Answer to Quiz 8 Fri 11 Mar 2005

Suppose we take f(x) = x^2 and sample f at x = 0,1,2,3.

1. Give the Newton form of the polynomial interpolating through the points (0,0), (1,1), (2,4), and (3,9).

   Answer:

   
     0   0
             1-0
             --- = 1
             1-0        2-1
     1   1              --- = 1
             4-0        2-1       1-1
             --- = 2              --- = 0
             2-0        3-1       3-2
     2   4              --- = 1
             9-0        3-1
             --- = 3
             3-0
     3   9
   
     p(x) = 0 + 1*(x-0) + 1*(x-0)*(x-1) + 0*(x-0)*(x-1)*(x-2)
   
   verify = x + x*(x-1) = x + x^2 - x = x^2
   
   

2. Explain why f[0,1,2,3] equals zero.

   Answer:

     The interpolating polynomial is unique.
     Because we interpolate at f = x^2 in 4 points,
     if f[0,1,2,3] were nonzero, then we would have
     two different interpolating polynomials.
   

   Alternative Answer:

     The error function E(x) = f(x) - p(x) equals
   
             f^(4)(a)
     E(x) = ------------ (x-0)*(x-1)*(x-2)*(x-3).
                4!
   
     For f = x^2, we have E(x) = 0, so p(x) = f(x),
     and thus f[0,1,2,3] = 0.