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I've recently come across this question and have no idea how to do it.

Let $P(x)$ be a 11-degree polynomial such that

$P(x)=\frac{1}{x+1}$ for $x=0,1,2,...11$

Find $P(12)$.

I've tried using the general form of a polynomial but it's turning to long. I think I'm missing some simple trick.

After this comes a general form of the problem.

If $P(x)$ is a polynomial of degree $n$ such that

$P(x)=\frac{x}{x+1}$ for $x=1,2,...,n$, find the value of $P(n+1)$.

R.D.
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    I'm afraid I don't understand the notation.. Why isn't $P(12)=1/13$? – Bobson Dugnutt Sep 03 '16 at 14:27
  • Is it $\frac 1{x+1}$ or $\frac x{x+1}$? – lulu Sep 03 '16 at 14:28
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    @Lovsos I assume the OP means that $P(i)=\frac 1{i+1}$ for $i\in {0,\cdots 11}$. Not that the functions are equal (it is stated that $P(x)$ is a degree $11$ polynomial). – lulu Sep 03 '16 at 14:30
  • @lulu it is as given in the question. – R.D. Sep 03 '16 at 16:59
  • @Fourier Transform That question does not have the second part for $n$ degree. – R.D. Sep 03 '16 at 17:00
  • In the header and in the first question you say $\frac 1{x+1}$ but in the generalization you call for $\frac x{x+1}$. Is that what you intended? – lulu Sep 03 '16 at 17:32
  • Actually...in the header and in the generalization you call for $\frac x{x+1}$ but in the specific question you call for $\frac 1{x+1}$. Doesn't really matter much...the same approach addresses both. – lulu Sep 03 '16 at 18:08
  • @lulu Actually, it doesn't matter much. – R.D. Sep 04 '16 at 06:47

1 Answers1

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Note that from the given conditions, we have that $P(x)(x+1)-1$ is a twelfth degree polynomial with solutions $0,1,2, \dots 11$. In other words, $$P(x)(x+1)-1=ax(x-1)(x-2) \dots (x-11)$$If $x=-1$, than $-1=12!\times a$. Thus $a=-\frac{1}{12!}$

If $x=12$, then $$P(12) \times 13 -1=-1$$ We have that $P(12)=0$. This can be generalized for $n$.

S.C.B.
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