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75 lines
2.3 KiB
C++
75 lines
2.3 KiB
C++
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/*------ NEWTON'S BACKWARD DIFFERENCES INTERPOLATION METHOD -----------*/
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/* THE PROGRAM GENERATES A BACKWARD DIFFERENCES TABLE FROM GIVEN
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DATA, & CALCULATES THE VALUE OF f(x) AT GIVEN VALUE OF x.
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INPUTS : 1) Number of entries of the data.
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2) Values of 'x' & corresponding y = f(x).
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3) Value of 'xr' at which y = f(x) to be calculated.
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OUTPUTS : Interpolated value f(x) at x = xr. */
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/*------------------------------ PROGRAM ---------------------------*/
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#include<stdio.h>
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#include<math.h>
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#include<stdlib.h>
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#include<conio.h>
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void main()
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{
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double y[20][20],x[20],xr,yr,h,r,sum,fy,facto;
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int i,j,k,n,m,t;
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clrscr();
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printf("\n\tNEWTON'S BACKWARD DIFFERENCES INTERPOLATION TECHNIQUE");
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printf("\n\nEnter the number of entries (max 20) = ");
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/* ENTER THE NUMBER OF ENTRIES IN THE TABLE */
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scanf("%d",&n);
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for(i = 0; i < n; i++)
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{
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/* LOOP TO GET x AND y = f(x) IN THE TABLE */
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printf("x%d = ",i);
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scanf("%lf",&x[i]);
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printf(" y%d = ",i);
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scanf("%lf",&y[i][0]);
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}
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printf("\nEnter the value of xr at which y = f(x)\nis to be "
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"interpolated, xr = ");
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scanf("%lf",&xr);
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h = x[1] - x[0];
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r = (xr - x[n-1])/h; /* CALCULATE VALUE OF 'r' */
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printf("\nThe value of h = %lf and value of r = %lf\n",h,r);
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k = 0;
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for(j = 1; j < n; j++)
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{
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/* LOOP TO CALCULATE BACKWARD DIFFERENCES */
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k++;
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for(i = n-1; i >= k; i--)
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{
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y[i][j] = y[i][j-1] - y[i-1][j-1];
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}
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}
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sum = 0;
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for(t = 1; t < n; t++)
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{
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/* LOOP FOR NEWTON'S BACKWARD DIFFERENCE INTERPOLATION FORMULA */
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fy = 1;
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facto = 1;
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for(m = 0; m < t; m++)
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{
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fy = fy * (r + m);
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facto = facto * (m + 1);
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}
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fy = fy * (y[n-1][t]/facto);
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sum = sum + fy;
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}
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yr = sum + y[n-1][0];
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printf("\nThe value of y = f(x) at xr = %lf is yr = %lf", xr,yr);
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}
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/*--------------------------- END OF PROGRAM ---------------------------*/
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