 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Follow the function (Posted on 2019-11-28) 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)?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) re: some other thoughts - Solution | Comment 4 of 6 | (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.

 Posted by Brian Smith on 2019-12-08 15:32:09 Please log in:

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