# Thread: Find polynomial if n is odd

1. ## Find polynomial if n is odd

Let p(x) be a polynomial of degree n and p(k) = $\frac{k}{k+1}, k = 0,1,2,3,..., n.$ Find p(n+1) if n is odd.

2. ## Re: Find polynomial if n is odd

We claim the degree n polynomial
$p(x) = \sum_{i=1}^n (-1)^{i+1} \frac{1}{i+1} \binom{x}{i}$
where
$\binom{x}{i} = \frac{x(x-1)(x-2) \cdots (x-i+1)}{i!}$
(a generalized binomial coefficient) satisfies
$p(k) = \frac{k}{k+1}$ for $k = 0, 1, 2, \dots n$
and that
$p(n+1) = 1$ for n odd.

The key to these claims is the identity
(*) ... $\sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = \frac{k}{k+1}$
$(1+k)^k = \sum_{i=0}^k \binom{k}{i} x^i$
Integrate both sides from 0 to x and then set x=-1, with the result
$\frac{-1}{k+1} = \sum_{i=0}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = -1 + \sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i}$
so
$\sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1}\binom{k}{i} = 1 - \frac{1}{k+1} = \frac{k}{k+1}$
which shows that $p(k) = \frac{k}{k+1}$ for $k = 0, 1, 2, \dots n$, as claimed.

To show that p(n+1) = 1, start with (*) for k = n+1, i.e.
$\sum_{i=1}^{n+1} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2}$
so
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} + (-1)^{n+2} \frac{1}{n+2}= \frac{n+1}{n+2}$
so for n odd
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} - \frac{1}{n+2}= \frac{n+1}{n+2}$
hence
$\sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2} + \frac{1}{n+2} = 1$