A textbook of Computer Based Numerical and Statiscal Techniques part 62 docx

Start of the program to interpolate the given data Step 2.. 13.28 ALGORITHM FOR TRAPEZOIDAL RULE Step 1.. Start of the program for numerical integration Step 2.. Input the number of subi

13.26 ALGORITHM FOR LAGRANGE’S INTERPOLATION METHOD

Step 1 Start of the program to interpolate the given data

Step 2 Input the number of terms n

Step 3 Input the array ax and ay

Step 4 For i = 0; i < n; i++

Step 5 nr = 1

Step 6 dr = 1

Step 7 for j = 0; j < n; j++

Step 8 if j !=1

(a) nr = nr * (x – ax[j])

(b) dr * (ax [i] – ax [j])

Step 9 End loop j

Step 10 y + =( nr / dr)*ay[i]

Step 11 end loop i

Step 12 print output x, y

Step 13 End of the program

13.27 PROGRAMMING FOR LAGRANGE’S INTERPOLATION METHOD

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<string.h>

#include<process.h>

void main()

{

int n;

int i, j;

float ax[100];

float ay[100];

float h;

float p;

float nr, dr;

float x = 0, y = 0;

clrscr();

printf(“enter the no of term–“);

scanf(“%d”, & n);

printf(“enter the value in the form of x–”);

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

printf(“enter the value of x%d”, i+1);

scanf(“%f”, & ax[i]);

}

printf(“enter the value in the form of y”);

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

{

printf(“enter the value of y%d “, i+1);

scanf(“%f”, & ay[i]);

}

printf(“enter the value of x for”);

printf(“which you want the value of y”);

scanf(“%f”, %x);

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

{

nr = 1;

dr = 1;

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

{

if(j != i)

{

nr = nr*(x–ax[j]);

dr = dr*(ax[i]–ax[j]);

}

y = y+(nr/dr)*ay[i];

}

printf(“when x = %5.2f, y = %6.8f”, x, y);

printf(“press enter to exit”);

getch();

}

OUTPUT

Enter the no of term –5

Enter the value in form of x–

Enter the value of x1 – 5

Enter the value of x2 – 7

Enter the value of x3 – 11

Enter the value of x4 – 13

Enter the value of x5 – 17

Enter the value in the form of y–

Enter the value of y1 – 150

Enter the value of y2 – 392

Enter the value of y3 – 1452

Enter the value of y4 – 2366

Enter the value of y5 – 5202

Enter the value of x for

Which you want the value of y – 9.0

When x = 9.0, y = 810.00

Press enter to exit

13.28 ALGORITHM FOR TRAPEZOIDAL RULE

Step 1 Start of the program for numerical integration

Step 2 Input the upper and lower limits a and b

Step 3 Obtain the number of subinterval by h = (b–a)/n

Step 4 Input the number of subintervals

Step 5 sum = 0

Step 6 sum = func(a) + func(b)

Step 7 for i = 1; i < n; i++

Step 8 sum += 2 * func(a + i)

Step 9 End loop i

Step 10 Result = sum *h/2

Step 11 Print output

Step 12 End of the program and start of section func

Step 13 temp = 1/(1+(x * x))

Step 14 Return temp

Step 15 End of section func

13.29 PROGRAMMING FOR TRAPEZOIDAL RULE

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

float h, a, b, s1 = 0, s2 = 0, s3 = 0, s = 0, c, f, y;

int i;

clrscr ( );

printf(“Integrate the equation 1/(1+x^2) with limit 0 to 2”);

printf(“\n enter the initial and final limits =:”);

scanf(“%f %f”, &a, &b);

printf(“\n enter the interval = ”);

scanf(“%f”, &h);

c = 0, y = 0;

for (i = 0; y <= b; i++)

{

s1 = 0;

s2 = 0;

c = h;

f = 1/ (1+(y*y));

if (i == a :: y == b)

s1 = f/2;

else

s2 = f;

s3 = s3+_(s1 + s2);

y = y + c;

}

s = s3*h;

printf(“the exact value of the function = %f”, s);

getch();

}

OUTPUT

Enter the interval = 0.5

The exact value of the function = 1.103846

13.30 ALGORITHM FOR SIMPSON’S 1/3 RULE

Step 1 Start of the program for numerical integration

Step 2 Input the upper and lower limits a and b

Step 3 Obtain the number of subinterval by h = (b–a)/n

Step 4 Input the number of subintervals

Step 5 sum = 0

Step 6 sum = func(a) + 4 * func(a + h) + func(b)

Step 7 for i = 3; i < n; i += 2

Step 8 sum += 2 * func(a + (i–1) * h) + 4 * func(a + i * h)

Step 9 End loop i

Step 10 Result = sum * h/3

Step 11 Print output

Step 12 End of the program and start of section func

Step 13 temp = 1/ (1+(x * x))

Step 14 Return temp

Step 15 End of section func

13.31 PROGRAMMING FOR SIMPSON’S 1/3 RULE

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

float h, a, b, s1 = 0, s2 = 0, s3 = 0, s4 = 0, s = 0, c, f, y;

int i = 0;

clrscr ( );

printf(“Integrate the equation 1/(1+x^2) with limit 0 to 2”);

printf(“\n enter the initial and final limits =:”);

scanf(“%f %f”, &a, &b);

printf(“\n enter the interval = ”);

scanf(“%f”, &h);

c = 0, y = 0;

while (y <= b)

{

s1 = 0;

s2 = 0;

s4 = 0;

c = h;

f = 1/(1+(y*y));

if (i == a :: y == b)

s1 = f;

else

if ((i! = a :: i! = b) && (i%2 == 1))

s2 = 4*f;

else if((i! = a :: i! = b) && (i%2 == 0))

s4 = 2*f;

s3 = s3 + (s1 + s2 + s4);

y = y + c;

i++;

}

s = s3*(h/3);

printf(“the exact value of the function = %f”, s);

getch();

}

OUTPUT

Enter the interval = 1

The exact value of the function = 1.066667

13.32 ALGORITHM FOR SIMPSON’S 3/8 RULE

Step 1 Start of the program for numerical integration

Step 2 Input the upper and lower limits a and b

Step 3 Obtain the number of subinterval by h = (b–a)/n

Step 4 Input the number of subintervals

Step 5 sum = 0

Step 6 sum = func(a) + func(b)

Step 7 for i = 1; i < n; i++

Step 8 if i%3 = 0

Step 9 sum += 2 * func(a + i * h)

Step 10 else

Step 11 sum += 3 * func(a + (i) * h)

Step 12 End loop i

Step 13 Result = sum * 3 * h/8

Step 14 Print output

Step 15 End of the program and start of section func

Step 16 temp = 1/(1+(x * x))

Step 17 Return temp

Step 18 End of section func

13.33 PROGRAMMING FOR SIMPSON’S 3/8 RULE

#include<stdio.h>

#include<conio.h>

float sim(float);

void main()

{

float res, a, b, h, sum;

int i, j, n;

clrscr ( );

printf(“Enter the Range\n”);

printf(“\n Enter the Lower limit a=”);

scanf(“%f”, &a);

printf(“\n Enter the Upper limit b=”);

scanf(“%f”, &b);

printf(“\n Enter the number of sub-intervals= ”);

scanf(“%d”, &n);

h = (b–a)/n;

sum = 0;

res = 1;

sum = sim(a) + sim(b);

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

{

if (i%3 == 0) sum += 2*sim(a + i*h);

else

sum += 3*sim(a + i*h);

}

res = sum*3*h/8;

printf(“\n value of the integral is: % 4f”, res);

getch();

}

float sim(float x)

{

float temp;

temp = 1/(1+(x*x));

return temp;

}

OUTPUT

Enter the Range

Lower limit a = 0

Upper limit b = 6

Enter the number of subintervals = 6

Value of the integral is: 1.3571

13.34 ALGORITHM FOR FITTING A STRAIGHT LINE OF THE FORM

Y = a + bX

Step 1 Start of the program for fitting straight line

Step 2 Input the number of terms observe

Step 3 Input the array ax and ay

Step 4 for i = 0 to observe

Step 5 sum1 + = x[i]

Step 6 sum2 + = y[i]

Step 7 xy[i] = x[i] * y[i]

Step 8 sum3 += xy[i]

Step 9 End of the loop i

Step 10 for i = 0 to observe

Step 11 x2[i] = x[i] * x[i]

Step 12 sum4 += x2[i]

Step 13 End of the loop i

Step 14 temp1 = (sum2 * sum4) – (sum3 * sum1)

Step 15 a = temp1/((observe * sum4) – (sum1 * sum1)

Step 16 b = (sum2 – observe * a)/sum1

Step 17 Print output a, b and line “a + bx”

Step 18 End of the program

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