使用C中的递归计算pi的值

Question

so I'm trying to calculate value of pi using recursion. My code looks like this:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <malloc.h>

double pi (int n){
    double s = 0;
    s = s + 4 * ( pow(-1,n+1 ) * (1/(2*n-1) ) );
    if( n > 0 ) {
        pi( n - 1 );
    }
    return s;
}

int main(void) {

    int n,i;
    float *A;
    scanf("%d", &n);
    A = (float*)malloc( n *sizeof(float) );
    for( i = 0 ; i < n; i++ ) {
        A[i] = pi( i + 1 );
    }
    for( i = 0; i < n; i++ ) {
        printf( "%f\n", A[i] );
    }

    return 0;
}

for value of n = 1 , it returns the expected answer, pi = 4 , but for any other value it computes that pi = 0 . Anyone care to explain why?

Answer 1

use s = s + 4 * ( pow(-1,n+1 ) * (1.0/(2*n-1) ) );
instead of s = s + 4 * ( pow(-1,n+1 ) * (1/(2*n-1) ) );
because if n=2 , the (1/(2*n-1)) part will give 1/3 , and as both 1 and 3 are integers , result will be converted to integer that is 0 , thats why u get a 0.

Answer 2

You can do it like this:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <malloc.h>


double pi (int n){
    double s = 0;
    if( n > 0 ) {
      s = s + 4 * ( pow(-1,n+1 ) * (1/(2*(double)n-1) ) );
      s +=  pi( n - 1 );
    }
    return s;
}

int main(void) {
    int n,i;
    float *A;
    scanf("%d", &n);
    A = (float*)malloc( n *sizeof(float) );
    for( i = 0 ; i < n; i++ ) {
        A[i] = pi( i + 1 );
    }
    for( i = 0; i < n; i++ ) {
        printf( "%f\n", A[i] );
    }

    return 0;
}

These are the things that you are doing wrong:

In the function pi , you are only returning the value of s from the first recursive call. The value of s from successive recursive calls is lost.

..

In this piece of code 1/(2*n-1) , since n is an integer, integer division takes place. You need to cast n to a double to avoid losing digits after the floating point

..

The piece of code, s = s + 4 * ( pow(-1,n+1 ) * (1/(2*n-1) ) ); should be placed inside the if condition if( n > 0 ) This is because, when n equals 0, you will simply be adding another 4 to the value of s .

s = s +  4 * ( pow(-1, 0+1 ) * (1/(2*0-1) ) );
s = s +  4 * ( -1 * -1)
s = s +  4