User Tools

Site Tools


math104-s21:s:vpak

This is an old revision of the document!


vpak

Hi welcome to my note page

Summary of Material

1. Numbers, Sets, and Sequences

Rational Zeros Theorem. For polynomials of the form cnxn + … + c0 = 0 , where each coefficient is an integer, then the only rational solutions have the form $\frac{c}{d}$ where c divides cn and d divides c0; rational root r must divide c0.

The maximum of a set S is the largest element in the set.
The minimum is the smallest element in the set.
The $\inf$ of S is the greatest lower bound.
The $\sup$ of S is the smallest upper bound.
S is bounded if $\forall$s $\in$ S, s$\leq$M for some M $\in$ $\reals$
Completeness Axiom. If S is a nonempty bounded set in $\reals$, then $\inf$ S and $\sup$ S exist.
Archimedean Property. If a, b $\gt$ 0, then $\exists$n such that na $\gt$ b.

A sequence (sn) is a function mapping from $\N$ to $\R$. It converges to s if $\forall$ $\epsilon$ > 0 there exists N such that N > n $\implies$ |(sn)-s| < $\epsilon$
In other words, $\lim$(sn) = s

Important limit theorems include:
$\lim$(sn)(tn) = ($\lim$(sn))($\lim$(tn))
$\lim$(sn)+(tn) = ($\lim$(sn)) + ($\lim$(tn))
$\lim$($\frac{1}{n^p}$) = 0 for p > 0
$\lim$ n(1/n) = 1

A subsequence (sn(k)) of (sn) is a sequence that is a subset of the elements in the original sequence with relative order preserved.
Bolzano-Weierstrass Theorem. Every bounded sequence has a convergent subsequence, having some subsequential limit.

Given any (sn) and let S be the set of subsequential limits of (sn). Define:
$\lim$ $\sup$ (sn) = $\lim\limits_{N \to \infin}$ $\sup${(sn): n > N} = $\sup$ S
$\lim$ $\inf$ (sn) = $\lim\limits_{N \to \infin}$ $\inf${(sn): n > N} = $\inf$ S

$\lim$ $\inf$ |sn+1| / |sn| $\leq$ $\lim$ $\inf$ |sn|^(1/n) $\leq$ $\lim$ $\sup$ |sn|^(1/n) $\leq$ $\lim$ $\sup$ |sn+1| / |sn|

2. Topology

math104-s21/s/vpak.1620450780.txt.gz · Last modified: 2026/02/21 14:44 (external edit)