All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
A Further Polynomial Problem (Posted on 2007-02-06) Difficulty: 3 of 5
Let P(x) denote a polynomial of degree n such that P(k)= k/(k+1) for k= 0,1,2,,n. Determine P(n+1).

  Submitted by K Sengupta    
No Rating
Solution: (Hide)
Let Q(x) = (x+1)P(x) x.

Then, since Q(x) is a polynomial (n+1)th degree and vanishes for x = 0,1,2,,n; we must have:

Q(x) = (x+1)P(x) x = Bx(x-1)(x-2).(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)*.*(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
Solutiona solutionDennis2007-02-06 13:34:19
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information