Polynomials sprint: How to solve a seemingly disgusting polynomial

On page 248, it is shown how to solve the polynomial x4+x3+x2+x+1=0x^4+x^3+x^2+x+1=0. In this note, I will explain how to solve the polynomial x8+x7+x6+x5+x4+x3+x2+x+1=0x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0 First we multiply both sides by (x1)(x-1) to get x91=0    x9=1x^9-1=0\implies x^9=1 Then since x9=1x^9=1, we can divide some terms by x9x^9 x8+x7+x6+x5+x4+x3+x2+x+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1= x8x9+x7x9+x6x9+x5x9+x4+x3+x2+x+1=\dfrac{x^8}{x^9}+\dfrac{x^7}{x^9}+\dfrac{x^6}{x^9}+\dfrac{x^5}{x^9}+x^4+x^3+x^2+x+1= 1x+1x2+1x3+1x4+x4+x3+x2+x+1\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+\dfrac{1}{x^4}+x^4+x^3+x^2+x+1 (x+1x)+(x2+1x2)+(x3+1x3)+(x4+1x4)+1=0\left(x+\dfrac{1}{x}\right)+\left(x^2+\dfrac{1}{x^2}\right)+\left(x^3+\dfrac{1}{x^3}\right)+\left(x^4+\dfrac{1}{x^4}\right)+1=0 Now we let y=x+1xy=x+\dfrac{1}{x} so we have (y)+(y22)+(y33y)+((y22)22)+1=0(y)+(y^2-2)+(y^3-3y)+((y^2-2)^2-2)+1=0 which after simplification becomes y4+y33y22y+1=0y^4+y^3-3y^2-2y+1=0 Using the Rational Root Theorem (discussed on page 246) we quickly find that y=1y=-1 is a root and factor the polynomial as (y+1)(y33y+1)(y+1)(y^3-3y+1) Now the rest is simple! We can use the cubic solving method discussed on pages 247-248 to find the roots of y33y+1y^3-3y+1 Then simple quadratic bashing gives us our roots!

Note by Nathan Ramesh
5 years ago

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Or, we can use the 9th roots of unity.

David Lee - 5 years ago

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Yes, what happened to the classic roots of unity? Roots are just e2kiπ/9e^{2ki\pi/9}with k=18k=1\to 8.

Daniel Liu - 5 years ago

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Except all roots of unity give you is e2πi9e^{\frac{2\pi i}{9}} and not a numerical answer with radicals. The method here allows you to compute sin40\sin {40^{\circ}}

Nathan Ramesh - 5 years ago

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@Nathan Ramesh Or you can also just do some Euler's formula with e2πi9 e^{\dfrac{2\pi i}{9}} .

David Lee - 5 years ago

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@David Lee ummmm which euler's formula?

Nathan Ramesh - 5 years ago

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@Nathan Ramesh Where i can find this book

Aman Sharma - 4 years, 6 months ago

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@Nathan Ramesh eiϕe^{i\phi} = cosϕ\phi + i{i}sinϕ\phi ... substituting π\pi gives eiπe^{i\pi} = -1, where i{i} = 1\sqrt{-1}

Curtis Clement - 4 years, 6 months ago

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@Nathan Ramesh eiθ=cis(θ)e^{i\theta}=\text{cis}(\theta)

Daniel Liu - 5 years ago

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@Daniel Liu How does that let you solve for sin40\sin{40^{\circ}}

Nathan Ramesh - 5 years ago

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wow

Daniel Liu - 5 years ago

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:o

Daniel Liu - 5 years ago

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2 Daniels

David Lee - 5 years ago

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@David Lee 2 Lius

Jubayer Nirjhor - 5 years ago

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@Jubayer Nirjhor 2 14-year-old-Daniel-Lius. Enough to get a conversation off-topic.

Ahaan Rungta - 5 years ago

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@Ahaan Rungta 2 14-year-old-Daniel-Lius-residing-in-USA. conversation closed.

Jubayer Nirjhor - 5 years ago

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@David Lee Oh God, this is not good. One Daniel is enough for me, but two? Double Trouble.

Sharky Kesa - 5 years ago

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I reshared your note. :D

Finn Hulse - 5 years ago

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@Nathan Ramesh Can you add this to the Roots of Unity Applications Wiki page?

Select "Write a summary", and then copy-paste your text into it (with minor formatting adjustments if relevant. Thanks!

Calvin Lin Staff - 4 years, 9 months ago

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Done! I put it at the bottom. Let me know if it is bugged (I posted it from my phone). Thanks!

Nathan Ramesh - 4 years, 9 months ago

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Yes I used the same to expand cosnx

U Z - 4 years, 6 months ago

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Where are all other pages?

U Z - 4 years, 6 months ago

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