Methods in C version Toma 0.0.1

by **BIT0107-Toma** Fri May 21, 2010 6:28 am

CHAPTER 6:

6.6. Bisection Method:

Code:: #include<stdio.h> #include<math.h> double f(double x); int main(void){ double a=-2, b=-1, c, fa = f(a), fb = f(b), fc, e = .000001; int count=0; if(fa*fb>0){ printf("Root is not in range\n"); return 0; } do{ c=(a+b)/2; fc = f(c); if(fa*fc b = c; fb = fc; } else{ a = c; fa = fc; } count++; }while(fabs(b-a)>e); printf("The root is %f & solved by %d iterations\n", c, count); return 0; } double f(double x){ return (x*x) - (4*x) -10; }

6.7. False Position Method:

Code:: #include<stdio.h> #include<math.h> double f(double x); int main(void){ double a=1, b=3, c, fa = f(a), fb = f(b), fc, e = .000001; int count=0; if(fa*fb>0){ printf("Root is not in range\n"); return 0; } do{ c = a - fa*((b-a)/(fb-fa)); fc = f(c); if(fa*fc b = c; fb = fc; } else{ a = c; fa = fc; } count++; }while(fabs(fc)>e); printf("The root is %f & solved by %d iterations\n", c, count); return 0; } double f(double x){ return (x*x) - x - 2; }

6.8. Newton-Raphson Method:

Code:: #include<stdio.h> #include<math.h> double f(double x); double f_(double x); int main(void){ double a=0, b, fa=f(a), f_a=f_(a), fb, e=.000001; int count=0; do{ b = a - (fa/f_a); fb = f(b); a = b; fa = fb; f_a = f_(b); count++; }while(fabs(fb)>e); printf("The root is %f & solved by %d iterations\n", b, count); return 0; } double f(double x){ return (x*x) - (3*x) + 2; } double f_(double x){ return (2*x) - 3; }

6.9. Secant Method:

Code:: #include<stdio.h> #include<math.h> double f(double x); int main(void){ double a=4, b=2, c, fa=f(a), fb=f(b), fc, e=.000001; int count=0; do{ c = ((fb*a)-(fa*b))/(fb-fa); a = b; fa = fb; b = c; fb = f(c); count++; }while(fabs(a-b)>e); printf("The root is %f & solved by %d iterations\n", c, count); return 0; } double f(double x){ return (x*x) - (4*x) - 10; }

6.10. Fixed-Point Method:

Code:: #include #include double f(double x); double g(double x); int main(void){ double a=0, b, c, e=.000001; int count=0; do{ b = g(a); c = a; a = b; count++; }while(fabs(b-c)>e); printf("The root is %f & solved by %d iterations\n", b, count); return 0; } double f(double x){//no need of this. only for understanding the function return (x*x) + x - 2; } double g(double x){ return 2 - (x*x); }

by **BIT0107-Toma** Fri May 21, 2010 6:33 am

CHAPTER 8:

8.2. Jacobi Iteration Method:

Code:: #include<stdio.h> #include<math.h> int main(void){ int n=3; double x0[3], x[3], b[3] = {5, 15, 8}, e=.000001; double a[3][3] = { 2, 1, 1, 3, 5, 2, 2, 1, 4 }; int count=0; int i, j; for(i=0; i<n; i++){ x0[i] = b[i]/a[i][i]; } int isError=0; do{ isError = 0; for(i=0; i<n; i++){ double sum = b[i]; for(j=0; j<n; j++) if(i!=j) sum = sum - a[i][j]*x0[j]; x[i] = sum/a[i][i]; if(isError==0 && fabs(x[i]-x0[i])>e){ isError = 1; x0[i] = x[i]; } } count++; }while(isError); printf("Roots are: "); for(i=0; i<n; i++) printf("%f ", x[i]); printf("& solved by %d iterations\n", count); return 0; }

8.3. Guass-Seidel Method:

Code:: #include<stdio.h> #include<math.h> int main(void){ int n = 3; double x[3], b[3] = {5, 15, 8}, e=.000001; double a[3][3] = { 2, 1, 1, 3, 5, 2, 2, 1, 4 }; int count=0; int i, j; for(i=0; i<n; i++) x[i] = b[i]/a[i][i]; int isError; do{ isError=0; for(i=0; i<n; i++){ double sum = b[i]; for(j=0; j<n; j++) if(i!=j) sum = sum - a[i][j]*x[j]; double dummy = sum/a[i][i]; if(isError==0 && fabs(dummy-x[i])>e) isError = 1; x[i] = dummy; } count++; }while(isError); printf("Roots are: "); for(i=0; i<n; i++) printf("%f ", x[i]); printf("& solved by %d iterations\n", count); return 0; }

8.4. Method of Relaxation:
I have not done it coz I dont know what to assume the value of r [You must be registered and logged in to see this image.] .