numerical method mdu programs

8/2/2019 numerical method mdu programs

http://slidepdf.com/reader/full/numerical-method-mdu-programs 1/36

/*to find the roots of non linear equation using bisection

method*/

#include<stdio.h>

#include<math.h>

float f (float x)

{

return (x*x*x-4*x-9);

}

void bisect (float *x,float a,float b,int *itr)

{

*x=(a+b)/2;

++(*itr);

printf("iteration no. %3d x=%7.5\n",*itr,*x);

}

main()

{

int itr=0,maxitr;

float x,a,b,x1,aerr;

printf("enter value of a,b,""allowed error,maximum iteration\n");

scanf("%f%f%f%d",&a,&b,&aerr,&maxitr);

bisect (&x,a,b,&itr);

do

{

if(f(a)*f(x)<0)



b=x;

else

a=x;

bisect (&x1,a,b,&itr);

if (fabs (x1-x)<aerr)

{

printf("after %d iteration ,root""=%6.4f\n",itr,x1);

return 0;

}

x=x1;

}

while (itr<maxitr);

printf("soln not converges,""iteration not sufficient\n");

return (1);

}



/*result*/

enter value of a,b,allowed error,maximum iteration

2 4 .0001 20

iteration no. 1 x=%7.5















after 15 iteration ,root=2.7065





{

printf("Newton algorithm didn't converge after %d steps.\n",iters);

printf("The final estimate was %19.16e\n", x);

printf("f(%19.16e) = %19.16e\n", x, f(x));

}

return 0;

} /* main */

double newton(double x_0, double allowed_error, int max_iterations,

int* iters_p, int* converged_p)

{

double x = x_0;

double x_prev;

int iter = 0;

do

{

iter++;

x_prev = x;

x = x_prev - f(x_prev)/f_prime(x_prev);

}

while (fabs(x - x_prev) > allowed_error && iter < max_iterations);

if (fabs(x - x_prev) <= allowed_error)

*converged_p = 1;

else

*converged_p = 0;



*iters_p = iter;

return x;

} /* newton algorithm */

double f(double x) {

return x*x-2;

} /* f */

double f_prime(double x) {

return 2*x; //the derivative

} /* f_prime */

/*result*/

Enter x_0, allowed_error, and max_iterations



2 0.000001 10

Newton algorithm converged after 5 steps.

The approximate solution is 1.4142135623730951e+00

f(1.4142135623730951e+00) = 2.7343578790084422e-16


3 0.000001 10



f(1.4142135623731118e+00) = 4.7376165490664590e-14


4 0.000001 10



f(1.4142135623730951e+00) = 2.7343578790084422e-16


5 0.000001 10



f(1.4142135623730951e+00) = 2.7343578790084422e-16


6 0.000001 10





/*curve fitting by least square approximation*/

#include <stdio.h>

main()

{

float augm[3][4]={{0,0,0,0},{0,0,0,0},{0,0,0,0}};

float t,a,b,c,x,y,xsq;

int i,j,k,n;

puts("enter the no. of pairs of "" observed values:");

scanf("%d",&n);

augm[0][0]=n;

for (i=0;i<n;i++)

{

printf("pair no.%d\n",i+1);

scanf("%f%f",&x,&y);

xsq=x*x;

augm[0][1]+=x;

augm[0][2]+=xsq;

augm[1][2]+=x*xsq;



augm[2][2]+=xsq*xsq;

augm[0][3]+=y;

augm[1][3]+=x*y;

augm[2][3]+=xsq*y;

}

augm[1][1]=augm[0][2];

augm[2][1]=augm[1][2];

augm[1][0]=augm[0][1];

augm[2][0]=augm[1][1];

puts("the augmented matrix is :-");

for (i=0;i<3;i++)

{

for (j=0;j<4;j++)

printf("%9.4f",augm[i][j]);

printf("\n");

}

for (j=0;j<3;j++)

for(i=0;i<3;i++)

if(i!=j)

{

t=augm[i][j]/augm[j][j];

for (k=0;k<4;k++)

augm[i][k]-=augm[j][k]*t;

}

a=augm[0][3]/augm[0][0];



b=augm[1][3]/augm[1][1];

c=augm[2][3]/augm[2][2];

printf("a = %8.4f b = %8.4f""c = %8.4f\n",a,b,c);}

/*result*/

enter the no. of pairs of observed values:

8

pair no.1

1 1.3

pair no.2

2 2.5

pair no.3

1.1 2.3

pair no.4

2.6 2.8

pair no.5

3.1 3.5

pair no.6

4.1 4.3

pair no.7

6.1 5.2

pair no.8

3.1 3.7

the augmented matrix is :-

8.0000 23.1000 86.2100 25.6000



23.1000 86.2100 383.3910 87.7800

86.2100 383.39101916.0260 367.9780

a = 0.6307 b = 1.1016 c = -0.0568

/*to solve system of linear equation using gauss elimination

method*/

#include<stdio.h>

#define N 4

main()

{

float a[N][N+1],x[N],t,s;

int i,j,k;

printf("enter the elements of the""argumented matrix rowwise\n");

for(i=0;i<N;i++)

for(j=0;j<N+1;j++)



scanf("%f",&a[i][j]);

for(j=0;j<N-1;j++)

for(i=j+1;i<N;i++)

{

t= a[i][j]/a[i][j];

for (k=0;k<N+1;k++)

a[i][k]-=a[j][k]*t;

}

printf("the upper triangular matrix""is :-\n");

for (i=0;i<N;i++)

{

for (j=0;j<N+1;j++)

printf("%8.4f",a[i][j]);

printf("\n");

}

for (i=N-1;i>=0;i--)

{

s=0;

for (j=i+1;j<N;j++)

s+=a[i][j]*x[j];

x[i]=(a[i][N]-s)/a[i][i];

}

printf("the soln is :-\n");

for (i=0;i<N;i++)

printf("x[%3d]=%7.4f\n",i+1,x[i]);



getch();

}

/*result*/

enter the elements of the argumented matrix rowwise

10 -3 7 5 9

-6 8 2 -1 -4

2 5 3 7 6 1

3 -1 -2 5 8

the upper triangular matrixis :-

10.0000 -3.0000 7.0000 5.0000 9.0000

-16.0000 11.0000 -5.0000 -6.0000 -13.0000

8.0000 -3.0000 1.0000 8.0000 10.0000

-1.0000 -2.0000 -4.0000 -9.0000 -1.0000

the soln is :-

x[ 1]=-4.6273



x[ 2]= 3.0202

x[ 3]= 9.1111

x[ 4]= 0.1111

/*to solve system of linear equation using Gauss Seidal

method*/

#include<stdio.h>



#include<math.h>

#define N 4

main()

{

float a[N][N+1],x[N],aerr,maxerr,t,s,err;

int i,j,itr,maxitr;

for(i=0;i<N;i++) x[i]=0;

printf("enter the elements of the""argumented matrix rowwise\n");

for(i=0;i<N;i++)

for(j=0;j<N+1;j++)

scanf("%f",&a[i][j]);

printf("enter the allowed error,""maximum iterations\n");

scanf("%f%d",&aerr,&maxitr);

printf("iterations x[1] x[2]"" x[3]\n");

for (itr=1;itr<=maxitr;itr++)

{

maxerr=0;

for (i=0;i<N;i++)

{

s=0;

for(j=0;j<N;j++)

{

if(j!=i) s+= a[i][j]*x[j];

t=(a[i][N]-s)/a[i][i];

err=fabs(x[i]-t);



if (err>maxerr) maxerr=err;

x[i]=t;

}

printf("%5d",itr);

for (i=0;i<N;i++)

printf("%9.4f",x[i]);

printf("\n");

if (maxerr<aerr)

{

printf("converges in %3d""iteration\n",itr);

for (i=0;i<N;i++)

printf("x[%3d]=%7.4f\n",i+1,x[i]);

return 0;

}

}

printf("the soln does not converges,""iteration not sufficient\n");

return(1);}}

/*RESULT*/

enter the elements of the argumented matrix rowwise

10 -2 1 -1 3

-2 10 -1 -1 15



-1 -1 10 -2 27

-1 -1 -2 10 -9

Enter the allowed error,maximum iterations

.0001 15

Iteration x[1] x[2] x[3] x[4]

1 0.30000 1.5600 2.8860 -0.1368

2 0.8869 1.9523 2.9566 -0.0248

3 0.9836 1.9899 2.9924 -0.0042

4 0.9968 1.9982 2.9987 -0.0008

5 0.9994 1.9997 2.9998 -0.0001

6 0.9999 1.9999 3.0000 -0.0000

7 1.0000 2.0000 3.0000 -0.0000

converges in 7 iterations

x[ 1]= 1.0000

x[ 2]= 2.0000

x[ 3]= 3.0000

x[ 4]= -0.0000



/*to solve system of linear equation using Gauss Jordan

method*/

#include<stdio.h>

int main()

{

double matrix[10][10],a,b;

int i, j, k, n;

printf("Enter the no of variables: ");

scanf("%d", &n);

printf("Enter the agumented matrix:\n");

for(i = 0; i < n ; i++)

{

for(j = 0; j < (n+1); j++)

{

scanf("%lf", &matrix[i][j]);

}

}

for(i = 0; i < n; i++)

{



for(j = 0; j < n; j++)

{

if(i != j)

{

a = matrix[j][i];

b = matrix[i][i];

for(k = 0; k < n+1; k++)

{

matrix[j][k] = matrix[j][k] - (a/b) * matrix[i][k];

}

}

}

}

for(i = 0; i < n; i++)

{

a = matrix[i][i];

for(j = 0; j < n+1; j++)

{

matrix[i][j] /= a;

}

}

printf("The required solution is: \n\n");

for(i = 0; i < n ; i++){

printf("%c => %.2f", i+97, matrix[i][n]);

printf("\n");



}

return 0;}

/*result*/

Enter the no of variables: 3

Enter the agumented matrix:

2 4 5 6

4 6 1 6

4 7 2 8

The required solution is:

a => -1.14

b => 1.71

c => 0.29

Enter the no of variables: 3

Enter the agumented matrix:

1 -2 -4 -1

2 -8 2 -7

-1 2 5 9

The required solution is:

a => 73.50

b => 21.25



c => 8.00

/*integration of given function using Trapezoidal rule*/

#include<stdio.h>

float y(float x)

{

return 1/(1+x*x);

}

int main()

{

float x0,xn,h,s;

int i,n;



printf("Enter x0, xn, no. of subintervals: ");

scanf("%f%f%d",&x0,&xn,&n);

h = (xn-x0)/n;

s = y(x0) + y(xn);

for(i = 1; i < n; i++)

{

s += 2*y(x0+i*h);

}

printf("Value of integral is %6.4f\n",(h/2)*s);

return 0;

}

/*RESULT*/

Enter x0, xn, no. of subintervals:



45 12 10

Value of integral is -0.0619


10 23 1.01

Value of integral is 0.0766


1.234 12 5




/*integration of given function using Simpson's 3/8 rule*/

#include<stdio.h>

float y(float x)

{

return 1/(1+x*x); //function of which integration is to be calculated

}

int main(){

float x0,xn,h,s;

int i,n,j,flag;

printf("Enter x0, xn, no. of subintervals: ");

scanf("%f%f%d",&x0,&xn,&n);

h = (xn-x0)/n;

s = y(x0)+y(xn);

for(i = 1; i<=n-1;i++)

{

for(j=1;j<=n-1;j++)

{

if(i==3*j)



{

flag = 1;

break;

}

else

flag = 0;

}

if(flag==0)

s += 3*y(x0+i*h);

else

s += 2*y(x0+i*h);

}

printf("Value of integral is %6.4f\n",(3*h/8)*s);

return 0;

}



/*RESULT*/


23 10 21

Value of integral is -0.0562


12 13 2



0.02 1.3 3.6



2.45 1.3 -1.2


/*TO FIND LARGEST EIGEN VALUE OF A MATRIX BY POWER

METHOD*/

#include<stdio.h>



#include<math.h>

#include<stdlib.h>

void matrix_mul(float matrix1[3][3],float matrix2[3][1],float matrix3[3][1])

{

int i,j,k;

for(i = 0; i < 3; i++)

{

for(j = 0; j < 1; j++)

{

matrix3[i][j] = 0;

for(k = 0; k < 3; k++)

{

matrix3[i][j] += matrix1[i][k]*matrix2[k][j];

}

}

}

}

float findmax(float array[3][1])

{

int i;

float maximum;

maximum = array[0][0];

for(i = 0; i < 3; i++){

if(maximum < array[i][0])

maximum = array[i][0];



}

return maximum;

}

int isEqual(float matrix1[3][1], float matrix2[3][1])

{

if(matrix1[0][0] == matrix2[0][0]

&& matrix1[1][0] == matrix2[1][0]

&& matrix1[2][0] == matrix2[2][0])

return 1;

return 0;

}

int main()

{

float matrix1[3][3], matrix2[3][1],result[3][1];

float eigenValue;

int i,j,k;

printf("Enter 3x3 matrix:\n");

for(i = 0; i < 3; i++)

{

for(j = 0; j < 3; j++)

{

scanf("%f",&matrix1[i][j]);

}

}

for(i = 0; i < 3; i++){



for(j = 0; j < 1; j++)

{

matrix2[i][j] = 1.0;

}

}

while(1)

{

matrix_mul(matrix1,matrix2,result);

eigenValue = findmax(result);

for(i = 0; i < 3; i++)

{

result[i][0]/=eigenValue;

}

if(isEqual(matrix2,result)==1)

{

break;

}

for(i = 0; i < 3; i++)

{

matrix2[i][0] = result[i][0];

}

}

printf("Greatest Eigen Value = %f", eigenValue);

printf("\nAny one of Eigen Vector: \n");

for(i = 0; i < 3; i++)



{

for(j = 0; j < 1; j++)

{

printf("%.2f",result[i][j]);

printf("\t");

}

printf("\n");

}

return 0;

}

/*RESULT*/

Enter 3x3 matrix:

2 5 8

2 -1 4

3 -5 3

Greatest Eigen Value = 6.274918



Any one of Eigen Vector:

1.00

0.42

0.27

/*TO FIND NUMERICAL SOLUTION OF DIFFERENTIAL

EQUATION BY RUNGE –KUTTA METHOD*/

#include <stdio.h>

float f(float x,float y)

{

return x+y*y;

}

main()

{

float x,x0,y0,h,xn,y,k1,k2,k3,k4,k;

printf("enter the value of x0,y0,""h,xn\n");

scanf("%f%f%f%f",&x0,&y0,&h,&xn);

x=x0;y=y0;

while(1)



{

if(x==xn) break;

k1=h*f(x,y);

k2=h*f(x+h/2,y+k1/2);

k3=h*f(x+h/2,y+k2/2);

k4=h*f(x+h,y+k3);

k=(k1+(k2+k3)*2+k4)/6;

x+=h;y+=k;

printf("when x = %8.4f""y = %8.4f\n",x,y);}}

/*RESULT*/

enter the value of x0,y0,h,xn

0.0 1.0 0.1 0.2

when x = 0.1000y = 1.1165

when x = 0.2000y = 1.2736

enter the value of x0,y0,h,xn



1.0 2.0 1.1 1.2

when x = 2.1000y = 12484.0469

Floating point error: Overflow.

Abnormal program termination







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numerical method mdu programs