Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 Canadian MO Problems/Problem 1"
Line 29: Line 29:
  
 
<math>\left( c_2-c_1+c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2</math>
 
<math>\left( c_2-c_1+c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2</math>
 +
 +
<math>\left( c_2+\sum_{i=2}^{n}\left((-1)^{i-2}\binom{i}{i-2}+(-1)^{i-1}\binom{i}{i-1} \right)c_i\right)x^2</math>
  
  

Revision as of 00:51, 27 November 2023

Problem

Determine all polynomials $P(x)$ with real coefficients such that \[(x+1)P(x-1)-(x-1)P(x)\] is a constant polynomial.

Solution

Let $F(x)=(x+1)P(x-1)-(x-1)P(x)$

$P(x)=\sum_{i=0}^{n}c_ix^i$

$F(x)=(x+1)\sum_{i=0}^{n}(x-1)^ic_i-(x-1)\sum_{i=0}^{n}c_ix^i$

$F(x)=\sum_{i=0}^{n}x(x-1)^ic_i+\sum_{i=0}^{n}(x-1)^ic_i-\sum_{i=0}^{n}c_ix^{i+1}+\sum_{i=0}^{n}c_ix^i$

$\sum_{i=0}^{n}(x-1)^ic_i=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^j$

$\sum_{i=0}^{n}x(x-1)^ic_i=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^{j+1}$

$F(x)=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^j+\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^{j+1}-\sum_{i=0}^{n}c_ix^{i+1}+\sum_{i=0}^{n}c_ix^i$

In order for the new polynomial $F(x)$ to be a constant, all the coefficients in front of $x^i$ for $i>1$ need to be zero.

So we start by looking at the coefficient in front of $x^2$:

$\left( c_2-c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2$

Since $\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i=c_1+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i$,

$\left( c_2-c_1+c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2$

$\left( c_2+\sum_{i=2}^{n}\left((-1)^{i-2}\binom{i}{i-2}+(-1)^{i-1}\binom{i}{i-1} \right)c_i\right)x^2$


~Tomas Diaz. orders@tomasdiaz.com

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_Canadian_MO_Problems/Problem_1&oldid=206053"