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

nth component
$$\wedge^m_n = \vartriangle^{m-1}_n$$ consists of the coefficients of multinomial expansion of a polynomial with m terms raised to the power of n:
 * $$|x|^n=\sum_{|k|=n}{\binom{n}{k}x^k};\ \ x\in\mathbb{R}^m,\ k\in\mathbb{N}^m_0,\ n\in\mathbb{N}_0,\ m\in\mathbb{N}$$

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 ⋀4
Pascal's 4-simplex, sliced along the k4. All points of the same color belong to the same nth component, from red (for n = 0) to blue (for n = 3).



Pascal's 1-simplex
$\wedge$1 is not known by any special name.



nth component
$$\wedge^1_n = \vartriangle^0_n$$ (a point) is the coefficient of multinomial expansion of a polynomial with 1 term raised to the power of n:
 * $$(x_1)^n = \sum_{k_1=n} {n \choose k_1} x_1^{k_1};\ \ k_1, n \in \mathbb{N}_0$$

Arrangement of $$\vartriangle^0_n$$

 * $$\textstyle {n \choose n}$$

which equals 1 for all n.

Pascal's 2-simplex
$$\wedge^2$$ is known as Pascal's triangle.



nth component
$$\wedge^2_n = \vartriangle^1_n$$ (a line) consists of the coefficients of binomial expansion of a polynomial with 2 terms raised to the power of n:
 * $$(x_1 + x_2)^n = \sum_{k_1+k_2=n} {n \choose k_1, k_2} x_1^{k_1} x_2^{k_2};\ \ k_1, k_2, n \in \mathbb{N}_0$$

Arrangement of $$\vartriangle^1_n$$

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

Pascal's 3-simplex
$$\wedge^3$$ is known as Pascal's tetrahedron.



nth component
$$\wedge^3_n = \vartriangle^2_n$$ (a triangle) consists of the coefficients of trinomial expansion of a polynomial with 3 terms raised to the power of n:
 * $$(x_1 + x_2 + x_3)^n = \sum_{k_1+k_2+k_3=n} {n \choose k_1, k_2, k_3} x_1^{k_1} x_2^{k_2} x_3^{k_3};\ \ k_1, k_2, k_3, n \in \mathbb{N}_0$$

Arrangement of $$\vartriangle^2_n$$


\begin{align} \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{align} $$

Inheritance of components
$$\wedge^m_n = \vartriangle^{m-1}_n$$ is numerically equal to each (m − 1)-face (there is m + 1 of them) of $$\vartriangle^m_n = \wedge^{m+1}_n$$, or:
 * $$\wedge^m_n = \vartriangle^{m-1}_n \subset\ \vartriangle^m_n = \wedge^{m+1}_n$$

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
For more terms in the above array refer to

Equality of sub-faces
Conversely, $$\wedge^{m+1}_n = \vartriangle^m_n$$ is (m + 1)-times bounded by $$\vartriangle^{m-1}_n = \wedge^m_n$$, or:
 * $$\wedge^{m+1}_n = \vartriangle^m_n \supset \vartriangle^{m-1}_n = \wedge^m_n$$

From this follows, that for given n, all i-faces are numerically equal in nth components of all Pascal's (m &gt; i)-simplices, or:
 * $$\wedge^{i+1}_n = \vartriangle^i_n \subset \vartriangle^{m>i}_n = \wedge^{m>i+1}_n$$

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):

2-simplex  1-faces of 2-simplex         0-faces of 1-face 1 3 3 1   1 . . .  . . . 1  1 3 3 1    1 . . .   . . . 1   3 6 3      3 . .    . . 3    . . .    3 3        3 .      . 3      . .     1          1        1.

Also, for all m and all n:
 * $$1 = \wedge^1_n = \vartriangle^0_n \subset \vartriangle^{m-1}_n = \wedge^m_n$$

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
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 k1, ..., km in m-dimensional space, vertices of component at n on each axis, the tip at [0, ..., 0] for n = 0.

Numeric construction
Wrapped nth power of a big number gives instantly the nth 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 $$\textstyle b^{dp} = (b^{dp_1},\cdots,b^{dp_m})\in\mathbb{N}^m,\ p\cdot k={\sum_{i=1}^m{p_i k_i}}\in\mathbb{N}_0$$.