- #1

swampwiz

- 571

- 83

_{0}} (and the monic degree term of course) is:

Δ

_{n:0}= σ

_{0}n

^{n}a

_{0}

^{( n - 1 )}

where n is the degree of the polynomial, and σ

_{k}is a sign (i.e., +1 or -1)

while the discriminant for a monic polynomial having only the x

^{1}term (and the monic degree term again) is:

Δ

_{n:1}= σ

_{1}( n - 1 )

^{n}a

_{1}

^{n}

This can be seen as the singular term for the discriminant that has only the free term for the known expressions for degree 2, 3 & 4 - and can be proven for any degree as well (simple arithmetic for a regular polygon having chords between all vertices). The familiar quadratic discriminant of { ( b

^{2}- 4 c ) is an example.

The solutions for the free-term polynomial is:

χ

_{n:0:k}= ω

_{n}

^{k}( - a

_{0})

^{( 1 / n )}= ω

_{n:k}( - { σ

_{0}Δ

_{n:0}

^{(1 / [ n - 1 ] )}} )

^{( 1 / n )}

where ω

_{n:k}are the n-th DeMoivre roots of unity

while the solutions (except for the singular solution of

*0*) for the polynomial having only the x

^{1}term are:

χ

_{n:1:k}= ω

_{[ n - 1 ]}

^{j}( - a

_{1})

^{( 1 / [ n - 1 ] )}= ω

_{[ n - 1 ]:j}( - { σ

_{1}Δ

_{n:1}

^{(1 / n )}} )

^{( 1 / [ n - 1 ] )}

and thus if there is proper continuous function for the solutions, it must have a term that is the nested DeMoivre roots of the discriminant

χ

_{n:*:j,k}= Ψ

_{k}+ C [ ω

_{n}

^{j}( Ψ

_{j}+ ω

_{[ n - 1 ]}

^{k}Δ

_{n}

^{( 1 / [ n - 1 ] )})

^{( 1 / n )}]

where C is some complex constant

Nested DeMoivre roots are both "commutative" (i.e., the order can be permutation) and "associative" (i.e., any root degree can be decomposed into further nesting of any factors of the root degree), and so the nesting is equivalent to a net DeMoivre root of:

χ

_{NEST:n:*:m}= ω

_{[ n { n - 1 } ]}

^{m}Δ

_{n}

^{( 1 / [ n { n - 1 } ] )}

so the nesting degree product {and prime factors} for various n are;

p

_{2}= 2 -> { 2 }

p

_{3}= 6 -> { 2 , 3 }

p

_{4}= 12 -> { 2 , 2 , 3 }

p

_{5}= 20 -> { 2 , 2 , 5 }

Now turning to how to generate a solution function, the technique is to first do a substitution of variable to eliminate the penultimate term and then do a bunch of finagling to get a resolvent polynomial of one degree less, which after being solved is plugged back into the finagled expression. The reason that the resolvent is one degree less (and not any more than that) is that the monic polynomial has

*n*different coefficients, and thus once the penultimate term is reduced out, there are

*{ n - 1 }*coefficients, and so it might be possible to finagle it into a general (monic) polynomial of one degree less.

Indeed, the solutions for degrees 2, 3 & 4 are of the form:

DEGREE 2 : a square root of the discriminant

DEGREE 3 : a cubic root of an expression that contains a resolvent solution of degree 2, and thus the cubic root of a square root of the discriminant

DEGREE 4: a square root of an expression that contains a resolvent solution of degree 3, and thus the square root of a cubic root of a square root of the discriminant

which is as per what has already been proven to be the case

Thus, a solution for degree 5 would need to be some root of an expression that contains a resolvent of degree 4, and since the quintic root must be part of the solution (i.e., because it is a prime component of the net DeMoivre root of 20), it must be the quintic root of either a square root of a square root, or a quartic root, of the discriminant (NOTE: the order of the nesting can be anything); however, the quintic solution must have a resolvent solution of the quartic, and so it would need to have the net DeMoivre root of { 2 , 2 , 3 , 5 }, which is impossible because it must be { 2 , 2 , 5 }.

It should also be said that the Arnold proof of the impossibility of the quintic - which is the motivation for this analysis - has the result that the number of permutations that cannot be commutated out is 60, which exactly matches the degree of the DeMoivre root that would be required for a solution.

QED