MHB Roots of polynomial equations 2

[Erfan1]

The roots of the equation x^3 - x - 1 = 0 are α β γ and S(n) = α^n + β^n + γ^n
(i) Use the relation y = x^2 to show that α^2, β^2 ,γ^2 are the roots of the equation y^3 - 2y^2 + y - 1 =0
(ii) Hence, or otherwise , find the value of S(4) .
(iii) Find the values of S(8) , S(12) and S(16)I have solved the first and second part . I found S(4) to be 2. Any idea how to do part 3 ?

 

[MarkFL]

Re: Roots of polynomial equations

I think if I was going to solve the third part of this problem, I would view $s(n)$ as the closed form of the recursion having the characteristic equation:

$$r^3-r+1=0$$

Hence:

$$s(n)=s(n-2)-s(n-3)$$

where (using Vieta and our previous results):

$$s(0)=3,\,s(1)=0,\,s(2)=2$$

Now you have enough information to compute successive values of $s(n)$.

Perhaps even simpler, we could use the result of part i) to write the recursion:

$$s(2n)=2s(2(n-1))-s(2(n-2))+s(2(n-3))$$

where:

$$s(2(0))=3,\,s(2(1))=2,\,s(2(2))=2$$

 

[anemone]

Re: Roots of polynomial equations

The roots of the equation x^3 - x - 1 = 0 are α β γ and S(n) = α^n + β^n + γ^n
(i) Use the relation y = x^2 to show that α^2, β^2 ,γ^2 are the roots of the equation y^3 - 2y^2 + y - 1 =0
(ii) Hence, or otherwise , find the value of S(4) .
(iii) Find the values of S(8) , S(12) and S(16)I have solved the first and second part . I found S(4) to be 2. Any idea how to do part 3 ?

For me, I'd approach this using purely algebraic method...

Part (iii):

For the polynomial equation $$x^3-x-1=0$$ with roots $\alpha$, $\beta$ and $\gamma$, we have

$$\alpha+\beta+\gamma=0$$

$$\alpha\beta+\alpha\gamma+\beta\gamma=-1$$

$$\alpha\beta\gamma=1$$

This makes it so easy to calculate $$\alpha^2+\beta^2+\gamma^2$$, $$\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2$$ and $$(\alpha\beta\gamma)^2$$ since

$$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma)=0+2=2$$---(1)

$$\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2= \alpha^2(\beta^2+\gamma^2)+\frac{1}{\alpha^2}= \frac{1}{\alpha^2}(\alpha^4(2-\alpha^2)+1)=\frac{-\alpha^6+2\alpha^4+1}{\alpha^2}=\frac{\alpha^2-1+1}{\alpha^2}=1$$---(2)

and

$$(\alpha\beta\gamma)^2=1$$---(3)

Now, if we let $a, b$ and $c$ be the roots of $y^3-2y^2+y-1=0$ where $a=\alpha^2$, $b=\beta^2$ and $c=\gamma^2$, equations (1), (2) and (3) become

$a+b+c=2$

$ab+ac+bc=1$

$abc=1$

Now, our mission is to find the value for $a^2+b^2+c^2$ where

$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=2^2-2(1)=2$$---(4)

Hence, $$a^2+b^2+c^2=(\alpha^2)^2+(\beta^2)^2+(\gamma^2)^2=S(4)=\alpha^4+\beta^4+\gamma^4=2$$$$a^2b^2+a^2c^2+b^2c^2= a^2(b^2+c^2)+\frac{1}{a^2}= \frac{1}{a^2}(a^4(2-a^2)+1)=\frac{-a^6+2a^4+1}{a^2}$$

$$=\frac{-2a^4+4a^3-5a^2+2a}{a^2}=\frac{-2a^3+4a^2-5a+2}{a}=\frac{2a-2-5a+2}{a}=-3$$---(5)

and

$$(abc)^2=1$$---(6)

Therefore, we have another cubic polynomial $u^3-2u^2-3u-1=0$ with roots $a^2, b^2$ and $c^2$.

Now, if we repeat the whole process all over again by letting $d, e$ and $f$ be the roots of $u^3-2u^2-3u-1=0$ where $d=a^2=(\alpha^2)^2=\alpha^4$, $e=b^2=(\beta^2)^2=\beta^4$ and $f=c^2=(\gamma^2)^2=\gamma^4$, equations (4), (5) and (6) become

$d+e+f=2$

$de+df+ef=-3$

$def=1$

Now, our mission is to find the value for $d^2+e^2+f^2$ where

$$d^2+e^2+f^2=(d+e+f)^2-2(de+df+ef)=2^2-2(-3)=10$$---(7)

Hence, $$d^2+e^2+f^2=(\alpha^4)^2+(\beta^4)^2+(\gamma^4)^2=S(8)=\alpha^8+\beta^8+\gamma^8=10$$

$$d^2e^2+d^2f^2+e^2f^2= d^2(e^2+f^2)+\frac{1}{d^2}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{d^2}(d^4(10-d^2)+1)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{-d^6+10d^4+1}{d^2}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{6d^4-12d^3-13d^2-6d}{d^2}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{6d^3-12d^2-13d-6}{d}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{5d}{d}=5$$---(8)

and

$$(def)^2=1$$---(9)

Therefore, we have another cubic polynomial $v^3-10v^2+5v-1=0$ with roots $d^2, e^2$ and $f^2$.

Repeat this process again will give us the value for $S(16)$...

This time we let $g, h$ and $i$ be the roots of $v^3-10v^2+5v-1=0$ where $g=d^2=a^4=\alpha^8$, $h=e^2=b^4=\beta^8$ and $i=e^2=c^4=\gamma^8$, equations (7), (8) and (9) become

$g+h+i=10$

$gh+gi+hi=5$

$ghi=1$

Now, our mission is to find the value for $g^2+h^2+i^2$ where

$$g^2+h^2+i^2=(g+h+i=)^2-2(gh+gi+hi)=10^2-2(5)=90$$

Hence, $$g^2+h^2+i^2=(\alpha^8)^2+(\beta^8)^2+(\gamma^8)^2=S(16)=\alpha^{16}+\beta^{16}+\gamma^{16}=90$$

and we don't really need to find the values for $g^2h^2+g^2i^2+h^2i^2$ because we have already found all the values that the part (iii) asked.

But I find it a bit strange because this method isn't so straightforward to determine the value for $S(12)$...

To find the value for $S(12)=\alpha^{12}+\beta^{12}+\gamma^{12}$, I multiply the equation $S(8)$ by $S(4)$ and get

$S(8)\times S(4)=(\alpha^{8}+\beta^{8}+\gamma^{8})(\alpha^{4}+\beta^{4}+\gamma^{4})$

$(10)(2)=(\alpha^{12}+\beta^{12}+\gamma^{12})+( \alpha^{4}\beta^{4}(\alpha^{4}+\beta^{4})+\alpha^{4} \gamma^{4}(\alpha^{4}+\gamma^{4})+\beta^{4}\gamma^{4}(\beta^{4}+\gamma^{4})$

$20=(\alpha^{12}+\beta^{12}+\gamma^{12})+\alpha^{4}\beta^{4}(2-\gamma^{4})+\alpha^{4}\gamma^{4}(2-\beta^{4})+\beta^{4}\gamma^{4}(2-\alpha^{4})$

$20=(\alpha^{12}+\beta^{12}+\gamma^{12})-3+2(\alpha^{4}\beta^{4}+\alpha^{4}\gamma^{4}+\beta^{4}\gamma^{4})$

$S(12)=\alpha^{12}+\beta^{12}+\gamma^{12}=20+3-2(de+df+ef)=20+3-2(-3)=29$

 

[kaliprasad]

Re: Roots of polynomial equations

Roots of equation f(x) = x^3 - x - 1 = 0 are α β γ

So α^2, β^2, γ^2 are roots of equation

f(x^(1/2)) = 0

or x^3/2 – x^(1/2) = 1

or x^(1/2)(x-1) =1

we need to make integer power of x so square both sides to get

x(x^2 – 2x + 1) = 1 or x^3 – 2x^2+ x – 1 = 0