macOS usage of powl() function - C programming

[tom1992]

Hi, this is probably a very simple problem however I cannot get a valid output from the powl() function. everytime I run it, it gives an output of 0.00. This is probably just a simple usage problem but I can't work out how to raise long double variables to the power of other long double variables. My program is an iterative calculation of pi using 4*(4arctan(1/5)-arctan(1/239). I know you could use the inbuilt arctan functions in math.h but i wanted to make it iterative. I have read the man pages on pow()/powf()/powl() and scoured google but can't get it to work.

sorry for if i'm being thick.

/*
 program works by working out pi as 4*(4arctan(1/5)-arctan(1/239)
it approximates arctan(z) as (z^1)/1-(z^3)/3+(z^5)/5-... 
*/

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, const char * argv[]) {
    long double pi, arctan1, arctan2, divisor;
	bool sign;
	int count;
	sign = 0;
	count = 0;
	divisor = 1;
	printf("\n");
	while (count < 100) {
		
		if (sign == 0) 
			
		{
			arctan1 += (( powl( (1/5) , divisor))/divisor );
			arctan2 += (( powl ( (1/239) , divisor))/divisor );
			sign = 1;
			
		}
		
		else 
			
		{
			arctan1 -= (( powl ( (1/5) , divisor))/divisor );
			arctan2 -= (( powl ( (1/239) , divisor))/divisor );
			sign = 0;
			
		}
		
		
		pi=4*((4*arctan1)-arctan2);
		
		printf("\r%.050Lf", pi);
		count++;
		divisor +=2;
		
	}
	
	return 0;
	
}

 

[ianray]

Two things:

1. arctan1 and arctan2 are not initialized
2. 1/5 and 1/239 are an integer expressions -- change to 1/5. (or 1/5.0 if you prefer) to force floating point

Good luck

 

[tom1992]

thanks. I got it working now

/*
 program works by working out pi as 4*(4arctan(1/5)-arctan(1/239)
it approximates arctan(z) as (z^1)/1-(z^3)/3+(z^5)/5-... 
*/

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, const char * argv[]) {
    long double pi, arctan1, arctan2, divisor, a, b;
	bool sign;
	int count;
	a = (1.0)/(5.0);
	b = (1.0)/(239.0);
	sign = 0;
	count = 0;
	divisor = 1;
	printf("\n");
	while (count < 20) {
		
		if (sign == 0) 
			
		{
			arctan1 += (( powl ( a , divisor))/divisor );
			arctan2 += (( powl ( b , divisor))/divisor );
			sign = 1;
			
		}
		
		else 
			
		{
			arctan1 -= (( powl ( a , divisor))/divisor );
			arctan2 -= (( powl ( b , divisor))/divisor );
			sign = 0;
			
		}
		
		
		pi=4*((4*arctan1)-arctan2);
		
		printf("\r%.061Lf", pi);
		count++;
		divisor +=2;
		
	}
	printf("\n");
	return 0;
	
}

it also requires many less iterations than I expected!

 

[ianray]

Great! Note that arctan1 and arctan2 are not initialized to zero, so your program will fail if the stack contains non-zero values.