- #1
swampwiz
- 571
- 83
The discriminant for a monic polynomial having only the free term { a0 } (and the monic degree term of course) is:
Δn:0 = σ0 nn a0( n - 1 )
where n is the degree of the polynomial, and σk is a sign (i.e., +1 or -1)
while the discriminant for a monic polynomial having only the x1 term (and the monic degree term again) is:
Δn:1 = σ1 ( n - 1 )n a1n
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 { ( b2 - 4 c ) is an example.
The solutions for the free-term polynomial is:
χn:0:k = ωnk ( - a0 )( 1 / n ) = ωn:k ( - { σ0 Δn:0(1 / [ n - 1 ] ) } )( 1 / n )
where ωn: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 x1 term are:
χn:1:k = ω[ n - 1 ]j ( - a1 )( 1 / [ n - 1 ] ) = ω[ n - 1 ]:j ( - { σ1 Δn: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
χn:*:j,k = Ψk + C [ ωnj ( Ψj + ω[ n - 1 ]k Δn( 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:
χNEST:n:*:m = ω[ n { n - 1 } ]m Δn( 1 / [ n { n - 1 } ] )
so the nesting degree product {and prime factors} for various n are;
p2 = 2 -> { 2 }
p3 = 6 -> { 2 , 3 }
p4 = 12 -> { 2 , 2 , 3 }
p5 = 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
Δn:0 = σ0 nn a0( n - 1 )
where n is the degree of the polynomial, and σk is a sign (i.e., +1 or -1)
while the discriminant for a monic polynomial having only the x1 term (and the monic degree term again) is:
Δn:1 = σ1 ( n - 1 )n a1n
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 { ( b2 - 4 c ) is an example.
The solutions for the free-term polynomial is:
χn:0:k = ωnk ( - a0 )( 1 / n ) = ωn:k ( - { σ0 Δn:0(1 / [ n - 1 ] ) } )( 1 / n )
where ωn: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 x1 term are:
χn:1:k = ω[ n - 1 ]j ( - a1 )( 1 / [ n - 1 ] ) = ω[ n - 1 ]:j ( - { σ1 Δn: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
χn:*:j,k = Ψk + C [ ωnj ( Ψj + ω[ n - 1 ]k Δn( 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:
χNEST:n:*:m = ω[ n { n - 1 } ]m Δn( 1 / [ n { n - 1 } ] )
so the nesting degree product {and prime factors} for various n are;
p2 = 2 -> { 2 }
p3 = 6 -> { 2 , 3 }
p4 = 12 -> { 2 , 2 , 3 }
p5 = 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
) that when a pair of solutions are swapped (i.e., and with no path crossing), the discriminant does a wind around the origin (I proved this on my own as well) - and also, this gave me a reason as to why commutators of permutations have to be used to unwind a nested root - although the permutations are not in strict inverse order (i.e., A B B-1 A-1 ) for a commutator (i.e., A B A-1 B-1 ), since each permutation induces some net number of winds (and this number can be any number, with the caveat that each permutation can only have an odd or even number of binary swaps), so the commutator concatenation nulls out the winds since winding is additive, and of course, addition is commutative. Without this idea, I could not understand what was so special about the commutator, even though it works.