一題有關多項式的問題

Question

設 f(x) 是一個2000 次多項式 .若 f(k)= - 1/k , k=1,2,3,.......,2001.

試求 f(2002) 的值 ?

Answer 1

k f(k)= - 1 , k=1,2,3,.......,2001
kf(k) + 1 = 0, k=1,2,3,.......,2001
xf(x) + 1 = 0 為2001次方程式,所有根為 1,2,3,.......,2001
所以 xf(x) + 1 = a(x-1)(x-2)......(x-2001),
比較常數項得 a= -1/ 2001!
==> f(x) = [ -(x-1)(x-2).....(x-2001)/2001! - 1 ] / x
f(2002) = -2/2002 = - 1/1001

Answer 2

f(k)=-1/k
f(2002)=-1/2002