Question

Use recursion to compute \(a^{n}\), where \(n\) is a positive integer. Hint: If \(n\) is 1 , then \(a^{N}=4\). If \(n\) is even, then \(a^{n}=\left(a^{n / 2}\right)^{2}\), Otherwise, \(a^{n}=a \times a^{n-1}\).

Short answer

Use recursion with base case \( n=1 \), even case \( a^n=(a^{n/2})^2 \), and odd case \( a^n = a\times a^{n-1} \).

Step-by-step solution

Base Case Analysis

In recursion, we first need to consider the base case to stop the recursion. Here, if \( n = 1 \), the function returns \( a \) because \( a^1 = a \). This is our base case.

Recursive Case for Even Exponent

For an even exponent \( n \), according to the hint, we use the formula \( a^n = (a^{n/2})^2 \). This means we compute \( a^{n/2} \) using recursion and then square the result to get \( a^n \).

Recursive Case for Odd Exponent

For an odd exponent \( n \), the formula to use is \( a^n = a \times a^{n-1} \). This means we multiply \( a \) with the result of \( a^{n-1} \), which is calculated recursively.

Putting It All Together

We create a recursive function to compute \( a^n \). With the base case \( n = 1 \) returning \( a \), we check if \( n \) is even or odd. If even, compute \( a^{n/2} \) and square it; if odd, compute \( a \times a^{n-1} \). This completes the recursion logic.

Key concepts

Base Case Analysis

Understanding base cases is crucial when dealing with recursive algorithms. In a recursive function, the base case determines when the recursion should stop. It serves as a simple condition that returns a straightforward result without further recursive calls.
For the problem of computing \( a^n \) using recursion, the base case is identified when \( n = 1 \). In this scenario, the function directly returns \( a \), because \( a^1 = a \).
This stops any further recursive calls and provides a concrete answer for this specific scenario.
Thus, the base case is the fundamental building block ensuring that the recursion does not run indefinitely, providing a safe endpoint for the function.

Even and Odd Exponents

The distinction between even and odd exponents is a key aspect of the recursive algorithm to compute \( a^n \). This separation simplifies the calculation process and reduces redundancy in computation:

• Even Exponents: If \( n \) is even, the expression \( a^n \) can be simplified using the relation \( a^n = (a^{n/2})^2 \).

This means that instead of multiplying \( a \) by itself \( n \) times, we can halve the exponent, compute the result recursively, and then square it.
This transformation is highly efficient and avoids recalculating the exponent from scratch.

• Odd Exponents: For odd \( n \), the expression changes to \( a^n = a \times a^{n-1} \).

In this case, one instance of \( a \) is multiplied by the recursive result of \( a^{n-1} \), effectively reducing the problem into a smaller, manageable fraction.
This step-by-step reduction into smaller instances ultimately leads to reaching the base case, ensuring that recursion eventually concludes.

Recursion in Programming

Recursion is a widely-used concept in programming that involves solving a problem by breaking it down into smaller instances of the same problem.
This method is particularly effective for tasks that can naturally be divided into subproblems of the same type, such as computing powers or traversing tree structures.
In the recursive computation of \( a^n \), two key components define recursion:

• Recursive Calls: Involves calling the function itself with updated parameters, progressively reducing the problem’s complexity until reaching a base case.


• Base Case: Provides a clear condition for termination, delivering a direct solution for simple instances of the problem.

The power of recursion lies in its ability to simplify and reduce the complexity of code. However, it also demands precise design, especially in defining base cases and ensuring each recursive call moves closer to a base case.
Understanding these elements is vital for leveraging recursion to solve problems efficiently and elegantly.