# Hello World

Lets test some math,

$$\boxed{ e^{i\pi}+1=0. }$$

For any $x \in \mathbb R$, we have

$$1 + x \le e^{x}.$$

To see this, observe that $e^x$ is a convex function and $1+x$ is a tangent to $e^x$ at $x=0.$ By a change of variable, $x = \log y$ we have $\log y \le y-1$ for any $y \ge 0.$

Here is a useful inequality involving binomial coefficients,

$$\left( \frac{n}{m} \right)^{m} \le \binom{n}{m} \le \sum_{k=0}^{m} \binom{n}{k} \le \left( \frac{en}{m} \right)^{m}.$$

The $\ell_1$ norm is a convex surrogate for $\ell_0$ norm.