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Chapter 5: Problem 29

Devise a recursive algorithm to find the \(n\) th term of the sequence defined by \(a_{0}=1, a_{1}=2,\) and \(a_{n}=a_{n-1} \cdot a_{n-2}\) for \(n=2,3,4, \ldots\)

Short Answer

Expert verified
Base cases: \(a_0\) and \(a_1\). Recursive relation: \(a_n = a_{n-1} \times a_{n-2}\). Function: `def find_nth_term(n)`.

Step by step solution

01

- Understand the Problem

We need to find a recursive algorithm to compute the nth term of the sequence where the first two terms are given and each subsequent term is the product of the previous two terms.
02

- Define the Base Cases

The base cases are already provided: \(a_0 = 1\) and \(a_1 = 2\)
03

- Establish the Recursive Relation

For the sequence, the recursive relation given is \(a_n = a_{n-1} \times a_{n-2}\). This should be used when \(n \geq 2\).
04

- Write the Recursive Function

To create the recursive function, we can write the following pseudo-code:```python# Recursive function to find the nth term in the sequencedef find_nth_term(n): if n == 0: return 1 elif n == 1: return 2 else: return find_nth_term(n-1) * find_nth_term(n-2)```This function checks the base cases first and then uses the recursive relation to calculate the nth term.
05

- Analyze the Function

To compute the value of \(a_n\), the function calls itself recursively with the values \(n-1\) and \(n-2\) until it reaches the base cases. This ensures that all previous terms are computed and multiplied correctly.

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Key Concepts

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

Sequence Definition
A sequence is an ordered list of numbers that usually follow a specific pattern or rule. In this case, the sequence begins with the two base values: \(a_{0}=1\) and \(a_{1}=2\). The rest of the sequence is generated using the rule \(a_{n}=a_{n-1} \cdot a_{n-2}\) for \(n \geq 2\). This means you multiply the two preceding terms to get the next term in the sequence.
Let's illustrate this with the first few terms:
  • \(a_0 = 1\)
  • \(a_1 = 2\)
  • \(a_2 = a_1 \cdot a_0 = 2 \cdot 1 = 2\)
  • \(a_3 = a_2 \cdot a_1 = 2 \cdot 2 = 4\)
Understanding these initial terms helps us grasp how the sequence progresses through repeated multiplications of previous terms.
Base Cases
Base cases are critical in defining a recursive algorithm. They establish the initial values from which all other values are derived. For the sequence in question, the base cases are already given: \(a_0 = 1\) and \(a_1 = 2\). These values do not need any further computation as they serve as starting points. Without these base cases, the recursive function wouldn't know where to begin the computations.
In any recursive problem, clearly defining the base cases is essential because:
  • They prevent infinite recursion.
  • They provide known values to build upon.
  • They help in breaking complex problems into manageable calculations.
Recursive Relation
The recursive relation is the backbone of any recursive algorithm. It defines how each term in the sequence is related to earlier terms. For this sequence, the recursive relation is given by: \(a_n = a_{n-1} \cdot a_{n-2}\) when \(n \geq 2\). This relation tells us how to compute any term from the two preceding terms.
In this problem, each term beyond the base cases (\(a_0\) and \(a_1\)) depends on the values of the two preceeding terms. For example:
  • \(a_2 = a_1 \cdot a_0 = 2 \cdot 1 = 2\)
  • \(a_3 = a_2 \cdot a_1 = 2 \cdot 2 = 4\)
This pattern of using previously computed terms to determine the current term is what makes the relation recursive.
Pseudo-Code
Pseudo-code is a high-level representation of your algorithm. It is not meant to run on a computer but rather to illustrate the logic in a way that is easy to understand. Here's the pseudo-code for our recursive function: ```python# Recursive function to find the nth term in the sequence def find_nth_term(n): if n == 0: return 1 elif n == 1: return 2 else: return find_nth_term(n-1) * find_nth_term(n-2)```
Let's break it down:
  • \(if n == 0\): returns the base case \(a_0 = 1\)
  • \(elif n == 1\): returns the base case \(a_1 = 2\)
  • \(else\): uses the recursive relationship to return \(a_n = a_{n-1} \cdot a_{n-2}\)
This pseudo-code accurately represents the logic needed to compute the nth term in this sequence.
Term Computation
Term computation refers to the actual process of calculating the terms of the sequence using the recursive function we defined. Here's how the function works step-by-step:
  • When you call \(find_nth_term(n)\), the function first checks if \(n\) is one of the base cases (0 or 1). If so, it returns the pre-defined value.
  • If \(n\) is not a base case, the function calls itself twice: once with \(n-1\) and once with \(n-2\).
  • These calls continue until the function hits one of the base cases, then it works its way back up, multiplying the results of each call along the way until it computes the term for the original \(n\).
For example, to compute \(a_3\):
  • The function calls \(find_nth_term(2)\) and \(find_nth_term(1)\).
  • \(find_nth_term(2)\) calls \(find_nth_term(1)\) and \(find_nth_term(0)\).
  • It returns 2 for \(find_nth_term(1)\) and 1 for \(find_nth_term(0)\), thus \(a_2 = 2 \cdot 1 = 2\).
This way, the term \(a_3\) is finally computed as \(4\).

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