The Mathematics of ECE

Analysis: roadmap

Last time: review of concepts from topology
- Key concepts: normed vector spaces, open, closed sets, convergence
- Key results: algebraic structure not enough to talk about convergence
- Useful machine learning
Today: More on convergence, sequences and series
- Key concepts: Cauchy sequences, convergence
- Useful in signal processing, machine learning
Website for slides: https://bloch.ece.gatech.edu/teaching/MofECEfa21/

A sequence ${x_{i}}_{i \in N}$ in vector space $V$ is Cauchy if $\forall ϵ > 0 \exists N \in N^{*} such that \forall m, n > N {‖ x_{m} - x_{n} ‖}_{} \leq ϵ .$
Intuition: the points get closer to each others
Every converging sequence is Cauchy
Not every Cauchy sequence is converging.
Requesting that every Cauchy sequence converges amounts to requiring a complete vector space (ECE 7750)
A Cauchy sequence is bounded
We will review a very special situation: sequence in $R$ , for which there is a total order

$R$ is a vector space and everything discussed earlier applies
An increasing bounded sequence ${x_{n}}_{n \geq 1}$ in $R$ converges to its least upper bound.
Let ${x_{n}}_{n \geq 1}$ be a bounded sequence in $R$ . Then $\underset{n \to \infty}{lim sup} x_{n} = lim_{n \to \infty} (sup {x_{k} : k \geq n}) \underset{n \to \infty}{lim inf} x_{n} = lim_{n \to \infty} (inf {x_{k} : k \geq n})$
A sequence ${x_{n}}_{n \geq 1}$ converges to $x \in R$ if and only if $\underset{n \to \infty}{lim sup} x_{n} = \underset{n \to \infty}{lim inf} x_{n}$
A Cauchy sequence ${x_{n}}_{n \geq 1}$ in $R$ converges. This proves that $R$ is complete.

Let ${v_{n}}_{n \geq 1}$ be sequence in a normed vector space. The series $\sum_{n = 1}^{\infty} v_{n}$ converges to if and only if $lim_{m \to \infty} \sum_{n = 1}^{m} v_{n} = s$ .
If the series $\sum_{n = 1}^{\infty} v_{n}$ converges, then $lim_{n \to \infty} v_{n} = 0$ .
This is not a sufficient condition for convergence
There are other tests of convergence
Let ${a_{n}}_{n \geq 1}$ be real valued and non-negative. The series $\sum_{n = 1}^{\infty} a_{n}$ converges if and only if the partial sums $s_{m} ≜ \sum_{n = 1}^{m} a_{n}$ are bounded.
Let ${a_{n}}_{n \geq 1}$ be real valued and non-negative.
1. If there exists $C \in (0; 1)$ and $N$ such that $a_{n + 1} \leq C a_{n}$ for all $n > N$ , then $\sum_{n = 1}^{\infty} a_{n}$ converges
2. If there exists $C > 1$ and $N$ such that $a_{n + 1} \geq C a_{n}$ for all $n > N$ , then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
Let $f : R \to R$ be continuous and positive and decreasing for all $x \geq 1$ . Then $\sum_{n = 1}^{\infty} f (n)$ converges if and only if $\int_{1}^{\infty} f (x) d x$ converges.