A textbook of Computer Based Numerical and Statiscal Techniques part 61 pdf

Start of the program to interpolate the given data Step 2.. Input the value of n number of terms Step 3.. Input the array ax for data of x Step 4.. Input the array ay for data of y Step

for(i = 0; i < n–j; i++)

{

diff[i][j] = diff[i+1][j–1]–diff[i][j–1];

}

}

i = 0;

do

{

i++;

}while(ax[i]<x);

i– –;

p = (x–ax[i])/h;

y1 = p*diff[i–1][1];

y2 = p*(p+1)*diff[i–1][2]/2;

y3 = (p+1)*p*(p–1)*diff[i–2][3]/6;

y4 = (p+1)*p*(p-1)*(p+2)*diff[i–3][4]/24

y = ay[i] + y1 + y2 + y3 + y4;

printf(“when x = %6.4f, y = %6.8f”, x, y);

printf(“press enter to exit”);

getch();

}

OUTPUT

Enter the no of term –7

Enter the value in form of x–

Enter the value of x1 – 1.00

Enter the value of x2 – 1.05

Enter the value of x3 – 1.10

Enter the value of x4 – 1.15

Enter the value of x5 – 1.20

Enter the value of x6 – 1.25

Enter the value of x7 – 1.30

Enter the value in the form of y–

Enter the value of y1 – 2.7183

Enter the value of y2 – 2.8577

Enter the value of y3 – 3.0042

Enter the value of y4 – 3.1582

Enter the value of y5 – 3.3201

Enter the value of y6 – 3.4903

Enter the value of y7 – 3.6693

Enter the value of x for

Which you want the value of y – 1.35

When x = 1.35, y = 3.8483

Press enter to exit

13.20 ALGORITHM FOR STIRLING’S METHOD

Step 1 Start of the program to interpolate the given data

Step 2 Input the value of n (number of terms)

Step 3 Input the array ax for data of x

Step 4 Input the array ay for data of y

Step 5 Compute h = ax[1] – ax[0]

Step 6 For i = 0; i < n–1; i++

Step 7 diff[i] [1] = ay[i+1]–ay[i]

Step 8 End of the loop i

Step 9 for j = 2; j <= 4; j++

Step 10 for i = 0; i < n–j; i++

Step 11 diff [i][j] = diff [i+1] [j–1] – diff [i] [j–1]

Step 12 End of the loop i

Step 13 End of the loop j

Step 14 i = 0

Step 15 Repeat step 16 until ax[i] < x

Step 16 i = i+1

Step 17 i = i–1

Step 18 p = (x – ax[i])/h

Step 19 y1 = p * (diff [i][1] + diff [i – 1][1])/2

Step 20 y2 = p * p * diff[i – 2][2]/2

Step 21 y3 = p * (p *p–1)* (diff[i – 1][3] + diff[i–2][3])/6

Step 22 y4 = p * p * (p * p–1) * diff [i – 2] [4]/24

Step 23 y = ay[i] + y1 + y2 + y3 + y4

Step 24 Print the output x, y

Step 25 End of the program

13.21 PROGRAMMING FOR STIRLING’S 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[10];

float ay[10];

float h;

float p;

float x, y;

float diff[20][20];

float y1, y2, y3, y4;

clrscr();

printf(“enter the no of term–“);

scanf(“%d”, & n);

printf(“enter the no 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);

h = ax[1] –ax[0];

for(i = 0; i < n–1; i++)

{

diff[i][1] = ay[i+1]–ay[i];

}

for(j = 2; j <= 4; j++)

{

for(i = 0; i < n–j; i++)

{

diff[i][j] = diff[i+1][j–1]–diff[i][j–1];

}

}

i = 0;

do

{

i++;

}while(ax[i] < x);

i– –;

p = (x–ax[i])/h;

y1 = p*(diff[i][1]+diff[i–1][1]/2;

y2 = p*(p)*diff[i–1][2]/2;

y3 = p*(p*p–1)*(diff[i–1][3]+diff[i–2][3])/6;

y4 = p*p*(p*p–1)*diff[i–2][4]/24;

y = ay[i] + y1 + y2 + y3 + y4;

printf(“when x = %6.4f, y = %6.8f”, x, y);

printf(“press enter to exit”);

getch();

}

OUTPUT

Enter the no of term –7

Enter the value in form of x

Enter the value of x1 – 61

Enter the value of x2 – 62

Enter the value of x3 – 63

Enter the value of x4 – 64

Enter the value of x5 – 65

Enter the value of x6 – 66

Enter the value of x7 – 67

Enter the value in the form of y–

Enter the value of y1 – 1.840431

Enter the value of y2 – 1.858928

Enter the value of y3 – 1.877610

Enter the value of y4 – 1.896481

Enter the value of y5 – 1.915541

Enter the value of y6 – 1.934792

Enter the value of y7 – 1.954237

Enter the value of x for

Which you want the value of y – 0.6440

When x = 0.6440, y = 1.90408230

Press enter to continue

13.22 ALGORITHM FOR BESSEL’S METHOD

Step 1 Start of the program to interpolate the given data

Step 2 Input the value of n (number of terms)

Step 3 Input the array ax for data of x

Step 4 Input the array ay for data of y

Step 5 Compute h = ax[1] – ax[0]

Step 6 For i = 0; i < n–1; i++

Step 7 diff[i] [1] = ay[i+1]–ay[i]

Step 8 End of the loop i

Step 9 for j = 2; j <= 4; j++

Step 10 for i = 0; i < n–j; i++

Step 11 diff [i][j] = diff [i+1] [j–1] – diff [i] [j–1]

Step 12 End of the loop i

Step 13 End of the loop j

Step 14 i = 0

Step 15 Repeat step 16 until ax[i] < x

Step 16 i = i+1

Step 17 i = i–1

Step 18 p = (x – ax[i])/h

Step 19 y1 = p * (diff [i][1])

Step 20 y2 = p * (p–1) * (diff [i] [2] + diff [i – 1] [2]/4

Step 21 y3 = p * (p-1) * (p–0.5) * (diff [i – 1] [3])/6

Step 22 y4 = p * (p+1) * (p–2) * (p–1) * (diff [i – 2] [4] + diff [i – 1][4])/48

Step 23 y = ay[i] + y1 + y2 + y3 + y4

Step 24 Print the output x, y

Step 25 End of the program

13.23 PROGRAMMING FOR BESSEL’S 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[10];

float ay[10];

float h;

float p;

float x, y;

float diff[20][20];

float y1, y2, y3, y4;

clrscr();

printf(“enter the no of term–“);

scanf(“%d”, & n);

printf(“enter the no 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);

h = ax[1]–ax[0];

for(i = 0; i < n–1; i++)

{

diff[i][1] = ay[i+1]–ay[i];

}

for(j = 2; j <= 4; j++)

{

for(i = 0; i < n–j; i++)

{

diff[i][j] = diff[i+1][j–1]–diff[i][j–1];

}

}

i=0;

do

{

}while(ax[i] < x);

i– –;

p = (x–ax[i])/h;

y1 = p*(diff[i][1]);

y2 = p*(p–1)*(diff[i][2]+diff[i–1][2])/4/2;

y3 = p*(p–1)*(p–.5)*(diff[i–1][3])/6;

y4 = (p+1)*p*(p–1)*(p–2)*(diff[i–2][4]+diff[i–1][4])/48;

y = ay[i] + y1 + y2 + y3 + y4;

printf(“when x = %6.4f, y = %6.8f”, x, y);

printf(“press enter to exit”);

getch();

}

OUTPUT

Enter the no of term –7

Enter the value in form of x–

Enter the value of x1 – 61

Enter the value of x2 – 62

Enter the value of x3 – 63

Enter the value of x4 – 64

Enter the value of x5 – 65

Enter the value of x6 – 66

Enter the value of x7 – 67

Enter the value in the form of y–

Enter the value of y1 – 1.840431

Enter the value of y2 – 1.858928

Enter the value of y3 – 1.877610

Enter the value of y4 – 1.896481

Enter the value of y5 – 1.915541

Enter the value of y6 – 1.934792

Enter the value of y7 – 1.954237

Enter the value of x for

Which you want the value of y – 0.6440

When x = 0.6440, y = 1.90408230

Press enter to continue

13.24 ALGORITHM FOR LAPLACE EVERETT METHOD

Step 1 Start of the program to interpolate the given data

Step 2 Input the value of n (number of terms)

Step 3 Input the array ax for data of x

Step 4 Input the array ay for data of y

Step 5 Compute h = ax[1] – ax[0]

Step 6 For i = 0; i < n–1; i++

Step 7 diff[i] [1] = ay[i+1]–ay[i]

Step 8 End of the loop i

Step 9 for j = 2; j <= 4; j++

Step 10 for i = 0; i < n–j; i++

Step 11 diff [i][j] = diff [i+1] [j-1] – diff [i] [j-1]

Step 12 End of the loop i

Step 13 End of the loop j

Step 14 i = 0

Step 15 Repeat step 16 until ax[i] < x

Step 16 i = i+1

Step 17 i = i–1

Step 18 p = (x –ax[i])/h

Step 19 q = 1–p

Step 20 y1 = q * (ay [i])

Step 21 y2 = q * (q * q–1) * (diff [i–1] [2])/6

Step 22 y3 = q * (q * q–1) * (q *q–4) * (diff [i–2] [4])/120

Step 23 py 1 = p * ay [i + 1]

Step 24 py2 = ay[i] + y1 + y2 + y3 + y4

Step 25 Print the output x, y

Step 26 End of the program

13.25 PROGRAMMING FOR LAPLACE EVERETT 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[10];

float ay[10];

float h;

float p, q;

float x, y = 0;

float nr, dr;

float diff[20][20];

float y1, y2, y3, y4;

float py1, py2, py3, py4;

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);

h = ax[1]–ax[0];

for(i = 0; i < n–1; i++)

{

diff[i][1] = ay[i+1]–ay[i];

}

for(j = 2; j < = 4; j++)

{

for(i = 0; i < n–j; i++)

{

diff[i][j] = diff[i+1][j–1]–diff[i][j–1];

}

}

i = 0;

do

{

i++;

}while(ax[i] < x);

i– –;

p = (x–ax[i])/h;

q = 1–p;

y1 = q*(ay[i]);

y2 = q*(q*q–1)*(diff[i–1][2])/6;

y3 = q*(q*q–1)*(q*q–4)*(diff[i–2][4])/120;

py1 = p*ay[i+1];

py2 = p*(p*p–1)*diff[i][2]/6;

py3 = p*(p*p–1)*(p*p–4)*(diff[i–1][4])/120

y = y1 + y2 + y3 + y4 + py1 + py2 + py3;

printf(“when x = %6.4f, y = %6.8f”, x, y);

printf(“press enter to exit”);

getch();

}

OUTPUT

Enter the no of term –7

Enter the value in form of x–

Enter the value of x1 – 1.72

Enter the value of x2 – 1.73

Enter the value of x3 – 1.74

Enter the value of x4 – 1.75

Enter the value of x5 – 1.76

Enter the value of x6 – 1.78

Enter the value of x7 – 1.79

Enter the value in the form of y–

Enter the value of y1 – 1790661479

Enter the value of y2 – 1772844100

Enter the value of y3 – 1755204006

Enter the value of y4 – 1737739435

Enter the value of y5 – 1720448638

Enter the value of y6 – 1703329888

Enter the value of y7 – 1686381473

Enter the value of x for

Which you want the value of y – 1.7475

When x = 1.7475, y = 17420892

Press enter to exit