 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Further Polynomial Problem (Posted on 2007-02-06) Let P(x) denote a polynomial of degree n such that P(k)= k/(k+1) for k= 0,1,2,Ö,n. Determine P(n+1).

 Submitted by K Sengupta No Rating Solution: (Hide) Let Q(x) = (x+1)P(x) ñ x. Then, since Q(x) is a polynomial (n+1)th degree and vanishes for x = 0,1,2,Ö,n; we must have: Q(x) = (x+1)P(x) ñ x = Bx(x-1)(x-2)ÖÖ.(x-n) Substituting x =-1, we obtain: 1 = B*(-1)^(n+1)*(n+1)!; giving: P(x) = 1/(x+1)[x + {(-1)^(n+1)*x*(x-1)*Ö.*(x-n)}/ (n+1)!] ; so that: P(n+1) = 1 if n is odd = n/(n+2) if n is even. Comments: ( You must be logged in to post comments.)
 Subject Author Date a solution Dennis 2007-02-06 13:34:19 Please log in:
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