How to show that the solution of following equation is $1$ under certain condition?

Question

I have following third order polynomial equation $$x^3+x^2-x\log(t)+1-\log(t)=0$$ I want to know how to show that if $|\log(t)|>>1$ then the solution of above equation is $x=1$. I will be very thankful to you for your help in this regard.

Answer 1

Divide through by $\log (t)$, and let $\epsilon = \frac{1}{\log t}$, so that the equation becomes

\begin{equation*} \epsilon (x^3 + x^2 + 1) - x - 1 = 0. \end{equation*}

For $\epsilon = 0$, this equation has solution $x = -1$. By the implicit function theorem, for small $\epsilon$, this equation has solution $x = -1 + O(\epsilon)$. So for sufficiently large $t$, we have a solution

\begin{equation*} x = -1 + O\left(\frac{1}{\log t}\right). \end{equation*}

Answer 2

You can't?

Let $\log t = 3$, then the real root of $$ x^3 + x^2 -3 x -2 = 0 $$ is $x = -2$.

Let $t = 10^{10}$, then $\ln t = 23.0\dots$ and the real roots of your polynomial are $-4.825\dots$, $-0.954\dots$, and $4.780\dots$.

In short, taking $t$ large makes $\ln t$ large, but you aren't going to get a root near $1$. (In fact, by Descartes rule of signs, there is only one positive real root and either zero or two negative real roots. The positive real root slowly increases as $t$ increases.)

On the other hand, if we go to tiny $t$s, so that $-\log(t) \gg 1$, we get one real root very near $x = -1$ (and two complex roots which are very nearly purely imaginary), but this also is not near $x = 1$. To show that there is a root near $x = -1$. Let $f(x) = x^3 + x^2 - x \ln t + 1 - \ln t$. Then (all these decimals are exact numbers, not approximations) $$f(-1.1) = 0.879 + \frac{1}{10}\ln t $$ and $$f(-0.9) = 1.081 - \frac{1}{10} \ln t \text{.} $$ If $\frac{1}{10}\ln t < -0.869$, $f(-1.1) < 0$. This implies $\ln t < -8.79$. If $\frac{1}{10}\ln t < 1.081$, $f(0.9) >0$. This implies $\ln t < 10.81$. So if both inequalities are satisfied, there is a sign change in $f$ between $-1.1$ and $-0.9$, so $f$ has a root near $-1$. The two inequalities for $\ln t$ are satisfied if $\ln t < -8.79 < 10.81$, which is true when $0 < t < \mathrm{e}^{-8.79} = 1.522\dots \times 10^{-4}$.

What is $f$ doing near $1$?
$$ f(1) = 1 + 1 - \ln t + 1 - \ln t = 3-2 \ln t \text{.} $$ This is near zero when $\ln t$ is near $3/2$, but not for extremely large or small values of $\ln t$ (or of $t$).