/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 For a positive integer \(t,\) de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

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

Step by step solution

01

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.
02

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, \( S(n, n+1) = 0 \).
03

Compute Stirling Numbers for n=1

\[S(1, 0) = 0 \S(1, 1) = 1\]
04

Compute Stirling Numbers for n=2

Using the recurrence relation:\[S(2, 0) = 0 \S(2, 1) = 1 \cdot S(1, 1) + S(1, 0) = 1 \S(2, 2) = 1\]
05

Compute Stirling Numbers for n=3

\[S(3, 0) = 0 \S(3, 1) = 1 \cdot S(2, 1) + S(2, 0) = 1 \S(3, 2) = 2 \cdot S(2, 2) + S(2, 1) = 3 \S(3, 3) = 1\]
06

Compute Stirling Numbers for n=4

\[S(4, 0) = 0 \S(4, 1) = 1 \cdot S(3, 1) + S(3, 0) = 1 \S(4, 2) = 2 \cdot S(3, 2) + S(3, 1) = 7 \S(4, 3) = 3 \cdot S(3, 3) + S(3, 2) = 6 \S(4, 4) = 1\]
07

Compute Stirling Numbers for n=5

\[S(5, 0) = 0 \S(5, 1) = 1 \cdot S(4, 1) + S(4, 0) = 1 \S(5, 2) = 2 \cdot S(4, 2) + S(4, 1) = 15 \S(5, 3) = 3 \cdot S(4, 3) + S(4, 2) = 25 \S(5, 4) = 4 \cdot S(4, 4) + S(4, 3) = 10 \S(5, 5) = 1\]
08

Compute Stirling Numbers for n=6

\[S(6, 0) = 0 \S(6, 1) = 1 \cdot S(5, 1) + S(5, 0) = 1 \S(6, 2) = 2 \cdot S(5, 2) + S(5, 1) = 31 \S(6, 3) = 3 \cdot S(5, 3) + S(5, 2) = 90 \S(6, 4) = 4 \cdot S(5, 4) + S(5, 3) = 65 \S(6, 5) = 5 \cdot S(5, 5) + S(5, 4) = 15 \S(6, 6) = 1\]
09

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:\[\begin{array}{c|cccccc}n \backslash t & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\hline1 & 0 & 1 \2 & 0 & 1 & 1 \3 & 0 & 1 & 3 & 1 \4 & 0 & 1 & 7 & 6 & 1 \5 & 0 & 1 & 15 & 25 & 10 & 1 \6 & 0 & 1 & 31 & 90 & 65 & 15 & 1 \\end{array}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A six-person committee is to be chosen from 16 university students, (4 from each class-first, second, third, and fourth years). Determine how many committees are possible if: (a) Each class is represented. (b) No class has more than two representatives, and each class has at least one representative.

The Old Town Softball League has 16 teams arranged in four groups of four teams each. How many different ways can these groups be made up?

Prove that $$ 1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+\cdots+n \cdot(n+1) \cdot(n+2)=n(n+1)(n+2)(n+3) / 4 $$ This problem should not be solved using a proof by induction.

Internet Addresses: IPv4 and IPv6. The Internet requires an address for each machine that is connected to it. The address space of the addressing architecture of Internet Protocol version 4 (IPv4) consists of a 32 -bit field. Since not every combination of bits can be used as an address, plans are underway to change the address space to a 128 -bit field in IPv6. The 32 -bit IPv4 addresses are usually written in a form called dotted decimal. The 32 bit address is broken up into four 8 -bit bytes, and these bytes are then converted to their equivalent decimal form and separated by dots. For example. $$ \begin{array}{ll} 1000000000000011 & 00000010000000011 \end{array} $$ is written as 128.3 .2 .3 , which is obviously more readable. The 128 -bit IPv6 addresses are divided into eight 16 -bit pieces. Each 16 -bit piece is converted to its equivalent hexadecimal value (each sequence of 4 bits is converted to one hexadecimal digit). The eight four-character hexadecimal strings are separated by colons. It is not prac. tical to list 128 bits and show the conversion to the final IPv6 address form. As an example of what you might end up with, however, we show one IPv6 address: \(\mathrm{FFDC} \cdot \mathrm{BA} 98: 7654 \cdot 3210: \mathrm{FEDC}: \mathrm{BA} 98: 7654 \cdot 3210\) How many IPv4 addresses are possible?

How many ways can you draw a club or a heart from an ordinary deck of cards? A spade or an ace? An ace or a jack? A card numbered 3 through 9 ? A numbered card (Aces are not numbered cards) or a king?

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.