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GXWeb Catalan Playground

Saltire Software, home of Geometry Expressions and GXWeb

Symbolic computations on this page use Nerdamer Symbolic JavaScript to complement the in-built CAS of GXWeb

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Introduction Catalan Polygon Model Catalan Number Applications Catalan Number Definitions Catalan Matrix Magic A Catalan Polynomial Hyper-Catalan Numbers Solving Polynomials with Hyper-Catalan Numbers References Mathematical ToolKit Construct Your Own Model with GXWeb

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Introduction

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If you are like me (until quite recently), you may never have heard of the Catalan numbers, but it seems that they crop up in a remarkable range of situations! They may even hold the record for the longest reference list in the Online Encyclopedia of Integer Sequences (OEIS)! The Catalan numbers have recently been brought to my attention through the work of NJ Wildberger and Dean Rubine in their paper, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, which proposes a solution to higher order polynomials without recourse to irrationals or Galois Theory. The Hyper-Catalan numbers are explored below. The Catalan numbers \(C_{n}\) were introduced by Leonhard Euler in 1751 to count the number of ways a convex polygon with \(n+2\) sides could be divided into \(n\) (non-overlapping) triangles. A hexagon, for example, can be divided \(C_{4}\) = 14 ways into 4 triangles, as shown.

\[x\] \[Catalan(x)\] \[0\] \[1\] \[1\] \[1\] \[2\] \[2\] \[3\] \[5\] \[4\] \[14\] \[5\] \[42\] \[6\] \[132\] \[7\] \[429\] \[8\] \[1430\] \[9\] \[4862\] \[10\] \[16796\]

You might also view the above as the number of ways of joining \(2n\) points on a circle to form \(n\) nonintersecting chords - or perhaps the number of ways that \(2n\) people sitting around a table can introduce themselves by shaking hands without reaching across in front of someone else!

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Catalan Polygon Model

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Catalan Number Applications

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The Catalan numbers have surprisingly many other applications, including...

• CHALLENGE 1: In how many ways can \(n\) pairs of parentheses be arranged so that they are correctly matched? Did you guess the \(n\)th Catalan number? For example, the arrangements for \(n = 3\) are

  ((())), (()()), (())(), ()(()), ()()()

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CHALLENGE 2: Now imagine replacing each opening parenthesis ( with X and each ) with Y: these are called Dyck words. \(C_{n}\) is the number of Dyck words of length \(2n\). A Dyck word is a string consisting of \(n\) X's and \(n\) Y's such that no initial segment of the string has more Y's than X's. The Dyck words for \(n = 3\) are XXXYYY, XXYXYY, XXYYXY, XYXXYY, XYXYXY

CHALLENGE 3

• CHALLENGE 4: \(C_{n}\) is also the number of ways to insert \((n-1)\) pairs of parentheses in a word of \((n+1)\) letters. For n = 3, a,b,c,d can be arranged with 2 parenthesis pairs in 5 ways:

  (ab)(cd), ((ab)c)d, (a(bc))d, a((bc)d), a(b(cd))

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Other applications include...

• The number of paths of length \(2n\) in a grid that never rise above the main diagonal.

  

• \(C_{n}\) is also the number of full binary trees with \(n + 1\) leaves.

  

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Catalan Number Definitions

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And quite a few different formulae...

\[C_{n} = \frac{(2n)!}{(n+1)!\cdot n!} = \frac{1}{n+1}\cdot \begin {pmatrix} 2\cdot n \\ n \end{pmatrix} \]

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\[\ C_0 = 1 \ and \ C_{n+1} = \sum_{i=0}^{n} C_i \cdot C_{n-i}\]

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\[(1 + x + 2\cdot x^2 + 5\cdot x^3 + ... + C_{n}\cdot x^n)^2 \]\[= 1 + 2\cdot x + 5\cdot x^2 + ... + C_{n}\cdot x^{n-1} + C_{n+1}\cdot x^{n} + ...\]

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Catalan Matrix Magic

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The Catalan Numbers have many unique and interesting properties, including some involving the \(n\) x \(n\) Hankel matrix

\[\begin{bmatrix} a & b & c & d & e \\ b & c & d & e & f \\ c & d & e & f & g \\ d & e & f & g & h \\ e & f & g & h & i \end{bmatrix}\]

This matrix, with Catalan number entries, has determinant 1 for any value of \(n\) - and this remains true for a second sequence of Catalan numbers! (This is actually a unique and defining property of the Catalan numbers.)

\[det \begin{bmatrix} 1 & 1 & 2 & 5 & 14 \\ 1 & 2 & 5 & 14 & 42 \\ 2 & 5 & 14 & 42 & 132 \\ 5 & 14 & 42 & 132 & 429 \\ 14 & 42 & 132 & 429 & 1430 \end{bmatrix}\]\[ = \]\[det\begin{bmatrix} 1 & 2 & 5 & 14 & 42 \\ 2 & 5 & 14 & 42 & 132 \\ 5 & 14 & 42 & 132 & 429 \\ 14 & 42 & 132 & 429 & 1430 \\ 42 & 132 & 429 & 1430 & 4862 \end{bmatrix} \]\[= 1\]

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The Catalan-Hankel matrix has another unique feature: Any \(n\)x\(n\) sub-matrix beginning from 2 will have determinant \(n+1\).

\[det\begin{bmatrix} 2 \end{bmatrix} = 2\]\[det\begin{bmatrix} 2 & 5 \\ 5 & 14 \end{bmatrix} = 3\]\[det\begin{bmatrix} 2 & 5 & 14 \\ 5 & 14 & 42 \\ 14 & 42 & 132 \end{bmatrix} = 4\]\[det\begin{bmatrix} 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132 \\ 14 & 42 & 132 & 429 \\ 42 & 132 & 429 & 1430 \end{bmatrix} = 5\]\[det\begin{bmatrix} 2 & 5 & 14 & 42 & 132 \\ 5 & 14 & 42 & 132 & 429 \\ 14 & 42 & 132 & 429 & 1430 \\ 42 & 132 & 429 & 1430 & 4862 \\ 132 & 429 & 1430 & 4862 & 16796 \end{bmatrix} = 6\]

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A Catalan Polynomial

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Consider a polynomial of degree \(n\) with coefficients equal to the first \((n+1)\) Catalan Numbers. For example, if \(n = 4\), the first \(five\) coefficients are...\[[1,1,2,5,14]\]

\[(1+x+2\cdot x^2+5\cdot x^3+14\cdot x^4)^2\]

Now square this polynomial: what do you notice?

\[(1+x+2\cdot x^2+5\cdot x^3+14\cdot x^4)^2\] \[= 1+2\cdot x+5\cdot x^2+14\cdot x^3+42\cdot x^4+...\]

In general,

\[(1 + x + 2\cdot x^2 + 5\cdot x^3 + ... + C_{n}\cdot x^n)^2 \]\[= 1 + 2\cdot x + 5\cdot x^2 + 14\cdot x^3 + 42\cdot x^4 \]\[+ ... + C_{n}\cdot x^{n-1} + C_{n+1}\cdot x^{n} + ...\]Square that polynomial and the \((n+1)\)th coefficient of the result will give the next Catalan Number, where \(C_{0} = 1, C_{1} = 1...\)

[1,1,2,5,14,42,...]

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Hyper-Catalan Numbers

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Why stop at triangles?

The Catalan Numbers define the number of ways a polygon with \(n+2\) sides can be completely divided by \(n\) diagonals into triangles.

Consider now dividing polygons, not only into triangles, but into lesser polygons, and you enter the realm of Hyper-Catalan Numbers!

\[C_{m} = C(m_{2}, m_{3}, m_{4},...)\]\[= \frac{(2\cdot m_{2} + 3\cdot m_{3} + 4\cdot m_{4} + ...)!}{(1 + m_{2} + 2\cdot m_{3} + 3\cdot m_{4} + ...)! \cdot m_{2}! \cdot m_{3}! \cdot m_{4}! ...}\]

Thanks to the work of NJ Wildberger and colleague Dean Rubine, we have a deeper understanding of the Hyper-Catalan numbers \(C_{m} = C[m2, m3, m4,...]\), which give the possible polygonal combinations of \(m_{2}\) triangles, \(m_{3}\) quadrilaterals, \(m_{4}\) pentagons, and so on. Note that for a single input, \(C_{m} = C(m_{2})\), the Hyper-Catalan numbers coincide with the Catalan numbers.

It is also important to note the following relationships for such a divided polygon:

\[Vertices (V) = 2 + m_{2} + 2\cdot m_{3} + 3\cdot m_{4} +...\]\[Edges (E) = 1 + 2\cdot m_{2} + 3\cdot m_{3} + 4\cdot m_{4} +...\]\[Faces (F) = m_{2} + m_{3} + m_{4} +...\]\[and \ V - E + F = 1!\]

For example,

• \(C_{1} = C(3) = 5 \): dividing a pentagon into 3 triangles in 5 ways (V = 5, E = 7 and F = 3)

• \(C_{2} = C(2,1) = 21 \): dividing a hexagon into 2 triangles and 1 quadrilateral in 21 ways (V = 6, E = 8 and F = 3), and

• \(C_{3} = C(2,0,1) = 28 \), dividing a 7-gon into 2 triangles, 0 quadrilaterals and 1 pentagon in 28 ways (V = 7, E = 9 and F = 3)!

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Solving Polynomials using Hyper-Catalan Numbers

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Wildberger and Rubin propose a general Hyper-Catalan polynomial solution

(where the Hyper-Catalan number \(C_{m} = C[m_{2}, m_{3}, m_{4}, m_{5},...]\))

\[c_{0} - c_{1}\cdot x + c_{2}\cdot x^2 + c_{3}\cdot x^3 + c_{4}\cdot x^4 + c_{5}\cdot x^5 +... = 0\]\[x = \]\[\sum_{m_{2}, m_{3}, m_{4}, m_{5},...\geq 0} C_{m}\cdot \frac{c_{0}^{(1+m_{2}+2m_{3}+3m_{4}+4m_{5},...)}}{c_{1}^{(1+2m_{2}+3m_{3}+4m_{4}+5m_{5},+...)}}\cdot c_{2}^{m_{2}}\cdot c_{3}^{m_{3}}\cdot c_{4}^{m_{4}}\cdot c_{5}^{m_{5}}...\]

Note that, for \(m_{2}\) triangles, \(m_{3}\) quadrilaterals, \(m_{4}\) pentagons, \(m_{5}\) hexagons, ... the number of edges\[E = 1+2m_{2}+3m_{3}+4m_{4}+5m_{5},+...\]and the number of vertices\[V = 2+m_{2}+2m_{3}+3m_{4}+4m_{5},...\]Then let \[m = m_{2}, m_{3}, m_{4}, m_{5},...\] and \[c^{m} = c_{2}^{m_{2}}\cdot c_{3}^{m_{3}}\cdot c_{4}^{m_{4}}\cdot c_{5}^{m_{5}}...\] and our solution can be further simplified to\[x = \sum_{m \geq 0} C_{m}\cdot \frac{c_{0}^{V-1}}{c_{1}^{E-1}}\cdot c^{m}\]

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Hyper-Catalan Solution Example

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For example, for degree 5:

\[x = C[0,0,0,0]\cdot \frac{c_{0}}{c_{1}} + C[1,0,0,0]\cdot \frac{c_{0}^2}{c_{1}^3}\cdot c_{2} + C[2,0,0,0]\cdot \frac{c_{0}^3}{c_{1}^5}\cdot c_{2}^2 +...\]\[+ C[1,1,0,0]\cdot \frac{c_{0}^4}{c_{1}^6}\cdot c_{2}\cdot c_{3} + C[2,1,0,0]\cdot \frac{c_{0}^5}{c_{1}^8}\cdot c_{2}^2\cdot c_{3} + C[1,2,0,0]\cdot \frac{c_{0}^6}{c_{1}^9}\cdot c_{2}\cdot c_{3}^2 +...\]\[+ C[1,1,1,0]\cdot \frac{c_{0}^7}{c_{1}^{10}}\cdot c_{2}\cdot c_{3}\cdot c_{4} + C[1,1,1,1]\cdot \frac{c_{0}^{12}}{c_{1}^{15}}\cdot c_{2}\cdot c_{3}\cdot c_{4}\cdot c_{5} +...\]

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\[x = \frac{c_{0}}{c_{1}} + \frac{c_{0}^2}{c_{1}^3}\cdot c_{2} + 2\cdot \frac{c_{0}^3}{c_{1}^5}\cdot c_{2}^2 +...\]\[+ 5\cdot \frac{c_{0}^4}{c_{1}^6}\cdot c_{2}\cdot c_{3} + 21\cdot \frac{c_{0}^5}{c_{1}^8}\cdot c_{2}^2\cdot c_{3} + 28\cdot \frac{c_{0}^6}{c_{1}^9}\cdot c_{2}\cdot c_{3}^2 +...\]\[+ 72\cdot \frac{c_{0}^7}{c_{1}^{10}}\cdot c_{2}\cdot c_{3}\cdot c_{4} + 2184\cdot \frac{c_{0}^{12}}{c_{1}^{15}}\cdot c_{2}\cdot c_{3}\cdot c_{4}\cdot c_{5} +...\]

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Begin with the Quadratic

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Initially, you might try the Catalan solution for the quadratic equation (where \(C_{n}\) represents the \(n\)th Catalan number)

\[c_{0} - c_{1}\cdot x + c_{2}\cdot x^2 = 0\]\[x = \sum_{n\geq 0} C_{n}\cdot \frac{c_{0}^{(1+n)}\cdot c_{2}^n}{c_{1}^{(1+2\cdot n)}}\]\[x = \frac{c_{0}}{c_{1}} + \frac{c_{0}^2}{c_{1}^3}\cdot c_{2} + 2\cdot \frac{c_{0}^3}{c_{1}^5}\cdot c_{2}^2 + 5\cdot \frac{c_{0}^4}{c_{1}^7}\cdot c_{2}^3 + 14\cdot \frac{c_{0}^5}{c_{1}^9}\cdot c_{2}^4 +...\]

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Given the initial equation, these Catalan solutions, like the Newton-Raphson method, require taking estimates of any solutions, shifting the initial equation by these estimates, and then shifting the resulting solutions back.

For example, given the initial equation

\[x^2-x-1=0\]

solution estimates might be \(-1\) and \(2\).

In this case, shift by \(2\) to give

\[(x+(2))^2-(x+(2))-1 \]\[= x^2+3x+1\] \[⇒ 0.38196337279071135\]

and by \(-1\) to give

\[(x+(-1))^2-(x+(-1))-1 \]\[= x^2-3x+1\] \[⇒ -0.38196337279071135\]

Then shifting back again:

\[2-(0.38196337279071135)\] \[= 1.6180366272092885\]\[and\] \[-1-(-0.38196337279071135)\] \[= -0.6180366272092886\]

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CAS solution:

\[[\frac{1+\sqrt{5}}{2},\] \[\frac{1-\sqrt{5}}{2}]\] \[[1.618033988749648,\] \[-0.618033988749648]\]

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Consider the Cubic

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We move next to the Hyper-Catalan solution for the cubic equation (where \(C_{m}\) represents the Hyper-Catalan number \(C_{m2,m3}\).

\[c_{0} - c_{1}\cdot x + c_{2}\cdot x^2 + c_{3}\cdot x^3 = 0\]\[x = \sum_{m2,m3\geq 0} C_{m}\cdot \frac{c_{0}^{(1+m2+2\cdot m3)}}{c_{1}^{(1+2\cdot m2 + 3\cdot m3)}}\cdot c_{2}^{m2} \cdot c_{3}^{m3}\]

While we know that the values for \(m2\) and \(m3\) must be greater than or equal to zero, we do not have certainty as to their optimal order, as we do with the quadratic form, which uses the linear Catalan sequence.

We can choose to attempt to build these values up simply and systematically (as show in the initial example above - \[C[0,0,0,0], C[1,0,0,0], C[2,0,0,0], ...\]\[C[1,1,0,0], C[2,1,0,0], C[1,2,0,0],...\]\[C[1,1,1,0],... C[1,1,1,1]\] and this seems to lead to reasonably accurate results after around 10 steps. But a more efficient sequence would be appreciated - leading to the Geode array. In the case of the cubic, this presents as the Geode Bi-Tri array G[m2,m3].

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What is the Geode?

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Geometrically, the Hyper-Catalan numbers describe the number of possible arrangements of subdigons (planar convex roofed polygons subdivided by non-crossing diagonals into polygonal faces), each composed of a given number of vertices (V), edges (E) and faces (F). Wildberger and Rubine (2025) define the set of all such polygons S as a subdigon polyseries layered according to the number of vertices, edges or faces.

To organise S according to vertices \(V = 2+m_{2}+2\cdot m_{3}+3\cdot m_{4}+...\) they introduce a variable v such that

\[S_{V} = v^2 + t_{2}\cdot v^3 + (2\cdot t_{2}^2 + t_{3})\cdot v^4 + (5\cdot t_{2}^3 + 5\cdot t_{2}\cdot t_{3} + t_{4})\cdot v^5\]\[+ (14\cdot t_{2}^4 + 21\cdot t_{2}^2\cdot t_{3} + 3\cdot t_{3}^2 + 6\cdot t_{2}\cdot t_{4} + t_{5})\cdot v^6 + ...\]

Figure 5 shows the number of hexagons, corresponding to \(v_{6}\). With \(t_{2}\) representing triangles, \(t_{3}\) quadrilaterals, and so on, we observe that a hexagon has 14 ways (\(C_{4}\)) to be divided into triangles, 21 ways (\(C[2,1]\)) to be divided into 2 triangles and 1 quadrilateral, 3 ways (\(C[0,2]\)) to be divided into 2 quadrilaterals, 6 ways (\(C[1,0,1]\)) to be divided into 1 pentagon and 1 triangle and 1 way (\(C[0,0,0,1]\)) to be divided into a hexagon - 45 variations in all!

In the same way, we can organise S according to edges \(E = 1+2\cdot m_{2}+3\cdot m_{3}+4\cdot m_{4}+...\), introducing a variable e such that

\[S_{E} = e + t_{2}\cdot e^3 + t_{3}\cdot e^4 + (2\cdot t_{2}^2 + t_{4})\cdot e^5 + (5\cdot t_{2}\cdot t_{3} + t_{5})\cdot e^6 \]\[+ (5\cdot t_{2}^3 + 3\cdot t_{3}^2 + 6\cdot t_{2}\cdot t_{4} + t_{6})\cdot e^7\]\[+ (21\cdot t_{2}^2\cdot t_{3} + 7\cdot t_{3}\cdot t_{4} + 7\cdot t_{2}\cdot t_{5} + t_{7})\cdot e^8 + ...\]

Figure 6 shows the number of subdigons with 7 edges: 5 pentagons divided into 3 triangles (\(C[3]\)), 3 hexagons divided into 2 quadrilaterals (\(C[0,2]\)), 6 hexagons (1 triangle and a pentagon (\(C[1,0,1]\))) and 1 heptagon (\(C[0,0,0,0,1]\)).

The geode, then, is defined to be the number of possible arrangements of subdigons according to their central face. Defining \(S_{1} = t_{2} + t_{3} + t_{4} + ...\)

\[S_{F} = 1 + S_{1}\cdot f + S_{1}\cdot (2\cdot t_{2} + 3\cdot t_{3} + 4\cdot t_{4} + ...)\cdot f^2 \]\[+ S_{1}\cdot (5\cdot t_{2}^2 + 16\cdot t_{2}\cdot t_{3} + 12\cdot t_{3}^2 + 23\cdot t_{2}\cdot t_{4} + 33\cdot t_{3}\cdot t_{4} + 22\cdot t_{4}^2 + ...)\cdot f^3 + ...\]

For example, subdigons may be thought of as constructed from simpler subdigons around a central face, as shown. \(Geode(m2,m3,m4)\) gives the number of possible combinations of subdigons composed of triangles, quadrilaterals and pentagons.

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\[Geode(m2,m3)\]

m3 ⇒ m2 ⇓	0	1	2	3	4
0	1	3	12	55	273
1	2	16	110	728	4760
2	5	70	702	6160	50388
3	14	288	3850	42432	418950
4	42	1155	19448	259350	3010700

\[Geode(m2,m3,m4,m5,...) = \]\[\frac{(3 + 2\cdot m2 + 3\cdot m3 + 4\cdot m4 + 5\cdot m5 + ...)!}{d1\cdot d2 \cdot d3\cdot d4}\]

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\[G(4,0,0,0...) = 42\]\[ where \]\[ (3+2\cdot m2+3\cdot m3+ 4\cdot m4 + 5\cdot m5 + ...)! \]\[= (11)! = 39916800\]\[d1 =\]\[3+2\cdot m2+2\cdot m3+ 2\cdot m4 + 2\cdot m5 + ... = 11\]\[d2 =\]\[1+m2+m3+m4+m5+...= 5\]\[d3 =\]\[(2+m2+2\cdot m3+ 3\cdot m4 + 4\cdot m5 + ...)! = 720\]\[d4 =\]\[(m2)! = 24 \ (m3)! = 1 \ (m4)! = 1 \ and \ (m5)! = 1\]\[\frac{39916800}{11\cdot 5\cdot 720\cdot 24\cdot 1\cdot 1\cdot 1} = 42\]

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Solving a Cubic Polynomial with the Geode

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Wildberger and Rubine (2025) define a cubic function:

\[Q(t_{2},t_{3}) = 1 + (t_{2}+t_{3}) + (1+2\cdot t_{2} + 3\cdot t_{3} + 5\cdot t_{2}^{2} + 16\cdot t_{2}\cdot t_{3} + 12\cdot t_{3}^{3})\]

(Can you see the relationship with the first few diagonals of our Geode Bi-Tri array?)

Then to solve a general cubic polynomial

\[c_{0} - c_{1}\cdot x + c_{2}\cdot x^2 + c_{3}\cdot x^3\]

we define

\[K(c_{0},c_{1},c_{2},c_{3}) = \frac{c_{0}}{c_{1}}\cdot Q(\frac{c_{0}\cdot c_{2}}{c_{1}^{2}},\frac{c_{0}^{2}\cdot c_{3}}{c_{1}^{3}})\]

Once again, we begin with an estimate of our solution, and generally in just a couple of iterations we achieve a fairly accurate result. For example,

\[f(x) = x^3-2\cdot x -5 \ with \ estimate \ x=2\] \[⇒ f(x+2) = -1+10\cdot x+6\cdot x^2 + x^3\] \[K(-1,-10,6,1) = 2.0954682345\]

A second application results in \(x = 2.0945514808\), compared with a CAS result of \(x = 2.094551481542323\): correct to 9 decimal places after only 2 iterations.

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References

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Online Encyclopedia of Integer Sequences (A000108): Explore Catalan Numbers further in this extensive collection.

Wikipedia: Catalan Numbers.

Numberphile: Catalan Numbers (YouTube 13:16)

Wildberger, NJ and Rubine, D (2025) A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode The American Mathematical Monthly, 132:5, 383-402, DOI: 10.1080/00029890.2025.2460966

N J Wildberger: Paper with Dean Rubine on Solving Polynomial Equations and the Geode I (YouTube 47:12)

GXWeb Catalan Playground (YouTube 13:08)

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QR Codes are a great way to share data and information with others, even when no Internet connection is available. Most modern devices either come equipped with QR readers in-built, or freely available. The default link here is the GXWeb Showcase, but you can use it as an alternative to sending your assessment data via email, web or share with others in your class. You can even use it to send your own messages!

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