STATUS: Draft

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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
pd.set_option('display.max_colwidth', None)

#

# 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:])))

Solving Polynomial Equations (15)

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Aim: Explore the quartic equation and bBiTriQuad numbers

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Ref: The BiTri array can be found on OEIS (A104978), seen in diagonals

Ref: Interesting related reference: John Zhau, Fat and Thin Emergent Geometries of Hermitial One-Matrix Models (math physics)

Observe: a solution to a general cubic equation

$$C(m_2, m_3) \equiv(-1)^{m_3 + 1} \frac{(2 m_{2} + 3 m_{3})!}{(1 + m_{2} + 2 m_{3})!m_2!m_3!} \frac{c_0^{1 + m_{2} + 2 m_{3}} c_2^{m_2} c_3^{m_3} }{c_1^{2 m_{2} + 3 m_{3} + 1}}$$

Create needed unknown types

c_0, c_1, c_2, c_3, c_4, 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, c_3, c_4, 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')

Let $P1$, $P2$, $P3$ be from notebook 6.

with open('./SharedOutputs/SPE6.pickle', 'rb') as input:
    imported_outputs_from_another_notebook = pickle.load(input)
    
P1 = imported_outputs_from_another_notebook['xSubstitution']
P2 = imported_outputs_from_another_notebook['generalQuarticEquation']
P3 = imported_outputs_from_another_notebook['solutionForXForQuartic']

Observe: $P2$ was presented in Notebook 6 as a general quartic equation. Subsitutions were made, replacing $c_0$ with $t$ and $x$ with $P1$

renderListToLatex([P2, P1])

$$c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} + c_{4} x^{4} = 0$$
$$a_{0} + a_{1} t + a_{2} t^{2} + a_{3} t^{3} + a_{4} t^{4} + a_{5} t^{5} + a_{6} t^{6} + a_{7} t^{7} + a_{8} t^{8}$$

Observe: A solution for $x$ was found, $P3$.

P3

$\displaystyle - \frac{45 c_{0}^{8} c_{2} c_{4}^{2}}{c_{1}^{11}} - \frac{45 c_{0}^{8} c_{3}^{2} c_{4}}{c_{1}^{11}} + \frac{495 c_{0}^{8} c_{2}^{2} c_{3} c_{4}}{c_{1}^{12}} + \frac{165 c_{0}^{8} c_{2} c_{3}^{3}}{c_{1}^{12}} - \frac{495 c_{0}^{8} c_{2}^{4} c_{4}}{c_{1}^{13}} - \frac{990 c_{0}^{8} c_{2}^{3} c_{3}^{2}}{c_{1}^{13}} + \frac{1287 c_{0}^{8} c_{2}^{5} c_{3}}{c_{1}^{14}} - \frac{429 c_{0}^{8} c_{2}^{7}}{c_{1}^{15}} - \frac{4 c_{0}^{7} c_{4}^{2}}{c_{1}^{9}} + \frac{72 c_{0}^{7} c_{2} c_{3} c_{4}}{c_{1}^{10}} + \frac{12 c_{0}^{7} c_{3}^{3}}{c_{1}^{10}} - \frac{120 c_{0}^{7} c_{2}^{3} c_{4}}{c_{1}^{11}} - \frac{180 c_{0}^{7} c_{2}^{2} c_{3}^{2}}{c_{1}^{11}} + \frac{330 c_{0}^{7} c_{2}^{4} c_{3}}{c_{1}^{12}} - \frac{132 c_{0}^{7} c_{2}^{6}}{c_{1}^{13}} + \frac{7 c_{0}^{6} c_{3} c_{4}}{c_{1}^{8}} - \frac{28 c_{0}^{6} c_{2}^{2} c_{4}}{c_{1}^{9}} - \frac{28 c_{0}^{6} c_{2} c_{3}^{2}}{c_{1}^{9}} + \frac{84 c_{0}^{6} c_{2}^{3} c_{3}}{c_{1}^{10}} - \frac{42 c_{0}^{6} c_{2}^{5}}{c_{1}^{11}} - \frac{6 c_{0}^{5} c_{2} c_{4}}{c_{1}^{7}} - \frac{3 c_{0}^{5} c_{3}^{2}}{c_{1}^{7}} + \frac{21 c_{0}^{5} c_{2}^{2} c_{3}}{c_{1}^{8}} - \frac{14 c_{0}^{5} c_{2}^{4}}{c_{1}^{9}} - \frac{c_{0}^{4} c_{4}}{c_{1}^{5}} + \frac{5 c_{0}^{4} c_{2} c_{3}}{c_{1}^{6}} - \frac{5 c_{0}^{4} c_{2}^{3}}{c_{1}^{7}} + \frac{c_{0}^{3} c_{3}}{c_{1}^{4}} - \frac{2 c_{0}^{3} c_{2}^{2}}{c_{1}^{5}} - \frac{c_{0}^{2} c_{2}}{c_{1}^{3}} - \frac{c_{0}}{c_{1}}$

Observe: The general form of each term in thesolution:

$$\frac{c_0^{m_0} c_2^{m_2} c_3^{m_3} c_4^{m_4}}{c_1^{m_1}} $$

Observe: Based on the data in the solution, 2 conditions for relationships between the $m_i$ terms can be seen:

Observe: the Degree of each term in the quartic satisfies: $m_0 + m_2 + m_3 + m_4 - m_1 = 0$

Observe: the Index Degree of each term in the quartic solution satisfied: $ m_1 + 2m_2 + 3m_3 + 4m_4 = -1$

Observe: It is possible to use these equations to create the following equivalencies:

$$m_0 = 1 + m_2 + 2m_3 + 3m_4 $$$$ m_1 = 1 + 2m_2 + 3m_3 + 4m_4 $$

Observe: these identities appear to be an extension of cubic case.

Observe: the General term is given by:

$$ F(m_2, m_3, m_4) = (-1)^{m_3 + 1} \frac{(2m_2 + 3m_3 + 4m_4)! }{(1 + m_2 + 2m_3 + 3m_4)!m_2!m_3!m_4!} \frac{c_0^{ 1 + m_2 + 2m_3 + 3m_4 } c_2^{m_2} c_3^{m_3} c_4^{m_4}}{c_1^{1 + 2m_2 + 3m_3 + 4m_4}} $$

Observe: This can be written equivalently in terms of $m, n, p$ to simplify subscripts in the variables.

$$ F(m, n, p) = (-1)^{n + 1} \frac{(2m + 3n + 4p)! }{(1 + m + 2n + 3p)!m!n!p!} \frac{c_0^{ 1 + m + 2n + 3p } c_2^{m} c_3^{n} c_4^{p}}{c_1^{1 + 2m + 3n + 4p}} $$

Let $F1$ be an implementation of $F$

def F1(m, n, p, returnCoefficientsOnly = False, returnCoefficientsOnlyWithoutSigns = False, returnCoefficientsAsFactorialStrings = False):
    c_0, c_1, c_2, c_3, c_4 = sp.symbols('c_0, c_1, c_2, c_3, c_4')
    
    
    s1 = (-1)**(n + 1)
    
    s2 = sp.factorial(2 * m + 3 * n + 4 * p)
    s3 = sp.factorial(1 + m +2 * n + 3 * p) * sp.factorial(m) * sp.factorial(n) * sp.factorial(p)

    
    s4 = c_0**(1 + m + 2 * n + 3 * p) * c_2**m *c_3**n * c_4**p
    s5 = c_1**(1 + 2 * m + 3 * n + 4 * p)
    
    s7 = str(2 * m + 3 * n + 4 * p) + "!"
    s8 = str(1 + 2 * m + 3 * n + 4 * p) + "!" + str(m) + "!"  + str(n) + "!" + str(p) + "!"
    
    if returnCoefficientsOnly:
        s6 = s1 * (s2 / s3)
    elif returnCoefficientsOnlyWithoutSigns:
        s6 = (s2 / s3)
    elif returnCoefficientsAsFactorialStrings:
        s6 = str(s7 + " | " + s8)
    else:
        s6 = s1 * (s2 / s3) * (s4 / s5)

    return(s6)

Let $P6$ be an $8 \times 8 \times 1$ array of values evaluated with $F1$

P4 = np.arange(8)
P5 = np.array([[F1(j, i, 1) for i in P1] for j in P4])
P6 = sp.Matrix(P5)
P6

$\displaystyle \left[\begin{matrix}- \frac{c_{0}^{4} c_{4}}{c_{1}^{5}} & \frac{7 c_{0}^{6} c_{3} c_{4}}{c_{1}^{8}} & - \frac{45 c_{0}^{8} c_{3}^{2} c_{4}}{c_{1}^{11}} & \frac{286 c_{0}^{10} c_{3}^{3} c_{4}}{c_{1}^{14}} & - \frac{1820 c_{0}^{12} c_{3}^{4} c_{4}}{c_{1}^{17}} & \frac{11628 c_{0}^{14} c_{3}^{5} c_{4}}{c_{1}^{20}} & - \frac{74613 c_{0}^{16} c_{3}^{6} c_{4}}{c_{1}^{23}} & \frac{480700 c_{0}^{18} c_{3}^{7} c_{4}}{c_{1}^{26}}\\- \frac{6 c_{0}^{5} c_{2} c_{4}}{c_{1}^{7}} & \frac{72 c_{0}^{7} c_{2} c_{3} c_{4}}{c_{1}^{10}} & - \frac{660 c_{0}^{9} c_{2} c_{3}^{2} c_{4}}{c_{1}^{13}} & \frac{5460 c_{0}^{11} c_{2} c_{3}^{3} c_{4}}{c_{1}^{16}} & - \frac{42840 c_{0}^{13} c_{2} c_{3}^{4} c_{4}}{c_{1}^{19}} & \frac{325584 c_{0}^{15} c_{2} c_{3}^{5} c_{4}}{c_{1}^{22}} & - \frac{2422728 c_{0}^{17} c_{2} c_{3}^{6} c_{4}}{c_{1}^{25}} & \frac{17760600 c_{0}^{19} c_{2} c_{3}^{7} c_{4}}{c_{1}^{28}}\\- \frac{28 c_{0}^{6} c_{2}^{2} c_{4}}{c_{1}^{9}} & \frac{495 c_{0}^{8} c_{2}^{2} c_{3} c_{4}}{c_{1}^{12}} & - \frac{6006 c_{0}^{10} c_{2}^{2} c_{3}^{2} c_{4}}{c_{1}^{15}} & \frac{61880 c_{0}^{12} c_{2}^{2} c_{3}^{3} c_{4}}{c_{1}^{18}} & - \frac{581400 c_{0}^{14} c_{2}^{2} c_{3}^{4} c_{4}}{c_{1}^{21}} & \frac{5148297 c_{0}^{16} c_{2}^{2} c_{3}^{5} c_{4}}{c_{1}^{24}} & - \frac{43743700 c_{0}^{18} c_{2}^{2} c_{3}^{6} c_{4}}{c_{1}^{27}} & \frac{360540180 c_{0}^{20} c_{2}^{2} c_{3}^{7} c_{4}}{c_{1}^{30}}\\- \frac{120 c_{0}^{7} c_{2}^{3} c_{4}}{c_{1}^{11}} & \frac{2860 c_{0}^{9} c_{2}^{3} c_{3} c_{4}}{c_{1}^{14}} & - \frac{43680 c_{0}^{11} c_{2}^{3} c_{3}^{2} c_{4}}{c_{1}^{17}} & \frac{542640 c_{0}^{13} c_{2}^{3} c_{3}^{3} c_{4}}{c_{1}^{20}} & - \frac{5969040 c_{0}^{15} c_{2}^{3} c_{3}^{4} c_{4}}{c_{1}^{23}} & \frac{60568200 c_{0}^{17} c_{2}^{3} c_{3}^{5} c_{4}}{c_{1}^{26}} & - \frac{580179600 c_{0}^{19} c_{2}^{3} c_{3}^{6} c_{4}}{c_{1}^{29}} & \frac{5322259800 c_{0}^{21} c_{2}^{3} c_{3}^{7} c_{4}}{c_{1}^{32}}\\- \frac{495 c_{0}^{8} c_{2}^{4} c_{4}}{c_{1}^{13}} & \frac{15015 c_{0}^{10} c_{2}^{4} c_{3} c_{4}}{c_{1}^{16}} & - \frac{278460 c_{0}^{12} c_{2}^{4} c_{3}^{2} c_{4}}{c_{1}^{19}} & \frac{4069800 c_{0}^{14} c_{2}^{4} c_{3}^{3} c_{4}}{c_{1}^{22}} & - \frac{51482970 c_{0}^{16} c_{2}^{4} c_{3}^{4} c_{4}}{c_{1}^{25}} & \frac{590539950 c_{0}^{18} c_{2}^{4} c_{3}^{5} c_{4}}{c_{1}^{28}} & - \frac{6309453150 c_{0}^{20} c_{2}^{4} c_{3}^{6} c_{4}}{c_{1}^{31}} & \frac{63867117600 c_{0}^{22} c_{2}^{4} c_{3}^{7} c_{4}}{c_{1}^{34}}\\- \frac{2002 c_{0}^{9} c_{2}^{5} c_{4}}{c_{1}^{15}} & \frac{74256 c_{0}^{11} c_{2}^{5} c_{3} c_{4}}{c_{1}^{18}} & - \frac{1627920 c_{0}^{13} c_{2}^{5} c_{3}^{2} c_{4}}{c_{1}^{21}} & \frac{27457584 c_{0}^{15} c_{2}^{5} c_{3}^{3} c_{4}}{c_{1}^{24}} & - \frac{393693300 c_{0}^{17} c_{2}^{5} c_{3}^{4} c_{4}}{c_{1}^{27}} & \frac{5047562520 c_{0}^{19} c_{2}^{5} c_{3}^{5} c_{4}}{c_{1}^{30}} & - \frac{59609309760 c_{0}^{21} c_{2}^{5} c_{3}^{6} c_{4}}{c_{1}^{33}} & \frac{660885825600 c_{0}^{23} c_{2}^{5} c_{3}^{7} c_{4}}{c_{1}^{36}}\\- \frac{8008 c_{0}^{10} c_{2}^{6} c_{4}}{c_{1}^{17}} & \frac{352716 c_{0}^{12} c_{2}^{6} c_{3} c_{4}}{c_{1}^{20}} & - \frac{8953560 c_{0}^{14} c_{2}^{6} c_{3}^{2} c_{4}}{c_{1}^{23}} & \frac{171609900 c_{0}^{16} c_{2}^{6} c_{3}^{3} c_{4}}{c_{1}^{26}} & - \frac{2755853100 c_{0}^{18} c_{2}^{6} c_{3}^{4} c_{4}}{c_{1}^{29}} & \frac{39118609530 c_{0}^{20} c_{2}^{6} c_{3}^{5} c_{4}}{c_{1}^{32}} & - \frac{506679132960 c_{0}^{22} c_{2}^{6} c_{3}^{6} c_{4}}{c_{1}^{35}} & \frac{6113193886800 c_{0}^{24} c_{2}^{6} c_{3}^{7} c_{4}}{c_{1}^{38}}\\- \frac{31824 c_{0}^{11} c_{2}^{7} c_{4}}{c_{1}^{19}} & \frac{1627920 c_{0}^{13} c_{2}^{7} c_{3} c_{4}}{c_{1}^{22}} & - \frac{47070144 c_{0}^{15} c_{2}^{7} c_{3}^{2} c_{4}}{c_{1}^{25}} & \frac{1012354200 c_{0}^{17} c_{2}^{7} c_{3}^{3} c_{4}}{c_{1}^{28}} & - \frac{18027009000 c_{0}^{19} c_{2}^{7} c_{3}^{4} c_{4}}{c_{1}^{31}} & \frac{281015317440 c_{0}^{21} c_{2}^{7} c_{3}^{5} c_{4}}{c_{1}^{34}} & - \frac{3965314953600 c_{0}^{23} c_{2}^{7} c_{3}^{6} c_{4}}{c_{1}^{37}} & \frac{51770019087072 c_{0}^{25} c_{2}^{7} c_{3}^{7} c_{4}}{c_{1}^{40}}\end{matrix}\right]$

Let $P8$ be an $8 \times 8 \times 1$ array of values evaluated with $F1$ returning coefficients only and ignoring signs

P7 = np.array([[F1(j, i, 1, returnCoefficientsOnlyWithoutSigns=True) for i in P1] for j in P4])
P8 = sp.Matrix(P7)
P8

$\displaystyle \left[\begin{matrix}1 & 7 & 45 & 286 & 1820 & 11628 & 74613 & 480700\\6 & 72 & 660 & 5460 & 42840 & 325584 & 2422728 & 17760600\\28 & 495 & 6006 & 61880 & 581400 & 5148297 & 43743700 & 360540180\\120 & 2860 & 43680 & 542640 & 5969040 & 60568200 & 580179600 & 5322259800\\495 & 15015 & 278460 & 4069800 & 51482970 & 590539950 & 6309453150 & 63867117600\\2002 & 74256 & 1627920 & 27457584 & 393693300 & 5047562520 & 59609309760 & 660885825600\\8008 & 352716 & 8953560 & 171609900 & 2755853100 & 39118609530 & 506679132960 & 6113193886800\\31824 & 1627920 & 47070144 & 1012354200 & 18027009000 & 281015317440 & 3965314953600 & 51770019087072\end{matrix}\right]$

Observe: these coefficients can be regarded as BiTriQuad and are also sub multinomial

Observe: It is a natural conjecture, based on the assumption that if $m$ and $n$ relate to the number of ways of subdividing triangles and quadrilaterals, $p$ could relate to the number of ways of subdividing petagons. These numbers numbers may provide insight into the number of ways of subdividing polygons into triangles, quadrilaterals and pentagrams.

Observe: If this function is evaluated at $(m, n, 0)$ the BiTri (Catalan and Fuss numbers) will be recovered. This suggests a relatinoship between polynomials of increasing degrees.

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Aim: Explore the BiTriQuad numbers in a similiar fashion to exploring the BiTri numbers

Method: Explore different combinations of inputs such as $m, m + 1, n, n + 1, p, p + 1$ and keeping one variables as this is a 3D array of numbers.

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Observe: In the case of $(m, n, 1)$, the result given by $P8$, the first column of values is the OEIS entry A236194

Observe: In the case of $(m, n, 1)$, the result given by $P8$, the first row of values is the OEIS entry A002694

Todo: Check other patterns against OEIS

Conjecture: The BiTriQuad Numbers, which is the coeffiencts of F(m, n, p) counts the number of a diagonal subidivisons of a $(2 + m + 2n + 3p)$ sided polygon into $m$ triangles, $n$ quadrialterals and $p$ pentagons. In particular, this shows that $F(m, n, p)$ must a an integer

Observe: To "count" sides, start with 2, and then add one for every triangle that is being created. Quadrilaterals will add 2 new vertices/ edges Petagrams will add 3 vertices/edges. This is conjecture is true, this number will always be an integer.

Todo: Go through combinations in a similar way to the cubic case, and look at recursive relations.