3 ms·
Gonna butcher this, but it's good practice, so I'm gonna take a stab at it. The roots of polynomials are intimately related to the Alternating symmetry groups.
by kortex 5y ago
Gonna butcher this, but it's good practice, so I'm gonna take a stab at it. The roots of polynomials are intimately related to the Alternating symmetry groups. These describe the ways you can permute a set, specifially A_n describes the even permutations of a set of n elements. The even permutations of a sequence are those obtained by making an even number of transpositions ("moves").
I can make 2 swaps to go 1,2,3,4,5 -> 1,4,3,2,5 -> 1,4,5,2,3. That's an even permutation.
The order (size) of A_n is the number of even permutations of n unique elements. The orders are as follows:
A2: 1
A3: 3
A4: 12
A5: 60
A6: 360
A_n: n! / 2
The "solvability" of a polynomial depends on the the degrees of freedom and number of constraints. Quintic is where the number of degrees of freedom blows up and you can no longer factor out roots in such a way that you can uniquely factor the polynomial. Quadratic is basically trivial. Cubic cannot be directly factored, because you have 3 variables and 2 equations, but you can "factor out" a quadratic that you can solve, and then solve the cubic. Quartic is the same, you end up with 4 variables and 3 unique equations, you can factor out a cubic, which you can solve, etc. Quartic, you have 4 equations to solve and...something like 30 variables (or maybe it's 60. It's some way more than we can factor!).
https://en.wikipedia.org/wiki/Alternating_group https://en.wikipedia.org/wiki/Alternating_group
https://math.stackexchange.com/questions/550401/intuitive-reasoning-why-are-quintics-unsolvable https://math.stackexchange.com/questions/550401/intuitive-re...
- paulpauper 5y agoSorta. There are obviously infinite quintics that can be solved despite this. For example any polynomial in which the roots are arranged on a circle on the real,complex plane