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