Bound on log

Dependencies: None

\[ \forall x > 0, \frac{x-1}{x} \le \ln x \le x-1 \]

Proof

You can plot a graph of $y = \ln x$ and $y = x - 1$ to see that $\ln x \le x-1$.

Alternatively, let $f(x) = x - 1 - \ln x$. Prove that $f(x)$ attains a minimum value of $0$ at $x = 1$. Therefore, $f(x) \ge 0$, which means that $\ln x \le x-1$.

\begin{align} & \forall x \in \mathbb{R}, \ln x \le x-1 \\ &\implies \forall x \in \mathbb{R}, \ln \left(\frac{1}{x}\right) \le \frac{1}{x} - 1 \\ &\implies \forall x \in \mathbb{R}, -\ln x \le \frac{1-x}{x} \\ &\implies \forall x \in \mathbb{R}, \ln x \ge \frac{x-1}{x} \end{align}

Dependency for:

  1. Bounds on (1+x)^a for 0 ≤ a ≤ 1
  2. Bound on harmonic sum
  3. Approximation algorithm for covering LPs

Info:

Transitive dependencies: None