Pascal's simplex

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

n^th component[]

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

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

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

Example for ${\displaystyle \wedge ^{4}}$[]

Pascal's 4-simplex (sequence A189225 in the OEIS), sliced along the k₄. 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[]

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

n^th component[]

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

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

Arrangement of ${\displaystyle \vartriangle _{n}^{0}}$[]

${\displaystyle \textstyle {n \choose n}}$

which equals 1 for all n.

Pascal's 2-simplex[]

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

n^th component[]

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

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

Arrangement of ${\displaystyle \vartriangle _{n}^{1}}$[]

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

Pascal's 3-simplex[]

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

n^th component[]

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

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

Arrangement of ${\displaystyle \vartriangle _{n}^{2}}$[]

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

Properties[]

Inheritance of components[]

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

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

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

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

Example[]

        ${\displaystyle \wedge ^{1}}$         ${\displaystyle \wedge ^{2}}$        ${\displaystyle \wedge ^{3}}$         ${\displaystyle \wedge ^{4}}$

${\displaystyle \wedge _{0}^{m}}$     1          1          1          1

${\displaystyle \wedge _{1}^{m}}$     1         1 1        1 1        1 1  1
                             1          1

${\displaystyle \wedge _{2}^{m}}$     1        1 2 1      1 2 1      1 2 1  2 2  1
                            2 2        2 2    2
                             1          1

${\displaystyle \wedge _{3}^{m}}$     1       1 3 3 1    1 3 3 1    1 3 3 1  3 6 3  3 3  1
                           3 6 3      3 6 3    6 6    3
                            3 3        3 3      3
                             1          1

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

Equality of sub-faces[]

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

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

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

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

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):

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:

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

Number of coefficients[]

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

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

(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 n^th component ((m − 2)-simplex) of Pascal's (m − 1)-simplex, or by a number of all possible partitions of an n^th power among m exponents.

Example[]

Number of coefficients of n^th 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 n^th component ((m − 1)-simplex) of Pascal's m-simplex has the (m!)-fold spatial symmetry.

Geometry[]

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

Numeric construction[]

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

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

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