 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.) some other thoughts | Comment 2 of 6 | if f(x) = 1/x at the multiple x values indicated, then g(x) = xf(x) - 1 should have zeros at those same x values.
So g(x) = (x-1)(x-2) ... (x-(2n+1)) * h(x)
which is a polynomial of degree 2n+1 times h(x)

But g(x) is already of degree 2n+1 since g(x) = xf(x) - 1 and f(x) is of degree 2n.
Thus h(x) must be a constant, at least with respect to x; call it "H".

g(x) = (x-1)(x-2) ... (x-(2n+1)) * H
f(x) = (1 + g(x))/x
f(x) =  [1 + (x-1)(x-2) ... (x-(2n+1)) * H] / x

At this point, I'm stuck.  I can't come up with a way to determine a value of H that is independent of n. The only values of x where I know something about f(x) are the positive integers less than or equal to 2n+1.  But plugging in those values multiplies H by 0 bringing me no closer to solving for H.

But, at x = 2n+3 we can say
f(2n+3) =  [1 + (2n+2)(2n+1) ... (2) * H] / (2n+3)
f(2n+3) =  [1 + (2n+2)! * H] / (2n+3)
 Posted by Larry on 2019-12-07 08:41:30 Please log in:

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