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

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==Solution==
 
==Solution==
−
Let <math>Q(x) = (x+1)P(x) - x</math>. Clearly, <math>Q(x)</math> has a degree of <math>n+1</math>.  
+
Let <math>Q(x) = (x+1)P(x) - x</math>, and clearly, <math>Q(x)</math> has a degree of <math>n+1</math>.  
  
−
Then, for <math>k=0,1,2,\ldots,n</math>, <math>Q(k) = (k+1)P(k) - k = (k+1)\dfrac{k}{k+1} - k = 0</math>.  
+
Then, for <math>k=0,1,2,\ldots,n</math>, <math>Q(k) = (k+1)P(k) - k = (k+1)\cdot \dfrac{k}{k+1} - k = 0</math>.  
  
 
Thus, <math>k=0,1,2,\ldots,n</math> are the roots of <math>Q(x)</math>.  
 
Thus, <math>k=0,1,2,\ldots,n</math> are the roots of <math>Q(x)</math>.  
  
−
Since these are all <math>n+1</math> of the roots, we can write <math>Q(x)</math> as: <math>Q(x) = K(x)(x-1)(x-2) \cdots (x-n)</math> where <math>K</math> is a constant.  
+
Since these are all <math>n+1</math> of the roots of the <math>n+1^{\text{th}}</math> degree polynomial, we can write <math>Q(x)</math> as <cmath>Q(x) = c(x)(x-1)(x-2) \cdots (x-n)</cmath> where <math>c</math> is a constant.  
  
−
Thus, <math>(x+1)P(x) - x = K(x)(x-1)(x-2) \cdots (x-n)</math>
+
Thus, <cmath>(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).</cmath>
  
−
Plugging in <math>x = -1</math> gives:  
+
We plug in <math>x = -1</math> to cancel the <math>(x+1)P(x)</math> and find <math>c</math>:  
  
−
<math>(-1+1)P(-1) - (-1) = K(-1)(-1-1)(-1-2) \cdots (-1-n)</math>
+
<cmath>\begin{align*}
 +
-(-1) &= c(-1)(-1-1)(-1-2) \cdots (-1-n) \\
 +
1 &= c(-1)^{n+1}(1)(2) \cdots (n+1) \\
 +
c &= (-1)^{n+1}\dfrac{1}{(n+1)!} \\
 +
\end{align*}</cmath>
  
−
<math>1 = K(-1)^{n+1}(1)(2) \cdots (n+1)</math>
+
Finally, plugging in <math>x = n+1</math> to find <math>P(n+1)</math> gives:
  
−
<math>K = (-1)^{n+1}\dfrac{1}{(n+1)!}</math>
+
<cmath>\begin{align*}
 +
Q(n+1)&=(n+2)P(n+1)-(n+1)\\
 +
(-1)^{n+1}\dfrac{1}{(n+1)!}\cdot(n+1)! &=(n+2)P(n+1)-(n+1)\\
 +
(-1)^{n+1}&=(n+2)P(n+1)-(n+1)\\
 +
(-1)^{n+1}+(n+1)&=(n+2)P(n+1)\\
 +
P(n+1) &= \dfrac{(-1)^{n+1} + (n+1)}{n+2}\\
 +
\end{align*}</cmath>
  
−
Finally, plugging in <math>x = n+1</math> gives:
+
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>. <math>\Box</math>
  
−
<math>(n+1+1)P(n+1) - (n+1) =</math> <math>(-1)^{n+1}\dfrac{1}{(n+1)!}(n+1)(n+1-1)(n+1-2) \cdots (n+1-n)</math>
+
~Edits by BakedPotato66
−
 
 
−
<math>(n+2)P(n+1) = (-1)^{n+1}\dfrac{1}{(n+1)!}\cdot(n+1)! + (n+1)</math>
 
−
 
 
−
<math>(n+2)P(n+1) = (-1)^{n+1} + (n+1)</math>
 
−
 
 
−
<math>P(n+1) = \dfrac{(-1)^{n+1} + (n+1)}{n+2}</math>
 
−
 
 
−
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==
 
==Solution 2==

Revision as of 16:17, 9 August 2021

Problem

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

Solution

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

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

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

Since these are all $n+1$ of the roots of the $n+1^{\text{th}}$ degree polynomial, we can write $Q(x)$ as \[Q(x) = c(x)(x-1)(x-2) \cdots (x-n)\] where $c$ is a constant.

Thus, \[(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).\]

We plug in $x = -1$ to cancel the $(x+1)P(x)$ and find $c$:

\begin{align*} -(-1) &= c(-1)(-1-1)(-1-2) \cdots (-1-n) \\ 1 &= c(-1)^{n+1}(1)(2) \cdots (n+1) \\ c &= (-1)^{n+1}\dfrac{1}{(n+1)!} \\ \end{align*}

Finally, plugging in $x = n+1$ to find $P(n+1)$ gives:

\begin{align*} Q(n+1)&=(n+2)P(n+1)-(n+1)\\ (-1)^{n+1}\dfrac{1}{(n+1)!}\cdot(n+1)! &=(n+2)P(n+1)-(n+1)\\ (-1)^{n+1}&=(n+2)P(n+1)-(n+1)\\ (-1)^{n+1}+(n+1)&=(n+2)P(n+1)\\ P(n+1) &= \dfrac{(-1)^{n+1} + (n+1)}{n+2}\\ \end{align*}

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

~Edits by BakedPotato66

Solution 2

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


\[P(n+1) = \sum_{k=0}^n \frac{k}{k+1} \prod_{j \ne k} \frac{n+1-j}{k-j}\] \[= \sum_{k=0}^n \frac{k}{k+1} \cdot \frac{\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \dots 1\cdot (-1)(-2) \dots (k-n)}\] \[= \sum_{k=0}^n \frac{k}{k+1} (-1)^{n-k}\cdot \frac{(n+1)!}{k!(n+1-k)!}\] \[= \sum_{k=0}^n (-1)^{n-k} \binom{n+1}{k} - \sum_{k=0}^n \frac{(n+1)!(-1)^{n-k}}{(k+1)!(n+1-k)!}\] \[= -\left(\sum_{k=0}^{n+1} (-1)^{n+1-k} \binom{n+1}{k} - 1\right) + \frac{1}{n+2} \cdot \sum_{k=0}^n (-1)^{n+1-k} \binom{n+2}{k+1}\] \[= 1 + \frac{1}{n+2} \left(\sum_{k=-1}^{n+1} (-1)^{n+2 - (k+1)} \binom{n+2}{k+1} - (-1)^{n+2} - 1\right)\] \[= \boxed{1 - \frac{(-1)^n + 1}{n+2}}\]


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.

See Also

1975 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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