To Construct the New Data Points Using Stirling's Method of Interpolation - C Data Structure

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Description

To Construct the New Data Points Using Stirling's Method of Interpolation

Demo Code

#include<stdio.h>
#define MAX 20/*from   w ww  .j  ava2 s  . co  m*/

void main()
{
  int i, j, terms = 10;
  float ax[MAX], ay[MAX], x, y, h, p;
  float diff[MAX][5], y1, y2, y3, y4;

  printf("\nEnter the values of x upto 2 decimal points.\n");

  for (i=0; i<terms; i++) {
    printf("Enter the value of x%d: ", i+1);
    scanf("%f",&ax[i]);
  }
  printf("\nNow enter the values of y upto 4 decimal points.\n");
  for (i=0; i < terms; i++) {
    printf("Enter the value of y%d: ", i+1);
    scanf("%f", &ay[i]);
  }

  x = 20;

  h = ax[1] - ax[0];

  for(i=0; i < terms-1; i++)
    diff[i][1] = ay[i+1] - ay[i];

  for(j=2; j <= 4; j++)
    for(i=0; i < terms-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("\n\nFor x = %6.2f,     y = %6.4f", x, y);
}

Result


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