A textbook of Computer Based Numerical and Statiscal Techniques part 60 docx

13.14 ALGORITHM FOR NEWTON’S BACKWARD INTERPOLATIONMETHOD 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 arr

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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] [1]

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

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

Step 22 y4 = p * (p + 1) * (p + 2) * (p−1) * diff [i – 3] [4]/24

Step 23 Print the output x, y

Step 24 End of the program

13.12 PROGRAM FOR CONSTRUCTING DIFFERENCE TABLE

//program for newton forward difference table

#include<conio.h>

#include<stdio.h>

#include<math.h>

int fact(int a)

{

if(a==0)

return 1;

else

return (a*fact(a–1));

}

void main()

{

float x[60], y, diff[5][5], fx[60], u, h, temp = 1.00, sum;

int n, i = 0, j = 0, k = 0;

clrscr();

printf(“enter the no of values”);

scanf(“%d”,&n);

printf(“\n\n enter the values of x having constant difference between them \n”); for (i = 0; i < n; i++)

scanf(“%f”, & x[i]);

printf(“\n enter the values of y = f(x)\n”);

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

scanf(“%f”, & fx[i]);

for(i = 0; i<n–1; i++)

diff[0][i] = fx[i+1]–fx[i];

for(i = 1; i < n–1; i++)

for(j = 0; j < n–1; j++)

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

printf(“n\n\t newton forward difference table is:\n”);

printf(“\nX Y –Y”);

for(k = 2; k < n; k++)

printf(“–^%dY”,k);

printf(“\n”);

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

{

printf(“\n”);

printf(“%f %f”, x[i], fx[i]);

for(j = 0; j < n–1–i; j++) printf(“%f ”,diff[j][i]);

}

getch();

}

13.13 PROGRAMMING FOR NEWTON’S FORWARD INTERPOLATION

METHOD

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<string.h>

int fac(int a)

{

if(a==0)

return (1);

else

return (a*fac(a–1));

}

void main()

{

int x[60], X;

float dif[5][5], fx[60];

float u, h, sum, temp = 1.00;

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nt n, i = 0, j = k = 0;

clrscr();

print(“Enter the no of values”);

scanf(“%d”,&n);

printf(“Enter the values of x having constant diff b/w them\n”);

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

scanf(“%d”, &x[i]);

printf(“Enter the values of Y = f(x)\n”);

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

scanf(“%f”, & fx[i]);

for(i = 0; i < n–1; i++) dif[0][i] = fx[i+1]–fx[i];

for(i = 1; i < n–1; i++) for(j = 0; j < n–1–i; j++) dif[i][j] = dif[i–1][j+1]–dif[i–1][j];

printf(“\n\t The Newton Forward Difference Table is given by:-\n”);

printf(“\nX\tY\t\t–Y”);

for(k = 2; k < n; k++) printf(“\t–^%dY”, k);

printf(“\n”);

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

{

printf(“\n”);

printf(“%d\t%f\t”, x[i], fx[i]);

for(j = 0; j < n–1–i; j++)

printf(“%f ”, dif[j][i]);

printf(“\n\n\tEnter the value of X for which u want F(X)\t”);

scanf(“0%d”, & X);

h = (x[1]–x[0]);

u = ((X–x[0])/h);

sum = fx[0];

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

for(j = 0; j <= i; j++) temp* = (u–j);

sum+ = ((temp/fac(i+1))*dif[i][0]);

temp = 1;

} printf(“\n\n\tThe value of F(%d) is %f\t”, X, sum);

getch();

}

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13.14 ALGORITHM FOR NEWTON’S BACKWARD INTERPOLATION

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 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 14 i = 0

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

Step 16 i = i+1

Step 17 x0= mx[i]

Step 18 Sum = 0, y0 = my[i]

Step 19 fun = 1

Step 20 p = (x – x0)/h

Step 21 Sum = 0

Step 22 for k = 1; k <= 4; k++

Step 23 fun = (fun * (p – (k – 1)))/k

Step 24 sum = sum + fun* diff[i] [k]

Step 25 End of the loop k

Step 26 Print the output x, sum

13.15 PROGRAMMING FOR NEWTON’S BACKWARD INTERPOLATION

METHOD

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<string.h>

int fac(int a)

{

if(a==0)

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return (1);

else

return (a*fac(a–1));

}

void main()

{

int x[60], X;

float dif[5][5], fx[60];

float u, h, sum, temp = 1.00;

int n, i = 0, j = 0, k = 0;

clrscr();

printf(“Enter the no of values”);

scanf(“%d”, &n);

printf(“Enter the values of x having constant diff b/w them\n");

for(i = 0; i < n; i++) scanf(“%d”, & x[i]);

printf(“Enter the values of Y = f(x)\n”);

for(i = 0; i < n; i++) scanf(“%f”, & fx[i]);

for(i = 0; i < n–1; i++) dif[0][i] = fx[i+1]–fx[i];

for(i = 1; i < n–1; i++) for(j =0; j < n–1–i; j++) dif[i][j] = dif[i–1][j+1]–dif[i–1][j];

printf(“\n\tThe Newton Backward Difference Table is given by:-\n”);

printf(“\nX\tY\t\tY”);

for(k = 2; k < n; k++) printf(“\t ^%dY”, k);

printf(“\n”);

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

printf(“\n”);

printf(“%d\t%f\t”, x[i], fx[i]);

for(j = 0; j < n–1–i; j++) printf(“%f ”, dif[j][i]);

}

printf(“\n\n\t Enter the value of X for which u want F(X)\t”);

scanf(“%d”, & X);

h = (x[1]–x[0]);

u = ((X–x[n–1])/h);

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sum = fx[n–1];

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

for(j = 0; j <= i; j++) temp* = (u + j);

sum+ = ((temp/fac(i+1)*dif[i][n–2–i]);

temp = 1;

}

printf(“\n\n\tThe value of F(%d) is %f\t”, X, sum);

getch();

}

13.16 ALGORITHM FOR GAUSS FORWARD INTERPOLATION METHOD

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 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 14 i = 0

Step 16 i = i+1

Step 17 i = i–1

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

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

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

Step 22 y4 = p * (p+1) * (p–1) * (p–2) * diff[i–3][4]/24

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

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13.17 PROGRAMMING FOR GAUSS FORWARD 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[10];

float ay[10];

float x;

float nr, dr;

float h;

float p;

float diff[20][20];

float y1, y2, y3, y4;

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++)

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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];

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 in the form of y–

Enter the value of y1 – 2.7183

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Enter the value of x for

Which you want the value of y – 1.17

When x = 1.17, y = 3.2221

Press enter to exit

13.18 ALGORITHM FOR GAUSS BACKWARD INTERPOLATION METHOD

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 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 14 i = 0

Step 16 i = i+1

Step 17 i = i–1

Step 19 y1 = p *diff[i–1] [1]

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

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

Step 22 y4 = p * (p+1) * (p+2) * (p–1) * diff[i – 3][4]/24

13.19 PROGRAMMING FOR GAUSS BACKWARD INTERPOLATION METHOD

#include<stdio.h>

#include<conio.h>

#include<math.h>

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#include<process.h>

void main()

{

int; n;

int i, j;

float ax[10];

float ay[10];

float x;

float y = 0;

float h;

float p;

float diff[20][20];

float y1, y2, y3, y4;

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++)

{

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