Polynomial Question (1 Viewer)

[sasquatch]

This question has several parts, many of which i have done.

If w = x + x^-1 prove that,

i) x² + x^-2 = w² - 2
ii) x⁴ + x³ + x² + x + 1
= x²(w² + w - 1)
= [x² + (1/2)x(1 + √5) + 1][x² + (1/2)x(1 - √5) + 1]

Show that the roots of x⁴ + x³ + x² + x + 1 = 0 are the four complex roots of x⁵ = 1.

Deduce that cos 72 = (1/4)(√5 -1), cos 36 = (1/4)(√5 + 1).

I can do all the parts except for the deducing the cos angle bits. Could any body help. Ill post up my working for the first bits so that anyone trying to help can continue from.


i) Prove: x² + x^-2 = w² - 2

RHS = (x + x^-1)² -2
=x² + 2 + x^-2 -2
= x² + x^-2
= LHS

ii) Prove: x⁴ + x³ + x² + x + 1
= x²(w² + w - 1)
= [x² + (1/2)x(1 + √5) + 1][x² + (1/2)x(1 - √5) + 1]

w² + w - 1 = 0

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

.:. w² + w - 1 = (w + 1/2 + √5/2)(w + 1/2 - √5/2)
= (x + x^-1 + 1/2 + √5/2)(x + x^-1 + 1/2 - √5/2)

and x²(w² + w - 1)
= x²(x + x^-1 + 1/2 + √5/2)(x + x^-1 + 1/2 - √5/2)
=[x² + 1 + (1/2)x + (√5/2)x][x² + 1 + (1/2)x - (√5/2)x]
= [x² + (1/2)x(1 + √5) + 1][x² + (1/2)x(1 - √5) + 1]

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Now to Show that the roots of x⁴ + x³ + x² + x + 1 = 0 are the four complex roots of x⁵ = 1:

x⁵ = 1
r⁵cis5θ = 1cis0

r = 1, 5θ = 2π k
.:. θ = (2π k) / 5

x₁ = 1
x₂ = cis(2π/5)
x₃ = cis(-2π/5)
x₄ = cis(4π/5)
x₅ = cis(-4π/5)

x⁵ - 1 = 0, where (x - 1) is a factor.

hence by polynomial division... (i cant really show this bit)

x⁵ - 1= (x-1)(x⁴ + x³ + x² + x + 1)

.:. x⁴ + x³ + x² + x + 1 = 0 holds the complex roots of x⁵ = 0.

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Now to prove that cos crap:

x⁴ + x³ + x² + x + 1
= [x² + (1/2)x(1 + √5) + 1][x² + (1/2)x(1 - √5) + 1]
= [x - cis(2π/5)][x - cis(-2π/5)][x - cis(4π/5)][x - cis(-4π/5)]
= [x² - 2xcos(2π/5) + 1][x² - 2xcos(4π/5) + 1]

(I Skipped a few steps cuz its too long)..

Now if i equate... i can no way prove the results the question asks me.. so this is where im stuck and i hope somebody can help...

 

[joesmith1975]

take you some time to write that :S

 

[sasquatch]

Damn yeah...

 

[Riviet]

Note that cos 72 = cos(2pi/5) and cos 36 = cos(pi/5).
Let w=cis(pi/5)
Then w+w^-1=2cos(pi/5)
.'. cos(pi/5)=1/2.(w+w^-1)
Also w+w^-2=2cos(2pi/5)
.'. cos(2pi/5)=1/2.(w+w^-2)
Now we want to form a quadratic equation with these roots, so use sum and product of roots to get a value from the sum and a value from the products. You should form an equation x²-Sx+P=0, where S is the sum of the two roots and P is their product. Use the quadratic formula to find x, but since x is the root and we know the roots as cos(2pi/5) and cos(pi/5) we can equate the roots.
I hope that helps. :]

 

[sasquatch]

Man i dont get that...... i guess ill have to ask my teacher.. fitzpatrick doesnt explain anything... nothing at all like that..

 

[Riviet]

There are various ways of doing it, this one is based on Terry Lee's textbook.
Basically you're trying to find a way to get the exact value of cos 72 or cos 36. The quadratic allows us to use the formula and find exact values of cos 72 and cos 36 by putting them together in an equation and solving it.

 

[sasquatch]

Terry lee's a dickhead