Question

求次数不高于3次的多项式使其在x0=1,x1=3,x2=6,x3=7与f(x)=x2的值相等

Answer 1

记不高于 3 次的多项式为 g(x) = ax^3+bx^2+cx+d, f(x) = x^2， 由题目条件得
g(1) = f(1), 得 a+b+c+d = 1 (1)
g(3) = f(3), 得 27a+9b+3c+d = 9 (2)
g(6) = f(6), 得 216a+36b+6c+d = 36 (3)
g(7) = f(7), 得 343a+49b+7c+d = 49 (4)
(4)-(3), (3-2), (2)-(1) 得
127a+13b+c = 13 (5)
189a+27b+3c = 27 即 63a+9b+c = 9 (6)
26a+8b+2c = 8 即 13a+4b+c = 4 (7)
(5)-(6), (6)-(7) 得
64a+4b = 4 即 16a+b = 1 (8)
50a+5b = 5 即 10a+b = 1 (9)
(8) - (9) 得 6a = 0, a = 0, b = 1, c = 0, d = 0
所求多项式 g(x) = x^2