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A Further Polynomial Problem (Posted on 20070206) 

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

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Solution:

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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(x1)(x2)…….(xn)
Substituting x =1, we obtain:
1 = B*(1)^(n+1)*(n+1)!; giving:
P(x) = 1/(x+1)[x + {(1)^(n+1)*x*(x1)*….*(xn)}/ (n+1)!] ; so that:
P(n+1)
= 1 if n is odd
= n/(n+2) if n is even.

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Subject 
Author 
Date 
 a solution  Dennis  20070206 13:34:19 


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