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 (26)

Ascertaining the uniqueness of algebraic solution to the geometric quintic equation

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Aim: Check if the proposed solution to a quintic equation in geometic form is unique.

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Observe: The previous notebook proposed a solution to a general quintic equation in geometric form, by utilising a power series to an abritrary degree.

Theorem: The quintic equation in geometric form

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

has the 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: In the previous notebooks it was seen that these numbers represent a power series and relate to roofed polygonal subdivision numbers.

Observe: To ascertain if the solution to the above general quintic equation in geometric form is unique, a good starting point is to consider a equation in with a lower degree first.

Observe: In the linear case, the geometric equation can be stated as $1 - x = 0$ and this is a unique solution, $x = 1$.

Let the following be variables.

x, a0, a1, a2, a3, a4, a5, a6, c, d = sp.symbols('x, a0, a1, a2, a3, a4, a5, a6, c, d')

Let $F1$ be the quadratic case of a quadratic equation in geometric form.

F1 = sp.Eq(1 - x + c * x**2, 0)
F1

$\displaystyle c x^{2} - x + 1 = 0$

Observe: It is possible to use a power series solution to this equation to ascertain whether there could be a unique solution, where $x$ is:

$$ x = \Sigma_{k = 0}^6 = a_0 + ca_1 + c^2a_2+ c^3a_3 + c^4a_4 + c^5a_5 + c^6a_6 $$

Let $F2$ be a equation setting $x$ to a power series of degree 6:

F2 = sp.Eq(x, a0 + c*a1 + c**2*a2+ c**3*a3 + c**4*a4 + c**5*a5 + c**6*a6)
F2

$\displaystyle x = a_{0} + a_{1} c + a_{2} c^{2} + a_{3} c^{3} + a_{4} c^{4} + a_{5} c^{5} + a_{6} c^{6}$

Let $F3$ be $F1$ witha subsution of $x$ using $F2$

F3 = F1.subs(x, F2.rhs)
F3

$\displaystyle - a_{0} - a_{1} c - a_{2} c^{2} - a_{3} c^{3} - a_{4} c^{4} - a_{5} c^{5} - a_{6} c^{6} + c \left(a_{0} + a_{1} c + a_{2} c^{2} + a_{3} c^{3} + a_{4} c^{4} + a_{5} c^{5} + a_{6} c^{6}\right)^{2} + 1 = 0$

Let $F4$ be a power series expansion of $F3$ up to degree 6.

F4 = sp.expand(F3.lhs).series(c, 0, 6).removeO()
F4

$\displaystyle - a_{0} + c^{5} \left(2 a_{0} a_{4} + 2 a_{1} a_{3} + a_{2}^{2} - a_{5}\right) + c^{4} \left(2 a_{0} a_{3} + 2 a_{1} a_{2} - a_{4}\right) + c^{3} \left(2 a_{0} a_{2} + a_{1}^{2} - a_{3}\right) + c^{2} \left(2 a_{0} a_{1} - a_{2}\right) + c \left(a_{0}^{2} - a_{1}\right) + 1$

Observe: Assuming $a_1$ to $a_5$ is $0$, it can dedeced that can deduce a0 = 1. It is subsequentyly possible to obtain values for all of $a_i$. which will be unique.

Observe: The uniquess of this solution is clear from inductive examination, however a more general examintation the equation with subsitution also shows the solutions are unique.

Let $F5$ be the the equation written such that each coeffiecient in $c^k$ is written on a seperate line.

$$1$$$$ - (a_{0} + a_{1} c + a_{2} c^{2} + a_{3} c^{3} + a_{4} c^{4} + a_{5} c^{5} + a_{6} c^{6}) $$$$ c(a_{0} + a_{1} c + a_{2} c^{2} + a_{3} c^{3} + a_{4} c^{4} + a_{5} c^{5} + a_{6} c^{6})^{2} $$$$ = 0 $$

Observe: The value of $a_k$ can be determined by looking at the total coefficient of $c^k$ on each line: for $k > 0$ there is no such coefffient in the first line, $a_k$ appears linearly in the second line, and in the the third line the coefficient of $c^k$ will involve only coefficients$a_i$ for $i < k$ and these by induction are already uniquely determined.

Observe: This allows the conclusion that $a_k$ will be uniquely determined.

Observe: It is possible to ascertain this for the cubic case and a general argument is provided below.

Let $F6$ be a general cubic equation in geometric form.

F6 = sp.Eq(1 - x + c * x**2 + d * x**3, 0)
F6

$\displaystyle c x^{2} + d x^{3} - x + 1 = 0$

Observe: To ascertain uniqueness in this case using a power series as a subsutition for $x$, the expected output would be:

$$ x = \alpha + (\beta c + \gamma d) + (\delta c^2 + \phi cd + \eta d^2) + \text{(cubic terms)} \ldots$$

Observe: The value of $\alpha$ can be ascertained by looking at the constant term, so $\alpha = 1$

Observe: The values of $\beta$ and $\gamma$ are determined by looking at degree 1 terms and, similiarly the quadratic terms $\delta, \phi, \eta$ appear linearly in the coefficents in $c^2, cd, d^2$ respectively, which satisfies uniqueness.

Observe: This can be extended to higher degrees in a straightforward fashion using the same geometric form:

$$ 1 - x + c_2x^2 + \ldots = 0 $$

Observe Recall from SP2, when solving examining this problem by using a power series in $t$, two possible solutions were found: $a_0 = 0$ or $a_0 = \frac{-c_1}{-c_2}$. However the geometric form allows x to be determined as a single solution.