This method uses the newton method to compute the interpolation function on a point x. It takes as parameter a vector representing the abscissae Xi and the ordinates Yi with (Xi, Yi) the interpolation points, the size of the array and the calculation value of the polynomial. It calculates and returns the value of the interpolation polynomial co…

newtonInterpolation.c

double newton(double *x, double *y, int n,double valeur){
  double **difDiv = malloc2(n,n);
  int i,j;
  //affectation et calcul des differences divisées
  for(i = 0; i < n; i++){
    difDiv[i][0] = y[i];
    printf("Dif[%d][0] ====> %f\n",i,difDiv[i][0]);
  }// A LA FIN DE CETTE BOUCLE D[3][0] VAUT 6

  for(j = 1; j < n;j++){

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

      difDiv[i][j] = (difDiv[i+1][j-1]-difDiv[i][j-1])/(x[i+j]-x[i]);
     
    }

  }

  double a = 0.0 ;
  for(int i = 0 ; i < n ; i++){ // les Dif[0][i]
    
    double terme = difDiv[0][i] ;
    for(int j = 0 ; j < i ; j++){ // calcul des facteurs (x-x[j])
    
      terme *= valeur - x[j] ;
    }
    printf("terme %d = %lf\n", i, terme) ;
    a += terme ;
  }

  free(difDiv);
  
  return a;
}