Let $u$ and $v$ be real numbers such that \[ (u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8. \] Determine, with proof, which of the two numbers, $u$ or $v$, is larger.
Problem
Source: USAMO 1989
Tags: function, geometric series, algebra unsolved, algebra
paladin8
28.12.2005 12:20
Consider $f(x) = 1+x+\cdots+x^8+10x^9$ and $g(x) = 1+x+\cdots+x^{10}+10x^{11}$.
We are given that $f(u) = g(v) = 9$.
We have $g(x)-f(x) = 10x^{11}+x^{10}-9x^9 = x^9(10x-9)(x+1)$.
$f\left(\frac{9}{10}\right) = 1+\frac{9}{10}+\cdots+\frac{9^8}{10^8}+10 \cdot \frac{9^9}{10^9}$.
By summing the geometric series,
$f\left(\frac{9}{10}\right) = \frac{1-\frac{9^9}{10^9}}{1-\frac{9}{10}} = 10-10 \cdot \frac{9^9}{10^9}+10 \cdot \frac{9^9}{10^9} = 10 > 9$.
Since $f$ is an increasing function, we have $u < \frac{9}{10}$.
Hence $g(u)-f(u) = u^9(10u-9)(u+1) < 0 \Rightarrow g(u) < f(u) = 9$.
But since $g$ is an increasing function as well, we know $g(u) < g(v) = 9 \Rightarrow u < v$.
djmathman
10.03.2018 05:57
$12\tfrac14$ year revive! Remark that trivially $0 < u < 1$. Note that $(u + u^2 + \cdots + u^8) + 10u^9 = 8$ rearranges to \[\frac{u^9-1}{u-1} + 10u^9 = 9\quad\Rightarrow\quad 10u^{10} - 9u^9 - 9u + 8 = 0.\]This rearranges to $(u^9-1)(9u-8) = u^9(1-u)$; combined with our earlier observation we actually get $u < \tfrac89$. Furthermore, remark that the given relation further implies \[(10u^{10} - 9u^9 - 9u + 8)(u+1) = 0\quad\Rightarrow\quad 10u^{11} + u^{10} - 9u^9 = 9u^2 + u - 8 = (9u-8)(u+1) < 0.\]Thus, \[(u+u^2 + \cdots + u^{10}) + 10u^{11} = 8 - 9u^9 + u^{10} + 10u^{11} < 8.\]It follows that, upon letting $f(x) = (x+x^2+\cdots+x^{10}) + 10x^{11}-8$, we deduce that $f(u) < 0$. But $f$ is monotonically increasing, so in fact its root $v$ must satisfy $v > u$, implying that $v$ is larger.