Generalizing polynomial formulas?

[imatfaal]

QP - I think you are misunderstanding what it is that is impossible. No one is claiming that the roots don't exist - the curve clearly goes through the axis when you plot a graph. But there is no method/protocol/rearrangement that will always give you an exact answer.

 

Claiming that there must be an exact and findable answer to all quintics is akin to saying that you can give an exact answer to the ratio of a circles diameter to its radius. The answer is 2pi - but numerically we cannot give an exact answern (but a calculator will give a damn nice approximation.

 

All quadratics/cubics/quartics will have roots that can be expressed as integers (x=1), ratios of integers (x=5/7), or radicals (x=sqrt(2).

 

Quintics and higher order equations do not have to have such simple roots. x^5-x+1=y has one real root (at about 1.16) but it CANNOT be expressed in terms of integers/ratios/radicals. Your calculator will use the reversal of a taylor expansion to calculate an approximate answer (to any desired degree of accuracy) - but it is not exact!

[questionposter]

QP - I think you are misunderstanding what it is that is impossible. No one is claiming that the roots don't exist - the curve clearly goes through the axis when you plot a graph. But there is no method/protocol/rearrangement that will always give you an exact answer.

 

Claiming that there must be an exact and findable answer to all quintics is akin to saying that you can give an exact answer to the ratio of a circles diameter to its radius. The answer is 2pi - but numerically we cannot give an exact answern (but a calculator will give a damn nice approximation.

 

All quadratics/cubics/quartics will have roots that can be expressed as integers (x=1), ratios of integers (x=5/7), or radicals (x=sqrt(2).

 

Quintics and higher order equations do not have to have such simple roots. x^5-x+1=y has one real root (at about 1.16) but it CANNOT be expressed in terms of integers/ratios/radicals. Your calculator will use the reversal of a taylor expansion to calculate an approximate answer (to any desired degree of accuracy) - but it is not exact!

 

So your saying the roots are irrational numbers like e, and since the decimal places go on forever, you can't have an exact numerical answer? But aren't there exact things that generate irrational numbers and instead of writing out the irrational number you just keep the simplified version of whatever generates them? Like the square root of 2 is irrational when written out, so don't you just keep it as "sqrt(2)" to make it exact?

Edited October 18, 2011 by questionposter

[DrRocket]

So your saying the roots are irrational numbers like e, and since the decimal places go on forever, you can't have an exact numerical answer? But aren't there exact things that generate irrational numbers and instead of writing out the irrational number you just keep the simplified version of whatever generates them? Like the square root of 2 is irrational when written out, so don't you just keep it as "sqrt(2)" to make it exact?

 

No. [math]\sqrt 2 [/math] is exact even if it is not representable by a finite decimal expression.

 

For the general case of polynomials of degree 5 or larger no solution as an algebraic expression in terms of nth roots exists. It is not that no one has found such a solution. It has been proved (see a text on Galois theory) that no such solution exists.

 

The fundamental theorem of algebra (see a text on complex analysis) shows that roots for any polynomial exist, at least as complex numbers. But Galois theory shows that you cannot them, in general, exactly (as radical expressions. This does not mean thjat some particularly simple 5th degree polynomials cannot be solved exactly, but no formula exists for an arbitrary 5th degree or higher polynomial.

 

"e" is not only irrational, but is also transcendental. That means that it is not a root of any polynomial with rational coefficients.

[questionposter]

No. [math]\sqrt 2 [/math] is exact even if it is not representable by a finite decimal expression.

 

For the general case of polynomials of degree 5 or larger no solution as an algebraic expression in terms of nth roots exists. It is not that no one has found such a solution. It has been proved (see a text on Galois theory) that no such solution exists.

 

The fundamental theorem of algebra (see a text on complex analysis) shows that roots for any polynomial exist, at least as complex numbers. But Galois theory shows that you cannot them, in general, exactly (as radical expressions. This does not mean thjat some particularly simple 5th degree polynomials cannot be solved exactly, but no formula exists for an arbitrary 5th degree or higher polynomial.

 

"e" is not only irrational, but is also transcendental. That means that it is not a root of any polynomial with rational coefficients.

 

So again, the reason for not generalizing them is there isn't a process that will generate an exact answer all the time since some times the roots are irrational transcendentals which go on infinitely and therefore can't have 100% accuracy just as you cannot calculate the exact diameter of a circle because you must use pi which is an irrational transcendental?

[Cap'n Refsmmat]

No. No formula exists because there's no general method to use radicals and algebraic operations to get to the roots, not because one can't represent the roots with infinite accuracy.

[questionposter]

No. No formula exists because there's no general method to use radicals and algebraic operations to get to the roots, not because one can't represent the roots with infinite accuracy.

 

So again, I then say "so you can find the exact roots of higher degree polynomials even if they are all imaginary, but you just can't use some general formula for ALL polynomials?".

[Cap'n Refsmmat]

You can find the exact roots of some higher degree polynomials through radicals and algebra, but there is no general formula for all polynomials.

[questionposter]

You can find the exact roots of some higher degree polynomials through radicals and algebra, but there is no general formula for all polynomials.

 

Hmm, well I suppose the main problem is then that I don't fully understand the Abel-Ruffini theorum or whatever proves that some polynomials can't have exact roots.

Edited October 20, 2011 by questionposter

[Cap'n Refsmmat]

It proves there's no general formula to find those roots, not that they don't have roots.

[imatfaal]

hmmm - are you sure Capn? My understanding of the galois and abel-ruffini was that it proved that non-radical solutions existed - and it thus followed from this that no general equation that used rational coefficients could exist and not the other way around - its is a subtle difference but what the hell...

 

if you know/understand the theory then I will back down cos it is all a bit beyond me (and I last learnt it a few years ago)

[questionposter]

It proves there's no general formula to find those roots, not that they don't have roots.

 

Ok, so then all polynomial formulas have calculable roots, but you can't use the same method to calculate all those roots, correct?

[DrRocket]

hmmm - are you sure Capn? My understanding of the galois and abel-ruffini was that it proved that non-radical solutions existed - and it thus followed from this that no general equation that used rational coefficients could exist and not the other way around - its is a subtle difference but what the hell...

 

if you know/understand the theory then I will back down cos it is all a bit beyond me (and I last learnt it a few years ago)

 

One more time.

 

The fundamental Theorem of Algebra, usually proved using complex analysis, shows that any polynomial with complex coefficients has a complex root (the complex numbers are algebraically complete). It then follows from simple division of polynomials (Euclidean algorithm) that an nth degree polynomial with complex coefficients has n roots, counting multiplicity.

 

The Abel-Ruffini, now usually proved using Galois theory shows that in general a polynomial of degree 5 or higher is not solvable by radicals -- i.e. there is no formula in terms of addition, subtraction, multiplication division and nth roots. This does not mean that some particularly simple polynomials of degree 5 or higher are not exactly solvable, but there is no generally applicable solution for quintic or higher degree polynomials.

 

Numerical methods will allow you to approximate roots as closely as you wish, but will not yield exact solutions.

 

If the polynomial has rational coefficients you can find a polynomial with integer coefficients having the same roots, and the rational root theorem will let you find any rational roots by checking all possibilities. But there may not be any rational roots.

[imatfaal]

Ok, so then all polynomial formulas have calculable roots, but you can't use the same method to calculate all those roots, correct?

 

Nah - per DocRock above - "The Abel-Ruffini, now usually proved using Galois theory shows that in general a polynomial of degree 5 or higher is not solvable by radicals" - ie some higher orders will have roots that can only ever be numerically approximated - ie no method can exist and they cannot be calculated exactly