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I can get my head around this so someone explain it please.

$(1)$ From Galois theory it is known there is no formula to solve a general quintic equation.

But it is known a general quintic can be solved for the 5 roots exactly. Back in 1858 Hermite and Kronecker independently showed the quintic can be exactly solved for (using elliptic modular function). Also I think they're maybe other solution for the quintic which means a formula for each of the 5 roots.

So why is the claim in Galois theory that there is no formula to solve it? I know I am missing something here because the above $(1)$ is an established result.

So what is the value in saying, using Galois theory we do not have a formula for the 5 roots? Since for practical purposes we can actually find the 5 roots each time using say for example the formula based on elliptic modular functions.

Developer
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    Galois Theory tells us that the roots of a general quintic can not be obtained by the use of radicals alone (that is, with the usual operations of arithmetic and taking $n$-th roots for various $n$). That does not mean that the roots can't be expressed using other types of functions. – Geoff Robinson Feb 01 '16 at 14:39
  • @GeoffRobinson But what is the point in knowing that if you can solve for each of the five roots (exactly and numerically) using the Hermite method? – Developer Feb 01 '16 at 14:44
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    @Developer, the value lies in the difference between $1.4142$ and $\sqrt2$ – Mariano Suárez-Álvarez Feb 01 '16 at 20:41
  • @Mariano Suárez-Alvarez word play on the word "value"for a numerical value compared to a algebraic number. If that what you mean then. It depends on the situation if all that is required is a numerical value (or an expression for infinite precision like $sqrt(2)$) for each of the five roots then you can find them using the Hermite metthod-it doesnt depend on the fact that Galois proved that the Quintic is not solvable by his method (in other words that fact here has no value) – Developer Feb 02 '16 at 02:39
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    The «infinite precision» is irrelevant, @Developer. – Mariano Suárez-Álvarez Feb 02 '16 at 03:02
  • @Mariano Suárez-Alvarez Or should I have said theres an expression that lets you computer the root to an the required precision-what ever that may be – Developer Feb 02 '16 at 03:05
  • That is also quite irrelevant, really. Galois theory tells you when something is possible and when it is not possible — a completely different ballpark. – Mariano Suárez-Álvarez Feb 02 '16 at 03:07
  • @Mariano Suárez-Alvarez Yes I went into the wrong ball park by accident. Should I have said then the expression exists. – Developer Feb 02 '16 at 03:09

4 Answers4

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Galois-theory only says that there is no general formula to solve a quintic equation in terms of radicals. That is, there is no formula only using the arithmetic operations "sum, multiplication etc. and taking the $n$-th root".

For instance for the polynomial $x^5 - 4x + 2$ it is known that it has a root that is not expressible in the above mentioned operations (as its Galois-group is $S_5$). (Edit: Another example is $x^5 - x + 1$ which also has Galois-group $S_5$. If you Wolframalpha this polynomial you see nicely how four of its roots can be expressed by radicals, but the fifth can't.)

The solution you mean is the solution using Bring radicals - wikipedia-article here: https://en.wikipedia.org/wiki/Bring_radical -, which is not a contradiction, as it is not expressed in form of radicals (in the sense of $n$-th roots of something).

Oscar Lanzi
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Steven
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  • So what is the value in knowing the Galois result above, if you can solve for each of the five roots (exactly or numerically) with a formula which is not restricted to nth roots and arithmetic operations? – Developer Feb 01 '16 at 14:55
  • solving things in radicals is an algebraic viewpoint. taking the $n$-th root of something can still be made sense of in any field. note that the solution of the problem is more generally: for any field $k$, a polynomial in $k[X]$ is solvable in radicals if and only if its Galois-group is solvable. in a general field you can't use analytic methods like you do with the real numbers (which is done with the Bring radical). – Steven Feb 01 '16 at 20:04
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    I cheated however a bit in the above comment. the result which was historically important was indeed the special case of the rational numbers. but these kind of questions are more of the style: "what is the point in constructing things only with ruler and compass?" or "why care about Fermat's Last Theorem?" or "why would anyone care about classical music like Schönberg's that can't be used in movies?" or "why do painters paint if we have photography?" – Steven Feb 01 '16 at 20:08
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    $x^5+x+1=(x^2+x+1)(x^3-x^2+1)$? – CO2 Jun 24 '18 at 17:07
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    @CO2 yep, that is a mistake, all 5 roots are in radicals, and wolframalpha shows that if you press exact solution for real root. He meant -x. – Валерий Заподовников Apr 17 '23 at 07:18
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    That sign has been corrected. Properly, the Galois group for $x^5+x+1=0$ would be $S_3×C_2$, the cross product form coming from the polynomial being reducible. – Oscar Lanzi Jul 13 '23 at 12:16
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    @OscarLanzi Thanks for correcting. Upvoted. – Валерий Заподовников Jul 13 '23 at 21:16
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The problem of solving algebraic equations algebraically is one of the oldest problems humans have considered —thousands of years old— arising from everything, going from the subdivision of inheritance to riddles by mithological figures to the construction of actual buildings. The fact that we know exactly when we can do it and, when it is possible, that we can in fact carry out the construction of solutions is one of the greatest achievements of mankind.

That's the value of it.

No one cares about infinite precision.

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Before answering the question of what is the point of saying that there is no general formula in radicals for the roots of a quintic I would pose a preliminary question:

What is the point in having any formula for the roots of any polynomial?

If the goal is to compute numeric approximations for the roots of a polynomial with known numeric coefficients then you do not need any formula for the roots at all. Instead square-free factorisation and numerical root-finding algorithms can compute the roots of polynomials of any degree more accurately than any general formula. There might be certain situations where it is more efficient to use something like the quadratic formula rather than a root-finding algorithm but certainly no formulae are necessary to be able to compute the roots numerically.

In practice the value of having explicit formulae for the roots of a polynomial comes from using them as part of a symbolic calculation where the coefficients of the polynomial are symbolic expressions rather than explicit numbers. For example the differential equation $$ m\ddot{x} + b\dot{x} + kx = 0 $$ could describe a mass on a spring using symbols $m$, $b$ and $k$ as the coefficients. Explicit expressions for the solutions to the differential equation using radicals can be found using the quadratic formula and the behaviour of the solutions can be analysed by looking at how they depend on these symbolic parameters. Since this differential equation is 2nd order we would want the quadratic formula but for 5th order we would need the roots of a quintic and for a system of $n$ coupled masses we would need the roots of a polynomial of degree $2n$. This differential equation is just one of many examples where a symbolic calculation might lead to wanting symbolic expressions for the roots of a polynomial and where increasingly complex problems lead to polynomials of higher degree.

In this context we can use the quadratic, cubic or quartic formulae for polynomials up to degree 4 and obtain expressions involving radicals. Further symbolic manipulation of these expressions involving radicals is possible but can be awkward. One difficulty is that radical functions are generally awkward to work with and another is that e.g. the quartic formula gives large complicated expressions for the roots. Galois theory and the Abel-Ruffini theorem tell us that for degree 5 or more we might not even be able to get a solution in radicals and would instead need to work with more complicated functions that would be significantly harder to manipulate in further symbolic calculations. This problem is compounded by the fact that non-radical formulae that do exist for unsolvable quintics are also significantly more complicated than the quartic formula which is already unwieldy enough to be of questionable usefulness.

In practice even for symbolic calculations it is usually better to avoid using radical formulae whenever possible. There are however classes of problems that can be handled symbolically via the use of explicit formulae although doing so becomes more difficult for higher degree polynomials. The Abel-Ruffini theorem represents one hard limit on the approaches that can be used when attempting to scale up from simple symbolic calculations using the quadratic formula to more complicated problems involving higher degree polynomials.

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I'd like to share that the principal quintic $x^5+ax^2+bx+c=0$ is algebraically solvable by the "Fórmula Luderiana para Equação Quíntica" (Luderian Formula for Quintic Equation) just check it out at slideshare for details.