- 设n次多项式函数满足f(k)=k/(k+1)(k=0,1,2,.?
- 设n次多项式满足f(k)=k/(k+1) (k=0,1,2,...,n),求f(n+1)
- 由题设有:
(k+1)f(k)-k=0 (k=0,1,2,...,n),
因而n+1次多项式g(x)=(x+1)f(x)-x...①
有n+1个根x=0,1,2,...,n.于是g(x)=ax(x-1)(x-2)...(x-n)...②
其中a为待定系数,在①②中令x=-1,得
1=a[(-1)^(n+1)](n+1)!,
即 a=[(-1)^(n+1)]/(n+1)!.
从而知道多项式函数为
f(x)=1/(x+1)*{[(-1)^(n+1)]/(n+1)!*x(x-1)(x-2)...(x-n)+x}.
由此得 f(n+1)=[(-1)^(n+1)+(n+1)]=1 (n为奇数)
f(n+1)=n/(n+2) (n为偶数).