Saturday, January 12, 2008

Vandermonde: Observations on the Eleventh Root of Unity

Alexandre-Theophile Vandermonde was able to solve the eleventh roots of unity in radicals by first observing relations between the roots of unity using trigonometric methods.

Lemma 1:

if:

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

then there exists y such that:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0

Proof:

(1) Let y = -(x + x-1) = -x + -x-1

(2) Then:

y2 = x2 + 2(-x)*(-x)-1 + x-2 = x2 + 2 + x-2

y3 = y*y2 = -x3 + -2x + -x-1 + -x + -2x-1 + -x-3 = -x3 + -3x + -3x-1 + -x-3

y4 = y*y3 = x4 + 3x2 + 3 + x-2 + x2 + 3 + 3x-2 + x-4 = x4 + 4x2 + 6 + 4x-2 + x-4

y5 = y*y4 = -x5 + -4x3 -6x - 4x-1 + -x-3 + -x3 + -4x + -6x-1 + -4x-3 + -x-5 =

= -x5 - 5x3 - 10x -10x-1 -5x-3 - x-5

(3) So:

y5 - y4 -4y3 + 3y2 + 3y - 1 =

= [-x5 - 5x3 - 10x -10x-1 -5x-3 - x-5] - [x4 + 4x2 + 6 + 4x-2 + x-4] - 4[-x3 + -3x + -3x-1 + -x-3] + 3[x2 + 2 + x-2] + 3[-x + -x-1] - 1 =

= -x5 - x4 + [-5x3 +4x3] + [-4x2 + 3x2] + [-10x +12x -3x] + [-6 + 6 - 1] + [-10x-1 + 12x-1 -3x-1] + [-4x-2 + 3x-2] + [-5x-3 + 4x-3] -x-4 -x-5 =

= -x5 -x4 -x3 -x2 -x - 1 -x-1 - x-2 -x-3 -x-4 -x-5

(4) We assume that:

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

(5) We divide each side of the equation by -x5 to get:

-x5 -x4 -x3 -x2 -x - 1 -x-1 - x-2 -x-3 -x-4 -x-5 = 0

(6) Now, we apply step #3 to get:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0

QED

Lemma 2: The eleventh roots of unity using trigonometric methods

Using trigonometric methods, the eleventh roots of unity are:

cos 0 + isin 0
cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

Proof:

This follows directly from a previous result [See Corollary 1.1, here].

QED

Corollary 2.1:

The 10 solutions for:

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

are:

cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

Proof:

(1) From Lemma 2 above, we can see that the solutions for:

x11 - 1 = 0

are:

cos 0 + isin 0
cos 2π/11 + isin 2π/11
cos 4π/11 + isin 4π/11
cos 6π/11 + isin 6π/11
cos 8π/11 + isin 8π/11
cos 10π/11 + isin 10π/11
cos 12π/11 + isin 12π/11
cos 14π/11 + isin 14π/11
cos 16π/11 + isin 16π/11
cos 18π/11 + isin 18π/11
cos 20π/11 + isin 20 π/11

(2) Using the Fundamental Theorem of Algebra (see Theorem, here), we can see that:

x11 - 1 = (x - cos0 - isin 0)(x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(3) Since, cos 0 - isin 0 = 1 - i*0 = 1, dividing both sides by (x - 1) gives us:

(x11-1)/(x - 1) = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

(4) From a previous result (see Lemma 1, here):

(x11-1)/(x-1) = x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1

(5) So, combining step #3 and step #4 gives us:

(x11-1)/(x-1) = x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 = (x - cos 2π/11 - isin 2π/11)*...*(x - cos 20π/11 - isin 20π/11)

QED

Lemma 3:

-2cos 12π/11 = -2cos(10π/11)

-2 cos 14π/11 = -2cos(8π/11)

-2 cos 16π/11 = -2cos(6π/11)

-2 cos 18π/11 = -2cos(4π/11)

-2cos 20π/11 = -2cos(2π/11)

Proof:

Using Property 5 here and the fact that 2π = 360 degrees [see Definition, here], we have:

-2cos 12π/11 = -2 cos (12π/11 - 2π) = -2cos([12π - 22π]/11) = -2cos(-10π/11) = -2cos(10π/11)

-2 cos 14π/11 = -2 cos (14π/11 - 2π) = -2cos([14π - 22π]/11) = -2cos(-8π/11) = -2cos(8π/11)

-2 cos 16π/11 = -2cos (16π/11 - 2π) = -2cos([16π - 22π]/11) = -2cos(-6π/11) = -2cos(6π/11)

-2 cos 18π/11 = -2cos (18π/11 - 2π) = -2cos([18π - 22π]/11) = -2cos(-4π/11) = -2cos(4π/11)

-2cos 20π/11 = -2cos(20π/11 - 2π) = -2cos([20π - 22π]/11) = -2cos(-2π/11) = -2cos(2π/11)

QED

Lemma 4

if:

x = cos u + isin u

then:

1/x = cos u - isin u

Proof:

(1) Let x = cos u + isin u

(2) 1/x = 1/[cos u + isin u] = [cos u - isinu]/[cos2 u - i2sin2 u] = [cos u - isin u]/[cos2 u + sin2 u] = cos u - isin u

QED

Corollary 4.1

if:

x = cos u + isin u

then:

-x + -x-1 = -2cos u

Proof:

(1) Let x = cos u + isin u

(2) Then x-1 = cosu - isin u

(3) x + x-1 = 2cos u

(4) -x + x-1 = -2cos u

QED

Corollary 4.2:

if:

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

and

y = -x + -x-1

then:

the possible values of y are:

-2cos 2π/11

-2cos 4π/11

-2cos 6π/11

-2cos 8π/11

-2cos 10π/11

Proof:

(1) Case 1: x = cos 2π/11 + isin 2π/11

-x + -x-1 = -2cos 2π/11

(2) Case 2: x = cos 4π/11 + isin 4π/11

-x + -x-1 = -2 cos 4π/11

(3) Case 3: x = cos 6π/11 + isin 6π/11

-x + -x-1 = -2 cos 6π/11

(4) Case 4: x = cos 8π/11 + isin 8π/11

-x + -x-1 = -2 cos 8π/11

(5) Case 5: x = cos 10π/11 + isin 10π/11

-x + -x-1 = -2cos 10π/11

(6) Case 6: x = cos 12π/11 + isin 12π/11

-x + -x-1 = -2cos 12π/11 = -2cos(10π/11) [See Lemma 3 above]

(7) Case 7: x = cos 14π/11 + isin 14π/11

-x + -x-1 = -2 cos 14π/11 = -2cos(8π/11) [See Lemma 3 above]

(8) Case 8: x = cos 16π/11 + isin 16π/11

-x + -x-1 = -2 cos 16π/11 = -2cos(6π/11) [See Lemma 3 above]

(9) Case 9: x = cos 18π/11 + isin 18π/11

-x + -x-1 = -2 cos 18π/11 = -2cos(4π/11) [See Lemma 3 above]

(10) Case 10: x = cos 20π/11 + isin 20π/11

-x + -x-1 = -2cos 20π/11 = -2cos(2π/11) [See Lemma 3 above]

QED

Lemma 5: 2cosαcosβ = cos(α + β) + cos(α - β)

Proof:

(1) cos(A+B) = cosA*cosB - sinA*sinB [See Theorem 2, here]

(2) Likewise:

cos (A - B) = cos (A + [-B]) = cosA*cos(-B) - sin(A)sin(-B) = cosAcosB + sinAsinB

(3) cos(A + B) + cos(A-B) = cosA*cosB - sinA*sinB + cosA*cosB + sinA*sinB = 2cosAcosB

QED

Corollary 5.1: 2cos2(α) = cos(2*α) + 1

Proof:

cos(2a) + 1 = cos(2a) + cos(0) = cos(a + a) + cos(a - a) = 2cos(α)cos(α) = 2cos2(α)

QED

Corollary 5.2:

if:

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

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

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

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) Using Corollary 5.1 above, we have:

2cos2(2π/11) = cos(4π/11) + 1

2cos2(4π/11) = cos(8π/11) + 1

2cos2(6π/11) = cos(12π/11) + 1

2cos2(8π/11) = cos(16π/11) + 1

2cos2(10π/11) = cos(20π/11) + 1

(3) Multiplying both sides by 2 gives us:

4cos2(2π/11) = 2cos(4π/11) + 2

4cos2(4π/11) = 2cos(8π/11) + 2

4cos2(6π/11) = 2cos(12π/11) + 2

4cos2(8π/11) = 2cos(16π/11) + 2

4cos2(10π/11) = 2cos(20π/11) + 2

(4) So that we have:

a2 = 4cos2(2π/11) = 2cos(4π/11) + 2 = -b + 2

b2 = 4cos2(4π/11) = 2cos(8π/11) + 2 = -d + 2

c2 = 4cos2(6π/11) = 2cos(12π/11) + 2 = 2cos(10π/11) + 2 = -e + 2

d2 = 4cos2(8π/11) = 2cos(16π/11) + 2 = 2cos(6π/11) + 2 = -c + 2

e2 = 4cos2(10π/11) = 2cos(20π/11) + 2 = 2cos(2π/11) + 2 = -a + 2

QED

Corollary 5.3:

if:

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

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

a2 = -b + 2

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: a2 = -b + 2 becomes b2 = -d + 2 { a --> b and b --> d } which becomes d2 = -c + 2 { b --> d and d -->c } which becomes c2 = -e + 2 { d --> c and c --> e } which becomes e2 = -a + 2 { c --> e and e --> a }]

Proof:

This follows directly from Corollary 6.2 above.

QED

Lemma 6:

if:

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

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

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

Proof:

(1) Using Corollary 4.2 above, we have:

a = -2cos 2π/11

b = -2cos 4π/11

c = -2 cos 6π/11

d = -2 cos 8π/11

e = -2 cos 10π/11

(2) So, we have:

-a -b = (2cos 2π/11) + (2cos 4π/11)

-a -c = (2cos 2π/11) + (2cos 6π/11)

-a -d = (2cos 2π/11) + (2cos 8π/11)

-a -e = (2cos 2π/11) + (2cos 10π/11)

-b -c = (2cos 4π/11) + (2cos 6π/11)

-b -d = (2cos 4π/11) + (2cos 8π/11)

-b -e = (2cos 4π/11) + (2cos 10π/11)

-c -d = (2cos 6π/11) + (2cos 8π/11)

-c -e = (2cos 6π/11) + (2cos 10π/11)

-d -e = (2cos 8π/11) + (2cos 10π/11)

(3) Which gives us:

ab = (-2cos 2π/11)(-2cos 4π/11) = 4cos(2π/11)*cos(4π/11) = 2[2cos(2π/11)*cos(4π/11)] = 2[cos(4π/11 + 2π/11) + cos(4π/11 - 2π/11) = 2[cos(6π/11) + cos(2π/11)] = 2cos(6π/11) + 2cos(2π/11) = -a -c

ac = (-2cos 2π/11)(-2 cos 6π/11) = 4cos(2π/11)*cos(6π/11) = 2[2cos(2π/11)*cos(6π/11)] = 2[cos(6π/11 + 2π/11) + cos(6π/11 - 2π/11) = 2[cos(8π/11) + cos(4π/11)] = 2cos(8π/11) + 2cos(4π/11) = -b -d

ad = (-2cos 2π/11)(-2 cos 8π/11) = 4cos(2π/11)*cos(8π/11) = 2[2cos(2π/11)*cos(8π/11)] = 2[cos(8π/11 + 2π/11) + cos(8π/11 - 2π/11) = 2[cos(10π/11) + cos(6π/11)] = 2cos(10π/11) + 2cos(6π/11) = -c -e

ae = (-2cos 2π/11)(-2 cos 10π/11) = 4cos(2π/11)*cos(10π/11) = 2[2cos(2π/11)*cos(10π/11)] = 2[cos(10π/11 + 2π/11) + cos(10π/11 - 2π/11) = 2[cos(12π/11) + cos(8π/11)] = 2cos(12π/11) + 2cos(8π/11) = 2cos(10π/11) + 2cos(8π/11) = -d -e

bc = (-2cos 4π/11)( -2 cos 6π/11) = 4cos(4π/11)*cos(6π/11) = 2[2cos(4π/11)*cos(6π/11)] = 2[cos(6π/11 + 4π/11) + cos(6π/11 - 4π/11) = 2[cos(10π/11) + cos(2π/11)] = 2cos(10π/11) + 2cos(2π/11) = -a -e

bd = (-2cos 4π/11)(-2 cos 8π/11) = 4cos(4π/11)*cos(8π/11) = 2[2cos(4π/11)*cos(8π/11)] = 2[cos(8π/11 + 4π/11) + cos(8π/11 - 4π/11) = 2[cos(12π/11) + cos(4π/11)] = 2cos(12π/11) + 2cos(4π/11) = 2cos(10π/11) + 2cos(4π/11) = -b -e

be = (-2cos 4π/11)(-2 cos 10π/11) = 4cos(4π/11)*cos(10π/11) = 2[2cos(4π/11)*cos(10π/11)] = 2[cos(10π/11 + 4π/11) + cos(10π/11 - 4π/11) = 2[cos(14π/11) + cos(6π/11)] = 2cos(14π/11) + 2cos(6π/11) = 2cos(8π/11) + 2cos(6π/11) = -c -d

cd = ( -2 cos 6π/11)(-2 cos 8π/11) = 4cos(6π/11)*cos(8π/11) = 2[2cos(6π/11)*cos(8π/11)] = 2[cos(8π/11 + 6π/11) + cos(8π/11 - 6π/11) = 2[cos(14π/11) + cos(2π/11)] = 2cos(14π/11) + 2cos(2π/11) = 2cos(8π/11) + 2cos(2π/11) = -a -d

ce = (-2 cos 6π/11)(-2 cos 10π/11) = 4cos(6π/11)*cos(10π/11) = 2[2cos(6π/11)*cos(10π/11)] = 2[cos(10π/11 + 6π/11) + cos(10π/11 - 6π/11) = 2[cos(16π/11) + cos(4π/11)] = 2cos(16π/11) + 2cos(4π/11) = 2cos(6π/11) + 2cos(4π/11) = -b -c

de = (-2 cos 8π/11)(-2 cos 10π/11) = 4cos(8π/11)*cos(10π/11) = 2[2cos(8π/11)*cos(10π/11)] = 2[cos(10π/11 + 8π/11) + cos(10π/11 - 8π/11) = 2[cos(18π/11) + cos(2π/11)] = 2cos(18π/11) + 2cos(2π/11) = 2cos(4π/11) + 2cos(2π/11) = -a -b

QED

Corollary 6.1:

if:

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

then:

y5 - y4 -4y3 + 3y2 + 3y - 1 = 0 with roots a,b,c,d,e

and

ab = -a -c
ac = -b -d

where at all times we can exchange each letter by one arrow:

a --> b --> d --> c --> e --> a

[That is, each letter can be changed to the next letter so that: ab = -a - c becomes bd = -b - e (a --> b and b--> d and c --> e) which becomes cd = -d -a ( b --> d and d --> c and e --> a) which becomes ce = -c -b ( c --> e and d --> c and a --> b) which becomes ae = -e -d etc.]

Proof:

This follows directly from Lemma 7 above.

QED

References
• Jean-Pierre Tignol, , World Scientific, 2001

Friday, January 11, 2008

Vandermonde: Vandermonde Equation

Alexandre-Theophile Vandermonde independently discovered what is today called the Lagrange Resolvent. Some mathematical historians have politely called it the Lagrange (Vandermonde) Resolvent. Even so, there is little doubt that Vandermonde discovered the idea independently of Joseph-Louis Lagrange and only Vandermonde applied the idea to solve the eleventh root of unity. The Lagrange Resolvent is the equation:

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 l1]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 l1]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 j)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:

j]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:

l1)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: k*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, , World Scientific, 2001

Tuesday, January 08, 2008

Alexandre-Theophile Vandermonde

Alexandre-Theophile Vandermonde was born on Feb 28, 1735 in Paris, France. From a young age, he showed a strong interest in music and this interest was encouraged by his father. Vandermonde played the violin and for these early years did not show any special interest in mathematics. Today, Vandermonde is remembered primarily for his mathematical achievements.

Vandermonde did not take up math until he was already 35 years old. At 35, he wrote a paper that created enough of a stir in the scientific community of France to get him selected to the Academie des Sciences in 1771. In total, he wrote only four mathematical papers.

In 1778, he chose to work on paper regarding musical theory. Many might have expected a mathematical analysis that extended the work of the Ancient Greeks. Instead, Vandermonde argued that there should be no theory of music as mathematical theory, but, instead, musicians should judge music solely based on their trained ear.

Throughout his life, Vandermonde collaborated with the greatest talents of his time. In 1777, he collaborated with Etienne Bezout and Antoine-Laurent Lavoisier to study the effects of the severe frost of 1776. In 1787, he worked with Gaspard Monge and Claude Louis Berthollet to research the manufacturing of steel with the aim of improving bayonets. In this effort, he investigated different combinations of iron and carbon.

In Vandermonde's first paper on mathematics, he presented a solution to the eleventh root of unity using ideas of symmetrical polynomials and permutations. These ideas would predate both the work of Carl Friedrich Gauss and Evariste Galois.

Henri Lebesgue wrote (quoted from Jean Tignol, see below):
Surely, any man who discovers something truly important is left behind by his own discovery; he himself hardly understands it, and only by pondering it for a long time. Vandermonde never came back to his algebraic investigations because he did not realize their importance in the first place, and if he did not understand them afterwards, it is precisely because he did not reflect deeply on them: he was interested in everything, he was busy with everything; he was not able to go slowly to the bottom of anything...
Today, Vandermonde is best known for something that may have been accidentally named after him: the Vandermonde determinant. The idea is not found in any of his papers. Still, the last of his four math papers did focus on determinants in which for the first time, a determinant is presented as a mathematical function and Vandermonde proceeds to investigate its mathematical properties. One mathematician, Thomas Muir, has gone so far to say that Vandermonde is "the only one fit to be viewed the founder of the theory of determinants."

References

Monday, January 07, 2008

I found an interesting blog posting about Andrew Wiles and the pressures of academia. The claim in the blog is that in order for Andrew Wiles to spend 10 years on Fermat's Last Theorem, he needed to create a storehouse of technical papers.

Sunday, January 06, 2008

De Moivre: Fifth Root of Unity and the Seventh Root of Unity

Abraham de Moivre was able to solve Φ5 and Φ7 [see here for Definition of Cyclotomic Polynomial if needed] in order to express the primitive 5th roots of unity and the primitive 7th roots of unity as radicals.

In today's blog, I will show de Moivre's solutions.

Theorem 1: fifth root of unity can be solved by radicals.

Proof:

(1) Since 5 is prime, from a previous result (see Lemma 1, here), we have:

Φ5 = x4 + x3 + x2 + x + 1 = 0

(2) Dividing both sides by x2, the gives us:

x2 + x + 1 + x-1 + x-2 = 0

(3) Let y = x + x-1

(3) Then:

y2 = (x + x-1)2 = x2 + 2x*x-1 + x-2 = x2 + 2 + x-2

(4) So,

y2 + y - 1 = x2 + x-2 + x + x-1 + 1

(5) Using the quadratic equation to solve for y2 + y - 1 = 0 gives us (see Theorem, here if needed):

y = (-1 ± √1 + 4)/2 = (-1 ± √5)/2

(6) Since y = x + x-1, we have:

x2 - [(-1 ± √5)/2]x + 1 = 0

(7) Again, using the quadratic equation, we have:

x = { [(-1 ± √5)/2] ± √[(-1 ± √5)/2]2 - 4}/2

(8) We can simplify this by noting that:

[(-1 ± √5)/2]2 = [1 ± 2√5 + 5]/4 = [6 ± 2√5]/4

(9) So,

x = { [(-1 ± √5)/2] ± √[(-1 ± √5)/2]2 - 4}/2 =

= [(-1 ± √5)/2 ± √(6 ± 2√5)/4 - 16/4]/2 =

= [(-1 ± √5)/2 ± √(-10 ± 2√5)/4]/2 =

= [(-1 ± √5) ± √(-10 ± 2√5)]/4

QED

Theorem 2: Seventh root of unity can be solved by radicals.

Proof:

(1) Since 7 is prime, from a previous result (see Lemma 1, here), we have:

Φ7 = x6 + x5 + x4 + x3 + x2 + x + 1 = 0

(2) Dividing both sides by x3, the gives us:

x3 + x2 + x + 1 + x-1 + x-2 + x-3 = 0

(3) Let y = x + x-1

(3) Then:

y2 = (x + x-1)2 = x2 + 2xx-1 + x-2 = x2 + 2 + x-2

y3 = (x + x-1)(x2 + 2 + x-2) = x3 + 2x + x-1 + x + 2x-1 + x-3 = x3 + 3x + 3x-1 + x-3

(4) So,

y3 + y2 - 2y - 1 = x3 + 3x + 3x-1 + x-3 + x2 + 2 + x-2 -2(x + x-1) - 1 =

= x3 + x2 + (3x - 2x) + (2 -1) + (3x-1 - 2x-1) + x-2 + x-3 =

= x3 + x2 + x + 1 + x-1 + x-2 + x-3

(5) So using the cubic equation (see Theorem, here), we can solve for y in:

y3 + y2 - 2y - 1 = 0

to get:

where

p = (-2)/(1) - (1)2/3(1)2 = -2 - (1/3) = -7/3

q = (-2(1)3)/(27(1)3) + (1)(-2)/(3(1)2) + 1/1 = -2/27 -2/3 + 1 = (-2 -18 + 27)/27 = 7/27

r = (1)/(3) = 1/3

(6) Once a solution for y is found, then it is possible to solve for x using the quadratic equation:

x2 - [(-1 ± √5)/2]y + 1 = 0

QED

It would have to wait until Alexandre-Theophile Vandermonde to solve for the eleventh root of unity in terms of radicals. Later, Carl Friedrich Gauss would demonstrate that all roots of unity are expressible as radicals.

References
• Jean-Pierre Tignol, , World Scientific, 2001