THE LEMNISCATIC FUNCTIONAND ABEL’S THEOREM

Constructions on the Lemniscate

What is the lemniscate?

Curve in the plane defined by ሺ𝑥2 + 𝑦2ሻ2 = 𝑥2 − 𝑦2

1 .0 0 .5 0 .5 1 .0

0 .3

0 .2

0 .1

0 .1

0 .2

0 .3

What is the lemniscate?

An oval of Cassini - ሺሺ𝑥− 𝑎ሻ2 + 𝑦2ሻሺሺ𝑥+ 𝑎ሻ2 + 𝑦2ሻ= 𝑏4

Who studied the lemniscate?

Niels Abel Jacob Bernoulli Ferdinand Eisenstein Leonhard Euler Carl Gauss

Arc Length

In polar coordinates, the lemniscate is defined by 𝑟2 = cos2𝜃

Note ሺ𝑥2 + 𝑦2ሻ2 = 𝑥2 − 𝑦2 ⇒𝑟4 = r2(cos2 𝜃− sin2 𝜃)

r

1 .0 0 .5 0 .5 1 .0

0 .3

0 .2

0 .1

0 .1

0 .2

0 .3

Arc Length

𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = න ඥ𝑑𝑟2 + 𝑟2𝑑𝜃2𝑃0 = න ඨ1+ 𝑟2൬𝑑𝜃𝑑𝑟൰2𝑟

0 𝑑𝑟

r

1 .0 0 .5 0 .5 1 .0

0 .3

0 .2

0 .1

0 .1

0 .2

0 .3

Arc Length

𝑟2 = cos2𝜃 ⇒2𝑟= −sin2𝜃∗2𝑑𝜃𝑑𝑟 ⇒𝑑𝜃𝑑𝑟 = − 𝑟sin2𝜃

𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = න ඨ1+ 𝑟2൬𝑑𝜃𝑑𝑟൰2 𝑑𝑟= න ඨ1+ 𝑟4sin2 2𝜃𝑟0

𝑟0 𝑑𝑟

Arc Length

sin2 2𝜃 = 1− cos2 2𝜃 = 1− 𝑟4

𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = න ඨ1+ 𝑟4sin2 2𝜃𝑑𝑟=𝑟0 න

1ξ1− 𝑟4𝑟

0 𝑑𝑟

𝜛= 2න 1ξ1− 𝑟4 𝑑𝑟10

The Lemniscatic Function

𝑦= sin𝑠⟺𝑠= sin−1 𝑦= න1ξ1− 𝑡2

𝑦0 𝑑𝑡

y

1 .0 0 .5 0 .5 1 .0

1 .0

0 .5

0 .5

1 .0

The Lemniscatic Function

𝑟 = 𝜑ሺ𝑠ሻ⟺𝑠= න1ξ1− 𝑡4

𝑟0 𝑑𝑡

r

1 .0 0 .5 0 .5 1 .0

0 .3

0 .2

0 .1

0 .1

0 .2

0 .3

The Sine FunctionThe Lemniscatic

Function

The Lemniscatic Function

1 2 3 4 5

1 .0

0 .5

0 .5

1 .0

1 2 3 4 5 6

1 .0

0 .5

0 .5

1 .0

The Sine FunctionThe Lemniscatic

Function

Basic Identities

1) 𝑓ሺ𝑥+ 2𝜋ሻ= 𝑓ሺ𝑥ሻ 2) 𝑓ሺ−𝑥ሻ= −𝑓ሺ𝑥ሻ 3) 𝑓ሺ𝜋− 𝑥ሻ= 𝑓ሺ𝑥ሻ 4) 𝑓′2ሺ𝑥ሻ= 1− 𝑓2(𝑥)

1) 𝑓ሺ𝑠+ 2𝜛ሻ= 𝑓ሺ𝑠ሻ 2) 𝑓ሺ−𝑠ሻ= −𝑓ሺ𝑠ሻ 3) 𝑓ሺ𝜛− 𝑠ሻ= 𝑓ሺ𝑠ሻ 4) 𝑓′2ሺ𝑠ሻ= 1− 𝑓4(𝑠)

Basic Identities

𝑓ሺ−𝑠ሻ= −𝑓ሺ𝑠ሻ 𝑓ሺ𝜛− 𝑠ሻ= 𝑓ሺ𝑠ሻ

1 .0 0 .5 0 .5 1 .0

0 .3

0 .2

0 .1

0 .1

0 .2

0 .3

Basic Identities

From the previous two identities, 𝜑′ሺ𝑠ሻ= 𝜑′(−𝑠) and 𝜑′ሺ𝑠+ 𝜛ሻ= −𝜑′ሺ𝑠ሻ

Differentiating the arc length integral, 1 = 1ඥ1−𝜑4ሺ𝑠ሻ𝜑′ሺ𝑠ሻ, 0 ≤ 𝑠< 𝜛2

Thus 𝜑′2ሺ𝑠ሻ= 1− 𝜑4(𝑠)

The Sine FunctionThe Lemniscatic

Function

The Addition Law

𝑓ሺ𝑥+ 𝑦ሻ= 𝑓ሺ𝑥ሻ𝑓′ሺ𝑦ሻ+ 𝑓′ሺ𝑥ሻ𝑓(𝑦) 𝜑ሺ𝑥+ 𝑦ሻ= 𝜑ሺ𝑥ሻ𝜑′ሺ𝑦ሻ+ 𝜑ሺ𝑦ሻ𝜑′(𝑥)1+ 𝜑2ሺ𝑥ሻ𝜑2(𝑦)

The Addition Law

In 1753, Euler proved that 1ξ1−𝑡4𝛼0 𝑑𝑡+ 1ξ1−𝑡4𝛽0 𝑑𝑡 = 1ξ1−𝑡4𝛾0 𝑑𝑡

where 𝛼,𝛽 ∈[0,1] and 𝛾 = 𝛼ඥ1−𝛽4+𝛽ξ1−𝛼41+𝛼2𝛽2 ∈[0,1] If we let x,y, and z equal these three integrals, respectively, then

𝜑ሺ𝑥+ 𝑦ሻ= 𝜑ሺ𝑧ሻ= 𝛾 = 𝛼ඥ1−𝛽4+𝛽ξ1−𝛼41+𝛼2𝛽2 , 0 ≤ 𝑥+ 𝑦≤ 𝜛2.

Noting that 𝜑ሺ𝑥ሻ= 𝛼 and 𝜑ሺ𝑦ሻ= 𝛽, we obtain

𝜑ሺ𝑥+ 𝑦ሻ= 𝜑ሺ𝑥ሻ𝜑′ሺ𝑦ሻ+ 𝜑ሺ𝑦ሻ𝜑′(𝑥)1+ 𝜑2ሺ𝑥ሻ𝜑2(𝑦) , 0 ≤ 𝑥+ 𝑦≤ 𝜛2

Multiplication by Integers

It is an easy consequence of the addition law that

𝜑ሺ2𝑥ሻ= 𝜑ሺ𝑥+ 𝑥ሻ= 𝜑ሺ𝑥ሻ𝜑′ሺ𝑥ሻ+ 𝜑ሺ𝑥ሻ𝜑′ሺ𝑥ሻ1+ 𝜑2ሺ𝑥ሻ𝜑2(𝑥) = 2𝜑ሺ𝑥ሻ𝜑′(𝑥)1+ 𝜑4(𝑥)

In fact, formulas 𝜑(𝑛𝑥) may be generalized for all positive integers

Multiplication by Integers

Theorem: Given an integer 𝑛 > 0, there exist relatively prime polynomials𝑃𝑛ሺ𝑢ሻ,𝑄𝑛(𝑢) ∈ℤ[𝑢] such that if 𝑛 is odd, then 𝜑ሺ𝑛𝑥ሻ= 𝜑(𝑥) 𝑃𝑛(𝜑4ሺ𝑥ሻ)𝑄𝑛(𝜑4ሺ𝑥ሻ) and if 𝑛 is even, then 𝜑ሺ𝑛𝑥ሻ= 𝜑ሺ𝑥ሻ𝑃𝑛൫𝜑4ሺ𝑥ሻ൯𝑄𝑛൫𝜑4ሺ𝑥ሻ൯𝜑′ሺ𝑥ሻ

Multiplication by Integers

For the case 𝑛 = 1, it is clear that 𝑃1ሺ𝑢ሻ= 𝑄1ሺ𝑢ሻ= 1

And for the case 𝑛 = 2, we know that 𝜑ሺ2𝑥ሻ= 𝜑ሺ𝑥ሻ 21+𝜑4ሺ𝑥ሻ𝜑′(𝑥). Thus 𝑃2ሺ𝑢ሻ= 2 and 𝑄2ሺ𝑢ሻ= 1+ 𝑢

Multiplication by Integers

Proceeding by induction, we obtain recursive formulas for our polynomials

If n is even: 𝑃𝑛+1ሺ𝑢ሻ= −𝑄𝑛2ሺ𝑢ሻ𝑃𝑛−1ሺ𝑢ሻ+ 𝑃𝑛ሺ𝑢ሻሺ1− 𝑢ሻ൫2𝑄𝑛ሺ𝑢ሻ𝑄𝑛−1ሺ𝑢ሻ− 𝑢𝑃𝑛ሺ𝑢ሻ𝑃𝑛−1ሺ𝑢ሻ൯ 𝑄𝑛+1ሺ𝑢ሻ= 𝑄𝑛−1ሺ𝑢ሻ൫𝑄𝑛2ሺ𝑢ሻ+ 𝑢𝑃𝑛2ሺ𝑢ሻሺ1− 𝑢ሻ൯ If n is odd: 𝑃𝑛+1ሺ𝑢ሻ= −𝑄𝑛2ሺ𝑢ሻ𝑃𝑛−1ሺ𝑢ሻ+ 𝑃𝑛ሺ𝑢ሻ൫2𝑄𝑛ሺ𝑢ሻ𝑄𝑛−1ሺ𝑢ሻ− 𝑢𝑃𝑛ሺ𝑢ሻ𝑃𝑛−1ሺ𝑢ሻ൯ 𝑄𝑛+1ሺ𝑢ሻ= 𝑄𝑛−1ሺ𝑢ሻ൫𝑄𝑛2ሺ𝑢ሻ+ 𝑢𝑃𝑛2ሺ𝑢ሻ൯

Constructions

Recall that a complex number 𝛼 is constructible if there is a finite sequence of straightedge and compass constructions that begins with 0 and 1 and ends with 𝛼.

Theorem: The point on the lemniscate corresponding to arc length 𝑠 can be constructed by straightedge and compass if and only if 𝑟 = 𝜑(𝑠) is a constructible number.

Constructions

Solving for x and y using 𝑟2 = 𝑥2 + 𝑦2 and 𝑟4 = 𝑥2 − 𝑦2, we have

𝑥= ±ට12ሺ𝑟2 + 𝑟4ሻ and 𝑦= ±ට12ሺ𝑟2 − 𝑟4ሻ.

Remember that the set of constructible numbers is closed under square roots

Constructions

Consider the arc length 𝑠= 𝜛3, which is one sixth of the entire arc length of

the lemniscate. We see that 𝜑ሺ3𝑠ሻ= 𝜑ሺ𝜛ሻ= 0, since the arc length 𝜛 corresponds to the origin.

Using the recursive formulas from multiplication by integers, 𝑃3ሺ𝑢ሻ= −𝑄22ሺ𝑢ሻ𝑃1ሺ𝑢ሻ+ 𝑃2ሺ𝑢ሻሺ1− 𝑢ሻ൫2𝑄2ሺ𝑢ሻ𝑄1ሺ𝑢ሻ− 𝑢𝑃2ሺ𝑢ሻ𝑃1ሺ𝑢ሻ൯= 3− 6𝑢 − 𝑢2

𝑄3ሺ𝑢ሻ= 𝑄1ሺ𝑢ሻቀ𝑄22ሺ𝑢ሻ+ 𝑢𝑃22ሺ𝑢ሻሺ1− 𝑢ሻቁ= 1+ 6𝑢− 3𝑢2 so that 𝜑ሺ3𝑥ሻ= 𝜑ሺ𝑥ሻ 3− 6𝜑4ሺ𝑥ሻ− 𝜑8ሺ𝑥ሻ1+ 6𝜑4ሺ𝑥ሻ− 3𝜑8ሺ𝑥ሻ

Constructions

𝜑8ሺ𝑠ሻ+ 6𝜑4ሺ𝑠ሻ− 3 = 0

Using the quadratic formula with 𝑥= 𝜑4(𝑠), we have the constructible solution

𝜑ሺ𝑠ሻ= ට2ξ3− 34

Abel’s Theorem

Theorem (Gauss): Let 𝑛 > 2 be an integer. Then a regular n-gon can be constructed by straightedge and compass if and only if 𝑛 = 2𝑠𝑝1 ⋅⋅⋅ 𝑝𝑟, where 𝑠≥ 0 is an integer and 𝑝1,…,𝑝𝑟 are 𝑟 ≥ 0 distinct Fermat primes.

Theorem (Abel): Let 𝑛 be a positive integer. Then the following are equivalent: a) The n-division points of the lemniscate can be constructed using straightedge

and compass.

b) 𝜑ቀ2𝜛𝑛 ቁ is constructible.

c) 𝑛 is an integer of the form 𝑛 = 2𝑠𝑝1 ⋅⋅⋅ 𝑝𝑟,

where 𝑠≥ 0 is an integer and 𝑝1,…,𝑝𝑟 are 𝑟 ≥ 0 distinct Fermat primes.

Abel’s Theorem

For a finite field extension 𝐹⊂ 𝐿, 𝐺𝑎𝑙ቀ𝐿𝐹ቁ= ሼ𝜎:𝐿→𝐿ȁI𝜎ሺ𝑎ሻ= 𝑎 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈𝐹ሽ

For example, 𝐺𝑎𝑙ቀℚሺ𝑖ሻℚ ቁ= {𝜎1,𝜎2} where 𝜎1ሺ𝑎+ 𝑏𝑖ሻ= 𝑎+ 𝑏𝑖 and 𝜎2ሺ𝑎+ 𝑏𝑖ሻ= 𝑎− 𝑏𝑖

Abel’s Theorem

Theorem: If 𝐿= ℚ൬𝑖,𝜑ቀ𝜛𝑛ቁ൰ and 𝑛 is an odd positive integer, then ℚ(𝑖) ⊂ 𝐿 is a

Galois extension and there is an injective group homomorphism 𝐺𝑎𝑙൬𝐿ℚሺ𝑖ሻ൰→ቆ

ℤሾ𝑖ሿ𝑛ℤሾ𝑖ሿቇ∗

and 𝐺𝑎𝑙ቀ 𝐿ℚሺ𝑖ሻቁ is Abelian.

Theorem: If 𝜁𝑛 = 𝑒2𝜋𝑖/𝑛, then 𝐺𝑎𝑙ቆℚሺ𝜁𝑛ሻℚ ቇ≃ ൬

ℤ𝑛ℤ൰∗

and 𝐺𝑎𝑙ቀℚሺ𝜁𝑛ሻℚ ቁ is Abelian.

Abel’s Theorem

Proposition: If 𝑝= 22𝑚 + 1 is a Fermat prime, then

ቤቆℤሾ𝑖ሿ𝑝ℤሾ𝑖ሿቇ∗

ቤ= 𝑎 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 2.

Remember that 𝐺𝑎𝑙ቌℚቆ𝑖,𝜑ቀ𝜛𝑝ቁቇℚሺ𝑖ሻ ቍ→ቀℤሾ𝑖ሿ𝑝ℤሾ𝑖ሿቁ∗

This proposition implies that 𝜑ቀ2𝜛𝑝 ቁ is constructible

Abel’s Theorem

Proof: If 𝑚 ≥ 1, then 𝑝= 22𝑚 + 1 = ൫22𝑚−1 + 𝑖൯൫22𝑚−1 − 𝑖൯= 𝜋𝜋ത, where 𝜋 and 𝜋ത are prime conjugates in ℤ[𝑖]. Now by the Chinese Remainder Theorem, ℤሾ𝑖ሿ𝑝ℤሾ𝑖ሿ= ℤሾ𝑖ሿ𝜋𝜋തℤሾ𝑖ሿ≃ ℤሾ𝑖ሿ𝜋ℤሾ𝑖ሿ× ℤሾ𝑖ሿ𝜋തℤሾ𝑖ሿ.

Abel’s Theorem

Now since 𝑁ሺ𝑝ሻ= 𝑁ሺ𝜋𝜋തሻ= 𝑁ሺ𝜋ሻ𝑁(𝜋ത) = 𝑝2 and 𝑁ሺ𝜋ሻ= 𝑁(𝜋ത), we have ℤሾ𝑖ሿ𝜋ℤሾ𝑖ሿ× ℤሾ𝑖ሿ𝜋തℤሾ𝑖ሿ≃ 𝔽𝑝 × 𝔽𝑝. Thus we see that

ቤቆℤሾ𝑖ሿ𝑝ℤሾ𝑖ሿቇ∗

ቤ= ห𝔽𝑝∗ × 𝔽𝑝∗ห= ሺ𝑝− 1ሻ2 = 22𝑚+1.

Abel’s Theorem

The only remaining case is m=0, or equivalently p=3.

Since N(3)=9, the distinct elements that compose ቀℤሾ𝑖ሿ3ℤሾ𝑖ሿቁ∗

are those of the

form 𝑎+ 𝑏𝑖, where 𝑎,𝑏> 0 and 𝑁(𝑎+ 𝑏𝑖) < 9.

So ቀℤሾ𝑖ሿ3ℤሾ𝑖ሿቁ∗= ሼ1,2,𝑖,2𝑖,1+ 𝑖,2+ 𝑖,1+ 2𝑖,2+ 2𝑖ሽ and ቚቀ

ℤሾ𝑖ሿ3ℤሾ𝑖ሿቁ∗ ቚ= 8

The End

Thank you for coming!

Special thanks to Dr. Hagedornfor his many valuable insights