Bound on exponential

Dependencies: None

\[ \forall x \in \mathbb{R}, e^x \ge 1+x \]

Proof

Let $f(x) = e^x - (1+x)$. Then $f'(x) = e^x - 1$ and $f''(x) = e^x$.

Since $f''(x) \ge 0$, $f$ is convex.

$f'(x) = 0 \implies x = 0$. So $f$ is minimum at $x = 0$. $f(0) = 0$. Therefore, $f(x) \ge 0$ for all $x$.

Dependency for:

  1. Bounds on (1+x)^a for 0 ≤ a ≤ 1
  2. Chernoff bound

Info:

Transitive dependencies: None