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As everybody knows, our reach for the roots themselves of a polynomial of any degree ends at degree 4, except in special cases. However, since the formula for the sum of the roots of a quadratic is considerably simpler than the formula for the roots themselves, this prompts the hope that one can obtained the sum of the roots for polynomials of even degree for degree 6, and somewhat beyond – perhaps all the way to infinity?

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  • $\begingroup$ Actually, it is possible to find closed form (but generally ungainly) expressions for polynomials of degree greater than four (the Abel-Ruffini caveat here is that you cannot do it in terms of the four basic operations and the taking of radicals alone). The general quintic has been solved since 1858 or so, and if you look at my question here: math.stackexchange.com/questions/32616/… you can see pointers to resources about the solution of the quintic. $\endgroup$ – graveolensa May 7 '11 at 22:56
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The formula for the sum of the roots of any polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ is $$-\frac{a_{n-1}}{a_n}$$ See here.

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    $\begingroup$ Correct, but why stop there? We can write the polynomial as $a_n(x-r_0)(x-r_1)\cdots(x-r_{n-1})$, where the $r_i$s are the roots (including repetition). Expanding this product and then matching coefficient of each $x^i$ with $a_i$, we find that $a_{n-1} = -a_n(\text{sum of roots, } r_i)$, and $a_{n-2} = a_n(\text{sum of pairwise products of roots, } r_i r_j)$, and $a_{n-3} = -a_n(\text{sum of triple-products of roots, } r_i r_j r_k)$, ..., and $a_0 = a_{n-n} = (-1)^n a_n(\text{product of all } n \text{ roots, } r_0 r_1 r_2 \cdots r_{n-1})$. $\endgroup$ – Blue May 7 '11 at 21:54
  • $\begingroup$ Wow. This is why MSE is so wonderful. Thanks! $\endgroup$ – Mike Jones May 7 '11 at 21:57
  • $\begingroup$ So, while we can easily express the coefficients of a polynomial in terms of its roots (and its leading coefficient), what "everybody knows" is that we can't go the other way when $n>4$ (without using tools more sophisticated than extracting roots). $\endgroup$ – Blue May 7 '11 at 21:59
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In general, if $\alpha_1,\alpha_2,\ldots, \alpha_n$ are the $n$ roots (possibly repeated roots) of a $n^{th}$ degree polynomial, then we have

$$\sum_{k=1}^{\binom{n}{r}} p_k = (-1)^r \frac{a_{n-r}}{a_n}$$ where each $p_k$ denotes a product of a unique subset of $r$ roots of the polynomial i.e.

$p_k = \alpha_1^{t_{1k}} \alpha_2^{t_{2k}} \cdots \alpha_n^{t_{nk}}$ where $t_{jk} \in \{0,1 \}$ and $\displaystyle \sum_{j=1}^{n} {t_{jk}} = r, \forall k \in \{1,2,\ldots,\binom{n}{r} \}$

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  • $\begingroup$ I appreciate the generality of your answer, and so I up-voted it, but I'm accepting the answer of Zev Chonoles as THE answer to my question as being spot-on, and with a supporting link. $\endgroup$ – Mike Jones May 7 '11 at 21:55
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HINT $\rm\ \ \ (x^k-r\ x^{k-1}+\cdots\:)\ (x^n - s\ x^{n-1}+\cdots\:)\ =\:\ x^{k\:+\:n} - (r+s)\ x^{k\:+\:n-1} + \cdots$

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