/*    
	Title: Demonstration of computing x=j*h

	Two methods are used to compute x=j*h,
	using (3.40) and (3.41) in the text.
	see Section 3.3 for more information.
*/

#include <stdio.h>

const int n = 20;
const float h = 0.1;

main()
{
	float xsum, xprod, errsum, errprd;
	double dx, dh;
	int j;
 	

	/* Initialize */

	xsum = 0.0;	
	dh = h;
	printf("\nFollowing are the values of j, x=j*h, and");
	printf("\nthe errors in the computation of x using");
	printf("\n(3.40) [product] and (3.41) [sum], respectively.\n\n");
      
	for(j = 1; j <= n; j++) {
		xsum = xsum + h;
		xprod = j*h;
		dx = j*dh;
		errsum = dx - xsum;
		errprd = dx - xprod;
		printf("\t%3d\t%13.10f\t%11.2e\t%11.2e\n", j,dx,errprd,errsum);
	}
	return 0;
}