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Chapter 5: Problem 9

Give a recursive algorithm for finding the sum of the first \(n\) odd positive integers.

Short Answer

Expert verified
Define base case as \( \text{sum}(1) = 1 \) and use recursion: \( \text{sum}(n) = \text{sum}(n-1) + (2n - 1) \).

Step by step solution

01

- Define the Problem

To find the sum of the first \(n\) odd positive integers recursively, we need to formulate a function that uses the results of smaller subproblems.
02

- Base Case

Identify the simplest possible case. When \(n = 1\), the sum of the first odd positive integer is the integer itself, which is 1. Thus, we can define the base case as: \[ \text{sum}(1) = 1 \]
03

- Recursive Step

Define the recursive step that builds upon the smaller subproblems. The \(n^{th}\) odd positive integer can be written as \(2n - 1\). Therefore, the sum of the first \(n\) odd positive integers can be written recursively as: \[ \text{sum}(n) = \text{sum}(n-1) + (2n - 1) \]
04

- Combine Base Case and Recursive Step

Combine the base case and recursive step to form the complete recursive function: \[ \text{sum}(n) = \begin{cases} 1 & \text{if } n = 1 \ \text{sum}(n - 1) + (2n - 1) & \text{if } n > 1 \ \text{end{cases}} \]
05

- Implement the Algorithm

Implement the recursive function in a programming language of your choice. For example, in Python: ```pythondef sum_odd_integers(n): if n == 1: return 1 else: return sum_odd_integers(n-1) + (2 * n - 1)```

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Key Concepts

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

recursive function
A recursive function is a function that calls itself to solve smaller instances of the same problem. This approach simplifies complex problems by breaking them down into more manageable subproblems.
Imagine a scenario where you need to solve a complicated puzzle. Instead of doing it all at once, you solve one small piece at a time and use those solutions to complete the bigger puzzle. A recursive function works this way.
For example, in our exercise, the sum of the first \(n\) odd positive integers is calculated using a recursive function. Each call to the function handles a smaller value of \(n\) until it reaches the simplest form, known as the base case.
base case
The base case in a recursive function is the simplest possible instance of the problem. It serves as the stopping point for recursion and prevents infinite loops.
Think of the base case as the foundation of a building. Without it, the building (or our recursive function) would collapse.
In our exercise, the base case is identified when \( n = 1 \). At this point, the sum of the first odd positive integer is simply 1. Therefore, we define it as: \[ \text{sum}(1) = 1 \] This condition ensures that when the recursion reaches \( n = 1 \), it does not call itself again and returns 1.
recursive step
The recursive step is where the function calls itself to solve smaller versions of the same problem. It uses the results from these smaller subproblems to solve the original problem.
In our exercise, we know that each odd positive integer can be represented as \( 2n - 1 \). The sum of the first \( n \) odd positive integers can be derived from the sum of the first \( n - 1 \) odd integers by adding \( 2n - 1 \).
Mathematically, this can be expressed as: \[ \text{sum}(n) = \text{sum}(n - 1) + (2n - 1) \] This step builds upon the base case and continues to call itself until it reaches the simplest form.
sum of odd integers
To find the sum of the first \( n \) odd positive integers using a recursive algorithm, we combine the base case and recursive step.
The complete recursive function can be written as: \[ \text{sum}(n) = \begin{cases} \ 1 & \text{if } n = 1 \ \ \text{sum}(n - 1) + (2n - 1) & \text{if } n > 1 \ \end{cases} \]
This function starts with \( n = 1 \) (our base case) and recursively calls itself, adding \( 2n - 1 \) each time until all odd integers up to the given \( n \) are summed.

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Most popular questions from this chapter

Give a recursive definition of \(w^{i},\) where \(w\) is a string and \(i\) is a nonnegative integer. (Here \(w^{i}\) represents the concatenation of \(i\) copies of the string \(w . )\)

Use strong induction to show that if a simple polygon with at least four sides is triangulated, then at least two of the triangles in the triangulation have two sides that border the exterior of the polygon.

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Show that we can prove the well-ordering property when we take strong induction as an axiom instead of taking the well-ordering property as an axiom.

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.
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