STATUS: Draft

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import numpy as np
import sympy as sp
import pickle
from IPython.display import HTML, Image
import ipywidgets as widgets
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
import pandas as pd
import itertools
pd.set_option('display.max_colwidth', None)
import treelib as tr


# function to print latex
def renderListToLatex(e):
    latex_rendering = []

    for i in range(len(e)):
        latex_rendering.append("$$" + sp.latex(e[i]) + "$$")
    
    return(HTML("".join(latex_rendering[0:])))

Solving Polynomial Equations (25)

The algebraic solution of the general quintic polynomial equation

Observe Recall the quintic equation in geometric form:

$ \displaystyle c_{2} x^{2} + c_{3} x^{3} + c_{4} x^{4} + c_{5} x^{5} - x + 1 = 0 $

Observe: Recall this has the proposed solution:

$$ x = \Sigma_{m = 0}^+ \Sigma_{n = 0}^+ \Sigma_{p = 0}^+ \Sigma_{q = 0}^+ \frac{(2m + 3n + 4p + 5q)!}{(1 + m + 2n + 3p + 4q)!m!n!p!q!} c_2^{m} c_3^{n} c_4^{p} c_5^q $$

Observe: This solution is a polynomial power series in $c2, c3, c4$ and $c5$.

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Aim: This solution should be computationally verified up to a certain degree.

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Observe: The verification to be undertaken is not intended to work with infite sums or infite computations, but should be truncated to an arbitrary number.

Observe: The proposed solution includes an extension of the BiTriQuad numbers.

Definition: The extension of the BiTriQuad numbers to degree 5 is defined as BiTriQuadQuintic numbers:

$$ \frac{(2m + 3n + 4p + 5q)!}{(1 + m + 2n + 3p + 4q)!m!n!p!q!} $$

Definition: These types of numbers (such as BiTri, BiTriQuad etc.) can be denoted as Roofed Polygonal Subdivision numbers (here in the special case of degree 5).

$$ r(m,n, p, q) = \frac{(2m + 3n + 4p + 5q)!}{(1 + m + 2n + 3p + 4q)!m!n!p!q!} $$

Let $P1$ be a function that implements $r$

Let the following be unknown types.

m, n, p, q, c2, c3, c4, c5, x = sp.symbols('m, n, p, q, c2, c3, c4, c5, x')

Let $P1$ be a general quintic equation.

P1 = sp.Eq(c0 - c1 * x + c2 * x**2 + c3 * x**3 + c4 * x**4 + c5 * x**5, 0)
P1

$\displaystyle c_{0} - c_{1} x + c_{2} x^{2} + c_{3} x^{3} + c_{4} x^{4} + c_{5} x^{5} = 0$

Let $P2$ be $P1$ converted to geometric form.

P2 = P1.subs({c0:1, c1:1})
P2

$\displaystyle c_{2} x^{2} + c_{3} x^{3} + c_{4} x^{4} + c_{5} x^{5} - x + 1 = 0$

Let $P3$ be a function that returns a roofed polygon subdivision number (to degree 5).

def P3(m, n, p, q):
    numerator = sp.factorial(2 * m + 3 * n + 4 * p + 5 * q)
    denominator = sp.factorial(1 + m + 2 * n + 3 * p + 4 * q) * sp.factorial(m) * sp.factorial(n) * sp.factorial(p) * sp.factorial(q)
    
    return(numerator / denominator)
    

Example: $P3$ evaluated at $(3, 2, 0, 0)$.

P3(3, 2, 0, 0)

$\displaystyle 990$

Let $P4$ be a function that returns a roofed polygon subdivision number for variables $c_2, c_3, c_4$ and $c_5$.

def P4(m, n, p, q, c2, c3, c4, c5):
    numerator = sp.factorial(2 * m + 3 * n + 4 * p + 5 * q)
    denominator = sp.factorial(1 + m + 2 * n + 3 * p + 4 * q) * sp.factorial(m) * sp.factorial(n) * sp.factorial(p) * sp.factorial(q)
   
    v = (numerator / denominator) * (c2**m * c3**n * c4**p * c5**q)

    return(v)

Example: Evaluate $P4$ at $m, n, p, q, c2, c3, c4, c5$.

P4(m, n, p, q, c2, c3, c4, c5)

$\displaystyle \frac{c_{2}^{m} c_{3}^{n} c_{4}^{p} c_{5}^{q} \left(2 m + 3 n + 4 p + 5 q\right)!}{m! n! p! q! \left(m + 2 n + 3 p + 4 q + 1\right)!}$

Let $P5$ be a function that implements:

$$ x = \Sigma_{m = 0}^+ \Sigma_{n = 0}^+ \Sigma_{p = 0}^+ \Sigma_{q = 0}^+ \frac{(2m + 3n + 4p + 5q)!}{(1 + m + 2n + 3p + 4q)!m!n!p!q!} c_2^{m} c_3^{n} c_4^{p} c_5^q $$

def P5(degree = 2):
    degree = 2
    F1 = []
    for m in range(0, degree):
        for n in range(0, degree):
            for p in range(0, degree):
                for q in range(0, degree):
                    F1.append(P4(m, n, p, q, c2, c3, c4, c5))   
    return(np.sum(F1))

Let $P6$ be $P5$ evaluated with an an degree of $2$.

P6 = P5(2)
P6

$\displaystyle 2184 c_{2} c_{3} c_{4} c_{5} + 72 c_{2} c_{3} c_{4} + 90 c_{2} c_{3} c_{5} + 5 c_{2} c_{3} + 110 c_{2} c_{4} c_{5} + 6 c_{2} c_{4} + 7 c_{2} c_{5} + c_{2} + 132 c_{3} c_{4} c_{5} + 7 c_{3} c_{4} + 8 c_{3} c_{5} + c_{3} + 9 c_{4} c_{5} + c_{4} + c_{5} + 1$

Let $P7$ be the geometric form of the the general quintic equation, with a subsution of $x$ for $P6$

P7 = P2.subs(x, P6)
P7

$\displaystyle - 2184 c_{2} c_{3} c_{4} c_{5} - 72 c_{2} c_{3} c_{4} - 90 c_{2} c_{3} c_{5} - 5 c_{2} c_{3} - 110 c_{2} c_{4} c_{5} - 6 c_{2} c_{4} - 7 c_{2} c_{5} + c_{2} \left(2184 c_{2} c_{3} c_{4} c_{5} + 72 c_{2} c_{3} c_{4} + 90 c_{2} c_{3} c_{5} + 5 c_{2} c_{3} + 110 c_{2} c_{4} c_{5} + 6 c_{2} c_{4} + 7 c_{2} c_{5} + c_{2} + 132 c_{3} c_{4} c_{5} + 7 c_{3} c_{4} + 8 c_{3} c_{5} + c_{3} + 9 c_{4} c_{5} + c_{4} + c_{5} + 1\right)^{2} - c_{2} - 132 c_{3} c_{4} c_{5} - 7 c_{3} c_{4} - 8 c_{3} c_{5} + c_{3} \left(2184 c_{2} c_{3} c_{4} c_{5} + 72 c_{2} c_{3} c_{4} + 90 c_{2} c_{3} c_{5} + 5 c_{2} c_{3} + 110 c_{2} c_{4} c_{5} + 6 c_{2} c_{4} + 7 c_{2} c_{5} + c_{2} + 132 c_{3} c_{4} c_{5} + 7 c_{3} c_{4} + 8 c_{3} c_{5} + c_{3} + 9 c_{4} c_{5} + c_{4} + c_{5} + 1\right)^{3} - c_{3} - 9 c_{4} c_{5} + c_{4} \left(2184 c_{2} c_{3} c_{4} c_{5} + 72 c_{2} c_{3} c_{4} + 90 c_{2} c_{3} c_{5} + 5 c_{2} c_{3} + 110 c_{2} c_{4} c_{5} + 6 c_{2} c_{4} + 7 c_{2} c_{5} + c_{2} + 132 c_{3} c_{4} c_{5} + 7 c_{3} c_{4} + 8 c_{3} c_{5} + c_{3} + 9 c_{4} c_{5} + c_{4} + c_{5} + 1\right)^{4} - c_{4} + c_{5} \left(2184 c_{2} c_{3} c_{4} c_{5} + 72 c_{2} c_{3} c_{4} + 90 c_{2} c_{3} c_{5} + 5 c_{2} c_{3} + 110 c_{2} c_{4} c_{5} + 6 c_{2} c_{4} + 7 c_{2} c_{5} + c_{2} + 132 c_{3} c_{4} c_{5} + 7 c_{3} c_{4} + 8 c_{3} c_{5} + c_{3} + 9 c_{4} c_{5} + c_{4} + c_{5} + 1\right)^{5} - c_{5} = 0$

Observe: To evaluate $P7$ and and ascertain if it verifies $P2$ (truncated to degree 2), a power series expansion is needed.

Let $P8$ be a power series expansion of $P7$ in variables $c_2, c_3, c_4$ and $c_5$.

P8  = sp.expand(P7.lhs).series(c2, 0, 2).removeO().series(c3, 0, 2).removeO().series(c4, 0, 2).removeO().series(c5, 0, 2).removeO()
P8

$\displaystyle 0$

Observe: this result verifies up to degree 2 that the solution found using a substitution of $P6$ for $x$ satisfies $ \displaystyle c_{2} x^{2} + c_{3} x^{3} + c_{4} x^{4} + c_{5} x^{5} - x + 1 = 0 $

Todo: Find a a faster implementation in Python for the multivariate series expansion.