/*! 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 8 For \(n \geq 0\), $$ b_{n}=\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 8

For \(n \geq 0\), $$ b_{n}=\left(\frac{1}{n+1}\right)\left(\begin{array}{c} 2 n \\ n \end{array}\right) $$ is the \(n\)th Catalan number. a) Show that for all \(n \geq 0\), $$ b_{n+1}=\frac{2(2 n+1)}{(n+2)} b_{n} $$ b) Use the result of part (a) to write a computer program (or develop an algorithm) that calculates the first 15 Catalan numbers,

Short Answer

Expert verified
a) The proof shows that \(b_{n+1}=\frac{2(2 n+1)}{(n+2)} b_{n}\) holds true for all \(n \geq 0\). b) The algorithm begins with \(b = 1\) then loops 14 times to calculate and print the next Catalan number by using the formula \(\frac{2(2n+1)}{(n+2)}b_n\), updating \(b\) each time.

Step by step solution

01

Rewrite the formula for \(b_{n+1}\)

Begin by replacing \(n+1\) in \(b_{n}\) with \(n\) to get an expression for \(b_{n+1}\), which will be \(\frac{1}{n+2}\) \(\binom{2(n+1)}{n+1}\).
02

Expand the binomial coefficients

Next, expand the binomial coefficient of \(b_{n+1}\) and \(b_{n}\) as factorials to allow manipulation of formula. Here, the binomial coefficient \(\binom{2(n+1)}{n+1}\) becomes \(\frac{(2n+2)!}{(n+1)!(n+1)!}\) and \(\binom{2n}{n}\) becomes \(\frac{(2n)!}{n!n!}\).
03

Manipulate the formula

Now, replace \(b_{n+1}\) and \(b_{n}\) in the formula \(b_{n+1}=\frac{2(2 n+1)}{(n+2)} b_{n}\) with their equivalent factorial expressions. Next, simplify the formula. If the formula matches, then the proof is correct.
04

Develop an algorithm

To calculate the first 15 Catalan numbers, an algorithm/program should be developed. Initialize \(b_0\) as 1 as the base case. To find the next Catalan number, use the formula \(b_{n+1} = \frac{2(2n+1)}{(n+2)}b_n\), where \(b_n\) is the current Catalan number.
05

Write the Algorithm

The algorithm can be written as follows:\n\n1. Set \(b=1\)\n2. Print \(b\)\n3. For \(n\) from 0 to 13\n a. Calculate \(b = \frac{2(2n+1)}{(n+2)}*b\)\n b. Print \(b\)\n\nThis algorithm will produce the first 15 Catalan numbers.

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 91影视!

Key Concepts

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

Discrete Mathematics
Discrete mathematics is a branch of mathematics that deals with discrete elements that can be enumerated by integers. It includes a wide range of topics such as logic, set theory, graph theory, and combinatorics. Combinatorial problems, like those involving the Catalan numbers, are a central part of discrete mathematics.

Understanding problems in discrete mathematics often involves identifying and working with mathematical structures that are inherently discrete rather than continuous. For example, counting the number of ways to arrange or combine items without regard to certain restrictions falls within the realm of discrete mathematics.

The Catalan numbers, a sequence of natural numbers with significant applications in enumerative combinatorics, exemplify a deep relationship between discrete structures and combinatorial mathematics. These numbers play a vital role in various counting problems, such as counting the number of different associative ways to multiply a sequence of matrices or the number of rooted binary trees with a certain number of nodes.
Combinatorial Mathematics
Combinatorial mathematics is the field of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It's closely related to discrete mathematics, as it also deals with discrete elements.

In the context of combinatorial mathematics, the exercise provided relates to the enumeration of specific structures - a problem type commonly addressed by the field. Combinatorial problems can often seem straightforward but require a deep understanding of the principles involved to solve efficiently.

The study of the Catalan numbers is an excellent example of a combinatorial problem. They are sequences used to count combinatorial structures of various types, becoming a fundamental part of the theory. These numbers have an extensive range of applications across different areas of mathematics and computer science, including sorting algorithms and the construction of combinatorial objects like partitions and paths within a lattice.
Binomial Coefficients
Binomial coefficients are the positive integers that occur as coefficients in the binomial theorem. Typically written in the form \( \binom{n}{k} \), they represent the number of ways to choose a subset of k elements from a larger set of n distinct elements without considering order. Binomial coefficients are fundamental to many areas in mathematics and play a prominent role in combinatorial mathematics.

Understanding binomial coefficients is essential for working with problems involving combinations, permutations, and probability. In the provided exercise, binomial coefficients appear within the definition of the Catalan numbers.

The step-by-step solution demonstrates how the recursive relationship between successive Catalan numbers can be proven by manipulating these coefficients, specifically by expanding them into factorial terms which then simplifies to show the pattern. This process is a beautiful illustration of how binomial coefficients can be dynamically used to establish mathematical relationships and properties in both theory and applied problems.

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

Let \(n, k \in \mathbf{Z}^{+}\), and define \(p(n, k)\) to be the number of partitions of \(n\) into exactly \(k\) (positive-integer) summands. Prove that \(p(n, k)=p(n-1, k-1)+p(n-k, k)\).

Consider ternary strings - that is, strings where \(0,1,2\) are the only symbols used. For \(n \geq 1\), let \(a_{n}\) count the number of ternary strings of length \(n\) where there are no consecutive 1 's and no consecutive \(2^{\prime} s\). Find and solve a recurrence relation for \(a_{n}\) -

Solve the following recurrence relations. a) \(a_{n+2}^{2}-5 a_{n+1}^{2}+6 a_{n}^{2}=7 n, \quad n \geq 0, \quad a_{0}=a_{1}=1\) b) \(a_{n}^{2}-2 a_{n-1}=0, \quad n \geq 1, \quad a_{0}=2\) (Let \(b_{n}=\log _{2} a_{n}\), \(n \geq 0 .)\)

In Corollary \(10.2\) we were concerned with finding the appropriate "big-Oh" form for a function \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}^{+} \cup\\{0\\}\) where \(\begin{aligned} f(1) \leq & c, \quad \text { for } c \in \mathbf{Z}^{+} \\\ f(n) \leq & a f(n / b)+c, \\ & \text { for } a, b \in \mathbf{Z}^{+} \text {with } b \geq 2 \text {, and } n=b^{k}, k \in \mathbf{Z}^{+} \end{aligned}\) Here the constant \(c\) in the second inequality is interpreted as he amount of time needed to break down the given problem of size \(n\) into \(a\) smaller (similar) problems of size \(n / b\) and to combine the \(a\) solutions of these smaller problems in order to set a solution for the original problem of size \(n\). Now we shall examine a situation wherein this amount of time is no longer a) Let \(a, b, c \in \mathbf{Z}^{+}\), with \(b \geq 2\). Let \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}^{+} \cup\\{0\\}\) be a monotone increasing function, where $$ \begin{aligned} &f(1) \leq c \\ &f(n) \leq a f(n / b)+c n, \quad \text { for } n=b^{k}, \quad k \in \mathbf{Z}^{+} \end{aligned} $$ Use an argument similar to the one given (for equalities) in Theorem \(10.1\) to show that for all \(n=1, b, b^{2}, b^{3}, \ldots\), $$ f(n) \leq c n \sum_{t=0}^{k}(a / b)^{\prime} $$ b) Use the result of part (a) to show that \(f \in O(n \log n)\), where \(a=b\). (The base for the log function here is any real number greater than 1.) c) When \(a \neq b\), show that part (a) implies that $$ f(n) \leq\left(\frac{c}{a-b}\right)\left(a^{k+1}-b^{k+1}\right) $$ d) From part (c), prove that (i) \(f \in O(n)\), when \(ab\). [Note: The "big-Oh" form for \(f\) here and in part (b) is for \(f\) on \(\mathbf{Z}^{+}\), not just \(\left\\{b^{k} \mid k \in \mathbf{N}\right\\} .\) ]

a) For \(n \in \mathbf{Z}^{+}\), determine the number of ways one can tile a \(1 \times n\) chessboard using \(1 \times 1\) white (square) tiles and \(1 \times 2\) blue (rectangular) tiles. b) How many of the tilings in part (a) use (i) no blue tiles; (ii) exactly one blue tile; (iii) exactly two blue tiles; (iv) exactly three blue tiles; and (v) exactly \(k\) blue tiles, where \(0 \leq k \leq\lfloor n / 2\rfloor ?\) c) How are the results in parts (a) and (b) related?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation
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.