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The roots of the equation $x^4 -3x^2 + 5x - 2 = 0$ are $\alpha$, $\beta$, $\gamma$ and $\delta$. $\alpha^n + \beta^n + \gamma^n + \delta^n$ is denoted by $S(n)$. Find values of $S(2)$ and $S(4)$ and of $S(3)$ and $S(5)$.

Hence, find the value of $\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$

This topic is a rather strange one, and I can't seem to locate the formulas I need, and I need them for my exam. If you do use any formula (except $\sum\alpha$, $\sum\alpha\beta$, $\sum\alpha\beta\gamma$ and $\alpha\beta\gamma\delta$), please state it. Thank you in advance.

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  • $\begingroup$ The coefficients of the equation are the elementary symmetric polynomials in the roots. The $S(n) are the power sum symmetric polynomials. It is a theorem that you can write the latter in terms of the former. The trick is to do so explicitly. The algorithm to write one symmetric polynomial in terms of a family of others is to construct a polynomial whose leading coefficient agrees with the first, subtract, and repeat. $\endgroup$ – Aaron Mar 13 '14 at 20:19
  • $\begingroup$ Newton's identities (en.wikipedia.org/wiki/Newton%27s_identities) are helpful here. $\endgroup$ – Oleg567 Mar 13 '14 at 20:46
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$$S(2)=\left(\sum\alpha\right)^2-2\sum\alpha\beta$$ $$S(3)=\left(S(2)-\sum\alpha\beta\right)\left(\sum\alpha\right)+3\sum\alpha\beta\gamma$$ See here to verify the second identity.Now, summing cyclically we have: $$S(4)-3S(2)+5S(1)-8=0$$ More generally, multiplying by $x^n$ and summing cyclicaclly we get the recurrence relation: $$S(n+4)-3S(n+2)+5S(n+1)-2S(n)=0$$


Finally, for your bizarre looking last expression, we have $$\sum a^2\left(S(3)-\alpha^3\right)=S(3)\sum a^2-\sum\alpha^5=S(2)S(3)-S(5)$$

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  • $\begingroup$ I cannot thank you enough for that general formula and answers. However, I don't understand that last expression. Why do you subtract $α^3$ from $s(3)$ ? $\endgroup$ – George Mar 13 '14 at 21:37
  • $\begingroup$ @George You're welcome :) That is because $\alpha^2(\beta^3+\gamma^3+\delta^3)=\alpha^2(\alpha^3+\beta^3+\gamma^3+\delta^3-\alpha^3)=\alpha^2(S(3)-\alpha^3)$ . I would also recommend reading the links in the comments, because it gives the general case. $\endgroup$ – chubakueno Mar 13 '14 at 21:41
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Denote

$A(1) = \alpha+\beta+\gamma+\delta = 0$;
$A(2) = \alpha\beta+...+\gamma\delta = -3$;
$A(3) = \alpha\beta\gamma +...+\beta\gamma\delta = -5$;
$A(4) = \alpha\beta\gamma\delta = -2$.

Applying Newton's identities, we obtain

$S(1) = A(1) = 0$;
$S(2) = A(1) S(1) - 2A(2) = 0+6=6$;
$S(3) = A(1) S(2) - A(2)S(1) + 3A(3) = 0+0-15=-15$;
$S(4) = A(1) S(3) - A(2)S(2) + A(3)S(1) - 4A(4) = 0+18+0+8=26$;
$S(5) = A(1) S(4) - A(2)S(3) + A(3)S(2) - A(4)S(1) + 5A(5) = 0-45-30+0+0=-75$
(one can obtain last identity, when consider equation $x^5-3x^3+5x^2-2x=0$ with $5$ roots $\alpha,\beta,\gamma,\delta,0$, where $A(5)=0$, and previous $A(n)$ are the same).


Easy to see, that

$$\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$$

$$=\alpha^2 (\alpha^3+\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \beta^3+\gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \gamma^3+\delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3 + \delta^3) - \alpha^5-\beta^5-\gamma^5-\delta^5$$

$$ = (\alpha^2+\beta^2 + \gamma^2 + \delta^2)(\alpha^3+\beta^3 + \gamma^3 + \delta^3) - \alpha^5-\beta^5-\gamma^5-\delta^5 = S(2)S(3)-S(5). $$

$$ = 6\cdot (-15) - (-75) = -90+75=-15. $$


Note: equation $x^4-3x^2+5x-2=0$ has $2$ real and $2$ complex roots:
$x_1\approx -2.344470$;
$x_2\approx 0.578277$;
$x_{3,4} \approx 0.883096 \pm 0.833866i$.

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  • $\begingroup$ Thank you, the last part was made clear with the second part. The link to Newton's identities isn't very clear (I am not too familiar with the language), and the comment above gave me another equation ($S(n+4)−3S(n+2)+5S(n+1)−2S(n)=0$) with which I solved using $S(-1)$. If you would be so kind as to say the general equation (as a sum) using your notation of $A(n)$ and $S(n)$ so that you could clear it up, please? $\endgroup$ – George Mar 13 '14 at 22:17
  • $\begingroup$ @George, Newton's identities (in my notation) will look like $$ kA(k) = \sum_{j=1}^k (-1)^{j-1} A(k-j)S(k), \tag{1} $$ or $$ kA(k) = \sum_{j=0}^{k-1} (-1)^{k-j+1} A(k)S(k-j),\tag{2} $$ where $A(0) = 1$. If you'll consider equation $$ (x^4-3x^2+ 5x-2)x^n=0, $$ you'll have $A(0)$, $...$, $A(4)$ as were earlier, but $A(5)=...=A(n+4)=0$. So, for $k=n+4$ you'll get (looking at $(2)$) $$ (n+4) \underbrace{A(n+4)}_{=0} = A(0)S(n+4) - A(1)S(n+3) + A(2)S(n+2) - A(3)S(n+1)+A(4)S(n) - 0+0-0+...; $$ so $$ 0 = A(0)S(n+4) - A(1)S(n+3) + A(2)S(n+2) - A(3)S(n+1)+A(4)S(n). $$ $\endgroup$ – Oleg567 Mar 14 '14 at 6:17

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