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| A Further Polynomial Problem (Posted on 2007-02-06) |
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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).
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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(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.
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Subject |
Author |
Date |
 | a solution | Dennis | 2007-02-06 13:34:19 |
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