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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