PLease explain code SAMPLE PROBLEM # 4 ON RECURSIVE FUNCTIONS A function f(n) is

Question

PLease explain code

SAMPLE PROBLEM # 4 ON RECURSIVE FUNCTIONS

A function f(n) is given to be f(n) = 2 + 4 + 6 + …+ 2n.

Show that this function f(n) can be represented as a recursive function.

Write down the sequence of steps which the compiler will go through in order to recursively compute the value of f(n) for a sample value n=3 of n.

Design a program in C using a prototype function and the recursive process to compute the value of f(n) for a user-specified value of n.

SOURCE CODE:

       int f(int n){

              if(n==0){                  // Baseline condition

                    return 0;

              }

              else{

                    return 2*n+f(n-1);

              }   

       }

       int main(){   

              int n;

              printf("Please enter an n: ");

              scanf_s("%d",&n);

              printf("f(%d)=%d ",n,f(n));

              return 0;

       }

Explanation / Answer

initially there should be a command to enter the code in "C"

#include<stdio.h>

this is called standard input and output header file.

comes to the programming

initially

the compiler went to the main function.

int n; means it declares an integer n.

then it will print please enter an n:

you have to give an integer

if you given n=3;

next we have a printf statement it needs n and f(n), we know n but not f(n);

f(n) is a function and also n=3; so it goes f(3);

so n is not equal to "0". so it will return "2*n+f(n-1)". but here n=3, so it will return "2*3+f(3-1)"=="6+f(2)".

now also f(n)is called with n=2. then it goes to the function again checks for n is again not equals to "0".

so it will return "2*n+f(n-1)" here n=2 so it will return "2*2+f(2-1)"="4+f(1)".

no again f(n) is called with n=1. it again goes to the function here n=1 which is not equal to "0".

so ti will return "2*1+f(1-1)"="2+f(0)". here again we call the function so it will goes to it

with n=0; here n=0 so it will return =0;

now total answer="2*3+2*2+2*1+0"

for n=3 returns"2*3+f(2)"=="2*3+2*2+2*1+0"=6+4+2+0=12.

for n=2 returns"2*2+f(1)"

for n=1 returns"2*1+f(0)"

for n=1 returns"0"

then the answer is 12.

what is you are given is the code for prototype recursive programme for f(n).

Thank You.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.