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Difference between revisions of "2013 Canadian MO Problems/Problem 1"
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<math>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</math>
 
<math>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</math>
 +
 +
In order for the new polynomial <math>F(x)</math> to be a constant, all the coefficients in front of <math>x^i</math> for <math>i>1</math> need to be zero.
 +
 +
So we start by looking at the coefficient in front of <math>x^2</math>
 +
  
  

Revision as of 23:42, 26 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 $x^2$


~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.

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