Anonymous 寫到:設p(x)表2000次多項式,滿足P(K)
= (K
= 1,2,....,2001),則P(2002)= ?
蠻有趣的題目。
Since p(K) = 1/K for K = 1, 2, ...,2001
so Kp(K) = 1 for K =1,2,3,...,2001
If we let
f(x) = xp(x) - 1
then
f(K) = 0 for K = 1, 2,3,...,2001.
This implies that
f(x) = A(x-1)(x-2)(x-3)...(x-2001) = xp(x) -1
We can determine A by calculating f(0) = 0 * p(0) - 1 = -1
So A = 1/2001!
and f(2002) = A(2001)(2000)(1999)...(1) = A * 2001! = 1.
Now we are ready to calculate p(2002).
p(x) = [ f(x) + 1 ]/x
p(2002) = [f(2002) + 1]/2002 = (1+1)/2002 =1/1001.