Difference between revisions of "2023 WSMO Speed Round Problems/Problem 5"
(Created page with "==Problem== There exists a rational polynomial <math>f(x)</math> such that for all <math>x</math> in the range <math>(0,1),</math> <math>f(x)=\sum_{n=1}^{\infty}nx^n.</math>...") |
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==Solution== | ==Solution== | ||
| + | We have | ||
| + | <cmath>\begin{align*} | ||
| + | f(x) &= x+2x^2+3x^3+4x^4+\ldots\text{and}\\ | ||
| + | xf(x) &= x^2+2x^3+3x^4+\ldots.\\ | ||
| + | \end{align*}</cmath> | ||
| + | So, <cmath>f(x)-xf(x) = x+x^2+x^3+\ldots = \frac{x}{1-x}\implies f(x) = \frac{x}{(1-x)^2}</cmath> for <math>x</math> in the range <math>(0,1)</math>. For <math>x>1,</math> <math>f(x)</math> is strictly decreasing, meaning <math>f(x)</math> is maximized at <math>x=6.</math> Thus, our answer is <cmath>\frac{6}{(1-6)^2} = \frac{6}{25}\implies6+25 = \boxed{31}.</cmath> | ||
| + | |||
| + | ~pinkpig | ||
Latest revision as of 11:13, 12 September 2025
Problem
There exists a rational polynomial 
such that for all 
in the range 
If the maximum of over ![$[6,9]$](http://latex.artofproblemsolving.com/b/f/5/bf55b52cd93ba729e9055c5ee114b15e72dc8f0c.png)
is 
for relatively prime positive integers 
and 
find 
Solution
We have
So, ![\[f(x)-xf(x) = x+x^2+x^3+\ldots = \frac{x}{1-x}\implies f(x) = \frac{x}{(1-x)^2}\]](http://latex.artofproblemsolving.com/b/5/6/b56567c9c79040a2ae63e1937cd70ca5a742b322.png)
for in the range 
. For 
is strictly decreasing, meaning is maximized at 
Thus, our answer is ![\[\frac{6}{(1-6)^2} = \frac{6}{25}\implies6+25 = \boxed{31}.\]](http://latex.artofproblemsolving.com/0/4/f/04f0f7182439bea381882bdc664bff7231032505.png)
~pinkpig
