Oxford Interview: Comparing 2^x and x^2

← 목록으로 돌아가기

Here's a clean way to compare $2^x$ and $x^2$ for real $x$.

Key idea (turn it into a sign test)

For $x \neq 0$,

$$\ln\!\left(\frac{2^x}{x^2}\right)= x\ln 2 - 2\ln|x| \;\;=: F(x).$$

Because $\ln$ is strictly increasing,
$\;2^x > x^2 \iff F(x)>0$, $\;2^x = x^2 \iff F(x)=0$, $\;2^x < x^2 \iff F(x)<0$.

Shape of $F(x)$

$$F'(x)=\ln 2 - \frac{2}{x},\qquad F''(x)=\frac{2}{x^2}>0 \text{ for } x\neq 0.$$

Thus $F$ is:

strictly increasing on $(-\infty,0)$ (since $F'(x)=\ln 2-2/x>0$ for $x<0$),
strictly decreasing on $(0,\;2/\ln 2)$,
strictly increasing on $(2/\ln 2,\;\infty)$,

with $2/\ln 2\approx 2.885$. Also $F(x)\to+\infty$ as $x\to 0^\pm$, and $F(x)\to -\infty$ as $x\to -\infty$.

Intersections $2^x=x^2$

Positive solutions

Obvious solutions: $x=2$ and $x=4$ (since $2^2=4=2^2$ and $2^4=16=4^2$).

On $(0,\infty)$, the behavior above implies these are the only positive solutions, with $F$ dipping below 0 between them.

Negative solution

On $(-\infty,0)$, monotonicity gives exactly one negative solution, found numerically (e.g., Newton's method on $f(x)=2^x-x^2$, $f'(x)=2^x\ln2-2x$) at

$$x\approx -0.7666646959.$$

Final comparison (all real $x$)

결론:

$2^x < x^2$ for $x< -0.7666646959\ldots$ or $2
$2^x = x^2$ at $x\approx -0.7666646959\ldots,\; 2,\; 4$
$2^x > x^2$ for $-0.7666646959\ldots < x<2$ or $x>4$

(Optional) How to get the negative root quickly

Newton iteration

$$x_{n+1}=x_n-\frac{2^{x_n}-x_n^2}{2^{x_n}\ln2-2x_n},\quad x_0=-1$$

converges to $-0.7666646959\ldots$ in a few steps.

Oxford Interview: Comparing $2^x$ and $x^2$