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Here's the problem statement: Given an $n$ degree polynomial $p(k)$ such that:

$$p(k)=\frac{k}{k+1}$$

for all integer $k$ from $0$ to $n$, determine $p(n+1)$. Any ideas on how to solve it?

sayantankhan
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    There's always Lagrange interpolation. – lhf Apr 10 '14 at 11:09
  • For specific cases, like for n=3 or 4, I can determine the polynomial by determining p(0), p(1), p(2) till p(n). I get n+1 simultaneous equations which I solve to get the coefficients of the polynomial and then plug in (n+1). But it's too long and time consuming. – sayantankhan Apr 10 '14 at 11:23
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    Duplicate of http://math.stackexchange.com/q/668108/115115 – Lutz Lehmann Apr 11 '14 at 17:17

1 Answers1

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The answer is $P(n+1)=1$.

Indeed, consider $Q(X)=(X+1)P(X)-X$. We have $\deg Q\leq n+1$ and $Q(k)=0$ for $k=0,1,2,\ldots,n$. Thus, there is a constant $\lambda$ such that $Q(X)=\lambda\prod\limits_{k=0}^n(X-k)$. To determine lambda, we note that $Q(-1)=1=\lambda(-1)^{n+1}(n+1)!$, therefore $$ XP(X+1)-X=\frac{1}{(n+1)!}\prod_{k=0}^n(k-X) $$ Substituting $X=n+1$ we get $P(n+1)=1$.

Omran Kouba
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