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?
How far can we reach for the sum of roots in closed form for a polynomial of even degree?
4
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algebra-precalculus
polynomials
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0Actually, 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: http://math.stackexchange.com/questions/32616/hermites-solution-of-the-general-quintic-in-terms-of-theta-functions you can see pointers to resources about the solution of the quintic. – 2011-05-07
3 Answers
10
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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0So, 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). – 2011-05-07
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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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0I 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. – 2011-05-07
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HINT $\rm\ \ \ (x^k-r\ x^{k-1}+\cdots\:)\ (x^n - s\ x^{n-1}+\cdots\:)\ =\:\ x^{k\:+\:n} - (r+s)\ x^{k\:+\:n-1} + \cdots$