CODEWITHSUNDEEP
cmathwhile-loopsum
infinite
infinite

Reputation: 9

how to evaluate pi in c using while loop

I'm trying to write a code in c to approximate the value of pi using a while loop. I know it is much easier to do so with a for loop but I'm trying to do so using while. the formula I'm using to do so is in link below: https://www.paulbui.net/wl/Taylor_Series_Pi_and_e and the code I wrote looks like this:

#include <stdio.h>
#include <math.h>
int main(){
   long n=10;
   while(n>0){
      double a=0;
      a+=((pow(-1,n))/((2*n)+1));
      n=n-1;
      printf("%ld",4*a);
   }
return 0;
}

the reason I used long and double type is that I wanted to do the approximation to a good preciseness but first I should do st for this problem. thanks in advance.

Upvotes: 0

Views: 464

Answers (2)

user3629249
user3629249

Reputation: 16540

The posted code does not cleanly compile!

gcc -ggdb3 -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled.c" -o "untitled.o" 

untitled.c: In function ‘main’:
untitled.c:7:19: warning: conversion from ‘long int’ to ‘double’ may change value [-Wconversion]
7 |       a+=((pow(-1,n))/((2*n)+1));
  |                   ^

untitled.c:7:22: warning: conversion from ‘long int’ to ‘double’ may change value [-Wconversion]
7 |       a+=((pow(-1,n))/((2*n)+1));
  |                      ^

untitled.c:9:17: warning: format ‘%ld’ expects argument of type ‘long int’, but argument 2 has type ‘double’ [-Wformat=]
9 |       printf("%ld",4*a);
  |               ~~^  ~~~
  |                 |   |
  |                 |   double
  |                 long int
  |               %f

Compilation finished successfully.

Note: when there are warnings, fix those warnings. Also, when there are warnings, the compiler outputs it's best guess which is not necessarily what you wanted.

Upvotes: 0

MBo
MBo

Reputation: 80187

You have to move a initialization before loop and make stop condition - for example, evaluating current summand. Also it is worth to calculate sign incrementally without using pow:

double a=0;
double eps= 1.0e-6; //note this series has rather slow convergence
n = 0;
double tx = 1.0;
double t = 1.0;
while(abs(tx)>eps){
   tx = t / (2*n+1)); 
   a+= tx;
   printf("%f",4*a);
   n++;
   t = - t; 
}

Upvotes: 1

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