Newton Polynomials with Equally Spaced Nodes Solution: x k f(x k ) 0.4 0.423 0.6 0.684 0.8 1.030 1.0 1.557 Use the polynomial to interpolate for f(0.73). Newton Backward-Difference Formula We ordinarily use the interpolating nodes x 0 , x 1 , x 2 , ...., x n . But what if we use: x n , x n-1 , ..... , x 1 , x 0 ? We obtain the interpolating polynomial: P n (x) = f [ x n ] +f [ x n , x n-1 ] (x-x n )+ f [ x n , x n-1 , x n-2 ] (x-x n ) (x-x n-1 )+ . . . +f [ x n , x n-1 , ... , x 0 ] (x-x n ) (x-x n-1 ) ... (x-x 1 )