To find the root of a equation using Secant Method
Program:
#include<stdio.h> #include<conio.h> #include<math.h> #define EPS 0.000001 #define MAXIT 50 #define F(x) (x) * (x) * (x) - 2 * (x) - 5 void main() { float a, b, root, x1, x2 ; int count, status ; clrscr() ; printf("Input two starting points : ") ; scanf("%f %f", &a, &b) ; sec(&a, &b, &x1, &x2, &root, &count, &status) ; if(status == 1) { printf("\nDivision by zero") ; printf("\nLast x1 = %f", x1) ; printf("\nLast x2 = %f", x2) ; printf("\nNo. of iterations = %d", count) ; } else if(status == 2) { printf("\nNo convergence in %d iterations", MAXIT) ; } else { printf("\nRoot = %f", root) ; printf("\n\nFunction value at root = %f", F(root)) ; printf("\n\nNo. of iterations = %d", count) ; } getch() ; } sec(float *a, float *b, float *x1, float *x2, float *root, int *count, int *status) { float x3, f1, f2, error ; *x1 = *a ; *x2 = *b ; f1 = F(*a) ; f2 = F(*b) ; *count = 1 ; begin : if(fabs(f1 - f2) <= 1.E-10) { *status = 1 ; return ; } x3 = *x2 - f2 * (*x2 - *x1) / (f2 - f1) ; error = fabs((x3 - *x2) / x3) ; if(error > EPS) { if(*count == MAXIT) { *status = 2 ; return ; } else { *x1 = *x2 ; } *x2 = x3 ; f1 = f2 ; f2 = F(x3) ; *count = *count + 1 ; goto begin ; } else { *root = x3 ; *status = 3 ; return ; } }
Output:
Input two starting points : 2 3 Root = 2.094552 Function value at root = 0.000001 No. of iterations = 6
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To find the root of a equation using Secant Method