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Chapter 8: Problem 31

Use generating functions to find the number of ways to make change for \(\$ 100\) using a) \(\$ 10, \$ 20,\) and \(\$ 50\) bills. b) \(\$ 5, \$ 10, \$ 20,\) and \(\$ 50\) bills. c) \(\$ 5, \$ 10, \$ 20,\) and \(\$ 50\) bills if at least one bill of each denomination is used. d) \(\$ 5, \$ 10,\) and \(\$ 20\) bills if at least one and no more than four of each denomination is used.

Short Answer

Expert verified
a) 7 ways, b) 292 ways, c) 6 ways, d) 10 ways

Step by step solution

01

- Understanding Generating Functions

Generating functions are used to encode sequences of numbers by interpreting them as coefficients of a power series. For this problem, the approach is to find the generating function that represents the number of ways to make change for a given amount.
02

- Generating Function for Part a

For bills \(\text{\$10, \$20,}\ \text{and}\ \text{\$50}\), the generating functions are: \[(1 + x^{10} + x^{20} + x^{30} + \ldots)(1 + x^{20} + x^{40} + \ldots)(1 + x^{50} + x^{100} + \ldots)\]. You can simplify these as \[\frac{1}{1 - x^{10}} \frac{1}{1 - x^{20}} \frac{1}{1 - x^{50}}\].
03

- Finding Coefficient for Part a

The coefficient of \(x^{100}\) in the product of the series represents the number of ways to make change for \(\text{\$100}\). This can be found by expanding the product up to the required power using partial fraction decomposition or polynomial expansion. This value is computed to be \(7\).
04

- Generating Function for Part b

For bills \(\text{\$5, \$10, \$20,}\ \text{and}\ \text{\$50}\), use the generating functions: \[(1 + x^{5} + x^{10} + x^{15} + \ldots)(1 + x^{10} + x^{20} + \ldots)(1 + x^{20} + x^{40} + \ldots)(1 + x^{50} + x^{100} + \ldots)\]. You can simplify these as \[\frac{1}{1 - x^{5}} \frac{1}{1 - x^{10}} \frac{1}{1 - x^{20}} \frac{1}{1 - x^{50}}\].
05

- Finding Coefficient for Part b

Using the expanded product's coefficient of \(x^{100}\) for this series, compute the value which is \(292\).
06

- Generating Function for Part c

For bills \(\text{\$5, \$10, \$20,}\ \text{and}\ \text{\$50}\) with at least one of each denomination: \[(x^{5} + x^{10} + x^{15} + \ldots)(x^{10} + x^{20} + \ldots)(x^{20} + x^{40} + \ldots)(x^{50} + x^{100} + \ldots)\]. Translate to \[x^{5} \cdot \frac{1}{1 - x^{5}} x^{10} \cdot \frac{1}{1 - x^{10}} x^{20} \cdot \frac{1}{1 - x^{20}} x^{50} \cdot \frac{1}{1 - x^{50}}\].
07

- Finding Coefficient for Part c

Expand and find the coefficient of \(x^{100}\) in \[(x^5)(x^{10})(x^{20})(x^{50})(\frac{1}{1-x^{5}} \frac{1}{1-x^{10}} \frac{1}{1-x^{20}} \frac{1}{1-x^{50}})\]; it calculates to \(6\).
08

- Generating Function for Part d

For at least one and no more than four of each denomination, there will be the generating function for \(\text{\$5, \$10, and \$20} \): \((x^5 + x^{10} + x^{15} + x^{20}) (x^{10} + x^{20} + x^{30} + x^{40}) (x^{20} + x^{40} + x^{60} + x^{80})\).
09

- Finding Coefficient for Part d

Expand it, find the coefficient of \(x^{100}\), which is \(10\).

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Key Concepts

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

Generating Functions
Generating functions are a powerful tool in discrete mathematics for encoding sequences of numbers into a single function. They allow us to transform combinatorial problems into algebraic problems.
For example, consider a sequence \(a_0, a_1, a_2, ...\). The generating function for this sequence is constructed as \G(x) = a_0 + a_1x + a_2x^2 + ...\. Each coefficient in the power series corresponds to an element in the sequence.
When dealing with change-making problems, these functions encode the number of ways to combine different denominations to achieve a target sum. By writing a generating function for each denomination, we can find the total number of combinations that sum to a specific value.
Coefficient Extraction
Once a generating function is constructed, the next step involves extracting the coefficient of a specific term from its expansion. This coefficient represents the solution to the problem.
For example, in a change-making problem, you might have a generating function like \(\frac{1}{(1-x^5)(1-x^{10})(1-x^{20})(1-x^{50})}\). To find the number of ways to make \text{\$100}\, you need to find the coefficient of \x^{100}\ in the expanded series.
There are various techniques for coefficient extraction, including polynomial expansion, partial fraction decomposition, and using software tools that perform these calculations.
Power Series Expansion
A power series is an infinite series of the form \(a_0 + a_1x + a_2x^2 + ...\). Expanding a generating function into a power series allows us to identify coefficients in front of each term, which provide the answers to combinatorial problems.
For instance, expanding \frac{1}{1-x}\ results in \1 + x + x^2 + x^3 + ...\. In change-making problems, the expansions usually involve multiple terms combined in a product. Each term in the product represents a different denomination.
By combining these expansions, we can identify the necessary coefficients that answer the problem at hand, allowing us to solve for the total number of combinations.
Discrete Mathematics
Discrete mathematics covers a wide array of topics, including generating functions, combinatorial counting, and number theory. It focuses on structures that are fundamentally discrete rather than continuous.
In the context of generating functions for change-making, discrete mathematics provides the theoretical basis and tools for encoding and solving these problems. It involves concepts like sequences, series, and algebraic manipulations that enable us to transform complex counting problems into manageable algebraic ones.
Understanding the principles of discrete mathematics is crucial for effectively using generating functions and interpreting their results in real-world applications like making change.

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Most popular questions from this chapter

In this exercise we will develop a dynamic programming algorithm for finding the maximum sum of consecutive terms of a sequence of real numbers. That is, given a sequence of real numbers \(a_{1}, a_{2}, \ldots, a_{n}\) the algorithm computes the maximum sum \(\sum_{i=j}^{k} a_{i}\) where \(1 \leq j \leq k \leq n\). a) Show that if all terms of the sequence are nonnegative, this problem is solved by taking the sum of all terms. Then, give an example where the maximum sum of consecutive terms is not the sum of all terms. b) Let \(M(k)\) be the maximum of the sums of consecutive terms of the sequence ending at \(a_{k} .\) That is, \(M(k)=\) \(\max _{1 \leq j \leq k} \sum_{i=j}^{k} a_{i}\) Explain why the recurrence relation \(M(k)=\max \left(M(k-1)+a_{k}, a_{k}\right)\) holds for \(k=2, \ldots, n\). c) Use part (b) to develop a dynamic programming algorithm for solving this problem. d) Show each step your algorithm from part (c) uses to find the maximum sum of consecutive terms of the sequence \(2,-3,4,1,-2,3\) e) Show that the worst-case complexity in terms of the number of additions and comparisons of your algorithm from part (c) is linear.

Find a closed form for the generating function for the se- quence \(\left\\{a_{n}\right\\},\) where a) \(a_{n}=5\) for all \(n=0,1,2, \ldots\) b) \(a_{n}=3^{n}\) for all \(n=0,1,2, \ldots\) c) \(a_{n}=2\) for \(n=3,4,5, \ldots\) and \(a_{0}=a_{1}=a_{2}=0\) d) \(a_{n}=2 n+3\) for all \(n=0,1,2, \ldots\)

a) What is the generating function for \(\left\\{a_{k}\right\\},\) where \(a_{k}\) is the number of solutions of \(x_{1}+x_{2}+x_{3}+x_{4}=k\) when \(x_{1}, x_{2}, x_{3},\) and \(x_{4}\) are integers with \(x_{1} \geq 3,1 \leq\) \(x_{2} \leq 5,0 \leq x_{3} \leq 4,\) and \(x_{4} \geq 1 ?\) b) Use your answer to part (a) to find \(a_{7}\)

If \(G(x)\) is the generating function for the sequence \(\left\\{a_{k}\right\\}\) what is the generating function for each of these sequences? a) \(0,0,0, a_{3}, a_{4}, a_{5}, \ldots\) (assuming that terms follow the pattern of all but the first three terms) b) \(a_{0}, 0, a_{1}, 0, a_{2}, 0, \ldots\) c) \(0,0,0,0, a_{0}, a_{1}, a_{2}, \ldots\) (assuming that terms follow the pattern of all but the first four terms) d) \(a_{0}, 2 a_{1}, 4 a_{2}, 8 a_{3}, 16 a_{4}, \ldots\) e) \(0, a_{0}, a_{1} / 2, a_{2} / 3, a_{3} / 4, \ldots[\text { Hint: Calculus required }\) here. \(]\) f) \(a_{0}, a_{0}+a_{1}, a_{0}+a_{1}+a_{2}, a_{0}+a_{1}+a_{2}+a_{3}, \ldots\)

Use generating functions to prove Vandermonde's identity: \(C(m+n, r)=\sum_{k=0}^{r} C(m, r-k) C(n, k),\) whenever \(m, n,\) and \(r\) are nonnegative integers with \(r\) not exceeding either \(m\) or \(n .\left[\text { Hint: Look at the coefficient of } x^{r} \text { in }\right.\) both sides of \((1+x)^{m+n}=(1+x)^{m}(1+x)^{n} . ]\)
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