/*! 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 60 Show that each of these proposed... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that each of these proposed recursive definitions of a function on the set of positive integers does not produce a well-defined function. a) \(F(n)=1+F(\lfloor n / 2\rfloor)\) for \(n \geq 1\) and \(F(1)=1\) b) \(F(n)=1+F(n-3)\) for \(n \geq 2, \quad F(1)=2, \quad\) and \(F(2)=3\) c) \(F(n)=1+F(n / 2)\) for \(n \geq 2, F(1)=1,\) and \(F(2)=2\) d) \(F(n)=1+F(n / 2)\) if \(n\) is even and \(n \geq 2, F(n)=1-\) \(F(n-1)\) if \(n\) is odd, and \(F(1)=1\) e) \(F(n)=1+F(n / 2)\) if \(n\) is even and \(n \geq 2, F(n)=$$F(3 n-1)\) if \(n\) is odd and \(n \geq 3,\) and \(F(1)=1\)

Short Answer

Expert verified
Parts (b), (c), (d), and (e) do not produce well-defined functions due to undefined recursion, non-integer results, alternating inconsistencies, and endless recursion, respectively.

Step by step solution

01

Definition of a Well-Defined Function

A function is well-defined if every input corresponds to exactly one output. Therefore, to show that the proposed recursive definitions do not produce well-defined functions, it must be demonstrated that for some inputs, the function does not produce a unique output.
02

Analyze Part (a)

Consider the function definition: \[F(n) = 1 + F(\lfloor n / 2 \rfloor)\text{ for } n \geq 1, \quad F(1)=1\]For each positive integer n, the floor function \(\lfloor n/2 \rfloor\) will eventually reach 1. Since the same recursive process is applied to each integer until 1 is reached, it can be concluded that each call will return to the base case F(1). Hence, part (a) does produce a well-defined function.
03

Analyze Part (b)

Consider the function definition: \[F(n) = 1 + F(n - 3)\text{ for } n \geq 2, \quad F(1)=2, \quad F(2)=3\]Analyzing iteratively, for any n, the steps will reduce n by 3 until F(1) or F(2) is reached. For instance:\(F(5) = 1 + F(2) = 1 + 3 = 4\)\(F(8) = 1 + F(5) = 1 + 4 = 5\)As the steps continue, F(n) will reference potentially negative values, which do not have a definition in this context. This results in undefined behaviors, showing part (b) does not produce a well-defined function.
04

Analyze Part (c)

Consider the function definition: \[F(n) = 1 + F(n / 2)\text{ for } n \geq 2, \quad F(1)=1, \quad F(2)=2\]For instance, if n=4, the function calls:\(F(4) = 1 + F(2) = 1 + 2 = 3\)However, for n=3:\(F(3) = 1 + F(1.5)\)The division may produce non-integer results, leading to ambiguities. Therefore, part (c) does not produce a well-defined function.
05

Analyze Part (d)

Consider the function definition: \[F(n) = 1 + F(n / 2)\text{ if } n\text{ is even and } n \geq 2, \quad F(n) = 1 - F(n-1)\text{ if } n\text{ is odd, and } \quad F(1) = 1\]Proceed by analyzing the recursive calls for both even and odd n. The recursion for even values reduces to n=2 quickly, which is manageable. However, for odd n, we encounter the alternating behavior:\(F(3) = 1 - F(2)\)\(F(5) = 1 - F(4)\)Continuing this pattern leads to inconsistencies, showing part (d) does not produce a well-defined function.
06

Analyze Part (e)

Consider the function definition: \[F(n) = 1 + F(n / 2)\text{ if } n\text{ is even and } n \geq 2, \quad F(n) = F(3n-1)\text{ if } n\text{ is odd and } n \geq 3, \quad F(1) = 1\]For even numbers, the recursion is manageable similar to part (a). However, consider odd n like n=3:\(F(3) = F(8)\)Odd input references another value which leads to continuous odd referencing, creating recursion without natural termination, thus showing part (e) does not produce a well-defined function.

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 91Ó°ÊÓ!

Key Concepts

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

Well-Defined Functions
A function is considered well-defined if it assigns a single, unambiguous output to each input. This means that for every input, there must be one and only one output. If a function can produce different outputs for the same input or if it fails to produce an output for some inputs, it is not well-defined. In the context of the given exercise, we need to determine if each recursive function consistently provides a unique and valid output for all positive integers.
Floor Function
The floor function, denoted by \(\floor{...}\), is a mathematical function that maps a real number to the largest integer less than or equal to that number. For instance, \(\floor{2.9} = 2\) and \(\floor{3} = 3\). This function is particularly useful in recursion as it enables breaking down a problem into more manageable parts. In the given exercise, for example, \(\floor{n/2}\) is employed to reduce the size of n until it reaches the base case.
Recursive Definitions
Recursive definitions specify a function in terms of itself. They include a base case, which is a simple, non-recursive definition, and a recursive step that reduces the complexity of the problem. A well-defined recursive function should ensure every call eventually reaches the base case. However, if expressions lead to cyclic or unresolvable calls, the recursion mechanism may fail, making the function not well-defined. Each option in the exercise deals with how the functions break down positive integers recursively and whether those breakdowns are logically consistent.
Positive Integers
Positive integers are the set of all natural numbers greater than zero: \(1, 2, 3, 4, \ldots\). These are the numbers that we commonly use for counting and order. The exercise focuses on defining functions for positive integers, examining how values evolve under the recursive rules provided. Proper handling of positive integers in recursion is crucial as they ensure every recursive call progresses towards a base case rather than undefined or infinite scenarios.
Function Analysis
Function analysis involves examining the behavior and properties of functions, especially in terms of their outputs based on given inputs. To determine if a function is well-defined, we analyze:
  • The correctness and feasibility of the base case(s).
  • The clarity and logical flow of recursive steps.
  • The ability to reach the base case from any valid input.
In our provided recursive functions, we scrutinize if each reduces the problem size and terminates correctly without leading to contradictions or undefined behavior.

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

a) Determine which amounts of postage can be formed using just 4 -cent and 11 -cent stamps. b) Prove your answer to (a) using the principle of math- ematical induction. Be sure to state explicitly your inductive hypothesis in the inductive step. c) Prove your answer to (a) using strong induction. How does the inductive hypothesis in this proof differ from that in the inductive hypothesis for a proof using mathematical induction?

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.

Use a merge sort to sort 4, 3, 2, 5, 1, 8, 7, 6 into increasing order. Show all the steps used by the algorithm.

Give a recursive algorithm for finding the string \(w^{i},\) the concatenation of \(i\) copies of \(w,\) when \(w\) is a bit string.

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 \(j .\)

See all solutions

Recommended explanations on Math Textbooks

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.