## A Bit About Manifolds

A manifold is a structure

## The Banach Contraction Principle

This theorem guarantees the existence of fixed points in complete metric spaces under a contraction mapping.

## A Tutorial on the REINFORCE (aka Monte-Carlo Policy Differentiation) Algorithm

The REINFORCE Algorithm

## Some Notes on Functional Analysis

Normed Vector Spaces

## Some Observations on Taylor Series

Often in mathematics, it is useful to be able to compute an approximation to a given function $f(x)$. One way of doing this is to find some $a$ for which $f(x)$ is easy to compute. We can then calculate the approximate value of $f(x)$ when $x$ is near $a$. We will call this approximation $f_{a}(x)$. More concretely, we might say $x$ is near $a$ when $% $ for some threshold value of $\varepsilon$ our choosing. One way of settling upon such an $\varepsilon$ might be by determining an acceptable upper bound $U$ on the error of our approximation $\left|f(x) - f_{a}(x)\right|$ and then finding the greatest $\varepsilon$ such that: $\left|f(x) - f_{a}(x)\right| < U \quad \forall x \in (a - \varepsilon,\ a + \varepsilon)$

## Solutions to Rudin's Principles of Mathematical Analysis

Exercises from Chapter 1 of the book.

Exercise 1: If $r$ is rational ($r \neq 0$) and $x$ is irrational, prove that $r + x$ and $rx$ are irrational.

Suppose $r \in \mathbb{Q}$ and $x \in \mathbb{I}$ but $r+x,\ rx \in \mathbb{Q}$. Then we have $r +x = \frac{a}{b}$ for some $a,b \in \mathbb{Z}$. This implies that $x = \frac{a}{b} - r$. Since $r \in \mathbb{Q}$, $r = \frac{c}{d}$ for some $c, d \in \mathbb{Z}$. Therefore $x = \frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}$ which implies $x \in \mathbb{Q}$: a contradiction. Therefore, $r + x$ is irrational. Now, consider $rx$. If $rx \in \mathbb{Q}$, then

\begin{align} rx &= \frac{a}{b}\ \text{for}\ a,b \in \mathbb{Z}
\implies x &= \frac{a}{b} \cdot \frac{d}{c}
\implies x &\in \mathbb{Q} \end{align}

which is a contradiction. Therefore, $rx$ is irrational.