STATUS: Draft

---

Solving Polynomial Equations: 3

import numpy as np
import sympy as sp
import pickle
from IPython.display import HTML
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

# This is commented out
#%matplotlib widget

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

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

Create required variables

c_0, c_1, c_2, x, t, a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, x_1, x_2, s_1, s_2 = sp.symbols('c_0, c_1, c_2, x, t, a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, x_1, x_2, s_1, s_2')

Recall that the original quadratic general equation sought to find solutions for $x_1$ and $x_2$ for a general quadratic

$$t + c_0x + c_1x^2 = 0$$

Recall that, using a power series in $t$, value of $x_1$ and $x_2$, the following solutions could be found:

x_1 = sp.Eq(x_1, -t/c_1 - c_2*t**2/c_1**3 - 2*c_2**2*t**3/c_1**5 - 5*c_2**3*t**4/c_1**7 - 14*c_2**4*t**5/c_1**9 - 42*c_2**5*t**6/c_1**11 - 132*c_2**6*t**7/c_1**13 - 429*c_2**7*t**8/c_1**15 - 1430*c_2**8*t**9/c_1**17)
x_1

$\displaystyle x_{1} = - \frac{t}{c_{1}} - \frac{c_{2} t^{2}}{c_{1}^{3}} - \frac{2 c_{2}^{2} t^{3}}{c_{1}^{5}} - \frac{5 c_{2}^{3} t^{4}}{c_{1}^{7}} - \frac{14 c_{2}^{4} t^{5}}{c_{1}^{9}} - \frac{42 c_{2}^{5} t^{6}}{c_{1}^{11}} - \frac{132 c_{2}^{6} t^{7}}{c_{1}^{13}} - \frac{429 c_{2}^{7} t^{8}}{c_{1}^{15}} - \frac{1430 c_{2}^{8} t^{9}}{c_{1}^{17}}$

x_2 = sp.Eq(x_2, -c_1/c_2 + t/c_1 + c_2*t**2/c_1**3 + 2*c_2**2*t**3/c_1**5 + 5*c_2**3*t**4/c_1**7 + 14*c_2**4*t**5/c_1**9 + 42*c_2**5*t**6/c_1**11 + 132*c_2**6*t**7/c_1**13 + 429*c_2**7*t**8/c_1**15 + 1430*c_2**8*t**9/c_1**17)
x_2

$\displaystyle x_{2} = - \frac{c_{1}}{c_{2}} + \frac{t}{c_{1}} + \frac{c_{2} t^{2}}{c_{1}^{3}} + \frac{2 c_{2}^{2} t^{3}}{c_{1}^{5}} + \frac{5 c_{2}^{3} t^{4}}{c_{1}^{7}} + \frac{14 c_{2}^{4} t^{5}}{c_{1}^{9}} + \frac{42 c_{2}^{5} t^{6}}{c_{1}^{11}} + \frac{132 c_{2}^{6} t^{7}}{c_{1}^{13}} + \frac{429 c_{2}^{7} t^{8}}{c_{1}^{15}} + \frac{1430 c_{2}^{8} t^{9}}{c_{1}^{17}}$

Note that $x_1$ and $x_2$ into generative functions. $$ x_1 = -\sum_{k=1}^9 C(k - 1)\frac{c_2^{k-1}}{c_1^{2k - 1}}t^k$$

$$ x_2 = -\frac{c_1}{c_2} + \sum_{k=1}^9 C(k - 1)\frac{c_2^{k-1}}{c_1^{2k - 1}}t^k $$

Note that solutions are not available $x_1$ and $x_2$ beyond degree 9. However, note the general case of generative functions:

$$ x_1 = -\sum_{k=1}^+ C(k - 1)\frac{c_2^{k-1}}{c_1^{2k - 1}}t^k$$$$ x_2 = -\frac{c_1}{c_2} + \sum_{k=1}^+ C(k - 1)\frac{c_2^{k-1}}{c_1^{2k - 1}}t^k $$

The cubic case

Goal: Test the cubic case

Create the required variables:

a_0, a_1, a_2, a_3, a_4, a_5, a_6, t, c_1, c_2, c_3, x = sp.symbols('a_0, a_1, a_2, a_3, a_4, a_5, a_6, t, c_1, c_2, c_3, x ')

Assume that $t^8 == 0$.

Let s1 be a general cubic equation.

s1 = sp.Eq(t + c_1 * x + c_2*x**2 + c_3 * x**3, 0)
s1

$\displaystyle c_{1} x + c_{2} x^{2} + c_{3} x^{3} + t = 0$

Let s2 be a substitution for x.

s2 = a_0 + a_1 * t + a_2*t**2 + a_3*t**3 + a_4*t**4 + a_5*t**5 + a_6*t**6 + a_7*t**7
s2

$\displaystyle a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + a_{5} t^{5} + a_{6} t^{6} + a_{7} t^{7}$

Let s3 be s1 with the s2 subsituted for x

s3 = s1.subs(x, s2)
s3

$\displaystyle c_{1} \left(a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + a_{5} t^{5} + a_{6} t^{6} + a_{7} t^{7}\right) + c_{2} \left(a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + a_{5} t^{5} + a_{6} t^{6} + a_{7} t^{7}\right)^{2} + c_{3} \left(a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + a_{5} t^{5} + a_{6} t^{6} + a_{7} t^{7}\right)^{3} + t = 0$

Let s4 be the expansion of s3

s4 = s3.expand()

Enforce assumption that $t^8=0$ in s4

order = 8
coeffs = sp.Poly(s3, t).coeffs()
s4= sum(t**n * coeffs[-(n+1)] for n in range(order))
s4 = sp.Eq(s4, 0)
s4

$\displaystyle a_{0}^{3} c_{3} + a_{0}^{2} c_{2} + a_{0} c_{1} + t^{7} \left(3 a_{0}^{2} a_{7} c_{3} + 6 a_{0} a_{1} a_{6} c_{3} + 6 a_{0} a_{2} a_{5} c_{3} + 6 a_{0} a_{3} a_{4} c_{3} + 2 a_{0} a_{7} c_{2} + 3 a_{1}^{2} a_{5} c_{3} + 6 a_{1} a_{2} a_{4} c_{3} + 3 a_{1} a_{3}^{2} c_{3} + 2 a_{1} a_{6} c_{2} + 3 a_{2}^{2} a_{3} c_{3} + 2 a_{2} a_{5} c_{2} + 2 a_{3} a_{4} c_{2} + a_{7} c_{1}\right) + t^{6} \left(3 a_{0}^{2} a_{6} c_{3} + 6 a_{0} a_{1} a_{5} c_{3} + 6 a_{0} a_{2} a_{4} c_{3} + 3 a_{0} a_{3}^{2} c_{3} + 2 a_{0} a_{6} c_{2} + 3 a_{1}^{2} a_{4} c_{3} + 6 a_{1} a_{2} a_{3} c_{3} + 2 a_{1} a_{5} c_{2} + a_{2}^{3} c_{3} + 2 a_{2} a_{4} c_{2} + a_{3}^{2} c_{2} + a_{6} c_{1}\right) + t^{5} \left(3 a_{0}^{2} a_{5} c_{3} + 6 a_{0} a_{1} a_{4} c_{3} + 6 a_{0} a_{2} a_{3} c_{3} + 2 a_{0} a_{5} c_{2} + 3 a_{1}^{2} a_{3} c_{3} + 3 a_{1} a_{2}^{2} c_{3} + 2 a_{1} a_{4} c_{2} + 2 a_{2} a_{3} c_{2} + a_{5} c_{1}\right) + t^{4} \left(3 a_{0}^{2} a_{4} c_{3} + 6 a_{0} a_{1} a_{3} c_{3} + 3 a_{0} a_{2}^{2} c_{3} + 2 a_{0} a_{4} c_{2} + 3 a_{1}^{2} a_{2} c_{3} + 2 a_{1} a_{3} c_{2} + a_{2}^{2} c_{2} + a_{4} c_{1}\right) + t^{3} \left(3 a_{0}^{2} a_{3} c_{3} + 6 a_{0} a_{1} a_{2} c_{3} + 2 a_{0} a_{3} c_{2} + a_{1}^{3} c_{3} + 2 a_{1} a_{2} c_{2} + a_{3} c_{1}\right) + t^{2} \left(3 a_{0}^{2} a_{2} c_{3} + 3 a_{0} a_{1}^{2} c_{3} + 2 a_{0} a_{2} c_{2} + a_{1}^{2} c_{2} + a_{2} c_{1}\right) + t \left(3 a_{0}^{2} a_{1} c_{3} + 2 a_{0} a_{1} c_{2} + a_{1} c_{1} + 1\right) = 0$

Let s5 be the case where $t = 0$ in s4

s5 = s4.subs(t, 0)
s5

$\displaystyle a_{0}^{3} c_{3} + a_{0}^{2} c_{2} + a_{0} c_{1} = 0$

Let s6 be the factorisation of of s5. Note there are three solutions

s6 = sp.solve(s5, a_0)
renderListToLatex(s6)

$$0$$
$$- \frac{c_{2}}{2 c_{3}} - \frac{\sqrt{- 4 c_{1} c_{3} + c_{2}^{2}}}{2 c_{3}}$$
$$- \frac{c_{2}}{2 c_{3}} + \frac{\sqrt{- 4 c_{1} c_{3} + c_{2}^{2}}}{2 c_{3}}$$

Consider the case where $a_0 = 0$.

Substitute $a_0 = 0$ into s4

s7 = sp.Poly(s4.subs(a_0, 0), t)
s7

$\displaystyle \operatorname{Poly}{\left( \left(3 a_{1}^{2} a_{5} c_{3} + 6 a_{1} a_{2} a_{4} c_{3} + 3 a_{1} a_{3}^{2} c_{3} + 2 a_{1} a_{6} c_{2} + 3 a_{2}^{2} a_{3} c_{3} + 2 a_{2} a_{5} c_{2} + 2 a_{3} a_{4} c_{2} + a_{7} c_{1}\right) t^{7} + \left(3 a_{1}^{2} a_{4} c_{3} + 6 a_{1} a_{2} a_{3} c_{3} + 2 a_{1} a_{5} c_{2} + a_{2}^{3} c_{3} + 2 a_{2} a_{4} c_{2} + a_{3}^{2} c_{2} + a_{6} c_{1}\right) t^{6} + \left(3 a_{1}^{2} a_{3} c_{3} + 3 a_{1} a_{2}^{2} c_{3} + 2 a_{1} a_{4} c_{2} + 2 a_{2} a_{3} c_{2} + a_{5} c_{1}\right) t^{5} + \left(3 a_{1}^{2} a_{2} c_{3} + 2 a_{1} a_{3} c_{2} + a_{2}^{2} c_{2} + a_{4} c_{1}\right) t^{4} + \left(a_{1}^{3} c_{3} + 2 a_{1} a_{2} c_{2} + a_{3} c_{1}\right) t^{3} + \left(a_{1}^{2} c_{2} + a_{2} c_{1}\right) t^{2} + \left(a_{1} c_{1} + 1\right) t, t, domain=\mathbb{Z}\left[a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, c_{1}, c_{2}, c_{3}\right] \right)}$

Let s8 be a system of equations to linearly solve $a_1, a_2, a_3, a_4, a_5, a_6$, and $a_7$. Append $a_0 = 0$ as a valid equation in this system.

s8 = [sp.Eq(s7.coeffs()[i], 0) for i in range(len(s7.coeffs()))]
s8.append(sp.Eq(a_0, 0))
s8

[Eq(3*a_1**2*a_5*c_3 + 6*a_1*a_2*a_4*c_3 + 3*a_1*a_3**2*c_3 + 2*a_1*a_6*c_2 + 3*a_2**2*a_3*c_3 + 2*a_2*a_5*c_2 + 2*a_3*a_4*c_2 + a_7*c_1, 0),
 Eq(3*a_1**2*a_4*c_3 + 6*a_1*a_2*a_3*c_3 + 2*a_1*a_5*c_2 + a_2**3*c_3 + 2*a_2*a_4*c_2 + a_3**2*c_2 + a_6*c_1, 0),
 Eq(3*a_1**2*a_3*c_3 + 3*a_1*a_2**2*c_3 + 2*a_1*a_4*c_2 + 2*a_2*a_3*c_2 + a_5*c_1, 0),
 Eq(3*a_1**2*a_2*c_3 + 2*a_1*a_3*c_2 + a_2**2*c_2 + a_4*c_1, 0),
 Eq(a_1**3*c_3 + 2*a_1*a_2*c_2 + a_3*c_1, 0),
 Eq(a_1**2*c_2 + a_2*c_1, 0),
 Eq(a_1*c_1 + 1, 0),
 Eq(a_0, 0)]

Let s9 be the solutions to this system.

variables =  [a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7]
s9 = sp.nonlinsolve(s8, variables)
interimSolutions  = list(list(s9)[0])
s9 = [sp.Eq(interimSolutions[i], variables[i]) for i in range(len(interimSolutions))]
s9

[Eq(0, a_0),
 Eq(-1/c_1, a_1),
 Eq(-c_2/c_1**3, a_2),
 Eq((c_3/c_1**3 - 2*c_2**2/c_1**4)/c_1, a_3),
 Eq(5*(c_2*c_3/c_1**5 - c_2**3/c_1**6)/c_1, a_4),
 Eq((-3*c_3**2/c_1**6 + 21*c_2**2*c_3/c_1**7 - 14*c_2**4/c_1**8)/c_1, a_5),
 Eq((-28*c_2*c_3**2/c_1**8 + 84*c_2**3*c_3/c_1**9 - 42*c_2**5/c_1**10)/c_1, a_6),
 Eq((12*c_3**3/c_1**9 - 180*c_2**2*c_3**2/c_1**10 + 330*c_2**4*c_3/c_1**11 - 132*c_2**6/c_1**12)/c_1, a_7)]

Visualise solutions for $a_1, a_2, a_3, a_4, a_5, a_6$, and $a_7$.

latex_rendering = []

for i in range(len(s9)):
    latex_rendering.append("$$" + sp.latex(s9[i].simplify()) + "$$")
    
HTML("".join(latex_rendering[1:]))

$$a_{1} = - \frac{1}{c_{1}}$$$$a_{2} = - \frac{c_{2}}{c_{1}^{3}}$$$$a_{3} = \frac{c_{1} c_{3} - 2 c_{2}^{2}}{c_{1}^{5}}$$$$a_{4} = \frac{5 c_{2} \left(c_{1} c_{3} - c_{2}^{2}\right)}{c_{1}^{7}}$$$$a_{5} = \frac{- 3 c_{1}^{2} c_{3}^{2} + 21 c_{1} c_{2}^{2} c_{3} - 14 c_{2}^{4}}{c_{1}^{9}}$$$$a_{6} = \frac{- 28 c_{1}^{2} c_{2} c_{3}^{2} + 84 c_{1} c_{2}^{3} c_{3} - 42 c_{2}^{5}}{c_{1}^{11}}$$$$a_{7} = \frac{6 \left(2 c_{1}^{3} c_{3}^{3} - 30 c_{1}^{2} c_{2}^{2} c_{3}^{2} + 55 c_{1} c_{2}^{4} c_{3} - 22 c_{2}^{6}\right)}{c_{1}^{13}}$$

Note that if $c_3 = 0$, the cubic term, then the quadratic case is recovered

s10 = [s9[i].subs(c_3, 0) for i in range(len(s9))]
latex_rendering = []

for i in range(len(s10)):
    latex_rendering.append("$$" + sp.latex(s10[i].simplify()) + "$$")
    
HTML("".join(latex_rendering[1:]))

$$a_{1} = - \frac{1}{c_{1}}$$$$a_{2} = - \frac{c_{2}}{c_{1}^{3}}$$$$a_{3} = - \frac{2 c_{2}^{2}}{c_{1}^{5}}$$$$a_{4} = - \frac{5 c_{2}^{3}}{c_{1}^{7}}$$$$a_{5} = - \frac{14 c_{2}^{4}}{c_{1}^{9}}$$$$a_{6} = - \frac{42 c_{2}^{5}}{c_{1}^{11}}$$$$a_{7} = - \frac{132 c_{2}^{6}}{c_{1}^{13}}$$

Recall the cubic case expression.

s1

$\displaystyle c_{1} x + c_{2} x^{2} + c_{3} x^{3} + t = 0$

Substitute back into x to find one of the solutions of this general cubic.

values_to_solve2 = variables[1:]
solutions_as_list2 = interimSolutions[1:]
solutions_as_list2
r1_5 = sp.Eq(x, s2)

r15 = r1_5.subs(values_to_solve2[0], solutions_as_list2[0])
r15 = r15.subs(values_to_solve2[1], solutions_as_list2[1].simplify())
r15 = r15.subs(values_to_solve2[2], solutions_as_list2[2].simplify())
r15 = r15.subs(values_to_solve2[3], solutions_as_list2[3].simplify())
r15 = r15.subs(values_to_solve2[4], solutions_as_list2[4].simplify())
r15 = r15.subs(values_to_solve2[5], solutions_as_list2[5].simplify())
r15 = r15.subs(values_to_solve2[6], solutions_as_list2[6].simplify())

# after going through all the solutions, subs in 0 for a_0
#r15 = r15.subs(a_0, 0)
r15

$\displaystyle x = a_{0} - \frac{t}{c_{1}} - \frac{c_{2} t^{2}}{c_{1}^{3}} + \frac{t^{3} \left(c_{1} c_{3} - 2 c_{2}^{2}\right)}{c_{1}^{5}} + \frac{5 c_{2} t^{4} \left(c_{1} c_{3} - c_{2}^{2}\right)}{c_{1}^{7}} + \frac{t^{5} \left(- 3 c_{1}^{2} c_{3}^{2} + 21 c_{1} c_{2}^{2} c_{3} - 14 c_{2}^{4}\right)}{c_{1}^{9}} + \frac{t^{6} \left(- 28 c_{1}^{2} c_{2} c_{3}^{2} + 84 c_{1} c_{2}^{3} c_{3} - 42 c_{2}^{5}\right)}{c_{1}^{11}} + \frac{6 t^{7} \left(2 c_{1}^{3} c_{3}^{3} - 30 c_{1}^{2} c_{2}^{2} c_{3}^{2} + 55 c_{1} c_{2}^{4} c_{3} - 22 c_{2}^{6}\right)}{c_{1}^{13}}$

Subsitute into original equation

Summary

There still appears to be a relationship to the Catalan numbers in the cubic case, but there are numbers present also. Note the the above is only one of the solutions to a general cubic equation. It is valid to degree 7, expressing the values as an on-series in powers in t, where the coefficients are rational functions of $c_i$

Save outputs.

# Write data that will be used again
data = {'generalCubicSolutions': s9,
       'solutionOfGeneralCubicInTermsOfx':r15}
f = open('./SharedOutputs/SPE3.pickle', 'wb')
pickle.dump(data, f)
f.close()