Lagrange: Lagrange's Formula

Shortly after Alexander-Theophile Vandermonde completed his paper on equations and his solution of the eleventh root of unity, Joseph-Louis Lagrange published a 200+ page work that covered much of the same material in a clearer and more authoritative manner.

Lagrange came up with the following function that returned its parameters as possible values:



where

t(ω) = x₁ + ωx₂ + ω²x₃ + ... + ω^n-1x_n

and

ω is an n-th root of unity

I will later use Lagrange's Formula in Gauss's general proof on cyclotomic equations.

Theorem: Lagrange's Formula

If:



where

t(ω) = x₁ + ωx₂ + ω²x₃ + ... + ω^n-1x_n

and

ω is an n-th root of unity

then:



Proof:

(1) From the above definition of t(ω), we have:

t(ω⁰) = x₁ + x₂ + x₃ + ... + x_n

t(ω¹) = x₁ + ωx₂ + ω²x₃ + ... + ω^n-1x_n

t(ω²) = x₁ + ω²x₂ + ω⁴x₃ + ... + ω^2(n-1)x_n

...

t(ω^n-1) = x₁ + ω^n-1x₂ + ω^2(n-1)x₃ + ... + ω^(n-1)2x_n

(2) In general, we can see that:



where i = 1 ... n

(3) Now:



















(4) Now, when j ≠ k, ω^(j-k) is an n-th root of unity ≠ 1 since:

[ω^(j-k)]ⁿ = ω^n(j-k) = [ωⁿ]^j-k = (1)^j-k = 1

(5) Since ω^(j-k) is an n-th root of unity ≠ 1, and from a previous result (see Lemma 1, here), we know that:

1 + ω^(j-k) + ω^(j-k)2 + ... + ω^(j-k)(n-1) = 0

(6) So, if j ≠ k,



And, if j = k,



(7) From step #6, it is clear that:



(8) So, that we have:



(9) But ∑ (i=1,n)ω^(i-1) is the same as ∑ (ω) where ω is each n-th root of unity (including 1).

(10) So that we have:



(11) Adjusting for k+1 → k, gives us:



QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Vandermonde: Eleventh Root of Unity expressed as radicals

In today's blog, I will show how Alexandre-Theophile Vandermonde solved the eleventh root of unity by radicals. Carl Friedrich Gauss would later generalize this result to show that all n-th roots of unity are expressible as radicals. I will go through Gauss' results in a future blog.

Lemma 1:

if a,b,c,d,e are the roots of:

x⁵ - x⁴ - 4x³ + 3x² + 3x - 1 = 0

then:

a + b + c +d + e = 1

Proof:

(1) Using the Fundamental Theorem of Algebra (see Theorem, here), we know that:

x⁵ - x⁴ - 4x³ + 3x² + 3x - 1 = (x - a)(x - b)(x - c)(x -d)(x -e)

(2) From this it is clear that the coefficient for x⁴ is the sum of -ax⁴ + -bx⁴ + -cx⁴ + -dx⁴ + -ex⁴

Thus, we have:

-1(x⁴) = -ax⁴ + -bx⁴ + -cx⁴ + -dx⁴ + -ex⁴

(3) Dividing -x⁴ from both sides gives us:

1 = a + b + c + d + e

QED

Lemma 2:

If:

a² = -b + 2
b² = -d + 2
c² = -e + 2
d² = -c + 2
e² = -a + 2

and

ab = -a -c
bc = -a -e
cd = -a -d
de = -a-b
ac = -b -d
bd = -b -e
ce = -b -c
ad = -c - e
be = -c -d
ae = -d -e

Then:

(ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe)ⁿ = Aa + Bb + Cc +Dd + Ee + F

where

A,B,C,D,E,F are functions consisting solely of combinations of ρ_kⁱ (it does not include a,b,c,d or e)

Proof:

(1) The statement is clearly true for n=1.

(2) Assume that it is true up to some n-1 so that we have:

(Aa + Bb + Cc +Dd + Ee) *(ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe) =

Aa[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Bb[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Cc[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Dd[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Ee[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe]

(3) Aa[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] =

Aρ₁ⁱa² + Aaρ₂ⁱb + Aaρ₃ⁱc + Aaρ₄ⁱd + Aaρ₅ⁱe

(4) Since a² = -b + 2, ab = -a -c, ac= -b -d, ad=-c - e, ae=-d -e

Aρ₁ⁱa² + Aaρ₂ⁱb + Aaρ₃ⁱc + Aaρ₄ⁱd + Aaρ₅ⁱe =

Aρ₁ⁱ[-b + 2] + Aρ₂ⁱ[-a -c] + Aρ₃ⁱ[ -b -d] + Aρ₄ⁱ[-c - e] + Aρ₅ⁱ[-d -e]

(5) We can make the same type of argument for:

Bb[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Cc[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Dd[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe] +
Ee[ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe]

so we are done.

QED

Theorem 3: The Eleventh Root of Unity is Solvable by Radicals

Proof:

(1) Using Lemma 1 here, we have:

x¹⁰ + x⁹ + x⁸ + x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0

where x is an eleventh root of unity and x ≠ 1.

(2) Using Lemma 1 here and setting y = -(x + x^-1), we can simplify the equation to:

y⁵ - y⁴ -4y³ + 3y² + 3y - 1 = 0

(3) Using the fundamental theorem of algebra (see Theorem, here), we know that there are 5 roots ot this equation which we can label: a,b,c,d,e

(4) The plan of attack for solving this equation is to use Vandermonde's equation (see Theorem, here) to return a,b,c,d,e as its answers.

(5) So for values: a,b,c,d,e, we have:



where

V_i = (ρ₁ⁱa + ρ₂ⁱb + ρ₃ⁱc + ρ₄ⁱd + ρ₅ⁱe) where each ρ_k is a distinct 5th root of unity.

(6) Let ω be a primitive a fifth root of unity. [See definition here]

(7) Then, we can choose ρ_k such that (see Theorem 3, here):

ρ₁ = 1, ρ₂ = ω, ρ₃ = ω³, ρ₄ = ω², and ρ₅ = ω⁴

(8) So,

V_i(a,b,c,d,e) = a + ωⁱb + ω²ⁱd + ω³ⁱc + ω⁴ⁱe

and

V_i(b,d,e,c,a) = b + ωⁱd + ω²ⁱc + ω³ⁱe + ω⁴ⁱa

(9) This then gives us that:

V_i(b,d,e,c,a) = ω^-1V_i(a,b,c,d,e)

(10) We can use the same logic to confirm that:

V_i(a,b,c,d,e)⁵ = V_i(b,d,e,c,a)⁵ = V_i(d,c,a,e,b)⁵ = V_i(c,e,b,a,d)⁵ =
V_i(e,a,d,b,c)⁵

since (ω^k)⁵ = (ω⁵)^k = (1)^k

(11) Now, applying Lemma 2 above, we get:

V_i(a,b,c,d,e)⁵ = Aa + Bb + Cc + Dd + Ee + F

where A,B,C,D,E,F are rational expressions in ρ₁, ..., ρ₅

(12) Reusing the same function in step #11 gives us:

V_i(b,d,e,c,a)⁵ = Ab + Bd + Ce + Dc + Ea + F
V_i(d,c,a,e,b)⁵ = Ad + Bc + Ca + De + Eb + F
V_i(c,e,b,a,d)⁵ = Ac + Be + Cb + Da + Ed + F
V_i(e,a,d,b,c)⁵ = Ae + Ba + Cd + Db + Ec + F

(13) From step #10, step #11, and step #12, we have:

5V_i(a,b,c,d,e)⁵ = (A + B + C +D +E)(a + b + c + d +e) + 5F

(14) Using Lemma 1 above, we know that a + b + c + d + e =1, so we have:

5V_i(a,b,c,d,e)⁵ = (A + B + C +D +E) + 5F

which is the same as:

V_i(a,b,c,d,e)⁵ = 1/5(A + B + C + D +E) + F

(15) Now, since (a + b + c + d + e) = 1, we have as our solution:


where

V_i(a,b,c,d,e)⁵ = 1/5(A + B + C + D + E) + F

and

A,B,C,D,E,F are functions of the fifth roots of unity which are expressible by radicals based on a previous result from de Moivre (see Theorem 1, here)

QED

Here is the solution that Vandermonde came up with:

x = 1/5[1 + Δⁱ + Δⁱⁱ + Δⁱⁱⁱ + Δ^iv]

where:









Please note that in Vandermonde's original work, there was a mistake. Professor Tignol sent me the corrections which I have updated above. This mistake was noticed by Oscar Luis Palacios-Velez.

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001