′′′′0

So that turning the question of finding coefficients of y into anEigenvalue problem, (A− yI5)x̄ = 0;

p0 − y p1 p2 p3 p4

p′0 p′1 − y p′2 p′3 p′4p′′0 p′′1 p′′2 − y p′′3 p′′4p′′′0 p′′′1 p′′′2 p′′′3 − y p′′′4

p′′′′0 p′′′′1 p′′′′2 p′′′′3 p′′′′4 − y

Solving the quintic

If the characteristic polynomial

det(yI5 − A) = y5 + b4y4 + b3y

3 + b2y2 + b1y + b0

has no multiple roots, we can solve for b4 = b3 = b2 = 0.

Solving the quintic

Solution due to Bring:

b4(p0, p1, p2, p3, p4) = 0 is a linear equation so solving for p0,we can write every other p0 in the other equations in pi fori = 1, . . . , 4.

b3(p1, p2, p3, p4) = 0 is a quadratic equation and can bewritten as u2

1 − v21 + u2

2 − v22 where WLOG u1(p1, p2, p3, p4),

v1(p2, p3, p4), u2(p3, p4) and v2(p4) are all linear so that onecan solve ui = vi resulting in writing every other variable in p4

and p2.

Choosing a suitable p2 we have all equations in p4.

b2(p4) = 0 is a cubic equation in p4 hence can be solved.

After finding p4 we can solve p3, p1 and p0 in this order.

Solving the quintic

Hence we can transform the general quintic to Bring-JerrardForm.

y5 + b1y + b0 = 0

We can transform the quintic Bring-Jerrard Form to

z5 + 5z + a = 0

by a linear change of variables y 7→ 4√

5/b1z , which we can solveusing theta functions.

Solving the quintic

This has the inverse function z 7→ 4√b1/5y which can be easily

computed once one finds roots for z5 + 5z + a = 0, thus giving thesolutions for y5 + b1y + b0 = 0.

Similarly the inverse map of the Tschirnhaus transformation canbe calculated as a rational function in variable y from the matrix:

A− yI =

p0 − y p1 p2 p3 p4

p′0 p′1 − y p′2 p′3 p′4p′′0 p′′1 p′′2 − y p′′3 p′′4p′′′0 p′′′1 p′′′2 p′′′3 − y p′′′4

p′′′′0 p′′′′1 p′′′′2 p′′′′3 p′′′′4 − y

Solving the quintic

One can find the inverse transformation by eliminating the matrixyI5 − A upto a certain level:

∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ 0∗ ∗ ∗ 0 0

k(y) h(y) 0 0 00 0 0 0 0

Then since multiplying this matrix with (1, x , x2, x3, x4)t gives zerothe inverse transformation is defined by x = −k(y)/h(y).

Moreover k(y), h(y) has degree both at most 4 in y sincey5 + b1y + b0 = 0.

Solving the quintic

Furthermore, one can replace the inverse map

x = −k(y)/h(y)

with a polynomial of degree at most 4 which shows that there is asymmetry between mapping a Bring-Jerrard form to a generalquintic and vice versa.

Solving the quintic

Example: Let us take the quintic function:

f := x5 + x4 − x3 + x2 + 2x + 1

and the transformation:

y := p4x4 + p3x

3 + p2x2 + p1x + p0

. Recall that we have the freedom of choosing one pi (except p4

being zero). Then the characteristic polynomial in y of matrixyI5 − A becomes:

y5 + b4y4 + b3y

3 + b2y2 + b1y + E0 = 0

where bi s are going to be chosen in order to make

b4 = b3 = b2 = 0

.Ali Uncu & Frank Patane Theta Functions and the Quintic

Solving the quintic

Here equations are:

b4 = 7p3 − 3p4 + p1 − 5p0 − 3p2

b3 = 13p4p1 − 5p2p3 − 4p1p0 − 29p24

. . .+ 10p20 + 12bp4b0

b2 = −12p3p0p1 − 42p23p0 + p3

1 + 2p3p22

· · · − 24p3p0p4 − 23p2p1p4

We can solve p0 using b4 in the other variables as:

p0 = (7/5)p3 − (3/5)p4 + (1/5)p1 − (3/5)p2

Solving the quintic

We can rewrite the second equation as:

b3 = u21 − v2

1 + u22 − v2

√(−7

)(p1 −

2p4 −

√−47√

47p4 +

√−1431

√188

(p3 −

1431p4

√−8705

√5724p4

and solve ui = vi .Ali Uncu & Frank Patane Theta Functions and the Quintic

Solving the quintic

Picking p2 = 1 as our free choice, we get1

p4 = −0.552 . . . , p3 = −0.475 . . . , p1 = −0.36 . . . , p0 = −1.006 . . .

in this order. More importantly the characteristic polynomial turnsinto Bring-Jerrard form:

y5 + By + C = 0

Now we can replace Y by ((1/5)B)(1/4) ∗ Z and normalize to getthe polynomial:

Z 5 + 5Z − 3.14924 · · · = 0

1Exact numbers are all algebraic, yet some exceed maybe 20 pages.Ali Uncu & Frank Patane Theta Functions and the Quintic

Solving the quintic

Intro. to Theta Functions

Before we move further, let us gain some motivation byconsidering one way to solve the cubic.

We may easily depress the general cubic with a linear transformtionto get x3 + px + q = 0.

The key to our argument lies in 4 cos3(θ)− 3 cos(θ)− cos(3θ) = 0

We may exploit the trig identity to write the roots of the cubic interms or cos and cos−1

Solving the quintic

The Jacobi theta functions are defined for two complex variablesν, τ with ν any complex number, and τ in the upper half plane H.

We first set q := eπiτ and z := e2πiν and see that Im(τ) > 0implies |q| < 1.

The main Jacobi theta function is defined by the series,

θ3(ν|τ) = θ3(ν) =∞∑

m=−∞qm

Note that |q| < 1 implies that θ3(ν) is an entire function.

Solving the quintic

θ3(ν|τ) = θ3(ν) =∞∑

m=−∞qm

Solving the quintic

θ3(ν|τ) = θ3(ν) =∞∑

m=−∞qm

Solving the quintic

The other Jacobi theta functions are a shift of θ3. We have:

θ0(ν) = θ3(ν +1

∑(−1)mqm

θ1(ν) = ie−πi(ν−τ4

)θ3(ν +1− τ

2) = ie−πiν

∑(−1)mq(m− 1

θ2(ν) = e−πi(ν−τ4

)θ3(ν − τ

2) = e−πiν

∑q(m− 1

Solving the quintic

θ0(ν) = θ3(ν +1

∑(−1)mqm

)θ3(ν +1− τ

2) = ie−πiν

∑(−1)mq(m− 1

)θ3(ν − τ

2) = e−πiν

∑q(m− 1

Solving the quintic

θ0(ν) = θ3(ν +1

∑(−1)mqm

)θ3(ν +1− τ

2) = ie−πiν

∑(−1)mq(m− 1

)θ3(ν − τ

2) = e−πiν

∑q(m− 1

Solving the quintic

θ0(ν) = θ3(ν +1

∑(−1)mqm

)θ3(ν +1− τ

2) = ie−πiν

∑(−1)mq(m− 1

)θ3(ν − τ

2) = e−πiν

∑q(m− 1

Solving the quintic

Much of the time we set ν = 0, and consider the properties ofθ3(0|τ) = θ3.

Similarly we set θi (0|τ) = θi .

The Jacobi Identity states: θ43 = θ4

2 + θ40

Solving the quintic

2 + θ40

Solving the quintic

2 + θ40

Solving the quintic

Product Representation of Theta

Valid for complex q, z with |q| < 1, z 6= 0 the Jacobi TripleProduct Identity gives:

θ3(ν|τ) =∞∑

m=−∞qm

∞∏m=1

(1−q2m)(1+zq2m−1)(1+z−1q2m−1).

With ν = 0 we get

θ3 =∞∏

(1− q2m)(1 + q2m−1)2.

Solving the quintic

Product Representation of Theta

Valid for complex q, z with |q| < 1, z 6= 0 the Jacobi TripleProduct Identity gives:

θ3(ν|τ) =∞∑

m=−∞qm

∞∏m=1

(1−q2m)(1+zq2m−1)(1+z−1q2m−1).

With ν = 0 we get

θ3 =∞∏

(1− q2m)(1 + q2m−1)2.

Solving the quintic

Definition of η, f , f1, f2

Using the product representation for θ3 we get productrepresentations for θ0 and θ2:

In particular, θ3 = η(τ)f 2(τ), θ0 = η(τ)f 21 (τ), and

θ2 = η(τ)f 22 (τ) where

η(τ) = q1

(1− q2k), f (τ) = q−124∏

(1 + q2k−1),

f1(τ) = q−124∏

(1− q2k−1) f2(τ) =√

(1 + q2k).

Solving the quintic

η(τ) = q1

(1− q2k), f (τ) = q−124∏

(1 + q2k−1),

f1(τ) = q−124∏

(1− q2k−1) f2(τ) =√

(1 + q2k).

Solving the quintic

η(τ) = q1

(1− q2k), f (τ) = q−124∏

(1 + q2k−1),

f1(τ) = q−124∏

(1− q2k−1) f2(τ) =√

(1 + q2k).

Solving the quintic

Using the identity θ43 = θ4

2 + θ40 we derive f 8 = f 8

1 + f 82 .

We will also need the property f1f2f =√

f1f2f =√

(1−q2k−1)(1+q2k)(1+q2k−1) =√

(1−q2k−1)(1+qk) =√

2∏ (1− qk)(1 + qk)

(1− q2k)

Solving the quintic

1 + f 82 .

f1f2f =√

(1−q2k−1)(1+q2k)(1+q2k−1) =√

(1−q2k−1)(1+qk) =√

2∏ (1− qk)(1 + qk)

(1− q2k)

Solving the quintic

1 + f 82 .

f1f2f =√

(1−q2k−1)(1+q2k)(1+q2k−1) =√

(1−q2k−1)(1+qk) =√

2∏ (1− qk)(1 + qk)

(1− q2k)

Solving the quintic

The Modular Equation

Let us set u = f (τ), v∞ = f (5τ) , vc = f ( τ+c5 ) , c ∈ Z.

Since f has period 48, there are only 5 · 48 distinct vc .

We only consider the 5 vc with c ≡ 0 mod 48. Reducing the indexmodulo 5 we may write

v±2 = f (τ ∓ 48

5); v±1 = f (

τ ± 96

Solving the quintic

v±2 = f (τ ∓ 48

5); v±1 = f (

τ ± 96

Solving the quintic

v±2 = f (τ ∓ 48

5); v±1 = f (

τ ± 96

Solving the quintic

Our first goal is to establish the modular equation:

v)3 + (

u)3 = (uv)2 − 4

or equivalently : v6 − u5v5 + 4uv + u6 = 0

with u = f (τ) and v = vc or v∞.

Solving the quintic

v)3 + (

u)3 = (uv)2 − 4

Solving the quintic

v)3 + (

u)3 = (uv)2 − 4

Solving the quintic

v)3 + (

u)3 = (uv)2 − 4

Solving the quintic

Modular to Quintic

Thus we will have shown that the coefficients of a sixth degreepolynomial with roots vc can be expressed in terms of u.

We will then show that the roots of the quinticw(w2 + 5)2 − u12 + 64u−12 are a polynomial in the vc .

After the substitution y =f 81 −f 8

2f 2(w2+5)

we will have y5 + 5y =f 81 −f 8

2f 2 .

Solving the quintic

Modular to Quintic

2f 2(w2+5)

2f 2 .

Solving the quintic

Modular to Quintic

2f 2(w2+5)

2f 2 .

Solving the quintic

Modular to Quintic

2f 2(w2+5)

2f 2 .

Solving the quintic

Transformations of u and v

We first consider the transformation τ → τ + 2 on u and v .

When seeing f has period 48 we noted η(τ + 1) = eπi12 η(τ) implies

f (τ + 2) = e−πi

12 f (τ)

Thus τ → τ + 2 sends u to e−πi

Now τ → τ + 2 takes v∞ = f (5τ) to f (5τ + 10) = e−5πi

12 f (5τ).

Solving the quintic

f (τ + 2) = e−πi

12 f (τ)

12 f (5τ).

Solving the quintic

f (τ + 2) = e−πi

12 f (τ)

12 f (5τ).

Solving the quintic

f (τ + 2) = e−πi

12 f (τ)

12 f (5τ).

Solving the quintic

In general the substitution τ → τ + 2 replaces vc with

f (τ + 2 + c

5) = f (

τ + 50− 48 + c

5) = e

−5πi12 f (

τ + c ′

where c ′ = c − 48 ≡ 0mod 48 and c ′ ≡ c + 2 mod 5.

In other words vc → e−5πi

12 vc+2.

Solving the quintic

f (τ + 2 + c

5) = f (

τ + 50− 48 + c

5) = e

−5πi12 f (

τ + c ′

12 vc+2.

Solving the quintic

f (τ + 2 + c

5) = f (

τ + 50− 48 + c

5) = e

−5πi12 f (

τ + c ′

12 vc+2.

Solving the quintic

f (τ + 2 + c

5) = f (

τ + 50− 48 + c

5) = e

−5πi12 f (

τ + c ′

12 vc+2.

Solving the quintic

Thus we have uv → e−πi

2 uv and uv → e

πi3uv .

where the index of v transforms like τ → τ + 2.

One can also show the τ → −1τ takes uv → uv , u

v →uv , and

τ → τ−1τ+1 takes uv → −2

uv , uv →

−vu with the index of v changing in

the same way as τ .

τ → τ + 2 e−πi

2 uv eπi3uv

τ → −1τ uv u

vτ → τ−1

τ+1−2uv

Solving the quintic

2 uv and uv → e

πi3uv .

v →uv , and

uv , uv →

τ → τ + 2 e−πi

2 uv eπi3uv

τ → −1τ uv u

vτ → τ−1

τ+1−2uv

Solving the quintic

2 uv and uv → e

πi3uv .

v →uv , and

uv , uv →

τ → τ + 2 e−πi

2 uv eπi3uv

τ → −1τ uv u

vτ → τ−1

τ+1−2uv

Solving the quintic

2 uv and uv → e

πi3uv .

v →uv , and

uv , uv →

τ → τ + 2 e−πi

2 uv eπi3uv

τ → −1τ uv u

vτ → τ−1

τ+1−2uv

Solving the quintic

2 uv and uv → e

πi3uv .

v →uv , and

uv , uv →

τ → τ + 2 e−πi

2 uv eπi3uv

τ → −1τ uv u

vτ → τ−1

τ+1−2uv

Solving the quintic

Functions invariant under τ + 2, −1τ , τ−1

The function f 24(τ) is fixed under the transformations τ → τ + 2and τ → −1

However τ → τ−1τ+1 is an involution for f (τ). We have

f (τ)→ 212

f 24(τ).

Hence the function F (τ) = f 24(τ) + 212

f 24(τ)is invariant under all 3

transformations.

Solving the quintic

f (τ)→ 212

f 24(τ).

transformations.

Solving the quintic

f (τ)→ 212

f 24(τ).

transformations.

Solving the quintic

Lemma: Let g(τ) be meromorphic in H, and lacks an essentialsingularity at τ = i∞. If g(τ) is invariant under τ → τ + 2,τ → −1

τ , and τ → τ−1τ+1 , then g(τ) is a rational function of F (τ).

Moreover, the equation F (τ) = c is solvable for any c 6= 0.

Compare the above to modular functions being a rationalexpression in j and j(τ) = c .

Solving the quintic

Thus if g(τ) = R(F (τ)) is finite for all τ with F (τ) 6=∞ then R isa polynomial.

If R(F (τ)) is finite for q = 0 (F =∞) as well as being finite for allτ with F (τ) 6=∞, then R is a constant

We are now in a position to deduce the modular equation.

Solving the quintic

Deducing the Modular Equation

Recall that we would like to show (uv )3 + ( vu )3 = (uv)2 − 4(uv)2

Let us first set Ac = ( uvc

)3 + ( vcu )3 and Bc = (uvc)2 − 4(uvc )2 .

Using the transformations for u and v we get:

A Bτ → τ + 2 −A −Bτ → −1

τ A Bτ → τ−1

τ+1 −A −B

Solving the quintic

)3 + ( vcu )3 and Bc = (uvc)2 − 4(uvc )2 .

A Bτ → τ + 2 −A −Bτ → −1

τ A Bτ → τ−1

τ+1 −A −B

Solving the quintic

)3 + ( vcu )3 and Bc = (uvc)2 − 4(uvc )2 .

A Bτ → τ + 2 −A −Bτ → −1

τ A Bτ → τ−1

τ+1 −A −B

Solving the quintic

)3 + ( vcu )3 and Bc = (uvc)2 − 4(uvc )2 .

A Bτ → τ + 2 −A −Bτ → −1

τ A Bτ → τ−1

τ+1 −A −B

Solving the quintic

By the previous table, we see that the function∏

c(Ac − Bc)2 isfixed under the mentioned transformations of τ .

Using the definition of f (τ) we may gather the first few terms ofthe expansions of A∞ and B∞.

A∞ = q−12 (1− 2q + . . .) and B∞ = q

−12 (1− 2q + . . .).

Thus A∞ − B∞ vanishes at q = 0.

Solving the quintic

A∞ = q−12 (1− 2q + . . .) and B∞ = q

−12 (1− 2q + . . .).

Solving the quintic

A∞ = q−12 (1− 2q + . . .) and B∞ = q

−12 (1− 2q + . . .).

Solving the quintic

A∞ = q−12 (1− 2q + . . .) and B∞ = q

−12 (1− 2q + . . .).

Solving the quintic

For c ≡ 0 mod 48 we have, v∞(τ) = f (5τ) = u(5τ − c) andvc(5τ − c) = f ( 5τ−c+c

5 ) = f (τ) = u(τ).

Moreover, A and B are symmetric in u and v , so we haveAc(5τ − c) = A∞(τ) and Bc(5τ − c) = B∞(τ).

Letting τ → i∞ (q → 0) we get that Ac − Bc vanishes at q = 0.

Solving the quintic

5 ) = f (τ) = u(τ).

Solving the quintic

5 ) = f (τ) = u(τ).

Solving the quintic

Hence the function∏

c(Ac − Bc)2 vanishes at q = 0 and by theLemma is a constant.

But then Ac − Bc = 0 for some c .

Hence Ac − Bc = 0 for all c since, we establishedAc(5τ − c) = A∞(τ) and Bc(5τ − c) = B∞(τ).

This establishes the modular equation.

Solving the quintic

Deriving the Quintic

We have seen that the coefficients of a sixth degree polynomialwith roots vc can be expressed in terms of u.

We now show that the coefficients of the fifth degree polynomialwith roots

wi =(v∞ − vi )(vi+1 − vi−1)(vi+2 − vi−2)√

can be expressed in terms of u and find the explicit form of thepolynomial.

Solving the quintic

wi =(v∞ − vi )(vi+1 − vi−1)(vi+2 − vi−2)√

Solving the quintic

wi =(v∞ − vi )(vi+1 − vi−1)(vi+2 − vi−2)√

Solving the quintic

wi =(v∞ − vi )(vi+1 − vi−1)(vi+2 − vi−2)√

Solving the quintic

Using our established transformations for u and v , we find thetransformations of wi :

w0 w1 w2 w3 w4

τ → τ + 2 −w2 −w3 −w4 −w0 −w1

τ → −1τ w0 w2 w1 w4 w3

τ → τ−1τ+1 −w0 −w3 −w4 −w2 −w1

Solving the quintic

Using our established transformations for u and v , we find thetransformations of wi :

w0 w1 w2 w3 w4

τ → τ + 2 −w2 −w3 −w4 −w0 −w1

τ → −1τ w0 w2 w1 w4 w3

τ → τ−1τ+1 −w0 −w3 −w4 −w2 −w1

Solving the quintic

Consider the polynomial∏(w − wi ) = w5 + A1w

4 + A2w3 + A3w

2 + A4w + A5.

It’s coefficients are finite at u 6= 0,∞.

Also A21,A2,A

23,A4,A

25 are invariant under τ → τ + 2, τ → −1

τ ,and τ → τ−1

τ+1 .

Thus they are polynomials inF (τ) = u24 + 212u−24 = q−1 + 24 + . . .

Such a polynomial is nonconstant only if its power series expansionin q begins with cqr , where r ≤ −1, c 6= 0.

Solving the quintic

4 + A2w3 + A3w

2 + A4w + A5.

Also A21,A2,A

23,A4,A

τ+1 .

Solving the quintic

4 + A2w3 + A3w

2 + A4w + A5.

Also A21,A2,A

23,A4,A

τ+1 .

Solving the quintic

4 + A2w3 + A3w

2 + A4w + A5.

Also A21,A2,A

23,A4,A

τ+1 .

Solving the quintic

4 + A2w3 + A3w

2 + A4w + A5.

Also A21,A2,A

23,A4,A

τ+1 .

Solving the quintic

If we calculate the first term in the expansion of wi , we would find

that the first term of the expansion of Ak is q−k10 .

Therefore A21,A2,A

23,A4 are constant while A2

5 depends linearly onF (τ) = u24 + 212u−24.

Comparing expansions we get A25 = u24 + 212u−24 + C .

To calculate the value of the constants A1,A2,A3,A4,C , wecalculate vc(i).

Solving the quintic

Therefore A21,A2,A

Solving the quintic

Therefore A21,A2,A

Solving the quintic

Therefore A21,A2,A

Solving the quintic

Since f1(τ) = f2(−1τ ) we have f1(i) = f2(i).

Using f1f2f =√

2 and f 8 = f 81 + f 8

2 we get u = f (i) = 214 .

Also f (τ) = f (−1τ ) implies v3 = f ( i+48

5 ) = f ( i−25 + 10) =

e−10πi

24 f ( i−25 ) = e

−10πi24 f (i + 2) = e

−πi2 f (i) = −i2

Similarly v2 = i214 = iδ, with δ defined as 2

Solving the quintic

Using f1f2f =√

2 and f 8 = f 81 + f 8

2 we get u = f (i) = 214 .

5 ) = f ( i−25 + 10) =

e−10πi

24 f ( i−25 ) = e

−10πi24 f (i + 2) = e

−πi2 f (i) = −i2

Solving the quintic

Using f1f2f =√

2 and f 8 = f 81 + f 8

2 we get u = f (i) = 214 .

5 ) = f ( i−25 + 10) =

e−10πi

24 f ( i−25 ) = e

−10πi24 f (i + 2) = e

−πi2 f (i) = −i2

Solving the quintic

Using f1f2f =√

2 and f 8 = f 81 + f 8

2 we get u = f (i) = 214 .

5 ) = f ( i−25 + 10) =

e−10πi

24 f ( i−25 ) = e

−10πi24 f (i + 2) = e

−πi2 f (i) = −i2

Solving the quintic

So for τ = i we have already found two roots of the modularequation v6 − δ5v5 + δ9v + δ6 = 0.

Dividing the modular equation by (v − v2)(v − v3) = v2 + δ2 weget:

v4 − δ5v3 + δ2v2 + δ7v + δ4 = (v − α)2(v − β)2,

where α + β = δ and αβ = −δ2.

Solving the quintic

v4 − δ5v3 + δ2v2 + δ7v + δ4 = (v − α)2(v − β)2,

Solving the quintic

v4 − δ5v3 + δ2v2 + δ7v + δ4 = (v − α)2(v − β)2,

Solving the quintic

v4 − δ5v3 + δ2v2 + δ7v + δ4 = (v − α)2(v − β)2,

Solving the quintic

Thus we can solve for α, β and we get α = δ 1+√

52 and β = δ 1−

Lastly we see v∞ = f (5i) = f (−15i ) = f ( i

5 ) = v0 and v∞ > 0implies

v0 = v∞ = δ 1+√

52 and v1 = v4 = δ 1−

Solving the quintic

52 and β = δ 1−

v0 = v∞ = δ 1+√

52 and v1 = v4 = δ 1−

Solving the quintic

52 and β = δ 1−

v0 = v∞ = δ 1+√

52 and v1 = v4 = δ 1−

Solving the quintic

Plugging in we find w0 = 0,w1 = w2 = i√

5,w3 = w4 = −i√

Thus our quintic∏(w − wi ) = w(w − i

√5)2(w + i

√5)2 = w(w2 + 5)2.

So our constant term A5(i) = 0 and we calculate C = −27.

We deduce A5 = −u12 + 64u−12.

Solving the quintic

5,w3 = w4 = −i√

√5)2(w + i

√5)2 = w(w2 + 5)2.

We deduce A5 = −u12 + 64u−12.

Solving the quintic

5,w3 = w4 = −i√

√5)2(w + i

√5)2 = w(w2 + 5)2.

We deduce A5 = −u12 + 64u−12.

Solving the quintic

5,w3 = w4 = −i√

√5)2(w + i

√5)2 = w(w2 + 5)2.

We deduce A5 = −u12 + 64u−12.

Solving the quintic

Thus we have w(w2 + 5)2 = u12 − 64u−12.

We use f 8 = f 81 + f 8

2 and f1f2f =√

2 to get

u12 − 64

f 24 − 64

f 12= (

f 81 − f 8

f 2)2.

Thus we have√w = ± f 8

1 −f 82

f 2(w2+5).

Solving the quintic

Thus we have w(w2 + 5)2 = u12 − 64u−12.

We use f 8 = f 81 + f 8

2 and f1f2f =√

2 to get

u12 − 64

f 24 − 64

f 12= (

f 81 − f 8

f 2)2.

1 −f 82

f 2(w2+5).

Solving the quintic

Thus we have w(w2 + 5)2 = u12 − 64u−12.

We use f 8 = f 81 + f 8

2 and f1f2f =√

2 to get

u12 − 64

f 24 − 64

f 12= (

f 81 − f 8

f 2)2.

1 −f 82

f 2(w2+5).

Solving the quintic

Thus we have w(w2 + 5)2 = u12 − 64u−12.

We use f 8 = f 81 + f 8

2 and f1f2f =√

2 to get

u12 − 64

f 24 − 64

f 12= (

f 81 − f 8

f 2)2.

1 −f 82

f 2(w2+5).

Solving the quintic

So we let y =f 81 −f 8

2f 2(w2+5)

Thenf 81 −f 8

2f 2 = y(w2 + 5) = y(y4 + 5) = y5 + 5y .

Hence to find the roots of the Bring-Jerrard quintic

y5 + 5y − a = 0 we only need to find a τ such that a =f 81 −f 8

2f 2 .

Solving the quintic

2f 2(w2+5)

Thenf 81 −f 8

2f 2 = y(w2 + 5) = y(y4 + 5) = y5 + 5y .

2f 2 .

Solving the quintic

2f 2(w2+5)

Thenf 81 −f 8

2f 2 = y(w2 + 5) = y(y4 + 5) = y5 + 5y .

2f 2 .

Solving the quintic

Solving for τ

Squaring the equation f 81 (τ)− f 8

2 (τ) = af 2(τ), and usingf 81 + f 8

2 = f 8, f1f2f =√

2 we get

f 24 − a2f 12 − 64 = 0 and can solve for f 12(τ).

Let j be Klein’s j-invariant. Then j = f 24−16f 24 .

The ability to solve j(τ) = c for any complex c shows that we cansolve for a τ such that f 8

1 (τ)− f 82 (τ) = af 2(τ).

Solving the quintic

Solving for τ

2 (τ) = af 2(τ), and usingf 81 + f 8

2 = f 8, f1f2f =√

2 we get

1 (τ)− f 82 (τ) = af 2(τ).

Solving the quintic

Solving for τ

2 (τ) = af 2(τ), and usingf 81 + f 8

2 = f 8, f1f2f =√

2 we get

1 (τ)− f 82 (τ) = af 2(τ).

Solving the quintic

Solving for τ

2 (τ) = af 2(τ), and usingf 81 + f 8

2 = f 8, f1f2f =√

2 we get

1 (τ)− f 82 (τ) = af 2(τ).

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