Order of Natural Logarithm Function

Theorem
Let $\epsilon > 0$ be a real number.

Then:


 * $\ln x = \map \OO {x^\epsilon}$

where $\OO$ is big-O notation.

Proof
We show that for $x \ge 1$, we have:


 * $\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$

We first show that for $t \ge 0$, we have:


 * $\ds t \le \frac 1 \epsilon e^{\epsilon t}$

The claim then follows taking $t = \ln x$.

Define a real function $f : \openint 0 \infty \to \R$ by:


 * $\ds \map f t = \frac 1 \epsilon e^{\epsilon t} - t$

for $t \ge 0$.

Then $f$ is differentiable with derivative:


 * $\map {f'} t = e^{\epsilon t} - 1$

by Derivative of Exponential Function.

So:


 * $\map {f'} t \ge 0$ for all $t \ge 0$.

So, from Real Function with Positive Derivative is Increasing:


 * $f$ is increasing for $t \ge 0$.

That is, for $t \ge 0$:


 * $\map f t \ge \map f 0 = 1 > 0$

So:


 * $\ds \frac 1 \epsilon e^{\epsilon t} - t > 0$

for $t \ge 0$.

So:


 * $\ds \frac 1 \epsilon e^{\epsilon t} \ge t \ge 0$

for $t \ge 0$ as required.

Hence, for $x \ge 1$ we have:


 * $\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$

From the definition big-O notation, we have:


 * $\ln x = \map \OO {x^\epsilon}$