Bounds on (1+x)^a for 0 ≤ a ≤ 1

Dependencies:

Let $0 \le a \le 1$. \begin{align} \forall x \ge -1, &\quad (1+x)^a \le 1+ax \\ \forall 0 \le x \le k, &\quad 1 + \frac{ax}{k+1} \le 1 + \frac{\ln(1+k)}{k}ax \le (1+x)^a \end{align}

Proof

Upper bound

Let $f(x) = (1+ax) - (1+x)^a$. We would like to prove that $f(x) \ge 0$ for all $x$.

\[ f'(x) = a - a(1+x)^{a-1} = a\left(1 - (1+x)^{-(1-a)}\right) \] \[ x \ge 0 \implies 1+x \ge 1 \implies (1+x)^{-(1-a)} \le 1 \implies f'(x) \ge 0 \] \[ -1 \le x \le 0 \implies 0 \le 1+x \le 1 \implies (1+x)^{-(1-a)} \ge 1 \implies f'(x) \le 0 \] Since $f(0) = 0$ and $f(x)$ increases as we go away from 0, $\forall x, f(x) \ge 0$.

Lower bound

Let $\gamma = \frac{1}{k}\ln(1+k)$. We'll prove that $\forall 0 \le x \le k, \; e^{\gamma x} \le 1+x$. That would imply (using the bound on exponential) that \[ 1+\gamma ax \le e^{\gamma ax} \le \left(e^{\gamma x}\right)^a \le (1+x)^a \]

Let $g(x) = e^{\gamma x} - (1+x)$. $g(0) = 0$. $g(k) = e^{\gamma k} - (1+k) = 0$.

$g''(x) = \gamma^2 e^{\gamma x} > 0$ and $g(0) = g(k) = 0$. Therefore, $g(x) \le 0$ for $x \in [0, k]$. Therefore, $e^{\gamma x} \le 1+x$ for $x \in [0, k]$.

By the bound on log, we get $\frac{k}{k+1} \le \ln(1+k)$. Therefore, \[ 1 + \frac{ax}{k+1} \le 1 + \frac{\ln(1+k)}{k}ax \]

Dependency for: None

Info:

Depth: 1
Number of transitive dependencies: 3