6. [Exam 2011] Apply Newton's divided difference method to construct a polynomial passing through all the following four points (1, 3), (2, 7), (3, 13) and (4,25). 6. We let To = 1, $1 = 2, 12 = 3, 13 = 4 and f(xo) = 3, f(x]) = 7, f(x2) = 13, f(13) = 25. Then we have f[zo, x]] = f(=]) - f(To) =4, F1 - To f(12) - f(21) = 6 2 - T1 and f[r2, 13] = f(13) - f(12) = 12 C3 - T2 and f [zo, 1,12] = = 1 C2 - To and f [z1, 12, 13] = f [x2, Ta] - f , 12] 12 - 6 = 3 2 and _ 3-1 I3 - To 4 - Therefore, we can obtain the following polynomial: pz(r) = f(ro) + (x - ro)f [ro, mi] + (x - zo)(x - x1)f [To, I1, 12] +(x- ro)(x- )(x -2)f [To, 1, 12, 23] 3+4(x- 1) + (x-1)(x- 2) + }(x-1)(x -2)(x -3) 2x3 - 3x' + "x -3. which is a degree three polynomial passing through all the given points (1, 3), (2, 7), (3, 13) and (4, 25).p2(x) = f(zo) + (2 - Co) f [zco, zi] + (2 - CO) (2 - 1) f [zco, 21, 202 +(x - To) (x - x1)(x - 2) f To, X1, X2, 23 = 3+4(x - 1) + (x - 1)(x-2)+ 3(x-1)(x -2)(x -3) 203 - 3x2 + 2x- 3