Given a sequence y0, y1, y2, . . . , define the backward difference operator by yi = yi yi1. Powers of are defined recursively by 0yi = yi, jyi = (j1yi), j = 1, 2, . . . . Thus,2yi = (yi yi1) = yi 2yi1 yi2,etc. Consider polynomial interpolation at equispaced points, xi = x0 ih, i = 0, 1, . . . , n. 1 (a) Show that f[xn, xn1, . . . , xnj ] = 1j!hj jf(xn). (Hint: Use mathematical induction.) (b) Show that the interpolating polynomial of degree at most n is given by the Newton backward difference formula pn(x) = nXj=0 (1)j sj jf(xn). where s = xn x h and js = s(s 1) (s j 1) j! (with s0 = 1)