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75 lines
2.2 KiB
C++
75 lines
2.2 KiB
C++
5 years ago
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/*------------ NEWTON'S DIVIDED DIFFERENCES INTERPOLATION METHOD --------*/
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/* THE PROGRAM GENERATES A DIVIDED DIFFERENCES TABLE FROM GIVEN
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DATA, AND IT CALCULATES THE VALUE OF f(x) AT GIVEN VALUE
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OF xr.
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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 f(x) is to be interpolated.
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VALUES OF x NEED NOT BE EQUALLY SPACED.
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OUTPUTS : Interpolated value of 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],sum,fy,xr;
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/* ARRAY OF y[n][n] ELEMENTS FOR DIVIDED DIFFERENCE TABLE */
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int i,j,k,n,t,m;
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clrscr();
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printf("\n NEWTON'S DIVIDED DIFFERENCES INTERPOLATION METHOD");
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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) is to be"
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" calculated xr = ");
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scanf("%lf",&xr);
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k = n;
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for(j = 1; j < n; j++)
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{
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/* LOOP TO CALCULATE DIVIDED DIFFERENCES IN THE TABLE */
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k = k - 1;
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for(i = 0; i < k; i++)
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{
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y[i][j] = (y[i+1][j-1] - y[i][j-1])/(x[i+j]-x[i]);
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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 TO CALCULATE INTERPOLATED VALUE OF 'y' */
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fy = 1;
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for(m = 0; m < t; m++)
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{
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fy = fy * (xr - x[m]);
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}
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sum = sum + (fy * y[0][t]);
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}
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sum = sum + y[0][0];
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printf("\nThe interpolated value of y at xr = %lf"
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" is yr = %lf\n",xr,sum);
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}
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/*-------------------- END OF PROGRAM ---------------------------------*/
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