Question

已知f(x)是四次多项式,且满足f(i)=1/i ,i=1,2,3,4,5,求f(6)的值

答案不是1/6

Answer 1

解：f(6)=-7/4
设f(x)=ax^4+bx³+cx²+dx+e，依题意有：
f(1)=a+b+c+d+e=1 (1)
f(2)=16a+8b+4c+2d+e=1/2 (2)
f(3)=81a+27b+9c+3d+e=1/3 (3)
f(4)=256a+64b+16c+4d+e=1/4 (4)
f(5)=625a+125b+25c+5d+e=1/5 (5)
(5)-(4),(4)-(3),(3)-(2),(2)-(1)，得：
369a+61b+9c+d=1/5-1/4 (6)
175a+37b+7c+d=1/4-1/3 (7)
65a+19b+5c+d=1/3-1/2 (8)
15a+7b+3c+d=1/2-1/1 (9)
(6)-(7),(7)-(8),(8)-(9)，得：
194a+24b+2c=1/5-1/4-1/4+1/3 (10)
110a+18b+2c=1/4-1/3-1/3+1/2 (11)
50a+12b+2c=1/3-1/2-1/2+1/1 (12)
(10)-(11),(11)-(12)，得：
84a+6b=1/5-1/2+1/3-1/4+2/3-1/2=1/5-1/4 (13)
60a+6b=1/4-2/3+1/2-1/3+1/1-1/1=1/4-1/2 (14)
(13)-(14)，得：
24a=1/5-1/4-1/4+1/2=1/5
a=1/120 (15)
将(15)代入(14)，得：
60*(1/120)+6b=1/4-1/2
6b=-1/2-1/2=-1
b=-1/6 (16)
将(16),(15)代入(12)，得：
50*(1/120)-12*(1/6)+2c=1/3
2c=1/3+2-5/12
c=23/24 (17)
将(17),(16),(15)代入(9)，得：
15*1(/120)-7*(1/6)+3*(23/24)+d=-1/2
d=-9/4 (18)
将(18),(17),(16),(15)代入(1)，得：
1/120-1/6+23/24-9/4+e=1
e=49/20 (19)
于是有：f(x)=(1/120)x^4-(1/6)x³+(23/24)x²-(9/4)x+(49/20)
f(6)=(1/120)*1296-(1/6)*216+(23/24)*36-(9/4)*6+(49/20)
=54/5-36+69/2-27/2+49/20=-7/4

Answer 2

f（6）=1/6