USA Mathematical Olympiad 1989 Problem

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 $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

The upper part of the solution below is mine and the lower part belongs to the math professor from Arizona, USA.

If we let

$\begin{align*}f(x)&=x+x^2+\cdots +x^8 +10x^9 - 8\\&=x(1+x+\cdots+x^7)+10x^9 - 8\\&=\dfrac{x(1-x)(1+x+\cdots+x^7)}{(1-x)}+10x^9 - 8\\&=\dfrac{x(1-x^8)}{(1-x)}+10x^9 - 8\\&=\dfrac{x-x^9}{(1-x)}+10x^9 - 8\end{align*}$

We prove that the first derivative of $f(x)$ is positive so it is an increasing function that has only one real root.

$\begin{align*}f'(x)&=\dfrac{(1-x)(1-9x^8)-(-1)(x-x^9)}{(1-x)^2}+90x^8\\&=\dfrac{90x^{10}-172x^9+81x^8+1}{(1-x)^2}\end{align*}$

If we can prove $90x^{10}-172x^9+81x^8+1\ge 0$ for all $x$, then we're done, since $(1-x)^2\gt 0$ for all $x\in R,\,x\ne 1$.

Let $y=90x^{10}-172x^9+81x^8+1$

$\begin{align*}y'=900x^9-9(172)x^8+8(81)x^7&=9x^7(x-0.72)(x-1)\end{align*}$

Note that $(0,\,1)$ and $(1,\,0)$ are the minimum point whereas $(0.72,\,1.276)$ is maximum point.

Therefore $y\ge 0$ and hence $f'(x)\ge 0$, so the function of $f$ is an increasing function with only one real root.

We do the same for $g(x)$ where we let

$g(x)=x+x^2+\cdots +x^{10}+10x^{11}-8$

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

$\begin{align*}g'(x)&=\dfrac{(1-x)(1-11x^{10})-(-1)(x-x^{11})}{(1-x)^2}+110x^{10}\\&=\dfrac{110x^{12}-210x^{11}+99x^{10}+1}{(1-x)^2}\end{align*}$

If we can prove $110x^{12}-210x^{11}+99x^{10}+1\ge 0$ for all $x$, then we're done, since $(1-x)^2\gt 0$ for all $x\in R,\,x\ne 1$.

Let $m=110x^{12}-210x^{11}+99x^{10}+1$

$\begin{align*}m'=1320x^{11}-2310x^{11}+990x^9&=x^9(x-0.75)(x-1)\end{align*}$

Note that $(0,\,1)$ and $(1,\,0)$ are the minimum point whereas $(0.75,\,1.19)$ is maximum point.

Therefore $m\ge 0$ and hence $g'(x)\ge 0$, so the function of $g$ is an increasing function with only one real root.

Note that we also have $f(0.8) \lt 0, \; f(0.9) \gt0, \;g(0.8) \lt0, \;g(0.9)\gt0$ meaning that both roots lie between $0.8$ and $0.9$.

Now consider the difference $h(x) = f(x)-g(x) =9x^8-x^{10}-10x^{11}= -(x+1)(10x-9)x^9 $.

On the interval $(0.8,\,0.9)$, $h(x)\gt 0$ meaning that $ f(x) \gt g(x)$ giving that the root of $g(x)$ is larger than the root of $f(x)$.