Question

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

Short answer

Stirling numbers table for \(n=1\) to 6 follows the given relations and recurrence formula.

Step-by-step solution

Initialize the Table

First, create a table with the rows representing different values of \(n\) ranging from 1 to 6 and the columns representing \(t\), ranging from 0 to \(n\). In the initial table, fill out columns for \(t=0\) with zeros for all \(n\), since \(s(n, 0) = 0\) for \(n \geq 1\).

Fill Diagonals with Ones

Fill the main diagonals of the table (where \(n=t\)) with ones because \(s(n, n) = 1\) for \(n = 0, 1, 2, \ldots\). This gives the rightmost values of each row as ones.

Apply Recurrence Relation

Use the recurrence relation \(s(n, t)=s(n-1, t-1)-(n-1)s(n-1, t)\) to fill out the rest of the table. For each \(n\) from 2 to 6, calculate \(s(n, t)\) for each \(t\) from 1 to \(n-1\).

Calculate for each n

For \(n=1\), you have already filled \(s(1, 0)=0\) and \(s(1, 1)=1\). For \(n=2\), fill \(s(2, 1)=0-(1)\times 1 = -1\) and \(s(2, 2)=1\). Continue similarly for each \(n\) from 3 to 6.

Complete the Table

Continue filling in the table using the calculations for \(s(n, t)\) until the table is complete for all \(n\) from 1 to 6.

Key concepts

Recurrence Relations

Recurrence relations are essentially equations that define sequences recursively. They rely on previous terms to calculate the next terms in the sequence. For instance, suppose you're working with numbers or functions and need to determine the current term based on past results. Recurrence relations help establish a rule for this practice, aiding in building the whole sequence without direct computation for each term.

In the context of Stirling numbers of the first kind, the recurrence relation is vital. It states: \[ s(n, t) = s(n-1, t-1) - (n-1) \times s(n-1, t) \]This formula says that any given Stirling number depends on two specific terms from the previous step:

• One term from one row above and one column left, representing \(s(n-1, t-1)\) which helps "build up" the current calculation.
• The other term, \((n-1) \times s(n-1, t)\), reflects a subtraction, critical for the balancing act inherent in these numbers.

Understanding how recurrence relations work is especially helpful in fields like computer science, physics, and more, offering a concise method to handle large sets of data.

Polynomial Sequences

Polynomial sequences are sequences where each term can be described by polynomials. These could be simple or complex, but they always maintain a consistent increase in powers. Within the polynomial sequences, each term is foundational for building cubic, quadratic, and higher-degree polynomials.

When discussing Stirling numbers, the sequence of factorial numbers \([x]_t = x(x-1)(x-2)\cdots(x-t+1)\) represents each polynomial sequence segment. It develops as a polynomial in \(x\), where each segment reflects a polynomial degree. This approach allows us to express these polynomial sequences as a sum involving powers of \(x\):\[ [x]_t = \sum_{t=0}^{n} s(n, t) x^t \] This sum uses Stirling numbers of the first kind as coefficients, turning what initially seems complex into a structured polynomial form. Each part of the polynomial builds on the power of \(x\), with Stirling numbers modulating or shaping how the polynomial looks.

The concept of polynomial sequences helps link number theory and algebra, supplying a framework by which we view numbers through a polynomial "lens," thus expanding their use and understanding in various mathematical fields.

Combinatorial Mathematics

Combinatorial mathematics explores how we count, arrange, and permute elements in sets. It's a fascinating field encompassing permutations, combinations, the binomial theorem, and much more. In daily life, combinatorics finds its place in problems involving arrangements and selections.

Within the realm of Stirling numbers of the first kind, combinatorics plays a critical role. These numbers help in counting permutations, where applying cycles contributes to an underlying structure. Each Stirling number represents permutations of a specific kind – those that can be arranged as a series of overlapping cycles.

• These numbers aren't just about counting; they structure data, dissecting arrangements within cyclical constraints.
• The beauty of combinatorial mathematics lies in its application across various disciplines, from solving puzzles to cracking codes.

Understanding Stirling numbers of the first kind leverages combinatorial methods, breaking down complex permutations into comprehensible parts while uncovering elegant mathematical truths in the process.