text: Lagrange 's Interpolation | C # program for implementation of Lagrange 's Interpolation ; To represent a data point corresponding to x and y = f ( x ) ; function to interpolate the given data points using Lagrange 's formula xi corresponds to the new data point whose value is to be obtained n represents the number of known data points ; double result = 0 ; Initialize result ; Compute individual terms of above formula ; Add current term to result ; Driver code ; creating an array of 4 known data points ; Using the interpolate function to obtain a data point corresponding to x = 3
code: using System ; class GFG { class Data { public int x , y ; public Data ( int x , int y ) { this . x = x ; this . y = y ; } } ; static double interpolate ( Data [ ] f , int xi , int n ) { for ( int i = 0 ; i < n ; i ++ ) { double term = f [ i ] . y ; for ( int j = 0 ; j < n ; j ++ ) { if ( j != i ) term = term * ( xi - f [ j ] . x ) / ( f [ i ] . x - f [ j ] . x ) ; } result += term ; } return result ; } public static void Main ( String [ ] args ) { Data [ ] f = { new Data ( 0 , 2 ) , new Data ( 1 , 3 ) , new Data ( 2 , 12 ) , new Data ( 5 , 147 ) } ; Console . Write ( " Value ▁ of ▁ f ( 3 ) ▁ is ▁ : ▁ " + ( int ) interpolate ( f , 3 , 4 ) ) ; } }