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Chapter 5: Problem 49

A partition of a positive integer \(n\) is a way to write \(n\) as a sum of positive integers where the order of terms in the sum does not matter. For instance, \(7=3+2+1+1\) is a partition of \(7 .\) Let \(P_{m}\) equal the number of different partitions of \(m,\) and let \(P_{m, n}\) be the number of different ways to express \(m\) as the sum of positive integers not exceeding \(n\). a) Show that \(P_{m, m}=P_{m}\) . b) Show that the following recursive definition for \(P_{m, n}\) is correct: $$P_{m, n}=\left\\{\begin{array}{ll}{1} & {\text { if } m=1} \\ {1} & {\text { if } n=1} \\ {P_{m, m}} & {\text { if } m< n} \\ {1+P_{m, m-1}} & {\text { if } m=n>1} \\ {P_{m, n-1}+P_{m-n, n}} & {\text { if } m>n>1}\end{array}\right.$$ c) Find the number of partitions of 5 and of 6 using this recursive definition.

Short Answer

Expert verified
P_{5} = 7, P_{6} = 11.

Step by step solution

01

Prove that P_{m, m} = P_{m}

To show that the number of partitions of a positive integer m with each part not exceeding m is equal to the total number of partitions of m. Any partition of m that has parts greater than m would be invalid. Thus, the number of partitions of m in which each part does not exceed m is exactly the same as the number of partitions of m.
02

Show Correctness of Recursive Definition

A recursive definition of P_{m, n} can be shown correct by analyzing the base cases and recursion step by step:1. For m=1, there is exactly one way to partition 1, which is 1 = 1.2. For n=1, there is exactly one way to partition m, which is m = 1+1+...+1 (m times).3. For m1, we can form the partition either by using m or breaking it down into smaller parts.5. For m>n>1, the partitions either include n or they don't. Hence, P_{m, n-1} corresponds to partitions not using n and P_{m-n, n} to partitions including n.
03

Calculate the number of partitions of 5

Apply the recursive definition:1. Calculate P_{5,5} as it simplifies to P_{5} by Step 1.2. Using the known values for smaller partitions, build up:P_{5,1} = 1, P_{5,2} = P_{5,1} + P_{3,2} = 1 + 2 = 3, P_{5,3} = P_{5,2} + P_{2,3} = 3 + 2 = 5, P_{5,4} = P_{5,3} + P_{1,4} = 5 + 1 = 6, P_{5} = P_{5,4} + P_{0,4} = 6 + 1 = 7.
04

Calculate the number of partitions of 6

Similarly apply the recursive definition to 6:1. Calculate P_{6,6} = P_{6}.2. Using known relations:P_{6,1} = 1, P_{6,2} = P_{6,1} + P_{4,2} = 1 + 3 = 4, P_{6,3} = P_{6,2} + P_{3,3} = 4 + 3 = 7, P_{6,4} = P_{6,3} + P_{2,4} = 7 + 2 = 9, P_{6,5} = P_{6,4} + P_{1,5} = 9 + 1 = 10, P_{6} = P_{6,5} + P_{0,5} = 10 + 1 = 11.

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Key Concepts

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

Recursive Definition
Recursive definitions are widely used in mathematics and computer science. They define a sequence or function in terms of itself, using previously computed terms to calculate new ones. For partition functions, the recursive definition helps us break down the problem into smaller, simpler cases that are easy to solve.
For instance, the recursive definition for the partition function \( P_{m,n} \) is:
  • 1, if \( m = 1 \) or \( n = 1 \)
  • \( P_{m,m} \), if \( m < n \)
  • 1 + \( P_{m,m-1} \), if \( m = n > 1 \)
  • \( P_{m,n-1} + P_{m-n,n} \), if \( m > n > 1 \)
By using these rules, we can successively find the partition of larger numbers step by step.
Positive Integers
Positive integers are the set of all natural numbers excluding zero. In this context, they are used as the building blocks for partitions.
A partition of a positive integer \( n \) is any way of writing \( n \) as a sum of other positive integers. For example, the partitions of 4 include: 4, 3+1, 2+2, 2+1+1, and 1+1+1+1.
Understanding how to break down an integer into different sums is crucial for solving problems involving integer partitions.
Number Theory
Number theory is a branch of pure mathematics focusing on the properties and relationships of numbers, especially integers. It includes concepts like divisibility, the distribution of primes, and the study of integer partitions.
Integer partitions are a part of number theory that deals with writing a positive integer as a sum of other positive integers. The study of these partitions helps us understand the deeper properties of numbers and their behavior.
Applications of integer partitions can be found in combinatorics, algebra, and even physics.
Partition Function
The partition function \( P_{m,n} \) tells us the number of ways to partition a positive integer \( m \) using integers not exceeding \( n \). This function has several properties and base cases:
  • \( P_{m,m} = P_{m} \) 鈥� The partitions of \( m \) without exceeding \( m \) are the total partitions of \( m \)
  • \( P_{m,n-1} \) 鈥� Partitions that do not use \( n \)
  • \( P_{m-n,n} \) 鈥� Partitions that include \( n \)
Understanding the partition function helps break down complex partition problems into manageable pieces by using the defined recursive relationships.

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Most popular questions from this chapter

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers \(n\) greater than \(1,\) which, together with a verification for the case where \(n=1\) , establishes the weaker inequality we originally tried to prove using mathematical induction.

Consider this variation of the game of Nim. The game begins with \(n\) matches. Two players take turns removing matches, one, two, or three at a time. The player removing the last match loses. Use strong induction to show that if each player plays the best strategy possible, the first player wins if \(n=4 j, 4 j+2,\) or \(4 j+3\) for some nonnega- tive integer \(j\) and the second player wins in the remaining case when \(n=4 j+1\) for some nonnegative integer \(j .\)

Pick's theorem says that the area of a simple polygon \(P\) in the plane with vertices that are all lattice points (that is, points with integer coordinates) equals \(I(P)+B(P) / 2-1\) where \(I(P)\) and \(B(P)\) are the number of lattice points in the interior of \(P\) and on the boundary of \(P,\) respectively. Use strong induction on the number of vertices of \(P\) to prove Pick's theorem. [Hint: For the basis step, first prove the theorem for rectangles, then for right triangles, and finally for all triangles by noting that the area of a tri- angle is the area of a larger rectangle containing it with the areas of at most three triangles subtracted. For the inductive step, take advantage of Lemma \(1 . ]\)

Exercises 22 and 23 present examples that show inductive loading can be used to prove results in computational geometry. Let \(P(n)\) be the statement that when non intersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two vertices of the polygon are not endpoints of any of these diagonals. a) Show that when we attempt to prove \(P(n)\) for all integers \(n\) with \(n \geq 3\) using strong induction, the in- ductive step does not go through. b) Show that we can prove that \(P(n)\) is true for all inte- gers \(n\) with \(n \geq 3\) by proving by strong induction the stronger assertion \(Q(n),\) for \(n \geq 4,\) where \(Q(n)\) states that whenever nonintersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two nonadjacent vertices are not endpoints of any of these diagonals.

Show that if \(a_{1}, a_{2}, \ldots, a_{n}\) are \(n\) distinct real numbers, exactly \(n-1\) multiplications are used to compute the product of these \(n\) numbers no matter how parentheses are inserted into their product. [Hint: Use strong induction and consider the last multiplication. \(]\)
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