Let f(x) be a polynomial of degree 2n for some natural number n such that f(x)=1/x for x=1,2,3,...2n+1. What is the value of f(2n+3)?
(In reply to
some other thoughts by Larry)
f(x) is a polynomial and f(x) = [1 + (x-1)(x-2) ... (x-(2n+1)) * H] / x for some constant H.
Then the constant term of 1 + (x-1)(x-2) ... (x-(2n+1)) * H must equal zero.
The constant term of the product is (-1)^(2n+1) * (2n+1)!.
Then 0 = 1 + (-1)^(2n+1) * (2n+1)! * H, which means H = 1/(2n+1)!,
At x = 2n+3, f(2n+3) = [1 + (2n+2)! * H] / (2n+3).
Then substitute H = 1/(2n+1)! to get f(2n+3) = [1 + (2n+2)! * 1/(2n+1)!] / (2n+3). This simplifies to f(2n+3) = [1 + (2n+2)] / (2n+3), which then makes f(2n+3) = 1.
re: some other thoughts - Solution
