Why is it so hard to find the roots of polynomial equations?

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

1.) Number of solutions, Number of coefficients

Let's consider polynomial equations with complex coefficients.
The increase of the number of solutions of polynomial equations with the degree of the equation is one reason for increased complexity of the solution formula.
The increase of the number of possible coefficients of polynomial equations with the degree of the equation is another reason for increased complexity of the solution formula.

2.) Possibility of closed-form solutions

The history of solving polynomial equations of one unknown began with solutions in radicals. But there are polynomial equations of degree > 4 that have no solutions in radicals.

Radicals are expressions with explicit algebraic operations. If you additionally allow all algebraic functions, $\exp$ and $\ln$, you allow all elementary expressions.
Chow [Chow 1999] gives his
Corollary 1:
"If Schanuel's conjecture is true, then the algebraic numbers in" the explicit elementary numbers "are precisely the roots of polynomial equations with integer coefficients that are solvable in radicals."
That means, the algebraic equations that are not solvable in radicals cannot be solved by elementary numbers (means by applying elementary functions).

There are different reasons for insolvability of polynomial equations in radicals (or in elementary numbers): e.g. algebraic independence, symmetry restrictions, topological restrictions.

All solutions of polynomial equations can be represented with help of transcendental functions, e.g. theta functions, elliptic functions, Bring radicals, exponential/elliptic modular functions, Siegel modular forms, hyperelliptic integrals.
$\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

Answer 5

This may only be tangentially related, since it refers specifically to Diophantine equations, but I think it speaks to the general type of complexity you start to invoke when you get up into quintic polynomials, and thus is worth mentioning.

Matiyasevich showed a while back that you can represent any recursively enumerable set as the set of solutions to a Diophantine equation. IIRC, this requires in the general case using polynomials with variables having at least degree 4. And what this means is that all of the complexity you could find in any computer program (including fundamentally unprovable / undecidable problems) can be packed into one of these equations.

For example, see this formula which enumerates the primes:

https://en.wikipedia.org/wiki/Formula_for_primes#Formula_based_on_a_system_of_Diophantine_equations