where
ρ1, ..., ρn are n of the n-th roots of unity.
Below I present the reasoning that led Vandermonde to the above formula:
Vandermonde sought to find a function on n values that returned its answer those same n different values. In other words, a function of n values that returns the n values themselves as the answer.
If we are only talking about two values, then such an equation is pretty easy to find:
F(x1, x2) = (1/2)[(x1 + x2) + √(x1 - x2)2
To get two possible answers, we use the √(x1 - x2)2. This has two possible answers ± (x1 - x2)
The two results then are:
(1/2)[(x1 + x2) + x1 - x2)] = (1/2)[2*x1] = x1
and
(1/2)[(x1 + x2) + -(x1 - x2)] = (1/2)[x1 + x2 -x1 + x2] = (1/2)[2*x2] = x2
Vandermonde was able to find an equation that worked for an equation of n values which I present below.
Lemma 1:
if [ρl1]i = [ρl2]i where:
ρl is an n-th root of unity and 1 ≤ i ≤ n-1 and 1 ≤ l1,l2 ≤ n-1
Then:
l1 = l2
Proof:
(1) Assume that l1 ≠ l2
(2) Then [ρl1]i ≠ [ρl2]i [since each of the ρl are distinct]
(3) But this is not the case so we reject our assumption in step #1.
QED
Lemma 2:
In the sum:
where ρ1, ..., ρn are n of the n-th roots of unity.
It follows that each product above is an n-th root of unity and further, each product represents a distinct n-th root of unity ≠ 1
Proof:
(1) Now: [(ρj)i]n-1*(ρl)i is itself an n-th root of unity, since:
{[(ρj)i]n-1*(ρl)i}n = [(ρj)n]i(n-1)*[(ρl)n]i =
= (1)i(n-1)*(1)i = 1
(2) Assume that [(ρj)i]n-1*(ρl)i = 1
(3) Now (ρj)n-1*(ρj) = 1 so that:
[(ρj)n-1*(ρj)]i = 1
and
[(ρj)n-1]i*(ρj)i = 1
(4) So,[(ρj)n-1]i*(ρj)i= [(ρj)i]n-1*(ρl)i
(5) Which gives us that:
[ρj]i = [ρl]i which is impossible since l ≠ j. [See Lemma 1 above]
(6) So we have a contradictoin and reject our assumptoin at step #2.
(7) Assume that each product is not distinct so that there exists two products such that:
[(ρj)i]n-1*(ρl1)i = [(ρj)i]n-1*(ρl2)i
where l1 ≠ l2.
(8) Then it follows that:
(ρl1)i = (ρl2)i .
(9) But this is impossible since l1 ≠ l2. [See Lemma 1 above]
(10) So, we have a contradiction and we reject our assumption in step #7.
QED
Theorem 3: Vandermonde Equation
If:
where
and
ρ1, ..., ρn are n of the n-th roots of unity.
Then:
this equation returns x1, ..., xn in its set of possible values
Proof:
(1) First, we note that there are n possible roots for:
which are:
ρ1*Vi, ρ2*Vi, ..., ρn*Vi
This follows since: (ρk*Vi)n = (ρk)n*(Vi)n = (1)*(Vi)n = (Vi)n
(2) Let's consider that we want to return xj as an answer.
(3) Consider ρk*Vi such that ρk = ([ρj]i)n-1
Note: We know that ([ρj]i)n-1 is a root of unity since:
[([ρj]i)n-1]n = [(ρj)n]i*[n-1] = (1)i*[n-1] = 1
(4) So, we have:
ρk*Vi = ([ρj]i)n-1*[Vi] =
(5) And:
(6) Now, combining step #5 with the main equation gets us:
(7) Using Lemma 2 above, we know that each of the n-1 n-th root of unity products is distinct and ≠ 1.
(8) From step #7, it follows that there is a mapping from each product to a distinct ρm
Thus, we have:
where ρm ≠ 1.
(9) Now, since each n-th root of unity is itself a power of a primitive n-root of unity (see Theorem 3, here), if we let ρ be any primitive n-th root of unity, then we have:
(10) Now, from a previous result (see Lemma 1, here), we know that:
1 + ρ + ρ2 + ... + ρn-1 = 0
(11) So, now we can greatly simplify the equation in step #6:
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001
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