My Codes: August 2016

PROGRAM FOR TRAPEZOIDAL RULE

/*PROGRAM FOR TRAPEZOIDAL RULE*/

#include<stdio.h>

#include<conio.h>

void main()

{

int n,i;

float x0,xf,y,h;

float f(float);

clrscr();

printf("\n\nEnter the values of x0 & xf::");

scanf("%f%f",&x0,&xf);

printf("\n\nEnter the values of n::");

scanf("%d",&n);

h=(xf-x0)/n;

y=f(x0)+f(xf);

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

{

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

}

y=y*h/2;

printf("\n\nThe result is::%f",y);

getch();

}

float f(float x)

{

return(1/(1+x*x));

}

/* ******************OUTPUT*****************

Enter the values of x0 & xf::0

Enter the values of n::6

The result is::0.784241*/

ALGORITHM OF TRAPEZOIDAL RULE

1-Enter value of lower limit (a) & upper limit (b)

2-Enter the no. of sub-intervals (n)

3-h = (b-a) / n

4- s = y(a) + y(b)

5- for (j=1 to n-1)

{

s = s+2*y(a+i*h)

}

6-Result = (h/2) * s

7- Print result

8- Exit ( )

ALGORITHM OF SIMPSON 3/8 RULE

1-Enter value of upper limit (a) and lower limit(b)

2-Enter value of sub-interval (n)

3-h = (b-a) / n

4-s = y(a) +y(b)

5-for (j = 1 to n-1)

if (j % 3 = = 0)

s = s+2*y(a+j*h)

else

s= s +3*y(a+j*h)

6- result = 3*(h/8) * s

7-Print result

8-exit ( )

ALGORITHM OF SIMPSON’S 1/3 RULE

1-Enter value of upper limit & lower limit

2-Enter value of sub-interval(n)

3-h=(b-a)/n

4-s=y(a)+y(b)

5-For (j=1 to n-1) dø

if( j%2==0)

{

s= s+2*y(a+j*h)

else

s=s+4*y(a+j*h)
}

6-Result = (h/3)*s

7-Print “result”

8- exit( )

PROGRAM FOR SIMPSON 3/8 RULE

/*PROGRAM FOR SIMPSON 3/8 RULE*/

#include<stdio.h>

#include<conio.h>

void main()

{

int n,i;

float x0,xf,y,h;

float f(float);

clrscr();

printf("\n\nEnter the values of x0 & xf::");

scanf("%f%f",&x0,&xf);

printf("\n\nEnter the values of n::");
scanf("%d",&n);

if(n%3==0)

{

h=(xf-x0)/n;

y=f(x0)+f(xf);

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

{

if(i%3==0)

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

else

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

}

y=(3*(y*h))/8;

printf("\n\nThe result is::%f",y);

}

else

goto X;

getch();

}

float f(float x)

{

return(1/(1+x*x));

}

/***************OUTPUT********************

Enter the values of x0 & xf::0

Enter the values of n::6

The result is::1.357081 */

PROGRAM FOR SIMPSON 1/3 RULE

/*PROGRAM FOR SIMPSON 1/3 RULE*/

#include<stdio.h>

#include<conio.h>

void main()

{

int n,i;

float x0,xf,y,h;

float f(float);

clrscr();

printf("\n\nEnter the values of x0 & xf::");

scanf("%f%f",&x0,&xf);

printf("\n\nEnter the values of n::");

scanf("%d",&n);

h=(xf-x0)/n;

y=f(x0)+f(xf);

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

{

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

}

y=y*h/2;

printf("\n\nThe result is::%f",y);

getch();

}

float f(float x)

{

return(1/(1+x*x));

}

/*************OUTPUT******************

Enter the values of x0 & xf::0 6

Enter the values of n::6

The result is::1.410799 */

PROGRAM FOR REGULA FALSI

/*PROGRAM FOR REGULA FALSI*/

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

float x0,x1,x2,e;

int n,i,g=0;

float f(float);

clrscr();

printf("\n enter the value of x0,x1,e,n:\n");

scanf("%f%f%f%d",&x0,&x1,&e,&n);

if((f(x0)*f(x1)<0)

{

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

{

x2=(x0*f(x1)-x1*f(x0)/(f(x1)-f(x0));

}

if(fabs(f(x2))<e)

{

g=1;

break;

}

if(f(x2)*f(x0)<0)

x1=x2;

else

{

x0=x2;

printf("Iteration: %f\n",x2);

}

if(g==0)

printf("\n Root does not found in n iteration upto desired accuracy");

else

printf("\n ANSWER:Root=%f",x2);

getch();

}

float f(float x)

{

return(x*x*x-2*x-5);

}

/****************************OUTPUT*************************/

Enter the value of x0,x1,e,n:

.000006

Iteration:1.545455

Iteration:1.859163

Iteration:2.002119

Iteration:2.059644

Iteration:2.081572

Iteration:2.089974

Iteration:2.092782

Iteration:2.093899

Iteration:2.094311

Iteration:2.094463

Iteration:2.094519

Iteration:2.094539

Iteration:2.094547

Iteration:2.094550

Iteration:2.094551

ANSWER ROOT IS=2.094551*/

PROGRAM FOR NEWTON’S FORWARD DIFF. FORMULA

/*PROGRAM FOR NEWTON’S FORWARD DIFF. FORMULA*/

#include<stdio.h>

#include<conio.h>

#define n 4

void main()

{

float nr=1,dr=1,p,x,y;

float ax[n],ay[n],delta[n-1][n-1];

int i=0,j=0,k;

clrscr();

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

{

printf("\n Enter the value of ax and ay:\n");

scanf("%f%f",&ax,&ay);

}

printf("\n Enter the value of x:");

scanf("%f",&x);

for(i=0;i<n-1;i++)

{

delta[i][0]=(ay[i+1]-ay[i]);

printf("\n %f",delta[i][0]);

}

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

{

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

{

delta[i][j]=delta[i+1][j-1]-delta[i][j-1];

printf("\n\n%f",delta[i][j]);

}

for(i=0;i<n-1;i++)

{

printf("\n\n");

}

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

{

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

}

p=(x-ax[0]/ax[1]-ax[0]);

y=ay[0];

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

{

nr=nr*(p-k);

dr=dr*(k+1);

y+=((nr/dr)*delta[0][k]);

}

printf("ANSWER: VALUE OF Y AT%f IS %f",x,y);

getch();

}

/*OUTPUT

Enter the value of ax and ay:

13229

Enter the value of ax and ay:

31368

Enter the value of ax and ay:

55593

Enter the value of ax and ay:

87089

Enter the value of x:50

18139.000000

24225.000000

31496.000000

6086.000000

7271.000000

1185.000000

ANSWER: VALUE OF y AT 50.000000 IS 42645.687500*/

Tuesday, August 2, 2016