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You can write the n roots of an n degree polynomial in terms of its n coefficients, i.e., "Vieta's" formulas.

You can solve this system of nonlinear equations using Newton's method and the Jacobian.

What I am missing is which part of this procedure violates n>5 unsolvable algebraically --aren't all the matrix operations algebraic?

Is it taking the derivative that is the non-algebraic step that lets us solve?

In what sense is a derivative an "operation", what characteristics does it share with the other 5? Is there a group with nth derivative as its binary operation? Where does derivative fit in some kind of framework of algebraic operations? Quotient is an operation but is obtaining the remainder also an "operation" or would this also be in a separate part of the framework to the other 5? What are some references that address the placement of different operations in some kind of theoretical framework?

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    $\begingroup$ Newton's method gives an approximation, it does not solve the system of non linear equations. Of course, the system is equivalent to solving the polynomial. $\endgroup$
    – J.C. Ottem
    Commented Nov 3, 2011 at 19:04
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    $\begingroup$ In case there is discussion as to the merits of this question, please take it to tea.mathoverflow.net/discussion/1198/… . $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 3, 2011 at 20:17
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    $\begingroup$ Concerning the edit; we don't discuss MO on MO. Take it to the meta site. $\endgroup$
    – Gerry Myerson
    Commented Nov 4, 2011 at 21:21
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    $\begingroup$ Those of you who want to know the questioner's opinions concerning the future direction of the mathematics community are welcome to view the edit history of this post. $\endgroup$
    – S. Carnahan ♦
    Commented Nov 5, 2011 at 1:50
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    $\begingroup$ or here tea.mathoverflow.net/discussion/1198/… $\endgroup$
    – Cris Stringfellow
    Commented Nov 5, 2011 at 6:03

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The proper notion is "unsolvability with respect to a certain set of operations"; in the case of Galois-Abel's result regarding the quintic equation, this means that there will be no nice algebraic formula using just nth-roots, addition, etc. (Use Some encyclopedia for the proper set.) There are formulas for solving the quintic in terms of more advanced operations; again there are many easily found sources showing how and with what functions.

Similar problems that are not solvable at one level become solvable with more powerful tools, from squaring the circle to determining which Turing machines halt on a blank tape. You need to adjust your perspective to the situation and decide what is appropriate.

This type of situation arises often in mathematical logic, especially universal algebra. Look up reverse mathematics to see what you need for proving certain theorems; look up clone theory, interpretability theory, and classifying your favorite kinds of algebra for examples of what you can or can't do if given extra operations or relations.

Gerhard "Ask Me About System Design" Paseman, 2011.11.03

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  • $\begingroup$ You need to adjust your perspective to the situation and decide what is appropriate -- totally agree. Lookups appreciated. $\endgroup$
    – Cris Stringfellow
    Commented Nov 4, 2011 at 14:40
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Furthermore, Viète's formulas for the n roots of a degree n polynomial actually return $n!$ answers, because it returns all permutations of the answers. Of course, they are all (algebraic) conjugates, but this is not inherently apparent, certainly not numerically, although it should be 'obvious' to a mathematician.

Furthermore, Newton's method is not guaranteed to converge. In fact, there are no generally convergent algorithms (in the family of rational maps) for finding roots of polynomials - this result can be found a 1987 of McMullen, Families of rational maps and iterative root-finding algorithms.

You can still use Newton's method to find all the roots - you just have to account for the fact that some of your starting points will, almost surely, get stuck. But if you are clever enough, you can get all the roots (with Newton's method), as Hubbard, Schleicher and Sutherland show. There are some recent variations which are even more clever and use fewer 'starting points' and still guarantee convergence to all roots, but that is getting too far afield of this particular question.

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  • $\begingroup$ Yes the technical detail that all the roots (in the Vieta formula) are cyclically permutable. It is obvious. But I don't see how that returns n! answers, if you plug in a particular order of coefficients that should fix the answer permutation....Any more detail here that is worth discussing? Will solving Jacobian in Newton's method produce n! with equal probability? Appreciate the pdfs. $\endgroup$
    – Cris Stringfellow
    Commented Nov 4, 2011 at 14:42
  • $\begingroup$ No, plugging in a particular order of coefficients does not fix it. The system of equations you get inherently has n! solutions. $\endgroup$
    – Jacques Carette
    Commented Nov 5, 2011 at 13:12
  • $\begingroup$ Sorry I was thinking over Z. Over Z how many solutions pop out? Is it still n!. I don't think so. By the way, how do you work out that there are n! solutions? $\endgroup$
    – Cris Stringfellow
    Commented Nov 5, 2011 at 13:55
  • $\begingroup$ If you have k integer roots, k! will pop out, I would think (i.e. all permutations). I got to n! first from working out n=2,3,4 explicitly (using a CAS), and extrapolating. Then I remembered Bezout's theorem. $\endgroup$
    – Jacques Carette
    Commented Nov 5, 2011 at 14:08

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