Pascal's simplex
# Pascal's simplex

In mathematics, Pascal's simplex is a generalisation of Pascal's triangle into arbitrary number of dimensions, based on the multinomial theorem.

## Generic Pascal's m-simplex

Let m (m > 0) be a number of terms of a polynomial and n (n ≥ 0) be a power the polynomial is raised to.

Let $\wedge ^{m}$ denote a Pascal's m-simplex. Each Pascal's m-simplex is a semi-infinite object, which consists of an infinite series of its components.

Let $\wedge _{n}^{m}$ denote its nth component, itself a finite (m − 1)-simplex with the edge length n, with a notational equivalent $\vartriangle _{n}^{m-1}$.

### nth component

$\wedge _{n}^{m}=\vartriangle _{n}^{m-1}$ consists of the coefficients of multinomial expansion of a polynomial with m terms raised to the power of n:

    $|x|^{n}=\sum _{|k|=n}{{\binom {n}{k}}x^{k}};\ \ x\in \mathbb {R} ^{m},\ k\in \mathbb {N} _{0}^{m},\ n\in \mathbb {N} _{0},\ m\in \mathbb {N}$

where $\textstyle |x|=\sum _{i=1}^{m}{x_{i}},\ |k|=\sum _{i=1}^{m}{k_{i}},\ x^{k}=\prod _{i=1}^{m}{x_{i}^{k_{i}}}$.

### Example for $\wedge ^{4}$

Pascal's 4-simplex (sequence A189225 in the OEIS), sliced along the $k_4.$ All points of the same color belong to the same n-th component, from red (for n = 0) to blue (for n = 3).

## Specific Pascal's simplices

### Pascal's 1-simplex

$\wedge ^{1}$ is not known by any special name.

**nth component**

$\wedge _{n}^{1}=\vartriangle _{n}^{0}$ (a point) is the coefficient of multinomial expansion of a polynomial with 1 term raised to the power of n:

    $(x_{1})^{n}=\sum _{k_{1}=n}{n \choose k_{1}}x_{1}^{k_{1}};\ \ k_{1},n\in \mathbb {N} _{0}$

**nth component**

**Arrangement of $\vartriangle _{n}^{0}$**


    $\textstyle {n \choose n}$

which equals 1 for all n.

### Pascal's 2-simplex

$\wedge ^{2}$ is known as Pascal's triangle (sequence A007318 in the OEIS).

**nth component**

$\wedge _{n}^{2}=\vartriangle _{n}^{1}$ (a line) consists of the coefficients of binomial expansion of a polynomial with 2 terms raised to the power of n:

    $(x_{1}+x_{2})^{n}=\sum _{k_{1}+k_{2}=n}{n \choose k_{1},k_{2}}x_{1}^{k_{1}}x_{2}^{k_{2}};\ \ k_{1},k_{2},n\in \mathbb {N} _{0}$

**nth component**

**Arrangement of $\vartriangle _{n}^{1}$**


    $\textstyle {n \choose n,0},{n \choose n-1,1},\cdots ,{n \choose 1,n-1},{n \choose 0,n}$

### Pascal's 3-simplex

$\wedge ^{3}$ is known as Pascal's tetrahedron (sequence A046816 in the OEIS).

**nth component**

$\wedge _{n}^{3}=\vartriangle _{n}^{2}$ (a triangle) consists of the coefficients of trinomial expansion of a polynomial with 3 terms raised to the power of n:

    $(x_{1}+x_{2}+x_{3})^{n}=\sum _{k_{1}+k_{2}+k_{3}=n}{n \choose k_{1},k_{2},k_{3}}x_{1}^{k_{1}}x_{2}^{k_{2}}x_{3}^{k_{3}};\ \ k_{1},k_{2},k_{3},n\in \mathbb {N} _{0}$

**nth component**

**Arrangement of $\vartriangle _{n}^{2}$**


    ${\begin{aligned}\textstyle {n \choose n,0,0}&,\textstyle {n \choose n-1,1,0},\cdots \cdots ,{n \choose 1,n-1,0},{n \choose 0,n,0}\\\textstyle {n \choose n-1,0,1}&,\textstyle {n \choose n-2,1,1},\cdots \cdots ,{n \choose 0,n-1,1}\\&\vdots \\\textstyle {n \choose 1,0,n-1}&,\textstyle {n \choose 0,1,n-1}\\\textstyle {n \choose 0,0,n}\end{aligned}}$

## Properties

### Inheritance of components

$\wedge _{n}^{m}=\vartriangle _{n}^{m-1}$ is numerically equal to each (m − 1)-face (there is m + 1 of them) of $\vartriangle _{n}^{m}=\wedge _{n}^{m+1}$, or:

    $\wedge _{n}^{m}=\vartriangle _{n}^{m-1}\subset \ \vartriangle _{n}^{m}=\wedge _{n}^{m+1}$

From this follows, that the whole $\wedge ^{m}$ is (m + 1)-times included in $\wedge ^{m+1}$, or:

    $\wedge ^{m}\subset \wedge ^{m+1}$

**Example**

| | $\wedge ^{1}$ | $\wedge ^{2}$ | $\wedge ^{3}$ | $\wedge ^{4}$ |
| - | - | - | - | - |
| $\wedge _{0}^{m}$ | 1 | 1 | 1 | 1 |
| $\wedge _{1}^{m}$ | 1 | 1 1 | 1 1 1 | 1 1 1 1 |
| $\wedge _{2}^{m}$ | 1 | 1 2 1 | 1 2 1 2 2 1 | 1 2 1 2 2 1 2 2 2 1 |
| $\wedge _{3}^{m}$ | 1 | 1 3 3 1 | 1 3 3 1 3 6 3 3 3 1 | 1 3 3 1 3 6 3 3 3 1 3 6 3 6 6 3 3 3 3 1 |

For more terms in the above array refer to (sequence A191358 in the OEIS)

### Equality of sub-faces

Conversely, $\wedge _{n}^{m+1}=\vartriangle _{n}^{m}$ is (m + 1)-times bounded by $\vartriangle _{n}^{m-1}=\wedge _{n}^{m}$, or:

    $\wedge _{n}^{m+1}=\vartriangle _{n}^{m}\supset \vartriangle _{n}^{m-1}=\wedge _{n}^{m}$

From this follows, that for given n, all i-faces are numerically equal in nth components of all Pascal's (m > i)-simplices, or:

    $\wedge _{n}^{i+1}=\vartriangle _{n}^{i}\subset \vartriangle _{n}^{m>i}=\wedge _{n}^{m>i+1}$

**Example**

The 3rd component (2-simplex) of Pascal's 3-simplex is bounded by 3 equal 1-faces (lines). Each 1-face (line) is bounded by 2 equal 0-faces (vertices):

Also, for all m and all n:

    $1=\wedge _{n}^{1}=\vartriangle _{n}^{0}\subset \vartriangle _{n}^{m-1}=\wedge _{n}^{m}$

### Number of coefficients

For the nth component ((m − 1)-simplex) of Pascal's m-simplex, the number of the coefficients of multinomial expansion it consists of is given by:

    ${(n-1)+(m-1) \choose (m-1)}+{n+(m-2) \choose (m-2)}={n+(m-1) \choose (m-1)}=\left({\binom {m}{n}}\right),$

(where the latter is the multichoose notation). We can see this either as a sum of the number of coefficients of an (n − 1)th component ((m − 1)-simplex) of Pascal's m-simplex with the number of coefficients of an nth component ((m − 2)-simplex) of Pascal's (m − 1)-simplex, or by a number of all possible partitions of an nth power among m exponents.

**Example**

Number of coefficients of nth component ((m − 1)-simplex) of Pascal's m-simplex :

| m-simplex | nth component | n = 0 | n = 1 | n = 2 | n = 3 | n = 4 | n = 5 |
| - | - | - | - | - | - | - | - |
| 1-simplex | 0-simplex | 1 | 1 | 1 | 1 | 1 | 1 |
| 2-simplex | 1-simplex | 1 | 2 | 3 | 4 | 5 | 6 |
| 3-simplex | 2-simplex | 1 | 3 | 6 | 10 | 15 | 21 |
| 4-simplex | 3-simplex | 1 | 4 | 10 | 20 | 35 | 56 |
| 5-simplex | 4-simplex | 1 | 5 | 15 | 35 | 70 | 126 |
| 6-simplex | 5-simplex | 1 | 6 | 21 | 56 | 126 | 252 |

The terms of this table comprise a Pascal triangle in the format of a symmetric Pascal matrix.

### Symmetry

An nth component ((m − 1)-simplex) of Pascal's m-simplex has the (m!)-fold spatial symmetry.

### Geometry

Orthogonal axes $k_{1}...k_{m}$ in m-dimensional space, vertices of component at n on each axis, the tip at [0,...,0] for $n=0$.

### Numeric construction

Wrapped n-th power of a big number gives instantly the n-th component of a Pascal's simplex.

    $\left|b^{dp}\right|^{n}=\sum _{|k|=n}{{\binom {n}{k}}b^{dp\cdot k}};\ \ b,d\in \mathbb {N} ,\ n\in \mathbb {N} _{0},\ k,p\in \mathbb {N} _{0}^{m},\ p:\ p_{1}=0,p_{i}=(n+1)^{i-2}$

where $\textstyle b^{dp}=(b^{dp_{1}},\cdots ,b^{dp_{m}})\in \mathbb {N} ^{m},\ p\cdot k={\sum _{i=1}^{m}{p_{i}k_{i}}}\in \mathbb {N} _{0}$.

