Question

For a positive integer \(t,\) define \([x]_{t}=x(x-1) \cdots(x-t+1) .\) We can represent \(x^{n}\) as a linear combination of \([x]_{t},\) where \(n=1,2,3, \ldots,\) and \(t=0,1,2, \ldots, n\). The coefficients for this expansion are denoted as \(S(n, t)\) and are known as the Stirling numbers of the second kind. Thus, for any \(n\), we can write $$x^{n}=\sum_{t=0}^{n} S(n, t)[x]_{t}$$ The numbers \(S(n, t)\) can be defined for \(n=1,2,3, \ldots\) as \(S(n, 0)=0 ; S(n, n)=1\) : and $$S(n, t)=t S(n-1, t)+S(n-1, t-1)$$ for \(1 \leq t \leq n-1\). Make a table of the Stirling numbers of the second kind for \(n=\) 1,2,3,4,5,6

Short answer

The Stirling numbers for \( n = 1, 2, 3, 4, 5, 6 \) are tabulated from steps 3 to 9.

Step-by-step solution

Understand the Stirling Numbers Definition

Stirling numbers of the second kind, denoted as \( S(n, t) \), represent the number of ways to partition a set of \( n \) objects into \( t \) non-empty subsets. The relationship \( S(n, t) = t \cdot S(n-1, t) + S(n-1, t-1) \) helps to compute these recursively.

Initialize Base Values

For any \( n \), the initial conditions are \( S(n, 0) = 0 \) for all \( n \geq 1 \) (no way to partition a set into zero subsets), and \( S(n, n) = 1 \) (only one way to partition a set of \( n \) elements into \( n \) subsets). In addition, \( S(n, n+1) = 0 \).

Compute Stirling Numbers for n=1

\[S(1, 0) = 0 \S(1, 1) = 1\]

Compute Stirling Numbers for n=2

Using the recurrence relation:\[S(2, 0) = 0 \S(2, 1) = 1 \cdot S(1, 1) + S(1, 0) = 1 \S(2, 2) = 1\]

Compute Stirling Numbers for n=3

\[S(3, 0) = 0 \S(3, 1) = 1 \cdot S(2, 1) + S(2, 0) = 1 \S(3, 2) = 2 \cdot S(2, 2) + S(2, 1) = 3 \S(3, 3) = 1\]

Compute Stirling Numbers for n=4

\[S(4, 0) = 0 \S(4, 1) = 1 \cdot S(3, 1) + S(3, 0) = 1 \S(4, 2) = 2 \cdot S(3, 2) + S(3, 1) = 7 \S(4, 3) = 3 \cdot S(3, 3) + S(3, 2) = 6 \S(4, 4) = 1\]

Compute Stirling Numbers for n=5

\[S(5, 0) = 0 \S(5, 1) = 1 \cdot S(4, 1) + S(4, 0) = 1 \S(5, 2) = 2 \cdot S(4, 2) + S(4, 1) = 15 \S(5, 3) = 3 \cdot S(4, 3) + S(4, 2) = 25 \S(5, 4) = 4 \cdot S(4, 4) + S(4, 3) = 10 \S(5, 5) = 1\]

Compute Stirling Numbers for n=6

\[S(6, 0) = 0 \S(6, 1) = 1 \cdot S(5, 1) + S(5, 0) = 1 \S(6, 2) = 2 \cdot S(5, 2) + S(5, 1) = 31 \S(6, 3) = 3 \cdot S(5, 3) + S(5, 2) = 90 \S(6, 4) = 4 \cdot S(5, 4) + S(5, 3) = 65 \S(6, 5) = 5 \cdot S(5, 5) + S(5, 4) = 15 \S(6, 6) = 1\]

Compile the Stirling Numbers in a Table

The computed Stirling numbers of the second kind for \( n = 1, 2, 3, 4, 5, 6 \) are as follows:\[\begin{array}{c|cccccc}n \backslash t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\hline1 & 0 & 1 \2 & 0 & 1 & 1 \3 & 0 & 1 & 3 & 1 \4 & 0 & 1 & 7 & 6 & 1 \5 & 0 & 1 & 15 & 25 & 10 & 1 \6 & 0 & 1 & 31 & 90 & 65 & 15 & 1 \\end{array}\]

Key concepts

Recurrence Relation

A recurrence relation is a mathematical formula that defines the elements of a sequence using preceding terms. This is particularly useful for calculating complex series where each term can be expressed as a function of its previous terms. In the context of Stirling numbers of the second kind, the recurrence relation is expressed as follows: \[ S(n, t) = t \cdot S(n-1, t) + S(n-1, t-1) \] This relation allows for the number of ways to partition a set of \( n \) objects into \( t \) non-empty subsets to be determined recursively. The first part, \( t \cdot S(n-1, t) \), represents adding one additional element to one of the existing partitions. The second part, \( S(n-1, t-1) \), accounts for creating a new subset with the additional element.

By using this recurrence relation, you get the computational advantage of breaking down complex problems into simpler, more manageable parts. It essentially reduces the number of calculations necessary by reusing previously computed values.

Linear Combination

A linear combination in mathematics refers to an expression constructed from a set of terms multiplied by coefficients. The Stirling numbers of the second kind utilize this concept in expressing powers of \( x \) as: \[ x^n = \sum_{t=0}^{n} S(n, t)[x]_t \] Here, \( S(n, t) \) are the coefficients, and \([x]_t\) denotes the falling factorial \( x(x-1) \cdots (x-t+1) \). This expression means that the power of \( x \) can be broken down into a sum of these falling factorials, each scaled by Stirling numbers.

Linear combination simplifies complex polynomials by representing them in terms of more basic elements. This provides a more flexible and intuitive mathematical toolset for analysis, particularly beneficial in areas like combinatorics, where partitioning and decomposing structures are frequent.

Partitioning Sets

Partitioning a set involves dividing a set into distinct, non-overlapping subsets such that the union of these subsets recovers the original set. Stirling numbers of the second kind specifically denote the number of ways to partition a set of \( n \) elements into \( t \) non-empty subsets.

Each partition is a distinct way of grouping objects where order does not matter, but the composition of different groups does. For example, the set \( \{ A, B, C \} \) can be partitioned into two subsets as \( \{ \{A, B\}, \{C\} \} \) or \( \{ \{A, C\}, \{B\} \} \), among others.

Understanding set partitions is crucial in combinatorics, as they form the basis for many counting problems and logical groupings. They are foundational across mathematics, particularly in fields dealing with probabilistic scenarios, statistics, and number theory.

Combinatorics

Combinatorics is a branch of mathematics focused on counting, understanding, and managing configurations of objects. One vital area within combinatorics is the study of permutations and combinations, which helps determine how elements can be arranged or selected. Stirling numbers of the second kind fall under this umbrella, providing a precise method to count partitions into non-empty subsets. They are key tools when dealing with problems that require partitioning a set, distributing items, or arranging entities in groupings according to defined rules.

In practical applications, combinatorics aids in problem-solving beyond pure mathematics:

• Designing experiments that need randomized group allocation.
• Calculating probabilities in games or scientific data.
• Optimizing resource distribution and scheduling tasks.

Through its methods and principles, combinatorics transforms complicated counting problems into structured and solvable equations.