USA Mathematical Olympiad 1989 Problem (Second Method)

USA Mathematical Olympiad 1989 Problem

Which is larger, the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$, or the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

Previously we checked for the maximum and minimum points to ascertain if the first derivative of $f(x)=x+x^2+\cdots +x^8 +10x^9 - 8$ and $g(x)=x+x^2+\cdots +x^{10}+10x^{11}-8$ are always greater than or equal to zero.

But do you know that we could use a much simpler method to prove those are the cases?

That is called the analytic method, but that required us to rewrite the functions for both $f(x)$ and $g(x)$, like follows:

Note that

$\begin{align*}f(x)&=x+x^2+\cdots +x^8 +10x^9 - 8\\&=1+x+x^2+\cdots +x^8 +x^9 +9x^9 - 9\\&=\dfrac{(x-1)(1+x+x^2+\cdots +x^8 +x^9)}{x-1}+9x^9-9\\&=\dfrac{x^{10}-1}{x-1}+9x^9-9\end{align*}$

$\begin{align*}\therefore f'(x)&=\dfrac{(x-1)(10x^9)-(x^{10}-1)}{(1-x)^2}+81x^8\\&=\dfrac{9x^{10}-10x^9+1}{(1-x)^2}+81x^8\\&=\dfrac{x^9(9x-10)+1}{(1-x)^2}+81x^8\end{align*}$

For $x\lt 0$, $f'(x)\gt 0$.

For $9x-10\lt 0\,\implies\,\,0\lt x \lt \dfrac{9}{8}$, note that the part of $y_1=9x^{10}-10x^9+1$ has its graph always greater than or equal to zero (since the examination of its first derivative where $y_1'=90x^9-90x^8=90x^8(x-1)$ tells us the minimum of $y_1$ is 0), so we also have $f'(x)\gt 0$.

For $x\gt \dfrac{10}{9}$, $f'(x)\gt 0$.

Therefore we have proved that $f(x)$ is an increasing function that it has only one root.

Similarly,

$\begin{align*}g(x)&=x+x^2+\cdots +x^{10} +x^{11}+9x^{11} - 8\\&=1+x+x^2+\cdots +x^{11} +9x^{11} - 9\\&=\dfrac{(x-1)(1+x+x^2+\cdots +x^8 +x^{11})}{x-1}+9x^{11} - 9\\&=\dfrac{x^{12}-1}{x-1}+9x^{11} - 9\end{align*}$

$\begin{align*}\therefore g'(x)&=\dfrac{(x-1)(12x^{11})-(x^{12}-1)}{(1-x)^2}+99x^{10}\\&=\dfrac{11x^{12}-12x^{11}+1}{(1-x)^2}+99x^{10}\\&=\dfrac{x^{11}(11x-12)+1}{(1-x)^2}+99x^{10}\end{align*}$

For $x\lt 0$, $g'(x)\gt 0$.

For $11x-12\lt 0\,\implies\,\,0\lt x \lt \dfrac{9}{8}$, note that the part of $y_2=11x^{12}-12x^{11}+1$ has its graph always greater than or equal to zero (since the examination of its first derivative where $y_2'=132x^{11}-132x^{10}=132x^{10}(x-1)$ tells us the minimum of $y_2$ is 0), so we also have $g'(x)\gt 0$.

For $x\gt \dfrac{12}{11}$, $g'(x)\gt 0$.

Therefore we have proved that $g(x)$ is an increasing function that it has only one root.

And the rest of the solution to prove for the real root of $g(x)$ is larger than that of $f(x)$ can follow from the previous blog post that the solution is provided by the math professor from Arizona, U.S.A.