Difference between revisions of "1975 USAMO Problems/Problem 3"

Revision as of 15:06, 13 August 2014

Problem

If $P(x)$ denotes a polynomial of degree $n$ such that $P(k)=k/(k+1)$ for $k=0,1,2,\ldots,n$ , determine $P(n+1)$ .

Solution

Let $Q(x) = (x+1)P(x) - x$ . Clearly, $Q(x)$ has a degree of $n+1$ .

Then, for $k=0,1,2,\ldots,n$ , $Q(k) = (k+1)P(k) - k = (k+1)\dfrac{k}{k+1} - k = 0$ .

Thus, $k=0,1,2,\ldots,n$ are the roots of $Q(x)$ .

Since these are all $n+1$ of the roots, we can write $Q(x)$ as: $Q(x) = K(x)(x-1)(x-2) \cdots (x-n)$ where $K$ is a constant.

Thus, $(x+1)P(x) - x = K(x)(x-1)(x-2) \cdots (x-n)$

Plugging in $x = -1$ gives:

$(-1+1)P(-1) - (-1) = K(-1)(-1-1)(-1-2) \cdots (-1-n)$

$1 = K(-1)^{n+1}(1)(2) \cdots (n+1)$

$K = (-1)^{n+1}\dfrac{1}{(n+1)!}$

Finally, plugging in $x = n+1$ gives:

$(n+1+1)P(n+1) - (n+1) =$ $(-1)^{n+1}\dfrac{1}{(n+1)!}(n+1)(n+1-1)(n+1-2) \cdots (n+1-n)$

$(n+2)P(n+1) = (-1)^{n+1}\dfrac{1}{(n+1)!}\cdot(n+1)! + (n+1)$

$(n+2)P(n+1) = (-1)^{n+1} + (n+1)$

$P(n+1) = \dfrac{(-1)^{n+1} + (n+1)}{n+2}$

If $n$ is even, this simplifies to $P(n+1) = \dfrac{n}{n+2}$ . If $n$ is odd, this simplifies to $P(n+1) = 1$ .

Solution 2

It is fairly natural to use Lagrange's Interpolation Formula on this problem:

$\begin{align} P(n+1) &= \sum_{k=0}^n \frac{k}{k+1} \prod_{j \text{not} k} \frac{n+1-j}{k-j} \\ &= \sum_{k=0}^n \frac{k}{k+1} \cdot \frac{\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \dots 1\cdot (-1)(-2) \dots (k-n)}\\ &= \sum_{k=0}^n \frac{k}{k+1} (-1)^{n-k}\cdot \frac{(n+1)!}{k!(n+1-k)!} \\ &= \sum_{k=0}^n (-1)^{n-k} \binom{n+1}{k} - \sum_{k=0}^n \frac{(n+1)!(-1)^{n-k}}{(k+1)!(n+1-k)!} \\ &= -(\sum_{k=0}^{n+1} (-1)^{n+1-k} \binom{n+1}{k} - 1) + \frac{1}{n+2} \cdot \sum_{k=0}^n (-1)^{n+1-k} \binom{n+2}{k+1} \\ &= 1 + \frac{1}{n+2} (\sum_{k=-1}^{n+1} (-1)^{n+2 - (k+1)} \binom{n+2}{k+1} - (-1)^{n+2} - 1) \\ &= \boxed{1 - \frac{(-1)^n + 1}{n+2}} \end{align}$ (Error compiling LaTeX. Unknown error_msg)

through usage of the Binomial Theorem.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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Difference between revisions of "1975 USAMO Problems/Problem 3"

Revision as of 15:06, 13 August 2014

Contents

Problem

Solution

Solution 2

See Also

@@ Line 33: / Line 33: @@
 If <math>n</math> is even, this simplifies to <math>P(n+1) = \dfrac{n}{n+2}</math>. If <math>n</math> is odd, this simplifies to <math>P(n+1) = 1</math>.
+==Solution 2==
+It is fairly natural to use Lagrange's Interpolation Formula on this problem:
+<math>\begin{align}
+P(n+1) &= \sum_{k=0}^n \frac{k}{k+1} \prod_{j \text{not} k} \frac{n+1-j}{k-j} \\
+&= \sum_{k=0}^n \frac{k}{k+1} \cdot \frac{\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \dots 1\cdot (-1)(-2) \dots (k-n)}\\
+&= \sum_{k=0}^n \frac{k}{k+1} (-1)^{n-k}\cdot \frac{(n+1)!}{k!(n+1-k)!} \\
+&= \sum_{k=0}^n (-1)^{n-k} \binom{n+1}{k} - \sum_{k=0}^n \frac{(n+1)!(-1)^{n-k}}{(k+1)!(n+1-k)!} \\
+&= -(\sum_{k=0}^{n+1} (-1)^{n+1-k} \binom{n+1}{k} - 1) + \frac{1}{n+2} \cdot \sum_{k=0}^n (-1)^{n+1-k} \binom{n+2}{k+1} \\
+&= 1 + \frac{1}{n+2} (\sum_{k=-1}^{n+1} (-1)^{n+2 - (k+1)} \binom{n+2}{k+1} - (-1)^{n+2} - 1) \\
+&= \boxed{1 - \frac{(-1)^n + 1}{n+2}}
+\end{align}</math>
+through usage of the Binomial Theorem.
 {{alternate solutions}}