End of lecture

week9/15-03-2022/15-03-2022.tex

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+%        File: 20-01-2022.tex
+%     Created: 12:28:35 Thu, 20 Jan 2022 EST
+% Last Change: 12:28:35 Thu, 20 Jan 2022 EST
+%
+\documentclass[letterpaper]{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{cancel}
+\usepackage{amssymb}
+\usepackage{listings}
+\usepackage[shortlabels]{enumitem}
+\usepackage{soul}
+%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{mathtools}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{03/15/2022}
+\title{%
+        Strong Induction\\
+            \large MATH--190--01 : Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+Previously, we talked on mathematical induction.
+Its used when we can't really use mathematical induction isn't really enough.
+In mathematical induction, the inductive step assumes $P(n)$ and use that to
+show that $P(n+1)$ is also true.
+In strong induction, we instead assume $P(k)$, where $base \le k\le n$, and use
+that to prove that $P(n+1)$ is true.\\
+
+\section{Proof Example 1}
+
+Lets say that every integer greater than $1$ can be written as the product of
+prime numbers.
+The proof follows as such.\\
+
+Let $P(n)$ be the predicate "$n$ can be written as a product of primes."
+
+Basis step: $P(2)$ is true, since $2$ is, itself, a prime.
+
+Inductive step: The inductive hypothesis is that $P(j)$ is true for every $2\le j\le k$.
+That is, for any $j$ between $2$ and $k$, we can write $j$ as the product of prime numbers.
+Now, we have to show that $P(k+1)$.
+Alternatively, we need to show that $k+1$ can be written as the product of primes.
+
+There are two cases to consider, when $k+1$ is prime, and when it is composite.
+
+Case 1: If $k+1$ is prime, then we're done.
+
+Case 2: If $k+1$ is composite, then we can write $k+1=a * b$ where $2\le a \le k+1$, and
+$2\le b \le k+1$. Since $b$ and $a$ are both less than $k$, through the inductive hypothesis
+they can also be written as a product of primes.
+Thus, if $k+1$ is composite, it can be written as a product of primes -- those primes
+in the factorizations of $a$ and $b$.\\
+
+\section{Proof Example 2}
+Recall that the Fibonnacci sequence is the recursively defined sequence given by:
+\begin{equation}
+    \begin{split}
+        a_0=1\\
+        a_1=1\\
+        a_{k+1}=a_k+a_{k-1}\ for\ k>1
+    \end{split}
+\end{equation}
+
+The claim to prove is that $a_k\ge (\frac{3}{2})^{k-2}$ for every $k\ge1$\\
+
+Let $P(n)$ be the predicate $a_{n}=a_k+a_{n-1}$.
+
+Basis Step: $P(1)$ is true, since:
+\begin{equation}
+    a_1=1\ge \frac{2}{3}=\left( \frac{3}{2} \right)^{-1}=\left( \frac{3}{2} \right)^{(1-2)}
+\end{equation}
+
+Inductive Step: Assume $P(j)$ is true for all $1\le j\le k$. Now we have to show $P(k+1)$ is true.
+That is, that $a_{k+1}\ge\left( \frac{3}{2} \right)^{k-1}$
+By definition, we know that $a_{k+1}=a_k+a_{k-1}$.
+By the inductive hypothesis, we know that $a_k\ge\left( \frac{3}{2} \right)^{k-2}$.
+Additionally, we know that $a_{k-1}\ge\left( \frac{3}{2} \right)^{k-3}$.
+Given this, we can do the following:
+\begin{equation}
+    \begin{split}
+        a_{k+1}=a_k+a_{k-1}\\
+        \ge \left( \frac{3}{2} \right) ^{k-2} + \left( \frac{3}{2} \right) ^{k-2}\\
+        =\left(1+\frac{3}{2}\right) \left( \frac{3}{2} \right)^{k-3}\\
+        =\frac{5}{2}\left( \frac{3}{2} \right)^{k-3}\\
+        \ge \frac{9}{4}\left( \frac{3}{2} \right)^{k-3}\\
+        = \frac{3}{2}^2\left( \frac{3}{2} \right)^{k-3}\\
+        =\left(\frac{3}{2}\right)^{k-1}
+    \end{split}
+\end{equation}
+
+\end{document}

week9/17-03-2022/17-03-2022.tex

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+%        File: 20-01-2022.tex
+%     Created: 12:28:35 Thu, 20 Jan 2022 EST
+% Last Change: 12:28:35 Thu, 20 Jan 2022 EST
+%
+\documentclass[letterpaper]{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{cancel}
+\usepackage{amssymb}
+\usepackage{listings}
+\usepackage[shortlabels]{enumitem}
+\usepackage{soul}
+%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{mathtools}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{03/15/2022}
+\title{%
+        Recursive Definitions and Structural Induction\\
+        \large MATH--190--01 : Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+
+Recursively defined functions have some initial value, and a defined rule for
+how to get to any other value.
+They can theoretically work for any integer, but for simplicity, we're only going to work
+with a domain of non-negative integers.
+
+Note that this is similar to induction.
+Example:
+\begin{equation}
+    \begin{split}
+        Basis: \sum_{k=0}^{n}a_k\\
+        Recursive: \sum_{k=0}^{n}a_k=(\sum_{k=0}^{n-1}a_k) + a_n
+    \end{split}
+\end{equation}
+
+$\Sigma$ is the alphabet (a set of characters.)
+$\Sigma$ does not necessarily mean the Latin alphabet.
+(i.e. $\Sigma=\{ 0,1\}$) is an acceptable definition.
+$\Sigma^*$ is the set of strings over the alphabet $\Sigma$
+You can define the length of a string recursively.
+
+\end{document}