Question

Write a recursive definition of x^n, where n≥0, similar to the recursive definition of the Fibonacci numbers. How does the recursion terminate?

Answer 1

Answer:

Following are the program to this question:

#include <iostream>//defining header file

using namespace std;

int recurs(int x, int n)//defining a method recurs that accepts two parameter

{

if(n==0)//defining if block that checks n value

{

return 1;//return value 1

}

else//defining else block

{

return x*recurs(x,n-1);//use return keyword that retun value

}

}

int main()//defining main method

{

cout<<recurs(5,3); //use print method to call recurs method

   return 0;

}

Output:

125

Explanation:

In the above-given program, the integer method "recurs" is declared which accepts, two integer variables, which are "x, n", inside the method the if conditional statement is used.

• In the if block, it checks the value of n is equal to "0" if this condition is true, it will return a value, that is 1.
• Otherwise, it will go to the else block, in this block, it will use the recursive method to print its value.    

Answer 2

Recursions are functions that execute itself from within.

The recursive definition x^n in Python, where comments are used to explain each line is as follows:

#This defines the function

def recursion(x, n):

   #This returns 1, if n is 0

   if(n==0):

       return 1

   #This calculates the power recursively, if otherwise

   else:

       return x*recursion(x,n-1);

The function terminates when the value of n is subtracted till 0

Read more about recursions at:

https://brainly.in/question/634885