View Full Version : Polynomial solutions

Enola Straight

09-02-2009, 05:18 PM

The Quadratic formula solves Ax^2+Bx+C=0

The ??? formula solves Ax^3+Bx^2+Cx+D=0

The Quartic formula solves Ax^4+Bx^3+Cx^2+Dx+E=0

Fifth degree and higher polynomials have no handy-dandy formula to solve for x.

Why not?

OldGuy

09-02-2009, 05:40 PM

The Quadratic formula solves Ax^2+Bx+C=0

The ??? formula solves Ax^3+Bx^2+Cx+D=0

The Quartic formula solves Ax^4+Bx^3+Cx^2+Dx+E=0

Fifth degree and higher polynomials have no handy-dandy formula to solve for x.

Why not?

Cubic

and it's the Abel Ruffini Theorem that proves what you want. Many years ago I sort of understood it, but it's been too long. I know it has to do with group theory.

http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

ultrafilter

09-02-2009, 06:00 PM

There is a general solution for the quintic equation, but not one that just involves arithmetic operators and square roots. It's a little tough to give intuitive meaning to the math behind this conclusion, but the basic idea is that there's a certain structure associated with any polynomial. If the polynomial is of degree at most four, this structure is small enough that it has to have a certain form, but once you get into higher degree polynomials the structures are larger, and other forms are possible.

Uncertain

09-02-2009, 06:47 PM

There is a general solution for the quintic equation, but not one that just involves arithmetic operators and square roots.

Just to be clear (I wasn't sure what you meant), my understanding is that although every quintic (indeed every polynomial) has roots, there is no general equation of any sort for these roots. Also, the general solutions to cubics and quartics involve cube roots and fourth roots.

Indistinguishable

09-02-2009, 07:15 PM

There's no equation that takes in the coefficients of the polynomial and, using only addition, subtraction, multiplication, division, and taking nth roots, produces the zeros of the polynomial. In fact, it's not just that there's no uniform such equation; sometimes, for some polynomials of degree five and higher, the zeros can't be expressed in such a form at all.

This doesn't mean such zero-finding formulas can't be written using other operators. It just means, well, exactly what it says (which is exactly what ultrafilter said).

ultrafilter

09-02-2009, 07:26 PM

There is a general formula, but it requires Bring radicals (http://en.wikipedia.org/wiki/Bring_radical).

Exapno Mapcase

09-02-2009, 07:29 PM

Symmetry: A Journey into the Patterns of Nature, by Marcus Du Sautoy (http://www.amazon.com/Symmetry-Journey-into-Patterns-Nature/dp/0060789417/ref=sr_1_4?ie=UTF8&s=books&qid=1251936961&sr=8-4) has the best explanation of the real deep meaning of why higher level equations have no general solution that I've read.

The reason, not surprisingly, is symmetry. Du Sautoy takes many chapters detailing the history of math to get to the explanation and I can't put it into a paragraph. But it's intimately involved with the creation of group theory, which explains and explores symmetries. And when Galois invented group theory he produced the explanation for the lack of a quintic. Here's the closest thing to a one paragraph answer:

In particular, a group of symmetries might be indivisible despite the fact that the number of symmetries of the object was not a prime number.

Galois was interested in this discovery because he realized that the indivisibility of the group of symmetries of an equation held the key to whether the underlying equation could be solved or not. If a group of symmetries could be broken down into rotational symmetries of prime-sided shapes, the equation could be solved. Otherwise, it couldn't be solved.

This probably doesn't make much sense by itself. But you didn't really expect an understandable answer to a question about math, did you?

DSYoungEsq

09-02-2009, 07:41 PM

There is a general formula, but it requires Bring radicals (http://en.wikipedia.org/wiki/Bring_radical).

Thus the famous expression: When solve quintic, bring radicals. :D

Thudlow Boink

09-02-2009, 08:04 PM

The Quadratic formula solves Ax^2+Bx+C=0

The ??? formula solves Ax^3+Bx^2+Cx+D=0

The Quartic formula solves Ax^4+Bx^3+Cx^2+Dx+E=0

Fifth degree and higher polynomials have no handy-dandy formula to solve for x.

Why not?Such a formula is one of those things that, like a way to "square the circle" or to trisect an angle with straightedge and compass, mathematicians sought for centuries before someone came along and proved that it was impossible (under the conditions laid down).

In this case, the "someone" who proved it impossible was the Norwegian mathematician Niels Henrik Abel in 1824, in the theorem OldGuy linked to earlier. Not long after, Evariste Galois developed the beginnings of what came to be known as Galois Theory, which provided further insight into the issue (and this before dying at age 20).

So there's not really a simple explanation of "why not," because if there were, it wouldn't have taken mathematicians so long to find it.

Symmetry: A Journey into the Patterns of Nature, by Marcus Du Sautoy has the best explanation of the real deep meaning of why higher level equations have no general solution that I've read.Mario Livio's The Equation That Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry (http://www.amazon.com/Equation-That-Couldnt-Solved-Mathematical/dp/0743258215/ref=sr_1_3?ie=UTF8&s=books&qid=1251938973&sr=1-3) is another popular-level book about the topic, though I found it only so-so as a pop math book.

Chronos

09-02-2009, 08:40 PM

This doesn't mean such zero-finding formulas can't be written using other operators. It just means, well, exactly what it says (which is exactly what ultrafilter said). Of course, it's kind of meaningless to say that there exist operators which allows you to find the solution. Just define a unary operator called "RootOf()" which takes the polynomial as input and whose value is defined as the root. It's not enough to just say that you can invent a new operator that'll solve your equation; your new operator has to be easy to use, or have applicability to other problems, or be interesting in some other way.

Indistinguishable

09-02-2009, 09:55 PM

Yes, I thought of pointing out that trivial possibility, but held off on it (even though it does best deliver the meaningless of "There is/is not a formula for..." without specifying what kind of formula one means). Anyway, the point was to stave off the common danger of misinterpreting the Abel-Ruffini Theorem as saying something too strong, to the point of ultimately being, as demonstrated by precisely the kind of reflection you've just given, incoherent (as the phrase "there is no general equation of any sort for these roots" which I was replying to appeared to do); to understand the theorem, it's important to note exactly what kind of formulas it does and does not prohibit.

tim314

09-03-2009, 11:18 AM

In this case, the "someone" who proved it impossible was the Norwegian mathematician Niels Henrik Abel in 1824, in the theorem OldGuy linked to earlier. Not long after, Evariste Galois developed the beginnings of what came to be known as Galois Theory, which provided further insight into the issue (and this before dying at age 20).Abel himself only lived to 26, before dying of tuberculosis. (Galois was shot in a duel.)

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