Quartic polynomial roots

Question

The equation $x^4 - x^3-1=0$ has roots α, β, γ, δ. By using the substitution $y=x^3$ find the exact value of $α^6+β^6+γ^6+δ^6$ .

The solution is

$x=y$^(1/3)

$y^4=(1+y)^3$

$y^4 -y^3 -3y^2-3y^2-1=0$

$S$_N+4 $=$ $S$_N + $S$_N+3

$S$_-1 = $\frac {0}{1} =0$

$S$₂ = $ 1^2 -2*0 =1$

$S$₃ = $0+1 =1$

$S$₄ = $1+4 =5$

$S$₅ = $5+1=6$

$S$₆ = $6+1 =7$

Therefore, $α^6+β^6+γ^6+δ^6 = 7$

Can someone explain what is this $S$_n is and how did they work out solution from that.

I know other way to solve it but I don't know this summation way — Who am I, Sep 01 '19 at 14:22
You could also continue the first approach, set $z=y^2$ and from $z^2-3z-1=y(z+3)$ after squaring get $$z^4-6z^3+7z^2+6z+1=z(z^2+6z+9)\implies 0=z^4-7z^3+z^2-3z+1$$ where you also read off $S_6=7$ from the cubic coefficient. — Lutz Lehmann, Sep 01 '19 at 14:38

Lutz Lehmann · Answer 1 · 2019-09-01T14:31:36.183

2

This looks like an application of the Newton identities which are a generalization of Viete's identities connecting roots and coefficients. If $S_k=\sum_{j=1}^d α_j^k$ is the degree $k$ power sum of the roots of the polynomial equation $0=x^4+c_1x^3+c_2x^2+c_3x+c_4$, then $$ c_1+S_1=0\\ 2c_2+c_1S_1+S_2=0\\ 3c_3+c_2S_1+c_1S_2+S_3=0\\ 4c_4+c_3S_1+c_2S_2+c_1S_3+S_4=0\\ c_4S_1+c_3S_2+c_2S_3+c_1S_4+S_5=0\\ c_4S_2+c_3S_3+c_2S_4+c_1S_5+S_6=0\\ $$

For the given polynomial with $(c_1,c_2,c_3,c_4)=(-1,0,0,-1)$ this gives $$ S_1=1,\\ S_2=S_1=1,\\ S_3=S_2=1,\\ S_4=4+S_4=5,\\ S_5=S_1+S_5=6,\\ S_6=S_2+S_6=7. $$

Or using the proposed approach, $S_6$ is the square sum of the roots of the polynomial for $y$ which was found as $$0=y^4-y^3-3y^2-3y-1=y^4+C_1y^3+C_2y^2+C_3y+C_4,$$ where again one finds $S_3=-C_1=1$ and $$S_6=-C_1S_3-2C_2=7.$$

edited Sep 01 '19 at 14:31

answered Sep 01 '19 at 14:16

Lutz Lehmann

126,666

Which ones, the first or second ones? In both cases just reading off the coefficient sequence in the indexing scheme as I defined. – Lutz Lehmann Sep 01 '19 at 14:29
Then what's up with this Sn+4 formula – Who am I Sep 01 '19 at 14:30
This is also true, but you still need the identities to get the initial values for the iteration. – Lutz Lehmann Sep 01 '19 at 14:33
Thanks, man it really helped me a lot :) – Who am I Sep 01 '19 at 16:04

score 0 · Accepted Answer · answered Sep 03 '19 at 13:25

0

Multiply each term by $x^n$

$\pmb S_n =x^n$

Then it becomes

$S$_N+4 $=$ $S$_N + $S$_N+3

answered Sep 03 '19 at 13:25

This is wrong. $S_n$ is the sum over the powers of the roots, at least write $$S_n=\sum_{x:f(x)=0}x^n.$$ To be absolutely precise, one has to include something for multiple roots. – Lutz Lehmann Sep 03 '19 at 17:01

Quartic polynomial roots

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