- #1

ATCG

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(If you get the equation solve 4x^3+2x^2+5x-2)

Cubic Formula

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- Thread starter ATCG
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- #1

ATCG

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(If you get the equation solve 4x^3+2x^2+5x-2)

Cubic Formula

- #2

Hurkyl

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The cubic formula was one of the most important discoveries in mathematics... not so much for what it was able to solve, but for the way it influenced attitudes.

First off, it was something the ancient mathematicians had not been able to do. This was a terrific ego boost to mathematicians of this era when they finally realized they weren't just struggling to understand and rediscover what the ancients already knew, but they really and truely were developing

Secondly, the use of negative and imaginary numbers are

- #3

HallsofIvy

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Very well said, Hurkyl, an important point.

- #4

MathematicalPhysicist

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thanks.

- #5

Hurkyl

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The proof is not an easy one; the one I know requires developing a fair amount of Galois theory...

- #6

mathwonk

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then the key idea is to observe that this number system has a certain symmetry that the rationals do not. I.e. you can interchange a+bsqrt(2) with a-bsqrt(2) and the structure is essentially preserved as before. i.e. sums go to sums and products go to products.

this is the only non trivial symmetry of this number system.

to solve a more complicated equation like the cubic x^3 = 2, one must throw in a cube root, a number like cubert(2), obtaining a number system consisting of numbers of form a+ b.cubert(2) + c.cubert(9). Notice this system is analogous to a 3 dimensiopnal vector space.

this new system has more symmetries than does the previous one.

Galois showed that every time you adjoin a root of a previously known number, then the symmetries of the new system can be described in a predictable way. the point is that symmetries can be composed to form a sort of multiplication. the key property is the commutativity of the composition, or "multiplication", in the family of symmetries.

Now one can also study the structure of the symmetries of any new system of numbers whether they arise from adding in roots or not, i.e. whether or not they arise from adding in the kinds of numbers that occur in solution formulas for equations.

the question was whether the set of solutions of some equations of degree 5 might not actually be expressible in terms of solution formulas involving only roots.

indeed he showed that the solutions of a general equation of degree 5 formed a number system whose symmetries were essentially non commutaive, hence could not have occurred from any ordinary solution formulas invoilving only radicals, i.e. from adding in successively just the roots of formerly known numbers.

it follws that a general polynomial of mdegree 5 has solutions which cannot be obtained from the complex numbers by adding in successively the roots of previously obtained numbers.

i.e. they cannot be expressed in terms of solution formulas of the usual form, involving roots and radicals and other arithmetic operations.

an example of such a polynomial is X^5 - 80X + 2.

the geometric realization of the symmetries of this equation is essentially set of the rotational symmetries of the icosahedron. the numerical properties of the edges, faces, and vertices, of an icosahedron imply that the set of symmetries is not composed of any smaller collections of simpler symmetries.

by the way, the cubic formula for the polynomial x^3+px+q =0, was published by Cardano in 1545, as follows.

x = (1/3)[{(-27q/2)+(3/2)(-3D)^1/2}^1/3 - 3p/{(-27q/2) + (3/2)(-3D)^1/2}^1/3].

where D = (x1- x2)^2(x1-x3)^2(x2-x3)^2 = -4p^3-27q^2, and we have fixed the square root (-3D)^1/2. Then vary the cube root to get all three solutions.

For example, in the equation x^3-1 = 0, p = 0, q = -1, D = -27, so we get x = (1/3){27/2 + 27/2}^1/3 = {1/2 + 1/2}1/3 = 1^1/3 = 1, as hoped.

Indeed, for x^3-a = 0, we have p = 0, q = -a, D = -27a^2, and hence

x = (1/3){27a/2 + (3/2)(81a2)1/2}^1/3 = (1/3){27a}^1/3 = a^1/3.

For x^3 - 4x = 0, we get p = -4, q = 0, D = 256, and so

x = (1/3) [{(3/2)(-768)1/2}^1/3 + 12/{(3/2)(-768)^1/2}^1/3]

= (1/3) [ {(-27)(64)}^1/6 + 12/{(-27)(64)}^1/6 ]

= (1/3) [ 2¯3 i1/3 + 12/{2¯3 i1/3} ] = (2/¯3 )( i1/3 + i-1/3)

= (4/¯3 )Re(i1/3). Varying the cube roots gives (4/¯3 )(cos(<pi>/6)) = (4/¯3 )(¯3 /2) = 2, (4/¯3 )(cos(5<pi>/6)) = (4/¯3 )(-¯3 /2) = -2, and (4/¯3 )(cos(9<pi>/6)) = (4/¯3 )(0) = 0.

Of course, factoring x^3-4x = x(x-2)(x+2) = 0 confirms these answers.

(Notice a point which fascinated earlier workers, who were not entirely happy with "imaginary" numbers: the solution formula involves imaginaries even though the final answer it gives is real! As mentioned above it can be proved that this cannot be avoided.)

- #7

matt grime

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commutativity is not important, though explaining what is might be the reason why you slipped in this white lie.

"the geometric realization of the symmetries of this equation is essentially set of the rotational symmetries of the icosahedron. the numerical properties of the edges, faces, and vertices, of an icosahedron imply that the set of symmetries is not composed of any smaller collections of simpler symmetries."

not sure i can agree with that: the set of symmetries is generated by the set of strictly smaller subgroups of order 3, but again this might be one of those white lies to hide the reader from the full ideas of normality and solvability.

- #8

mathwonk

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As to the next item, I used the word "composed" very loosely, to mean there is no normal subgroup of A5. It was poorly said though. Thanks

- #9

mathwonk

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In a little more detail, we may try to understand a group G in terms of another smaller or simpler group H by finding a surjective homomorphism from G onto H.

If there exists such a homomorphism G-->H with kernel K, i.e. such that K consists of those elements which go to zero in H, then we may say in some sense that G is composed of K and H, combined in some way.

(Actually I do not know how to make the way G is obtained from K and H any more precise, except in the case where there is also some subgroup H' of G which actually maps isomorphically onto H by this map. In that case G is the "semi direct product" of K and H.)

For example, we can let the group G of rotations of a cube centered at the origin, act on the three mutually perpendicular axes through its center, by permuting them. Then the group G of 24 rotations, is represented in this way on the permutation group H of order 6 of the three axes, with kernel of order 4.

A "simple" group is a group (other than {0}) that cannot be broken down in this way, i.e. such that every homomorphism of G onto any other group H is either an isomorphism, or is identically zero.

All (finite) groups G may be thought of as "composed" of simple groups in the following sense.

Find a simple group H1 and a homomorphism of G onto H1, with kernel K1.

Put H1 aside as one of the simple constituents of G, and focus now on K1.

In the same way find another simple group H2 and a homomorphism of K1 onto H2, with kernel K2.

Set H2 aside as a second simple constituent of H, and focus on K2.

etc.....

In this way one obtains a finite set of simple constituents of G, which are actually unique, i.e. independent of the choices made in finding them.

E.g. the group of 24 rotations of a cube breaks down in this way to three copies of the cyclic group of order 2 and one copy of the cyclic group of order 3, all commutative.

Then what was said above about solving polynomials, more precisely was that a polynomial with rational coefficients, has a classical solution formula, if and only if the simple constituents of its Galois group are all commutative.

In the case of the general polynomial of degree 5, the Galois group has two simple constituents, namely the cyclic group of order 2, and the rotation group of the icosahedron which is not commutative, so the polynomial is not solvable.

The "reason" solution formulas, were always found in lower degrees than 5, from this point of view is that the group of the icosahedron is the first (i.e. smallest) non commutative simple group.

In this sense, I guess to me commutativity is the essential point underlying the theory, but I confess that was not at all clear from my failure to make precise my use of the words "composed of".

Matt's following post is quite right in correcting an error I made in my previous post. This time I think I got it right, eh Matt?

If there exists such a homomorphism G-->H with kernel K, i.e. such that K consists of those elements which go to zero in H, then we may say in some sense that G is composed of K and H, combined in some way.

(Actually I do not know how to make the way G is obtained from K and H any more precise, except in the case where there is also some subgroup H' of G which actually maps isomorphically onto H by this map. In that case G is the "semi direct product" of K and H.)

For example, we can let the group G of rotations of a cube centered at the origin, act on the three mutually perpendicular axes through its center, by permuting them. Then the group G of 24 rotations, is represented in this way on the permutation group H of order 6 of the three axes, with kernel of order 4.

A "simple" group is a group (other than {0}) that cannot be broken down in this way, i.e. such that every homomorphism of G onto any other group H is either an isomorphism, or is identically zero.

All (finite) groups G may be thought of as "composed" of simple groups in the following sense.

Find a simple group H1 and a homomorphism of G onto H1, with kernel K1.

Put H1 aside as one of the simple constituents of G, and focus now on K1.

In the same way find another simple group H2 and a homomorphism of K1 onto H2, with kernel K2.

Set H2 aside as a second simple constituent of H, and focus on K2.

etc.....

In this way one obtains a finite set of simple constituents of G, which are actually unique, i.e. independent of the choices made in finding them.

E.g. the group of 24 rotations of a cube breaks down in this way to three copies of the cyclic group of order 2 and one copy of the cyclic group of order 3, all commutative.

Then what was said above about solving polynomials, more precisely was that a polynomial with rational coefficients, has a classical solution formula, if and only if the simple constituents of its Galois group are all commutative.

In the case of the general polynomial of degree 5, the Galois group has two simple constituents, namely the cyclic group of order 2, and the rotation group of the icosahedron which is not commutative, so the polynomial is not solvable.

The "reason" solution formulas, were always found in lower degrees than 5, from this point of view is that the group of the icosahedron is the first (i.e. smallest) non commutative simple group.

In this sense, I guess to me commutativity is the essential point underlying the theory, but I confess that was not at all clear from my failure to make precise my use of the words "composed of".

Matt's following post is quite right in correcting an error I made in my previous post. This time I think I got it right, eh Matt?

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- #10

matt grime

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For a group to be solvable/soluble it is not the subgroups that must be abelian but the quotients by the (normal) subgroups that need to be abelian (error in post 8). (Counter) Example

1 < V < A_4 <S_4

where V is the vier gruppe.

zero being the identity implies your are talking about (additively written) abelian groups when you need not be.

The composition factors are only unique up to reordering (eg, C_6 has two towers with distinct terms whose quotients are abelian and simple)

1 < V < A_4 <S_4

where V is the vier gruppe.

zero being the identity implies your are talking about (additively written) abelian groups when you need not be.

The composition factors are only unique up to reordering (eg, C_6 has two towers with distinct terms whose quotients are abelian and simple)

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- #11

mathwonk

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- #12

matt grime

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Yep, realized that after I posted, was about to remove that one, but instead I'll post an apology.

- #13

mathwonk

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I think I owe you one Matt. Doesn't the word "set" also ignore repetitions? I need those, but not ordering.

So what is the right word for an unordered set with repetitions?

Just what you said I guess. Thanks again!

So what is the right word for an unordered set with repetitions?

Just what you said I guess. Thanks again!

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- #14

matt grime

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- #15

mathwonk

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maybe a little over the edge from my usual sloppy style of writing.

- #16

matt grime

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it need only be indexed by the set of prime numbers, after all most simple groups are not soluble.

- #17

mathwonk

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- #18

matt grime

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[tex]1=G_0 < G_1 < .... G_n = G[/tex] and each quotient [tex]G_i/G_{i-1}[/tex] is abelian. It is certainly not necessary that the quotients are "essentially unique"

- #19

HallsofIvy

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- #20

matt grime

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