Express the sum of three roots as combination of quotients

Question

I'm trying to do an exercise relative to symmetrical polynomials.

We're given the following polynomial:

$ X^3 + pX +q = 0$

With $x_1, x_2, x_3 $ its roots.

We're asked to give an expression with $p, q$ of the following sum :

$x_1^8 + x_2^8 + x_3^8 $

The obvious way to do this is way to messy, but I recall our teacher saying there was a trick when working with high exponents like these ones.

Please notice that this expression is symmetric, so we can use Cardano-Vieta to solve this.

Thank you for reading.

Answer 1

I usually use this trick. Let's call:

$$S_j=x_1^j+x_2^j+x_3^j$$

We are searching $S_8$. Trivially:

$$\left\{\begin{matrix} x_1^3+px_1+q=0\\ x_2^3+px_2+q=0\\ x_3^3+px_3+q=0 \end{matrix}\right.$$

If we multiply both sides for the monomial $x^j$ the sum of the powers of the roots doesn't change cause we are adding $n$ roots that are equal 0. So: $$\left\{\begin{matrix} x_1^{j+3}+px_1^{j+1}+qx_1^{j}=0\\ x_2^{j+3}+px_2^{j+1}+qx_1^{j}=0\\ x_3^{j+3}+px_3^{j+1}+qx_1^{j}=0 \end{matrix}\right.$$

Sum all of the terms and you get:

$$S_{j+3}=-pS_{j+1}-qS_{j}$$

So now we must calculate $S_2$ and $S_3$ and apply the recursion some times. $S_2$ is a standard calculus:

$$S_2=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)$$

Applying Vieta formulas:

$$x_1+x_2+x_3=0$$ $$x_1x_2+x_1x_3+x_2x_3=p$$

So: $$S_2=-2p$$ Moreover for the recursion: $$S_{3}=-pS_{1}-qS_{0}=-3q$$

And for the recursive formula: $$S_5=-pS_3-qS_2=3pq+2pq=5pq$$ $$S_8=-pS_6-qS_5$$

A last effort for $S_7$: $$S_4=-pS_2-qS_1=2p^2$$ $$S_6=-pS_4-qS_3=-2p^3+3q^2$$

And finally:

$$S_8=-p(-2p^3+3q^2)-q(5pq)$$ $$S_8=2p^4-8pq^2 $$

:)

Answer 2

Hint:

Denote $\;\sigma_1=\sum_i x_i$, $\;\sigma_2=\sum_{i\ne j}x_ix_j$, $\; \sigma_3=x_1x_2x_3$. From the form of the equation, we know that $$\sigma_1=0,\enspace \sigma_2=p,\quad \sigma_3=-q.$$

Now rewrite the equation as $X^3=-pX-q$. We deduce \begin{cases} x_i^3=-px_i-q, \quad\text{ hence} \\ x_i^6=(px_i+q)^2=p^2x_i^2 +2pqx_i+q^2 \\ x_i^8=p^2x_i^4 +2pqx_i^3+q^2x_i^2 =\dotsm \end{cases} Can you continue to express $x_i^8$ as a quadratic polynomial in $x_i$, and, from there, express $\sum_i x_i^8$ with $\sigma_1, \sigma_2,\sigma_3$?

Answer 3

Use Newton's Identities (to relate sums of powers of roots to coefficients of the polynomial).

Using the notation at the cited page, you want $p_8(x_1,x_2,x_3)$, where (henceforth suppressing "$(x_1,x_2,x_3)$") $$ e_0 = 1 , e_1 = 0 , e_2 = p , e_3 = -q \text{, and } e_{\geq 4} = 0 \text{.} $$ Then\begin{align*} p_1 &= e_1 = 0 \\ p_2 &= e_1 p_1 - 2 e_2 \\ &= 0 \cdot 0 - 2 \cdot p = -2 p \\ p_3 &= e_1 p_2 - e_2 p_1 + 3 e_3 \\ &= 0 - 0 + 3 (-q) = -3 q \\ p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1 - 0 \\ &= 0 - p(-2p) + 0 - 0 = 2 p^2 \\ &\vdots \end{align*}

\begin{align*} p_5 &= e_1 p_4 - e_2 p_3 + e_3 p_2 - 0 \\ &= 0 - p(-3q) - q(-2p) = 5 pq \\ p_6 &= e_1 p_5 - e_2 p_4 + e_3 p_3 - 0 \\ &= 0 - p(2p^2) - q(-3q) = -2 p^3 + 3 q^2 \\ p_7 &= e_1 p_6 - e_2 p_5 + e_3 p_4 - 0 \\ &= 0 - p(5pq) - q(2p^2) = -7 p^2 q \\ p_8 &= e_1 p_7 - e_2 p_6 + e_3 p_5 - 0 \\ &= 0 - p(-2 p^3 + 3 q^2) - q(5 pq) \\ &= 2 p^4 - 8 p q^2 \text{.} \end{align*}

Note that, since $e_{\geq 4} = 0$, once we get to $p_4$ and forever after, we are performing the same dot product with the previous three terms, so this can be rolled up into a matrix multiplication and made much faster, especially for $p_\text{huge number}$.

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An entirely different way to go is this. We want to reduce $x_1^8 + x_2^8 +x_3^8$ by the relations $x_1^3 + p x_1+ q = 0$, $x_2^3 + p x_2+ q = 0$, $x_3^3 + p x_3+ q = 0$, $x_1 + x_2 + x_3 = 0$, $x_1 x_2 + x_1 x_3 + x_2 x_3 = p$, and $x_1 x_2 x_3 = -q$. Constructing a Groebner basis of those relations with variable order $x_1, x_2, x_3$, gives $$ \{x_3^3 + p x_3 + q, x_2^2 + x_2 x_3 + x_3^2 + p, x_1 + x_2 + x_3\} $$ and reducing $x_1^8 + x_2^8 + x_3^8$ by this basis gives

$$ 2p^4 - 8 p q^2 \text{.} $$

We can make Wolfram Alpha do all the tedious work (and use a shorter basis specification since the relations are redundant). The result is the last expression, the (canonical) remainder on dividing $x_1^8 + x_2^8 + x_3^8$ by the relations.

Answer 4

Denoting the polynomial by $P(X)$, we can consider the expansion of $\log\left[P(X)/X^3\right]$ in powers of $X^{-1}$. We have:

$$\frac{P(X)}{X^3} =\prod_{k=1}^3\left(1-\frac{x_k}{x}\right)$$

Therefore:

$$\log\left[\frac{P(X)}{X^3}\right] =-\sum_{r=1}^\infty\frac{S_r}{rX^r}$$

where

$$S_r = \sum_{k=1}^3 x_k^r$$

So, we need to compute the series expansion of

$$\log\left(1 + \frac{p}{X^2} + \frac{q}{X^3}\right)$$

the sum of the 8th powers of the roots is then the coefficient of $X^{-8}$ times $-8$. In this particular case, extracting this coefficient requires very little effort as the only contributions come from the third and fourth power of $\frac{p}{X^2} + \frac{q}{X^3}$. The fourth power yields a contribution

$$(-8)\times \left(-\frac{1}{4}\right) p^4= 2 p^4$$

The third power yields a contribution via the square of the second term multiplied by the first, this is

$$(-8)\times \frac{1}{3} \times 3 p q^2=-8 p q^2$$

The result is therefore:

$$S_8 = 2 p^4 - 8 p q^2$$