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The question that follows was inspired by this question:

When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula.

  1. That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it?
  2. Why is it that with each successive degree in a polynomial equation, the solution becomes so much more complex? Can I gain some intuition about what makes finding the roots so hard?
  3. According to the Abel-Ruffini theorem: "there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher." What is so special about the quintic that makes it the cut-off for finding a general algebraic solution?
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I think this should be closed - it's mentioned twice in the question you link to that there is no formula for polynomials of degree > 4, so asking how fast the complexity of the formulas grows is meaningless –  BlueRaja - Danny Pflughoeft Jul 27 '10 at 19:33
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it's not so hard to find roots numerically... –  Jason S Jul 27 '10 at 19:33
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possible duplicate of Is there a general formula for solving 4th degree equations? –  Akhil Mathew Jul 27 '10 at 19:37
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The article you linked to in 3 answers your question 3, and possibly your question 1 if I am understanding it correctly. If there is something about the proof you don't understand, that should be part of your question. Right now I don't really know what you're asking. –  Larry Wang Jul 27 '10 at 19:54
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@Kaestur: the question is about gaining an intuition. Intuition about why this problem is so hard is very different than a proof that this problem is hard. I recognize that one possible answer to my question might be: it is impossible to gain any intuition about why quintic is unique, all we know how to do is prove it. I'm just curious to see if anyone can surprise me with a particularly elegant way of thinking about this problem. –  Ami Jul 27 '10 at 19:59

3 Answers 3

up vote 14 down vote accepted

The idea is basically:

Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.

Now if we expand such a product:

$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$ $(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$

And so on. The pattern should be clear.

This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:

For a quadratic polynomial $x^2 + px + q$, find $a_1,a_2$, such that

$p = a_1 + a_2$

$q = a_1a_2$

For a cubic polynomial $x^3 + px^2 + qx + r$, find $a_1,a_2,a_3$, such that

$p = a_1 + a_2 + a_3$

$q = a_1a_2 + a_1a_3 + a_2a_3$

$r = a_1 a_2 a_3$

And similarly for higher degree polynomials.

Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.

The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).

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When you try to solve a degree $n$ equation, there are $n$ roots you have to find (in principle) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an $n$-fold symmetry in order to write down the roots.

Now the symmetry group of the n roots becomes more and more complicated the larger $n$ is. For $n = 2$, it is abelian (and very small!); for $n = 3$ and $4$ it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for $n = 5$ or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals.

Summary: The complexity of the symmetry group of the $n$ roots leads to a corresponding complexity in explicitly solving the equation.

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For a different take, some practical problems are discussed in Wilkinson's classic article The Perfidious Polynomial. If you can't access it, check what Wikipedia has to say on the subject.

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Jim Wilkinson's prize-winning article can be read here; briefly, one frequent source of trouble in numerical polynomial root finding is our habit of expressing polynomials in the monomial basis, and it happens that there are polynomials like $\prod_i (x-i)$ that numerically behave very poorly in rootfinding when expressed in the monomial basis. –  J. M. Apr 20 '11 at 8:51
    
@J.M. This link is broken. –  I. J. Kennedy Oct 25 at 23:48

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