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Use generating functions to explain why the number of partitions of an integer in which each part is used an even number of times equals the generating function for the number of partitions of an integer in which each part is even. (h)

Short Answer

Expert verified
Both generating functions have the form \(\prod_{{k=1}}^{\infty} \frac{1}{{1 - x^{2k}}}\). Thus, they are equal.

Step by step solution

01

- Define the Generating Functions

Start by defining the generating functions. Let the generating function for the number of partitions of an integer in which each part is used an even number of times be denoted as \(P(x)\), and let the generating function for the number of partitions of an integer in which each part is even be \(Q(x)\).
02

- Express Generating Functions for Each Condition

Express \(P(x)\). For each integer part \(k\), the contribution to the generating function is \(\sum_{{m=1}}^{\infty}x^{2mk} = \frac{x^{2k}}{{1 - x^{2k}}}\). Hence, \(P(x) = \prod_{{k=1}}^{\infty} \frac{1}{{1 - x^{2k}}}\).
03

- Express the Function for Even Parts

Now, express \(Q(x)\). For the generating function where each part is even, we only consider even numbers: \(Q(x) = \prod_{{k=1}}^{\infty} \frac{1}{{1 - x^{2k}}}\).
04

- Compare the Two Generating Functions

Observe that both \(P(x)\) and \(Q(x)\) have the same form: \[P(x) = \prod_{{k=1}}^{\infty} \frac{1}{{1 - x^{2k}}} = Q(x)\]. Therefore, the generating function for the number of partitions in which each part is used an even number of times is the same as the generating function for the number of partitions in which each part is even.

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Key Concepts

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

Partitions of Integers
In combinatorics, the concept of partitions of integers is an important topic. A partition of an integer is a way to write the integer as a sum of other integers. For example, the partitions of the integer 4 are:
  • 4
  • 3 + 1
  • 2 + 2
  • 2 + 1 + 1
  • 1 + 1 + 1 + 1
Each partition represents a different grouping of numbers that sum up to the original integer. This helps in understanding how numbers can be broken down and studied in different ways. Partitions are a building block for understanding more complex mathematical concepts like generating functions.
Even Parts
Partitions can also be categorized based on the properties of their parts. One such category is partitions of integers in which each part is even. For example, the partitions of the integer 6 into even parts are:
  • 6
  • 4 + 2
  • 2 + 2 + 2
Here, all parts are even numbers. This special kind of partitioning is useful in various mathematical contexts and serves as a prerequisite for understanding related generating functions. When solving problems involving even parts, it’s essential to focus on the conditions that define the parts of each partition.
Generating Functions
Generating functions are powerful tools in combinatorics used to encode sequences or sets of numbers. They help simplify the process of finding solutions to problems involving partitions and other combinatorial objects. A generating function is typically expressed in the form of a power series:
\[ G(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ... \]
where each coefficient represents a count of a particular combinatorial object, like partitions.
For example, the generating function for the number of partitions of an integer in which each part is even is given by:
\ P(x) = \prod_{{k=1}}^{\infty} \frac{1}{{1 - x^{2k}}} \
This product accounts for each even integer part we're considering. Generating functions offer a compact and elegant way to manage and manipulate large sets of combinatorial data, making them invaluable in advanced mathematical problems.

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Most popular questions from this chapter

In Problem 203, we multiplied together infinitely many power series. Here are two notations for infinite products that look rather similar: $$ \prod_{i=1}^{\infty} 1+x+x^{2}+\cdots+x^{i} \quad \text { and } \quad \prod_{i=1}^{\infty} 1+x^{i}+x^{2 i}+\cdots+x^{i^{2}} $$ However, one makes sense and one doesn't. Figure out which one makes sense and explain why it makes sense and the other one doesn't. If we want a product of the form $$ \prod_{i=1}^{\infty} 1+p_{i}(x) $$ where each \(p_{i}(x)\) is a nonzero polynomial in \(x\) to make sense, describe a relatively simple assumption about the polynomials \(p_{i}(x)\) that will make the product make sense. If we assumed the terms \(p_{i}(x)\) were nonzero power series, is there a relatively simple assumption we could make about them in order to make the product make sense? (Describe such a condition or explain why you think there couldn't be one.) \((\mathrm{m})\)

When studying partitions of integers, it is inconvenient to restrict ourselves to partitions with at most \(m\) parts or partitions with maximum part size \(m\). (a) Give the generating function for the number of partitions of an integer into parts of any size. Don't forget to use \(q\) rather than \(x\) as your variable. (h) (b) Find the coefficient of \(q^{4}\) in this generating function. (h) (c) find the coefficient of \(q^{5}\) in this generating function. (d) This generating function involves an infinite product. Describe the process you would use to expand this product into as many terms of a power series as you choose. (h) (e) Rewrite any power series that appear in your product as quotients of polynomials or as integers divided by polynomials.

We are going to choose a subset of the set \([n]=\\{1,2, \ldots, n\\}\). Suppose we use \(x_{1}\) to be the picture of choosing 1 to be in our subset. What is the picture enumerator for either choosing 1 or not choosing \(1 ?\) Suppose that for each \(i\) between 1 and \(n,\) we use \(x_{i}\) to be the picture of choosing \(i\) to be in our subset. What is the picture enumerator for either choosing \(i\) or not choosing \(i\) to be in our subset? What is the picture enumerator for all possible choices of subsets of \([n] ?\) What should we substitute for \(x_{i}\) in order to get a polynomial in \(x\) such that the coefficient of \(x^{k}\) is the number of ways to choose a \(k\) -element subset of \(n ?\) What theorem have we just reproved (a special case of)? (h)

In Problem 216 you see that when we added numerical multiples of the reciprocals of first degree polynomials we got a fraction in which the denominator is a quadratic polynomial. This will always happen unless the two denominators are multiples of each other, because their least common multiple will simply be their product, a quadratic polynomial. This leads us to ask whether a fraction whose denominator is a quadratic polynomial can always be expressed as a sum of fractions whose denominators are first degree polynomials. Find numbers \(c\) and \(d\) so that $$ \frac{5 x+1}{(x-3)(x+5)}=\frac{c}{x-3}+\frac{d}{x+5} $$

What is the generating function (using \(q\) for the variable) for the number of partitions of an integer in which each part is even? (h)

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