Quintic Equations
Moderators: gmalivuk, Moderators General, Prelates
Quintic Equations
$$x^5x+1$$
Apparently ones like this cannot be solved.
So, I felt like having a good scare, so I decided to go ahead and google the cubic formula (like the quadratic formula, except for cubics). Then, the quartic formula. Apparently that was not frightening enough, so I went ahead and googled ''quintic formula.'' Obviously, I discovered there is no such thing, because quintic equations cannot be solved with radicals.
Old news to you guys, most likely. What I would like to ask is a couple things:
First, if you can't express the roots with radicals, what in Zeus' name can you express them with?
Second, why can you not do this with quintics? Why are quintics the lowest polynomials that you cannot use radicals to express? Why not quartics, or sexics? That is special about polynomials of degrees of five or higher that makes them not able to be expressed with radicals?
Apparently ones like this cannot be solved.
So, I felt like having a good scare, so I decided to go ahead and google the cubic formula (like the quadratic formula, except for cubics). Then, the quartic formula. Apparently that was not frightening enough, so I went ahead and googled ''quintic formula.'' Obviously, I discovered there is no such thing, because quintic equations cannot be solved with radicals.
Old news to you guys, most likely. What I would like to ask is a couple things:
First, if you can't express the roots with radicals, what in Zeus' name can you express them with?
Second, why can you not do this with quintics? Why are quintics the lowest polynomials that you cannot use radicals to express? Why not quartics, or sexics? That is special about polynomials of degrees of five or higher that makes them not able to be expressed with radicals?
 intimidat0r
 Posts: 73
 Joined: Thu Aug 02, 2007 6:32 am UTC
Re: Quintic Equations
Questions such as this and many other interesting ones are addressed by Galois theory. That is the summary of what I know about Galois theory, that and it involves something with complex conjugates.
The packet stops here.
Re: Quintic Equations
To start things off, by radicals, I'm assuming you mean real numbers.
To answer the first question, you can express them with complex numbers (essentially it means it has imaginary numbers).
Now I don't know exactly what you've been looking at and everything, but every polynomial is solvable. You're going by it using cubic/quartic/quintic formulas, which personally I don't know. But I think you should look at it at a different approach. A polynomial will have as many roots as it's highest degree. So a quintic equation will have five roots. Now, if the function crosses the xaxis 5 times, it will have 5 real roots. But if it crosses the xaxis fewer times than that, the rest of the roots are in the complex range. So if it has 3 real roots, there are 2 complex roots. There is nothing special about quintics that gets you into complex roots but it depends on the specific function you are looking at (x^2 + 1 = 0 is an example of a quadratic with only complex roots).
Also the function you gave is solvable. I put the function you gave into wolfram alpha and it came with 1 real root and 4 complex roots.
Anyways my suggestion is to don't bother with cubic/quartic/etc. formulas. You won't ever need them, and if you need to find the roots of any equation with a degree higher than 2, just factor it.
As for finding the complex roots of a function, I have no idea. I don't know much about complex numbers algebra, just a few basics with i.
Edit: Ok so I'm way off. Disregard what I posted
To answer the first question, you can express them with complex numbers (essentially it means it has imaginary numbers).
Now I don't know exactly what you've been looking at and everything, but every polynomial is solvable. You're going by it using cubic/quartic/quintic formulas, which personally I don't know. But I think you should look at it at a different approach. A polynomial will have as many roots as it's highest degree. So a quintic equation will have five roots. Now, if the function crosses the xaxis 5 times, it will have 5 real roots. But if it crosses the xaxis fewer times than that, the rest of the roots are in the complex range. So if it has 3 real roots, there are 2 complex roots. There is nothing special about quintics that gets you into complex roots but it depends on the specific function you are looking at (x^2 + 1 = 0 is an example of a quadratic with only complex roots).
Also the function you gave is solvable. I put the function you gave into wolfram alpha and it came with 1 real root and 4 complex roots.
Anyways my suggestion is to don't bother with cubic/quartic/etc. formulas. You won't ever need them, and if you need to find the roots of any equation with a degree higher than 2, just factor it.
As for finding the complex roots of a function, I have no idea. I don't know much about complex numbers algebra, just a few basics with i.
Edit: Ok so I'm way off. Disregard what I posted
Last edited by 314man on Mon Oct 25, 2010 3:11 pm UTC, edited 1 time in total.
 phlip
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Re: Quintic Equations
314man wrote:To start things off, by radicals, I'm assuming you mean real numbers.
No, a "radical" is a squareroot sign [imath]\sqrt{\,}[/imath] or, more generally, the nthroot sign [imath]\sqrt[n]{\,}[/imath].
For instance, every root of a quadratic with rational coefficients can be written as rational numbers, your usual +*/, and square roots. And the quadratic formula tells you how.
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Re: Quintic Equations
Note that wolfram gives answers using [imath]\approx[/imath] rather than [imath]=[/imath]. I would suspect these were done numerically. If you're only interesting in numeric solutions then that's fine. And of course many such equations will be easily factorisable, so we can solve those cases too. However the OP wasn't saying these equations couldn't be solved. They were saying that a general equation for the solution, of a similar form to the quadratic, cubic and quartic equation, doesn't exist, and that this is an interesting observation in its own right. It also implies to me that many such equations are going to be difficult to 'just factor'.314man wrote:Also the function you gave is solvable. I put the function you gave into wolfram alpha and it came with 1 real root and 4 complex roots.
Anyways my suggestion is to don't bother with cubic/quartic/etc. formulas. You won't ever need them, and if you need to find the roots of any equation with a degree higher than 2, just factor it.
All I really have to add to the discussion is a wiki link to Abel's impossibility theorem.
 imatrendytotebag
 Posts: 152
 Joined: Thu Nov 29, 2007 1:16 am UTC
Re: Quintic Equations
As someone mentioned before, the proof involves what is known as Galois theory, which relates group theory to field theory.
The basic reason for "why 5?" is that the symmetric group on 5 elements is unsolvable, while the symmetric groups on 1,2,3,4 elements ARE solvable. Of course, if you've never seen a group before, or don't know what it means for a group to be "solvable", this answer doesn't really help.
The rough idea is as follows: You can think of taking the rational numbers Q and adding all of the roots of the polynomial, along with their sums/products. You'll get a new field (basically any set where you can do addition, subtraction, multiplication, and division by nonzero elements), let's call it E. This process is called "field extension" and you can kind of think of it in the same way you get the complex numbers from the real numbers by adding the square root of 1.
On the other hand, we can think of starting with Q and adding, say, the fifth root of 2, (and all sums and products), and we get some field say F1. We can continue by adding some nth root of some element in F1 to get a bigger field F2, and then again to get a bigger field F3, etc. So it is easy to see that the F1,F2, F3, and so on consist of elements which can be written with radical expressions.
The question is, if we continue adding, will the sequence F1, F2, F3, ... ever contain all of E? That is, will be be able to express the roots of the polynomial using radical expressions? In order to answer this question, you need to look at groups of symmetries. What does this mean? Well you can look at functions from say, F1 to F1, which preserve the operations and fix all the elements of Q. For instance, complex conjugation is a function from C to C (the complex numbers) which fixes all the elements of R. All of these functions form what is called a "group", in that you can compose them and reverse them. Now it turns out that when you look at the groups you get when adding a radical to a field, you get a very special class of groups, and the question is, sort of, can you "build up" groups like this to get the group of symmetries of E? This is the power of Galois theory: translating statements about fields into statements about groups.
Now think about the symmetries of E, fixing Q. It should seem reasonable that all the roots of the quintic should be sent to other roots of the quintic, and that where these roots get sent determines the map completely. Now there are up to 5 roots of this quintic, and so there are 5! permutations of these roots. So we can think of matching up symmetries of E to permutations of 5 elements. For some quintic polynomials, each of these permutations yields a valid symmetry of E fixing Q, so you have 5! of these symmetries, and the group they form is called the symmetry group on 5 elements (Also called S5). The statement that S5 is unsolvable is exactly what is necessary to show that E cannot be "built up" from groups of symmetries obtained by adding radical expressions.
Kudos if you made it through all of that without getting put off by my vagueness. Galois theory is a very deep and beautiful topic, and I couldn't hope to really explain it in just a few paragraphs. But hopefully it inspires you to dig deeper into the subject!
The basic reason for "why 5?" is that the symmetric group on 5 elements is unsolvable, while the symmetric groups on 1,2,3,4 elements ARE solvable. Of course, if you've never seen a group before, or don't know what it means for a group to be "solvable", this answer doesn't really help.
The rough idea is as follows: You can think of taking the rational numbers Q and adding all of the roots of the polynomial, along with their sums/products. You'll get a new field (basically any set where you can do addition, subtraction, multiplication, and division by nonzero elements), let's call it E. This process is called "field extension" and you can kind of think of it in the same way you get the complex numbers from the real numbers by adding the square root of 1.
On the other hand, we can think of starting with Q and adding, say, the fifth root of 2, (and all sums and products), and we get some field say F1. We can continue by adding some nth root of some element in F1 to get a bigger field F2, and then again to get a bigger field F3, etc. So it is easy to see that the F1,F2, F3, and so on consist of elements which can be written with radical expressions.
The question is, if we continue adding, will the sequence F1, F2, F3, ... ever contain all of E? That is, will be be able to express the roots of the polynomial using radical expressions? In order to answer this question, you need to look at groups of symmetries. What does this mean? Well you can look at functions from say, F1 to F1, which preserve the operations and fix all the elements of Q. For instance, complex conjugation is a function from C to C (the complex numbers) which fixes all the elements of R. All of these functions form what is called a "group", in that you can compose them and reverse them. Now it turns out that when you look at the groups you get when adding a radical to a field, you get a very special class of groups, and the question is, sort of, can you "build up" groups like this to get the group of symmetries of E? This is the power of Galois theory: translating statements about fields into statements about groups.
Now think about the symmetries of E, fixing Q. It should seem reasonable that all the roots of the quintic should be sent to other roots of the quintic, and that where these roots get sent determines the map completely. Now there are up to 5 roots of this quintic, and so there are 5! permutations of these roots. So we can think of matching up symmetries of E to permutations of 5 elements. For some quintic polynomials, each of these permutations yields a valid symmetry of E fixing Q, so you have 5! of these symmetries, and the group they form is called the symmetry group on 5 elements (Also called S5). The statement that S5 is unsolvable is exactly what is necessary to show that E cannot be "built up" from groups of symmetries obtained by adding radical expressions.
Kudos if you made it through all of that without getting put off by my vagueness. Galois theory is a very deep and beautiful topic, and I couldn't hope to really explain it in just a few paragraphs. But hopefully it inspires you to dig deeper into the subject!
Hey baby, I'm proving love at nth sight by induction and you're my base case.

 Posts: 165
 Joined: Wed Apr 11, 2007 10:17 am UTC
Re: Quintic Equations
Not being solveable by radicals means that there is no way to express the number using addition, multiplication and iterated roots (of any degree) in any combination. A simple way to still express the solutions are to say "a root of this polynomial" which defines it just fine. Now if you're wondering "how is that a good definition, aren't there several roots for the polynomial?", then this is actually a good remark. However, if the polynomial is irreducible, in laymens terms, if you cannot write it as the product of two smaller polynomials, then there isn't any way to distinguish the different roots (over the field your polynomial lives in, in many cases this is the RATIONAL NUMBERS, for polynomials over the real numbers the largest irreducible polynomials are quadratic). For example, the polynomial x^2+1 is irreducible over the real numbers, and it has the well known imaginary unit i as its root (in fact, it's a simple way to DEFINE i). However, the other root is i, and if you play around with it some you'll see that i has all the same properties that i seems to have. In fact, you can't find a way to distinguish the two when expressing things in real numbers (like, for instance, the root of this polynomial). So which one is i and which one is i then? Well, I don't know, and neither can you. The thing is, it doesn't make a difference, they behave the same.
Now the issue with quintic polynomials is a case of galoistheory as was mentioned before, although I don't consider it a property of the polynomials per se. The topic of Galoistheory, in short, is considering fields (which, in case you don't know the term, are sets that behave like numbers, in a sense, where you can add and multiply the same way you can in the real numbers, with the known properties of distributivity, commutativity etc.) where irreduicble polynomials have solutions, for instance considering the Complex Numbers over the Real Numbers as x^2+1 has the two roots i and i in it. Certain field extensions of this kind have an associated finite group (basically, something where you have one operation, think of the integers only with addition as an example of a group) called the Galoisgroup which states a lot about the extension you consider. The ability to express a number in such an extension with radicals is equal to the abstract property of this group being solvable, which I can't express in simple terms. This was first discovered by Galois, hence the name. All small groups have this property, (iirc all with 59 or less elements) but once you consider polynomials of degree 5 or larger the field extensions you need to have all of their roots in them get big enough that the corresponding Galoisgroup is large enough that it can possibly be unsolvable. That's the reason there cannot be a formula for quintic polynomials: Some of them result in Galoisgroups that have too complex of a structure to be solvable. There is nothing particularly special about the number 5.
I don't think this is the right topic to go into details about all of this, although you are free to ask further questions. The reason for this is pretty abstract, but rest assured that there is a very specific fundamental reason you cannot express all roots of quintic polynomials in radicals.
Bah, now I got ninja'd, guess you have several explanations now to get an idea of the topic.
Now the issue with quintic polynomials is a case of galoistheory as was mentioned before, although I don't consider it a property of the polynomials per se. The topic of Galoistheory, in short, is considering fields (which, in case you don't know the term, are sets that behave like numbers, in a sense, where you can add and multiply the same way you can in the real numbers, with the known properties of distributivity, commutativity etc.) where irreduicble polynomials have solutions, for instance considering the Complex Numbers over the Real Numbers as x^2+1 has the two roots i and i in it. Certain field extensions of this kind have an associated finite group (basically, something where you have one operation, think of the integers only with addition as an example of a group) called the Galoisgroup which states a lot about the extension you consider. The ability to express a number in such an extension with radicals is equal to the abstract property of this group being solvable, which I can't express in simple terms. This was first discovered by Galois, hence the name. All small groups have this property, (iirc all with 59 or less elements) but once you consider polynomials of degree 5 or larger the field extensions you need to have all of their roots in them get big enough that the corresponding Galoisgroup is large enough that it can possibly be unsolvable. That's the reason there cannot be a formula for quintic polynomials: Some of them result in Galoisgroups that have too complex of a structure to be solvable. There is nothing particularly special about the number 5.
I don't think this is the right topic to go into details about all of this, although you are free to ask further questions. The reason for this is pretty abstract, but rest assured that there is a very specific fundamental reason you cannot express all roots of quintic polynomials in radicals.
Bah, now I got ninja'd, guess you have several explanations now to get an idea of the topic.

 Posts: 224
 Joined: Tue Jun 17, 2008 11:04 pm UTC
Re: Quintic Equations
Some of this might be inaccurate or oversimplified but here's the idea. In Galois theory, you consider fields and their extensions and automorphisms.
A field is simply a space where you can do addition and multiplication and where they obey most of the rules you'd expect them to (every nonzero number has a multiplicative inverse, multiplication distributes over addition, etc). Examples of fields are the rational numbers, the real numbers, or the complex numbers. You can also have finite fields, such as [imath]Z_3[/imath], which is a field with 3 elements {0,1,2}, where you add and multiply as usual, but every time you get a number 3 or greater, you subtract 3 until you are less than 3 again. So, 2x2 = 1 in [imath]Z_3[/imath]. Another example of a less orthodox field is the set of all ratios of polynomials, such as (x^2+1)/(3x^31).
Given a field, one thing you can do is take a polynomial p and ask "does this p have a root in this field?" For example, x^22 does not have a root in [imath]Q[/imath], the rational numbers, because there is no rational number whose square is 2.
What if we just add it by brute force? Let's add an new element "a" to the rationals and declare by fiat that a^2  2 = 0. This is called a field extension. Notationally, we write it [imath]Q[a]/(a^22)[/imath]. So, we can now do our usual field operations, but we have this new element a that we can play with. So, we can do things like (a+4)*a = a^2+4a, and because a^2 =2, so (a+4)*a = 2 + 4a.
We may notice some funny things about this field extension. It turns out that if we replace every "a" with "a", everything remains the same! That is, every possible operation we could do adding and multiplying things involving a will behave the same as if we just used a. For example a^4 = 4, but also (a)^4 = 4.
This is not always the case. For instance, if we look at [imath]Q[b]/(b^3+1)[/imath], then b^3 = 1, but (b)^3 = 1 , so swapping b and b here actually affects things. However, in this case, we might observe that replacing every b with b^2 doesn't affect anything! For example, b^3 = 1, and also (b^2)^3 = b^6 = b^3*b^3 = 1
These replacements that don't affect anything are called "automorphisms" of the field. We might ask: Given a field extension, what are the possible automorphisms? Clearly they differ in different cases.
It turns out that every automorphism will be a permutation of the roots of your polynomial (this is not obvious). For instance, the roots of b^3+1 are 1, b, b^2, and indeed swapping b with b^2 was one of our automorphisms. Similarly, both a and a are roots of a^22, and swapping them was an automorphism. Of course, not every possible permutation gives an automorphism  swapping 1 and b will affect things. But in any case, especially with higher degree polynomials, we will in general have many possible automorphisms, giving us many possible permutations of the roots.
Now, let's back away and consider permutations. We can combine permutations by composing them. For instance, if we have the permutation that sends four elements (a,b,c,d) > (d,a,b,c), and we have the permutation that swaps the last two elements, then their composition is another permutation (d,a,c,b).
A set of permutations, also called a "group" is "cyclic" if there is one single permutation that if repeatedly composed with itself will give you every permutation in that group. For example, if we have 3 elements (a,b,c), then the group consisting of {0 = doing nothing, 1 = rotating a>b>c>a, 2 = rotating c>b>a>c} is cyclic, because doing 1 twice will get you 2, and doing it 1 three times will get you 0.
For an example of a group that is not cyclic, take 4 elements (a,b,c,d), and the permutations consisting of {doing nothing, swapping (a,b), swapping (c,d), swapping both (a,b) and (c,d)}. Then, this is not cyclic  no single permutation when repeatedly applied will get you all of the others.
However, in this case, it happens to be "built" of cyclic groups. We can think of it "built" out of the two separate groups {doing nothing, swapping (a,b)} and {doing nothing, swapping (c,d)}, both of which are individually cyclic. There is a precise mathematical sense, which I won't go into, defining what exactly what this means, but hopefully you get the intuitive idea.
Can every group be built out of smaller cyclic groups? No! The smallest group that cannot involves 60 permutations of 5 elements, and is called the "Alternating group on 5 elements", or "A5". But every group of permutations of 4 elements CAN be built out of smaller cyclic groups.
How does this relate to solving polynomials with radicals? Well, notice that when you take square roots or cube roots, or nth roots, you are essentially making a field extension. When you write down an expression like [imath]\sqrt{2}/2 + 6\sqrt[3]{5}[/imath], you are essentially working with an element in the field extension [imath]Q[a,b]/(a^22)/(b^35)[/imath], and talking about the element a/2 + 6b.
It turns out that when you take an nth root and look at the automorphisms of the resulting field extension, the group of permutations for those automorphisms will *always* be cyclic. So by taking repeated square, cube roots, etc, you can only ever build the groups of permutations that can be built out of cyclic sets of permutations.
However, some field extensions have automorphisms whose permutation set is not buildable this way. In particular, [imath]Q[x]/(x^5x+1)[/imath] will give you a group of permutations that cannot be built from cyclic groups. It gives you a group called "S5" with 120 permutations, and the best you can do is break it down into a cyclic group (with 2 permutations) combined with "A5" (which has 60 permutations), and then you're stuck, because A5 can't be broken down any further. Hence, it is impossible to express any solution to x^5x+1 in terms of radicals.
This only happens once you get to degree 5 and higher, because only degree 5 and higher polynomials have 5 or more roots, and you need at least 5 things to permute before you start getting groups of permutations that can't be built from cyclic groups.
A field is simply a space where you can do addition and multiplication and where they obey most of the rules you'd expect them to (every nonzero number has a multiplicative inverse, multiplication distributes over addition, etc). Examples of fields are the rational numbers, the real numbers, or the complex numbers. You can also have finite fields, such as [imath]Z_3[/imath], which is a field with 3 elements {0,1,2}, where you add and multiply as usual, but every time you get a number 3 or greater, you subtract 3 until you are less than 3 again. So, 2x2 = 1 in [imath]Z_3[/imath]. Another example of a less orthodox field is the set of all ratios of polynomials, such as (x^2+1)/(3x^31).
Given a field, one thing you can do is take a polynomial p and ask "does this p have a root in this field?" For example, x^22 does not have a root in [imath]Q[/imath], the rational numbers, because there is no rational number whose square is 2.
What if we just add it by brute force? Let's add an new element "a" to the rationals and declare by fiat that a^2  2 = 0. This is called a field extension. Notationally, we write it [imath]Q[a]/(a^22)[/imath]. So, we can now do our usual field operations, but we have this new element a that we can play with. So, we can do things like (a+4)*a = a^2+4a, and because a^2 =2, so (a+4)*a = 2 + 4a.
We may notice some funny things about this field extension. It turns out that if we replace every "a" with "a", everything remains the same! That is, every possible operation we could do adding and multiplying things involving a will behave the same as if we just used a. For example a^4 = 4, but also (a)^4 = 4.
This is not always the case. For instance, if we look at [imath]Q[b]/(b^3+1)[/imath], then b^3 = 1, but (b)^3 = 1 , so swapping b and b here actually affects things. However, in this case, we might observe that replacing every b with b^2 doesn't affect anything! For example, b^3 = 1, and also (b^2)^3 = b^6 = b^3*b^3 = 1
These replacements that don't affect anything are called "automorphisms" of the field. We might ask: Given a field extension, what are the possible automorphisms? Clearly they differ in different cases.
It turns out that every automorphism will be a permutation of the roots of your polynomial (this is not obvious). For instance, the roots of b^3+1 are 1, b, b^2, and indeed swapping b with b^2 was one of our automorphisms. Similarly, both a and a are roots of a^22, and swapping them was an automorphism. Of course, not every possible permutation gives an automorphism  swapping 1 and b will affect things. But in any case, especially with higher degree polynomials, we will in general have many possible automorphisms, giving us many possible permutations of the roots.
Now, let's back away and consider permutations. We can combine permutations by composing them. For instance, if we have the permutation that sends four elements (a,b,c,d) > (d,a,b,c), and we have the permutation that swaps the last two elements, then their composition is another permutation (d,a,c,b).
A set of permutations, also called a "group" is "cyclic" if there is one single permutation that if repeatedly composed with itself will give you every permutation in that group. For example, if we have 3 elements (a,b,c), then the group consisting of {0 = doing nothing, 1 = rotating a>b>c>a, 2 = rotating c>b>a>c} is cyclic, because doing 1 twice will get you 2, and doing it 1 three times will get you 0.
For an example of a group that is not cyclic, take 4 elements (a,b,c,d), and the permutations consisting of {doing nothing, swapping (a,b), swapping (c,d), swapping both (a,b) and (c,d)}. Then, this is not cyclic  no single permutation when repeatedly applied will get you all of the others.
However, in this case, it happens to be "built" of cyclic groups. We can think of it "built" out of the two separate groups {doing nothing, swapping (a,b)} and {doing nothing, swapping (c,d)}, both of which are individually cyclic. There is a precise mathematical sense, which I won't go into, defining what exactly what this means, but hopefully you get the intuitive idea.
Can every group be built out of smaller cyclic groups? No! The smallest group that cannot involves 60 permutations of 5 elements, and is called the "Alternating group on 5 elements", or "A5". But every group of permutations of 4 elements CAN be built out of smaller cyclic groups.
How does this relate to solving polynomials with radicals? Well, notice that when you take square roots or cube roots, or nth roots, you are essentially making a field extension. When you write down an expression like [imath]\sqrt{2}/2 + 6\sqrt[3]{5}[/imath], you are essentially working with an element in the field extension [imath]Q[a,b]/(a^22)/(b^35)[/imath], and talking about the element a/2 + 6b.
It turns out that when you take an nth root and look at the automorphisms of the resulting field extension, the group of permutations for those automorphisms will *always* be cyclic. So by taking repeated square, cube roots, etc, you can only ever build the groups of permutations that can be built out of cyclic sets of permutations.
However, some field extensions have automorphisms whose permutation set is not buildable this way. In particular, [imath]Q[x]/(x^5x+1)[/imath] will give you a group of permutations that cannot be built from cyclic groups. It gives you a group called "S5" with 120 permutations, and the best you can do is break it down into a cyclic group (with 2 permutations) combined with "A5" (which has 60 permutations), and then you're stuck, because A5 can't be broken down any further. Hence, it is impossible to express any solution to x^5x+1 in terms of radicals.
This only happens once you get to degree 5 and higher, because only degree 5 and higher polynomials have 5 or more roots, and you need at least 5 things to permute before you start getting groups of permutations that can't be built from cyclic groups.
Last edited by lightvector on Mon Oct 25, 2010 7:34 am UTC, edited 1 time in total.
Re: Quintic Equations
Quintic equations in general can't be solved by radicals, this is correct. But there's no law saying you have to solve equations by radicals.
If, in addition to radicals, you admit (some branch of) the function [imath]\rho[/imath] whose inverse is given by [imath]\rho^{1}(x) = x^5 + x[/imath], you can solve any quintic equation. For example, one solution to your equation, [imath]x^5  x + 1 = 0[/imath], is [math]x = e^{i\pi/4}\rho\left(e^{i\pi/4}\right),[/math] or equivalently [math]\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\rho\left(\frac{\sqrt{2}}{2}i\frac{\sqrt{2}}{2}\right).[/math] And of course once you have one solution, you can factor: letting [imath]\alpha[/imath] be the solution above, [math]x^5 + x  1 = \left(x  \alpha\right) \left(x^4 + \alpha x^3 + \alpha^2 x^2 + \alpha^3 x + \left(\alpha^4  1\right)\right).[/math] At that point, if you know how to deal with quartics, you can find the rest of the solutions.
If, in addition to radicals, you admit (some branch of) the function [imath]\rho[/imath] whose inverse is given by [imath]\rho^{1}(x) = x^5 + x[/imath], you can solve any quintic equation. For example, one solution to your equation, [imath]x^5  x + 1 = 0[/imath], is [math]x = e^{i\pi/4}\rho\left(e^{i\pi/4}\right),[/math] or equivalently [math]\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\rho\left(\frac{\sqrt{2}}{2}i\frac{\sqrt{2}}{2}\right).[/math] And of course once you have one solution, you can factor: letting [imath]\alpha[/imath] be the solution above, [math]x^5 + x  1 = \left(x  \alpha\right) \left(x^4 + \alpha x^3 + \alpha^2 x^2 + \alpha^3 x + \left(\alpha^4  1\right)\right).[/math] At that point, if you know how to deal with quartics, you can find the rest of the solutions.
Last edited by SunAvatar on Mon Oct 25, 2010 9:43 am UTC, edited 3 times in total.
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neque defensori dominus,
nec pater nec pater,
nihil supernum.
 jestingrabbit
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Re: Quintic Equations
Just to put a link to what SunAvatar said.
http://en.wikipedia.org/wiki/Bring_radical
And really, given that things like Aberth's algorithm work, exact isn't practically better, just theoretically interesting.
http://en.wikipedia.org/wiki/Aberth_method
http://en.wikipedia.org/wiki/Bring_radical
And really, given that things like Aberth's algorithm work, exact isn't practically better, just theoretically interesting.
http://en.wikipedia.org/wiki/Aberth_method
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Re: Quintic Equations
SunAvatar wrote:And of course once you have one solution, you can factor: letting [imath]\alpha[/imath] be the solution above, [math]x^5 + x  1 = \left(x  \alpha\right) \left(x + \alpha x + \alpha^2 x^2 + \alpha^3 x^3 + \left(\alpha^4  1\right)\right).[/math] At that point, if you know how to deal with quartics, you can find the rest of the solutions.
Neat trick, factoring a quintic into a cubic times a linear factor. Actually, it should be:
[math]x^5  x + 1 = \left(x  \alpha\right) \left(x^4 + \alpha x^3 + \alpha^2 x^2 + \alpha^3 x + \left(\alpha^4  1\right)\right).[/math]
You can see this is correct because if you do the multiplication, you get
[math]\left(x^5 + \alpha x^4 + \alpha^2 x^3 + \alpha^3 x^2 + \left(\alpha^4  1\right)x\right)\left(\alpha x^4 + \alpha^2 x^3 + \alpha^3 x^2 + \alpha^4 x + \left(\alpha^5  \alpha\right)\right)=x^5x(\alpha^5\alpha),[/math]
and \(\alpha^5\alpha=1\) since \(\alpha\) is a root of \(x^5x+1\).
Also, e^{iπ/4}≠√2+i√2.
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"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
Re: Quintic Equations
Oh for Christ's sake.
Maybe I should proofread for more than TeX syntax before I post, once in a while.
Maybe I should proofread for more than TeX syntax before I post, once in a while.
Non est salvatori salvator,
neque defensori dominus,
nec pater nec pater,
nihil supernum.
neque defensori dominus,
nec pater nec pater,
nihil supernum.

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Re: Quintic Equations
Many posters above have given beautiful introductions to Galois theory. If you want a short (but not terribly illuminating) answer, it's that the alternating group on five elements is simple.
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