我与大家分享我家庭作业的一个练习的代码。我需要一个横向部分，因为页面上没有足够的空间容纳我写的单词。但是当我使用环境时，landscape从左到右的空间太大了。以下是这种情况的图片：
所以这已经是错误的了。你能帮我找出我的错误吗？
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%PREAMBLE%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[12pt, letterpaper]{article}
%%%%%%%%%%%%%%%%%%%%%%%%PAQUETES%%%%%%%%%%%%%%%%%%%%%%%%%%
%Avoid problems with accents and special charracters
\usepackage[utf8]{inputenc}
%document Format
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
%Colors along the docuemnt.
\usepackage{xcolor}
%Format for mathematical assertions
\usepackage{amsthm}
%More useful symbols
\usepackage{mathtools}
%Even more useful symbols
\usepackage{amssymb}
%Language of Expresions
\usepackage[english]{babel}
%Format Sections
\usepackage{sectsty}
%Font for canonical sets
\usepackage{dsfont}
%Images inclusion
\usepackage{graphicx}
%Long comments
\usepackage{comment}
%Caritas
\usepackage{wasysym}
%Referencias
\usepackage{hyperref}
%Enumeración con Color
\usepackage{enumitem}
%Hoja Volteada
\usepackage{pdflscape}
%%%%%%%%%%%%%%%%%%%%%%%%%COLORS%%%%%%%%%%%%%%%%%%%%%%%%%%%
\definecolor{blue}{RGB}{0,176,246}
\definecolor{white}{RGB}{255,255,255}
%%%%%%%%%%%%%%%%%%%%%ENTORNOS%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Exercise Enviroment
\renewcommand{\thesection}{\Roman{section}}
\newcounter{exercise}
\renewcommand{\theexercise}{\arabic{exercise}}
\newenvironment{exercise}[1][]{\refstepcounter{exercise}
\par\medskip
\noindent \textcolor{blue}{\textbf{Exercise~\theexercise}. #1}} \rmfamily
{\medskip}
%Proofs
\renewenvironment{proof}{{\noindent\bfseries Proof:\\}}{\rightline{$\blacksquare$}}
%Mathematical Staments
\theoremstyle{definition}
\newtheorem{Theo}{Theorem}
\newtheorem{Lem}[Theo]{Lemma}
\newtheorem{Def}[Theo]{Definition}
\newtheorem{Not}[Theo]{Notation}
\newtheorem{Cor}{Corollary}[Theo]
\newtheorem{Propo}[Theo]{Proposition}
\newtheorem{Prope}[Theo]{Property}
\newtheorem{Obs}[Theo]{Observation}
%%%%%%%%%%%%%%%%%%%%%MY COMMANDS%%%%%%%%%%%%%%%%%%%%%%%%%
%New
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\I}{\mathbb{I}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
%Renewed
\renewcommand{\sup}{\operatorname{Sup}}
\renewcommand{\inf}{\operatorname{Inf}}
\renewcommand{\limsup}{\operatorname{LimSup}}
\renewcommand{\liminf}{\operatorname{LimInf}}
\renewcommand{\min}{\operatorname{Min}}
\renewcommand{\max}{\operatorname{Max}}
%%%%%%%%%%%%%%%%%%%ESTYLE%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsectionfont{\color{blue}}
\sectionfont{\color{blue}}
%Hyperlinks
\hypersetup{
colorlinks=true,
linkcolor=red,
filecolor=magenta,
urlcolor=red
}
\urlstyle{same}
%%%%%%%%%%%%%%%%%%%%%%%&TÍTULO&%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{\textcolor{blue}{\textbf{Homework \#5\\
Analysis I (English)}}}
\author{\textcolor{blue}{\textit{Germán Felipe López Díaz}}}
\date{\textit{\textcolor{blue}{March 9\textsuperscript{th}, 2021}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%DOCUMENTO%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\setcounter{exercise}{5}
\begin{exercise}
\textcolor{blue}{The Fibonacci sequence $(a_{n})_{n \in \N}$ is defined recursively by:
\[a_{0} = 1 \hspace{0.5cm} a_{1} = 1 \hspace{0.5cm} a_{n+1} = a_{n} + a_{n-1}\hspace{0.5cm}n \in \N\]
Moreover, let $\sigma < \tau$ the solutions of $x^{2} - x - 1 = 0$ and
\[x_{n} = \frac{a_{n+1}}{a_{n}}, n \in \N\]}
For its exercise we need some lemmas:
\begin{Lem}
\label{L1}
For any $n \in \N_{0}$. In Fibonacci sequence context, $a_{n} > 0$.
\end{Lem}
\begin{proof}
\textbf{Bases Cases:}
\begin{itemize}
\item[$\star$] $\mathit{n = 0}:$ Suppose by contradiction that not. That is that $a_{0} \leq 0$. By definition $a_{0} = 1$. Therefore:
\[1 \leq 0\]
A contradiction! with the order of natural numbers. In conclusion it must be true that $a_{0} > 0$.
\item[$\star$] $\mathit{n = 1}:$ Suppose by contradiction that not. That is that $a_{1} \leq 0$. By definition $a_{1} = 1$. Therefore:
\[1 \leq 0\]
A contradiction! with the order of natural numbers. In conclusion it must be true that $a_{1} > 0$.
\end{itemize}
\textbf{Inductive Step:}
Suppose the result for any $k \in \N$ with $k \leq n$. It is that:
\[\forall_{k \in \N}\left[(k \leq n) \rightarrow (a_{k} > 0)\right]\hspace{2cm}\textbf{(HI)}\]
We want to verify the result for $n+1$. By the recursive definition we have that:
\[a_{n+1} = a_{n} + a_{n-1}\]
By \textbf{(HI)} we have that the right side is a non negative number due to $n \leq n$ and $n-1 \leq 1$ and the addition of to non negative numbers is again a non negative number. Call it $P$:
\[a_{n+1} = P\]
By the Strong Induction Principle we have that the result is true for all $n \in \N_{0}$.
\end{proof}
\begin{Lem}
\label{L2}
Let $\sigma$ and $\tau$, $\sigma < \tau$, solutions for the equation $x^{2} - x - 1 = 0$. Then:
\begin{itemize}
\item[(i)] $\tau - \sigma = \sqrt{5}$
\item[(ii)] $\tau^{2} - \sigma^{2} = \tau - \sigma$
\item[(iii)] $\tau^{-1} + \tau^{-2} = 1 = \sigma^{-1}+ \sigma^{-2}$
\end{itemize}
\begin{proof}
Initially by definition of $\sigma$ and $\tau$ we have the following equality:
\begin{equation}
\label{Eq1.1}
\tau^{2} - \tau - 1 = 0
\end{equation}
\begin{equation}
\label{Eq1.2}
\sigma^{2} - \sigma - 1 = 0
\end{equation}
On the other hand, remember The Quadratic Formula:
\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
We use it to solve the polynomial $x^{2} - x - 1$ using the with $a = 1$, $b = -1$ and $c = -1$:
\[x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\]
\[x = \frac{1 \pm \sqrt{1 + 4}}{2}\]
\[x = \frac{1 \pm \sqrt{5}}{2}\]
According to Precalculus course we have that its polynomial has two roots:
\[x_{1} = \frac{1 - \sqrt{5}}{2}\hspace{1cm}x_{2} =\frac{1 + \sqrt{5}}{2}\]
Both of them are real! Since $\sigma$ and $\tau$ are roots of it, necesarily, we must have that:
\begin{equation}
\label{Eq1.3}
\sigma = \frac{1 - \sqrt{5}}{2}
\end{equation}
\begin{equation}
\label{Eq1.4}
\tau = \frac{1 + \sqrt{5}}{2}
\end{equation}
Because $\sigma < \tau$
\begin{itemize}
\item[(i)]
\[\begin{array}{rclr}
\tau - \sigma & = & \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} & \left(\text{Equation \textcolor{red}{(\ref{Eq1.3})} and \textcolor{red}{(\ref{Eq1.4})}}\right)\\
& = & \frac{1 + \sqrt{5} -1 + \sqrt{5}}{2}& \left(\text{Subtraction of homogeneous fractions}\right)\\
& = & \frac{2\sqrt{5}}{2}& \left(\text{Adding similar terms}\right)\\
& = & \sqrt{5} & \left(\text{Canceling}\right)\\
\end{array}\]
\item[(ii)] Using equation $\textcolor{red}{(\ref{Eq1.1})}$ and $\textcolor{red}{(\ref{Eq1.2})}$ finding the quadratic term we get that:
\[\tau^{2} = \tau + 1 \hspace{1cm}\sigma^{2} = \sigma + 1\]
Thus we get:
\[\begin{array}{rclr}
\tau^{2} - \sigma^{2} & = & \tau + 1 - (\sigma + 1) &\left(\text{Substraction of last equations}\right)\\
& = & \tau + 1 - \sigma - 1 & \left(\text{Distributing the minus}\right)\\
& = & \tau - \sigma & \left(\text{Adding similar terms}\right)\\
\end{array}\]
\item[(iii)] Let $s$ be solution of the equation $x^{2} - x - 1 = 0$. from the solution of the Quadratic Formula, that we did at the beginning, we have that $s = \frac{1 \pm \sqrt{5}}{2}$. There fore we have the following equality:
\[s^{-1} = \frac{2}{1 \pm \sqrt{5}} \hspace{2cm} s^{-2} = \frac{4}{(1 \pm \sqrt{5})^{2}}\]
Let´s see what is $s^{-1} + s^{-2}$:
\[s^{-1} + s^{-2} = \frac{2}{1 \pm \sqrt{5}} + \frac{2}{1 \pm \sqrt{5}}\]
A common denominator is $(1 \pm \sqrt{5})^{2}$. Thus we have that:
\[s^{-1} + s^{-2} = \frac{2 \pm 2\sqrt{5} + 4}{(1 \pm \sqrt{5})^{2}}\]
We can rewrite the expression as:
\[s^{-1} + s^{-2} = \frac{1 \pm 2\sqrt{5} + 5}{(1 \pm \sqrt{5})^{2}}\]
Notice the numerator is the same denominator but solved:
\[s^{-1} + s^{-2} = \frac{(1 \pm \sqrt{5})^{2}}{(1 \pm \sqrt{5})^{2}}\]
So canceling we get:
\[s^{-1} + s^{-2} = 1\]
Since it is true for any solution of the equation we have what we wanted:
\[\tau^{-1} + \tau^{-2} = 1 =\sigma^{-1} + \sigma^{-2}\]
\end{itemize}
\end{proof}
\end{Lem}
\begin{Lem}
\label{L3}
Let $p \in \R$. Define $a_{n} = p^{n}$. If $|p| < 1$ then, the sequence $(a_{n})_{n \in \N}\subseteq\R$ converges to zero.
\begin{proof}
It is enough to verify that the sequence is bounded above and is monotonic decreasing:
\textbf{The sequence is bounded above by $\boldsymbol{0}$: }We want to see that for any $n \in \N_{0}$ $, 0 \leq a_{n}$. Proceed by induction over $n \in \N_{0}$:
\textit{Base Case: $\mathit{n = 1}$}
Notice that $a_{0} = p^0 = 1$ and it clearly satisfies that:
\[0 \leq 1\]
So we have that $0 \leq a_{n}$
\textit{Inductive Step:}
Suppose the result for $n = k$. It is that:
\[0 \leq a_{k}\hspace{2cm}\textbf{(HI)}\]
We want to verify the inequality for $n = k+1$ (i.e. $0 \leq a_{k+1}$)
\[\begin{array}{rclr}
a_{k+1} & = & p^{k+1} & \left(\text{$a_{k+1}$ definition}\right)\\
& = & pp^{k} & \left(\text{Powers Properties}\right)\\
& = & pa_{k} & \left(\text{$a_{k}$ definition}\right)\\
& \geq & p0 & \left(\textbf{(HI)}\right)\\
& \geq & 0 & \left(\text{$0$ definition}\right)
\end{array}\]
So by the induction principle we have the result for any $n \in \N_{0}$. Thus $(a_{n})_{n \in \N}$ is bounded above by $0$.
\textbf{$\boldsymbol{{a_{n})_{n \in \N}}}$ is monotonically decreasing:} We want to verify that $a_{n+1} \leq a_{n}$. We proceed by induction over $n$:
\textbf{Base Case:}\textit{$n = 0$}
Notice that $a_{0 +1} = a_{1} = p$ and $a_{0} = p^{0} = 1$. Since $|p| < 1$, $-1 < p < 1$. Thus, in particular, we get that:
\[p \leq 1\]
Therefore $a_{1} \leq a_{0}$
\textbf{Inductive Step:}
Suppose the result for $n = k$. It is that:
\[a_{k+1} \leq a_{k}\hspace{2cm}\textbf{(HI)}\]
We want to verify the inequality for $n = k+1$ (i.e. $a_{k+2} \leq a_{k+1}$)
We have to cases:
\begin{enumerate}[label=(\alph*)]
\item $0 \leq p < 1$
\[\begin{array}{rclr}
a_{k+1} & \leq & a_{k} & \left(\textbf{(HI)}\right)\\
pa_{k+1} & \leq & pa_{k} & \left(\text{$p \geq 0$}\right)\\
pp^{k+1}& \leq & pp^{k} & \left(\text{$a_{k+1}$ and $a_{k}$ definition}\right)\\
p^{k+2}& \leq & p^{k+1} & \left(\text{Power properties}\right)\\
a_{k+2}& \leq & a_{k+1} & \left(\text{$a_{k+2}$ and $a_{k+1}$ definition}\right)
\end{array}\]
As we wanted!
\item $1 < p < 0$
\[\begin{array}{rclr}
a_{k+2} & = & p^{k+2} & \left(\text{$a_{k+2}$ definition}\right)\\
& = & pp^{k+1} & \left(\text{Power properties}\right)\\
& \leq & p^{k+1} & \left(\text{$p < 0$}\right)\\
& \leq & a_{k+1} & \left(\text{$a_{k+1}$ definition}\right)
\end{array}\]
\end{enumerate}
So by the induction principle we have the result for any $n \in \N_{0}$. Thus $a_{n +1} \leq a_{n}$.
\end{proof}
\end{Lem}
\begin{Lem}
\label{L4}
Let $k \in \N$ and we define $x_{n} = \frac{1}{n^{k}}$ for $n \in \N$. Then $(x_{n})_{n \in \N}$ converges to zero.
\begin{proof}
See Homework 4 Lemma 4.
\end{proof}
\end{Lem}
\begin{enumerate}[label=\textcolor{blue}{(\alph*)}]
\item \textcolor{blue}{Show that $(a_{n})_{n \in \N}$ does not converge in $\R$.}
Suppose by contradiction that it does. So we have that for any $n \in \N$ $\displaystyle{\lim_{n \to \infty}}{a_{n}} = r$ for some $r \in \R$. Using recursive definition of the sequence we get that:
\[a_{n+1} = a_{n} + a_{n-1}\]
Passing by the limit:
\[\lim_{n \to \infty}{a_{n+1}} = \lim_{n \to \infty}{a_{n} a_{n-1}}\]
\[\lim_{n \to \infty}{a_{n+1}} = \lim_{n \to \infty}{a_{n}} + \lim_{n \to \infty}{a_{n-1}}\]
Using our hypothesis:
\[r = r + r\]
\[r = 2r\]
Therefore, divide both sides by $r$. It is fair because of the lemma \textcolor{red}{\ref{L1}}:
\[1 = 2\]
A contradiction! It is a false statement. Thus we must have that $(a_{n})_{n \in \N}$ does not converge in $\R$.
\item \textcolor{blue}{$a_{n} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1}), n \in \N$.}
We are going to demonstrate it by Strong Induction o over $n \in \N_{0}$:
\textbf{Bases Cases:}
\begin{itemize}
\item[\textasteriskcentered] $\mathit{n = 0}:$ Let $R = \frac{1}{\sqrt{5}}(\tau^{0+1} - \sigma^{0+1})$. Recall that by definition we have that $a_{0} = 1$, so it is enough to verify that $R = 1$:
\[R = \frac{1}{\sqrt{5}}(\tau - \sigma)\]
Using the lemma \textcolor{red}{\ref{L2}.(i)} we get that:
\[R = \frac{1}{\sqrt{5}}\sqrt{5}\]
Canceling:
\[R = 1\]
\item[\textasteriskcentered] $\mathit{n = 1}:$ Let $R = \frac{1}{\sqrt{5}}(\tau^{1+1} - \sigma^{1+1})$. Recall that by definition we have that $a_{1} = 1$, so it is enough to verify that $R = 1$:
\[R = \frac{1}{\sqrt{5}}(\tau^{2} - \sigma^{2})\]
Using the lemma \textcolor{red}{\ref{L2}.(ii)} we get that:
\[R = \frac{1}{\sqrt{5}}(\tau - \sigma)\]
Using the lemma \textcolor{red}{\ref{L2}.(i)} we get that:
\[R = \frac{1}{\sqrt{5}}\sqrt{5}\]
Canceling:
\[R = 1\]
\end{itemize}
\textbf{Inductive Step:}
Suppose the result for $k \in \N$ with $k \leq n$. It is that:
\[\textbf{(HI)} \hspace{2cm} \forall_{k \in \N_{0}}{\left\{\left(k \leq n\right) \rightarrow \left[a_{k} = \frac{1}{\sqrt{5}}(\tau^{k+1} - \sigma^{k+1})\right]\right\}}\]
We prove it for $n+1$. It is that:
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})\]
By the recursive definition of Fibonacci Sequence:
\[a_{n+1} = a_{n} + a_{n-1}\]
By \textbf{(HI)} to $n-1$ and $n$ (they are less or equal to $n$):
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1}) + \frac{1}{\sqrt{5}}(\tau^{n-1+1} - \sigma^{n-1+1})\]
Use Common Factor factorization with $\frac{1}{\sqrt{5}}$:
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1} + \tau^{n} - \sigma^{n})\]
Use Common Factor factorization with $\tau^{n+2}$ and $\sigma^{n+2}$:
\[a_{n+1} = \frac{1}{\sqrt{5}}[\tau^{n+2}(\tau^{-1} + \tau^{-2}) - \sigma^{n+2}( \sigma^{-1} + \sigma^{-2})]\]
Using the lemma \textcolor{red}{\ref{L2}.(iii)} we can change the parenthesis:
\[a_{n+1} = \frac{1}{\sqrt{5}}[\tau^{n+2}(1) - \sigma^{n+2}(1)]\]
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})\]
So by the Strong Induction Principle we have that for any $n \in \N_{0}$:
\[a_{n} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1})\]
\begin{landscape}
\item[\textcolor{blue}{(c) }]\textcolor{blue}{$\displaystyle{\lim_{n \to \infty}x_{n} = \tau}$.}
\[\begin{array}{rclr}
\displaystyle{\lim_{n \to \infty}{x_{n}}} & = & \displaystyle{\lim_{n \to \infty}{\frac{a_{n+1}}{a_{n}}}} & \left(\text{$x_{n}$ definition}\right) \\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{\frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})}{\frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1})}}} & \left(\text{Exercise $(b)$ formula}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\cfrac{\frac{\tau^{n+2}}{\tau^{n+2}} - \frac{\sigma^{n+2}}{\tau^{n+2}}}{\frac{\tau^{n+1}}{\tau^{n+1}}\frac{1}{\tau} - \frac{\sigma^{n+1}}{\tau^{n+1}}\frac{1}{\tau}}}} & \left(\text{Canceling and dividing by $\tau^{n+2}$ up and down}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n+2}}{\frac{1}{\tau} - \frac{1}{\tau}\left(\frac{\sigma}{\tau}\right)^{n+1}}}} & \left(\text{Power properties}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n+2}}{\frac{1}{\tau}\left[ 1 - \left(\frac{\sigma}{\tau}\right)^{n+1}\right]}}} & \left(\text{Common Factor Factorization}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)^{2}}{\frac{1}{\tau}\left[ 1 - \left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)\right]}}} & \left(\text{Power properties}\right)\\
&&&\\
& = & \frac{ \displaystyle{\lim_{n \to \infty}{1} - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\lim_{n \to \infty}\left(\frac{\sigma}{\tau}\right)^{2}}}}{\displaystyle{\lim_{n \to \infty}{\left(\frac{1}{\tau}\right)}\left[ \lim_{n \to \infty}{1} - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\lim_{n \to \infty}\left(\frac{\sigma}{\tau}\right)}\right]}} & \left(\text{Limit properties}\right)\\
&&&\\
& = & \frac{\displaystyle{1 - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)^{2}}}}{\displaystyle{\left(\frac{1}{\tau}\right)\left[1 - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}}\left(\frac{\sigma}{\tau}\right)\right]}} & \left(\text{Solving constant limits}\right)\\
&&&\\
& = & \frac{\displaystyle{1 - 0\left(\frac{\sigma}{\tau}\right)^{2}}}{\displaystyle{\left(\frac{1}{\tau}\right)\left[1 - 0\left(\frac{\sigma}{\tau}\right)\right]}} & \left(\text{Using lemmas \textcolor{red}{\ref{L2}.(iv)} and \ref{L3}}\right)\\
&&&\\
\end{array}\]
\[\begin{array}{rclr}
\displaystyle{\lim_{n \infty}{x_{n}}} & = & \displaystyle{\frac{\frac{1}{1}}{\frac{1}{\tau}}} & \left(\text{0 definition}\right)\\
&&&\\
& = & \tau & \left(\text{Means and extremes law}\right)\\
&&&
\end{array}\]
As we wanted!
\end{landscape}
\end{enumerate}
\end{exercise}
答案1
从
到这里
可以使用包来完成nicematrix。
\Block提供可作为和的命令multicolumn，并且还允许在单元格内部multirow使用 。\\
最简单的用法\Block{}{<content1> \\ <content2>}是创建一个带有两条独立线的 1x1 单元格。
最后一个数组代码中更改的五行被标记为 %<<<< changed
这是完整的代码。
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%PREAMBLE%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[12pt, letterpaper]{article}
%%%%%%%%%%%%%%%%%%%%%%%%PAQUETES%%%%%%%%%%%%%%%%%%%%%%%%%%
%Avoid problems with accents and special charracters
\usepackage[utf8]{inputenc}
%document Format
\usepackage[left=2.5cm,top=2.5cm,right=2.5cm,bottom=2.5cm]{geometry}
%Colors along the docuemnt.
\usepackage{xcolor}
%Format for mathematical assertions
\usepackage{amsthm}
%More useful symbols
\usepackage{mathtools}
%Even more useful symbols
\usepackage{amssymb}
%Language of Expresions
\usepackage[english]{babel}
%Format Sections
\usepackage{sectsty}
%Font for canonical sets
\usepackage{dsfont}
%Images inclusion
\usepackage{graphicx}
%Long comments
\usepackage{comment}
%Caritas
\usepackage{wasysym}
%Referencias
\usepackage{hyperref}
%Enumeración con Color
\usepackage{enumitem}
%Hoja Volteada
\usepackage{pdflscape}
\usepackage{nicematrix} % <<<<<<<<<<<<<< added
%%%%%%%%%%%%%%%%%%%%%%%%%COLORS%%%%%%%%%%%%%%%%%%%%%%%%%%%
\definecolor{blue}{RGB}{0,176,246}
\definecolor{white}{RGB}{255,255,255}
%%%%%%%%%%%%%%%%%%%%%ENTORNOS%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Exercise Enviroment
\renewcommand{\thesection}{\Roman{section}}
\newcounter{exercise}
\renewcommand{\theexercise}{\arabic{exercise}}
\newenvironment{exercise}[1][]{\refstepcounter{exercise}
\par\medskip
\noindent \textcolor{blue}{\textbf{Exercise~\theexercise}. #1}} \rmfamily
{\medskip}
%Proofs
\renewenvironment{proof}{{\noindent\bfseries Proof:\\}}{\rightline{$\blacksquare$}}
%Mathematical Staments
\theoremstyle{definition}
\newtheorem{Theo}{Theorem}
\newtheorem{Lem}[Theo]{Lemma}
\newtheorem{Def}[Theo]{Definition}
\newtheorem{Not}[Theo]{Notation}
\newtheorem{Cor}{Corollary}[Theo]
\newtheorem{Propo}[Theo]{Proposition}
\newtheorem{Prope}[Theo]{Property}
\newtheorem{Obs}[Theo]{Observation}
%%%%%%%%%%%%%%%%%%%%%MY COMMANDS%%%%%%%%%%%%%%%%%%%%%%%%%
%New
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\I}{\mathbb{I}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
%Renewed
\renewcommand{\sup}{\operatorname{Sup}}
\renewcommand{\inf}{\operatorname{Inf}}
\renewcommand{\limsup}{\operatorname{LimSup}}
\renewcommand{\liminf}{\operatorname{LimInf}}
\renewcommand{\min}{\operatorname{Min}}
\renewcommand{\max}{\operatorname{Max}}
%%%%%%%%%%%%%%%%%%%ESTYLE%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsectionfont{\color{blue}}
\sectionfont{\color{blue}}
%Hyperlinks
\hypersetup{
colorlinks=true,
linkcolor=red,
filecolor=magenta,
urlcolor=red
}
\urlstyle{same}
%%%%%%%%%%%%%%%%%%%%%%%&TÍTULO&%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{\textcolor{blue}{\textbf{Homework \#5\\
Analysis I (English)}}}
\author{\textcolor{blue}{\textit{Germán Felipe López Díaz}}}
\date{\textit{\textcolor{blue}{March 9\textsuperscript{th}, 2021}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%DOCUMENTO%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\setcounter{exercise}{5}
\begin{exercise}
\textcolor{blue}{The Fibonacci sequence $(a_{n})_{n \in \N}$ is defined recursively by:
\[a_{0} = 1 \hspace{0.5cm} a_{1} = 1 \hspace{0.5cm} a_{n+1} = a_{n} + a_{n-1}\hspace{0.5cm}n \in \N\]
Moreover, let $\sigma < \tau$ the solutions of $x^{2} - x - 1 = 0$ and
\[x_{n} = \frac{a_{n+1}}{a_{n}}, n \in \N\]}
For its exercise we need some lemmas:
\begin{Lem}
\label{L1}
For any $n \in \N_{0}$. In Fibonacci sequence context, $a_{n} > 0$.
\end{Lem}
\begin{proof}
\textbf{Bases Cases:}
\begin{itemize}
\item[$\star$] $\mathit{n = 0}:$ Suppose by contradiction that not. That is that $a_{0} \leq 0$. By definition $a_{0} = 1$. Therefore:
\[1 \leq 0\]
A contradiction! with the order of natural numbers. In conclusion it must be true that $a_{0} > 0$.
\item[$\star$] $\mathit{n = 1}:$ Suppose by contradiction that not. That is that $a_{1} \leq 0$. By definition $a_{1} = 1$. Therefore:
\[1 \leq 0\]
A contradiction! with the order of natural numbers. In conclusion it must be true that $a_{1} > 0$.
\end{itemize}
\textbf{Inductive Step:}
Suppose the result for any $k \in \N$ with $k \leq n$. It is that:
\[\forall_{k \in \N}\left[(k \leq n) \rightarrow (a_{k} > 0)\right]\hspace{2cm}\textbf{(HI)}\]
We want to verify the result for $n+1$. By the recursive definition we have that:
\[a_{n+1} = a_{n} + a_{n-1}\]
By \textbf{(HI)} we have that the right side is a non negative number due to $n \leq n$ and $n-1 \leq 1$ and the addition of to non negative numbers is again a non negative number. Call it $P$:
\[a_{n+1} = P\]
By the Strong Induction Principle we have that the result is true for all $n \in \N_{0}$.
\end{proof}
\begin{Lem}
\label{L2}
Let $\sigma$ and $\tau$, $\sigma < \tau$, solutions for the equation $x^{2} - x - 1 = 0$. Then:
\begin{itemize}
\item[(i)] $\tau - \sigma = \sqrt{5}$
\item[(ii)] $\tau^{2} - \sigma^{2} = \tau - \sigma$
\item[(iii)] $\tau^{-1} + \tau^{-2} = 1 = \sigma^{-1}+ \sigma^{-2}$
\end{itemize}
\begin{proof}
Initially by definition of $\sigma$ and $\tau$ we have the following equality:
\begin{equation}
\label{Eq1.1}
\tau^{2} - \tau - 1 = 0
\end{equation}
\begin{equation}
\label{Eq1.2}
\sigma^{2} - \sigma - 1 = 0
\end{equation}
On the other hand, remember The Quadratic Formula:
\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
We use it to solve the polynomial $x^{2} - x - 1$ using the with $a = 1$, $b = -1$ and $c = -1$:
\[x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\]
\[x = \frac{1 \pm \sqrt{1 + 4}}{2}\]
\[x = \frac{1 \pm \sqrt{5}}{2}\]
According to Precalculus course we have that its polynomial has two roots:
\[x_{1} = \frac{1 - \sqrt{5}}{2}\hspace{1cm}x_{2} =\frac{1 + \sqrt{5}}{2}\]
Both of them are real! Since $\sigma$ and $\tau$ are roots of it, necesarily, we must have that:
\begin{equation}
\label{Eq1.3}
\sigma = \frac{1 - \sqrt{5}}{2}
\end{equation}
\begin{equation}
\label{Eq1.4}
\tau = \frac{1 + \sqrt{5}}{2}
\end{equation}
Because $\sigma < \tau$
\begin{itemize}
\item[(i)]
\[\begin{array}{rclr}
\tau - \sigma & = & \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} & \left(\text{Equation \textcolor{red}{(\ref{Eq1.3})} and \textcolor{red}{(\ref{Eq1.4})}}\right)\\
& = & \frac{1 + \sqrt{5} -1 + \sqrt{5}}{2}& \left(\text{Subtraction of homogeneous fractions}\right)\\
& = & \frac{2\sqrt{5}}{2}& \left(\text{Adding similar terms}\right)\\
& = & \sqrt{5} & \left(\text{Canceling}\right)\\
\end{array}\]
\item[(ii)] Using equation $\textcolor{red}{(\ref{Eq1.1})}$ and $\textcolor{red}{(\ref{Eq1.2})}$ finding the quadratic term we get that:
\[\tau^{2} = \tau + 1 \hspace{1cm}\sigma^{2} = \sigma + 1\]
Thus we get:
\[\begin{array}{rclr}
\tau^{2} - \sigma^{2} & = & \tau + 1 - (\sigma + 1) &\left(\text{Substraction of last equations}\right)\\
& = & \tau + 1 - \sigma - 1 & \left(\text{Distributing the minus}\right)\\
& = & \tau - \sigma & \left(\text{Adding similar terms}\right)\\
\end{array}\]
\item[(iii)] Let $s$ be solution of the equation $x^{2} - x - 1 = 0$. from the solution of the Quadratic Formula, that we did at the beginning, we have that $s = \frac{1 \pm \sqrt{5}}{2}$. There fore we have the following equality:
\[s^{-1} = \frac{2}{1 \pm \sqrt{5}} \hspace{2cm} s^{-2} = \frac{4}{(1 \pm \sqrt{5})^{2}}\]
Let´s see what is $s^{-1} + s^{-2}$:
\[s^{-1} + s^{-2} = \frac{2}{1 \pm \sqrt{5}} + \frac{2}{1 \pm \sqrt{5}}\]
A common denominator is $(1 \pm \sqrt{5})^{2}$. Thus we have that:
\[s^{-1} + s^{-2} = \frac{2 \pm 2\sqrt{5} + 4}{(1 \pm \sqrt{5})^{2}}\]
We can rewrite the expression as:
\[s^{-1} + s^{-2} = \frac{1 \pm 2\sqrt{5} + 5}{(1 \pm \sqrt{5})^{2}}\]
Notice the numerator is the same denominator but solved:
\[s^{-1} + s^{-2} = \frac{(1 \pm \sqrt{5})^{2}}{(1 \pm \sqrt{5})^{2}}\]
So canceling we get:
\[s^{-1} + s^{-2} = 1\]
Since it is true for any solution of the equation we have what we wanted:
\[\tau^{-1} + \tau^{-2} = 1 =\sigma^{-1} + \sigma^{-2}\]
\end{itemize}
\end{proof}
\end{Lem}
\begin{Lem}
\label{L3}
Let $p \in \R$. Define $a_{n} = p^{n}$. If $|p| < 1$ then, the sequence $(a_{n})_{n \in \N}\subseteq\R$ converges to zero.
\begin{proof}
It is enough to verify that the sequence is bounded above and is monotonic decreasing:
\textbf{The sequence is bounded above by $\boldsymbol{0}$: }We want to see that for any $n \in \N_{0}$ $, 0 \leq a_{n}$. Proceed by induction over $n \in \N_{0}$:
\textit{Base Case: $\mathit{n = 1}$}
Notice that $a_{0} = p^0 = 1$ and it clearly satisfies that:
\[0 \leq 1\]
So we have that $0 \leq a_{n}$
\textit{Inductive Step:}
Suppose the result for $n = k$. It is that:
\[0 \leq a_{k}\hspace{2cm}\textbf{(HI)}\]
We want to verify the inequality for $n = k+1$ (i.e. $0 \leq a_{k+1}$)
\[\begin{array}{rclr}
a_{k+1} & = & p^{k+1} & \left(\text{$a_{k+1}$ definition}\right)\\
& = & pp^{k} & \left(\text{Powers Properties}\right)\\
& = & pa_{k} & \left(\text{$a_{k}$ definition}\right)\\
& \geq & p0 & \left(\textbf{(HI)}\right)\\
& \geq & 0 & \left(\text{$0$ definition}\right)
\end{array}\]
So by the induction principle we have the result for any $n \in \N_{0}$. Thus $(a_{n})_{n \in \N}$ is bounded above by $0$.
\textbf{$\boldsymbol{{a_{n})_{n \in \N}}}$ is monotonically decreasing:} We want to verify that $a_{n+1} \leq a_{n}$. We proceed by induction over $n$:
\textbf{Base Case:}\textit{$n = 0$}
Notice that $a_{0 +1} = a_{1} = p$ and $a_{0} = p^{0} = 1$. Since $|p| < 1$, $-1 < p < 1$. Thus, in particular, we get that:
\[p \leq 1\]
Therefore $a_{1} \leq a_{0}$
\textbf{Inductive Step:}
Suppose the result for $n = k$. It is that:
\[a_{k+1} \leq a_{k}\hspace{2cm}\textbf{(HI)}\]
We want to verify the inequality for $n = k+1$ (i.e. $a_{k+2} \leq a_{k+1}$)
We have to cases:
\begin{enumerate}[label=(\alph*)]
\item $0 \leq p < 1$
\[\begin{array}{rclr}
a_{k+1} & \leq & a_{k} & \left(\textbf{(HI)}\right)\\
pa_{k+1} & \leq & pa_{k} & \left(\text{$p \geq 0$}\right)\\
pp^{k+1}& \leq & pp^{k} & \left(\text{$a_{k+1}$ and $a_{k}$ definition}\right)\\
p^{k+2}& \leq & p^{k+1} & \left(\text{Power properties}\right)\\
a_{k+2}& \leq & a_{k+1} & \left(\text{$a_{k+2}$ and $a_{k+1}$ definition}\right)
\end{array}\]
As we wanted!
\item $1 < p < 0$
\[\begin{array}{rclr}
a_{k+2} & = & p^{k+2} & \left(\text{$a_{k+2}$ definition}\right)\\
& = & pp^{k+1} & \left(\text{Power properties}\right)\\
& \leq & p^{k+1} & \left(\text{$p < 0$}\right)\\
& \leq & a_{k+1} & \left(\text{$a_{k+1}$ definition}\right)
\end{array}\]
\end{enumerate}
So by the induction principle we have the result for any $n \in \N_{0}$. Thus $a_{n +1} \leq a_{n}$.
\end{proof}
\end{Lem}
\begin{Lem}
\label{L4}
Let $k \in \N$ and we define $x_{n} = \frac{1}{n^{k}}$ for $n \in \N$. Then $(x_{n})_{n \in \N}$ converges to zero.
\begin{proof}
See Homework 4 Lemma 4.
\end{proof}
\end{Lem}
\begin{enumerate}[label=\textcolor{blue}{(\alph*)}]
\item \textcolor{blue}{Show that $(a_{n})_{n \in \N}$ does not converge in $\R$.}
Suppose by contradiction that it does. So we have that for any $n \in \N$ $\displaystyle{\lim_{n \to \infty}}{a_{n}} = r$ for some $r \in \R$. Using recursive definition of the sequence we get that:
\[a_{n+1} = a_{n} + a_{n-1}\]
Passing by the limit:
\[\lim_{n \to \infty}{a_{n+1}} = \lim_{n \to \infty}{a_{n} a_{n-1}}\]
\[\lim_{n \to \infty}{a_{n+1}} = \lim_{n \to \infty}{a_{n}} + \lim_{n \to \infty}{a_{n-1}}\]
Using our hypothesis:
\[r = r + r\]
\[r = 2r\]
Therefore, divide both sides by $r$. It is fair because of the lemma \textcolor{red}{\ref{L1}}:
\[1 = 2\]
A contradiction! It is a false statement. Thus we must have that $(a_{n})_{n \in \N}$ does not converge in $\R$.
\item \textcolor{blue}{$a_{n} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1}), n \in \N$.}
We are going to demonstrate it by Strong Induction o over $n \in \N_{0}$:
\textbf{Bases Cases:}
\begin{itemize}
\item[\textasteriskcentered] $\mathit{n = 0}:$ Let $R = \frac{1}{\sqrt{5}}(\tau^{0+1} - \sigma^{0+1})$. Recall that by definition we have that $a_{0} = 1$, so it is enough to verify that $R = 1$:
\[R = \frac{1}{\sqrt{5}}(\tau - \sigma)\]
Using the lemma \textcolor{red}{\ref{L2}.(i)} we get that:
\[R = \frac{1}{\sqrt{5}}\sqrt{5}\]
Canceling:
\[R = 1\]
\item[\textasteriskcentered] $\mathit{n = 1}:$ Let $R = \frac{1}{\sqrt{5}}(\tau^{1+1} - \sigma^{1+1})$. Recall that by definition we have that $a_{1} = 1$, so it is enough to verify that $R = 1$:
\[R = \frac{1}{\sqrt{5}}(\tau^{2} - \sigma^{2})\]
Using the lemma \textcolor{red}{\ref{L2}.(ii)} we get that:
\[R = \frac{1}{\sqrt{5}}(\tau - \sigma)\]
Using the lemma \textcolor{red}{\ref{L2}.(i)} we get that:
\[R = \frac{1}{\sqrt{5}}\sqrt{5}\]
Canceling:
\[R = 1\]
\end{itemize}
\textbf{Inductive Step:}
Suppose the result for $k \in \N$ with $k \leq n$. It is that:
\[\textbf{(HI)} \hspace{2cm} \forall_{k \in \N_{0}}{\left\{\left(k \leq n\right) \rightarrow \left[a_{k} = \frac{1}{\sqrt{5}}(\tau^{k+1} - \sigma^{k+1})\right]\right\}}\]
We prove it for $n+1$. It is that:
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})\]
By the recursive definition of Fibonacci Sequence:
\[a_{n+1} = a_{n} + a_{n-1}\]
By \textbf{(HI)} to $n-1$ and $n$ (they are less or equal to $n$):
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1}) + \frac{1}{\sqrt{5}}(\tau^{n-1+1} - \sigma^{n-1+1})\]
Use Common Factor factorization with $\frac{1}{\sqrt{5}}$:
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1} + \tau^{n} - \sigma^{n})\]
Use Common Factor factorization with $\tau^{n+2}$ and $\sigma^{n+2}$:
\[a_{n+1} = \frac{1}{\sqrt{5}}[\tau^{n+2}(\tau^{-1} + \tau^{-2}) - \sigma^{n+2}( \sigma^{-1} + \sigma^{-2})]\]
Using the lemma \textcolor{red}{\ref{L2}.(iii)} we can change the parenthesis:
\[a_{n+1} = \frac{1}{\sqrt{5}}[\tau^{n+2}(1) - \sigma^{n+2}(1)]\]
\[a_{n+1} = \frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})\]
So by the Strong Induction Principle we have that for any $n \in \N_{0}$:
\[a_{n} = \frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1})\]
% \begin{landscape}
\item[\textcolor{blue}{(c) }]\textcolor{blue}{$\displaystyle{\lim_{n \to \infty}x_{n} = \tau}$.}
% <<<<<<<< changed from here
\[\begin{NiceArray}{rclr} %<<<< changed
\displaystyle{\lim_{n \to \infty}{x_{n}}} & = & \displaystyle{\lim_{n \to \infty}{\frac{a_{n+1}}{a_{n}}}} & \left(\text{$x_{n}$ definition}\right) \\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{\frac{1}{\sqrt{5}}(\tau^{n+2} - \sigma^{n+2})}{\frac{1}{\sqrt{5}}(\tau^{n+1} - \sigma^{n+1})}}} & \left(\text{Exercise $(b)$ formula}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\cfrac{\frac{\tau^{n+2}}{\tau^{n+2}} - \frac{\sigma^{n+2}}{\tau^{n+2}}}{\frac{\tau^{n+1}}{\tau^{n+1}}\frac{1}{\tau} - \frac{\sigma^{n+1}}{\tau^{n+1}}\frac{1}{\tau}}}} & \Block{}{\text{(Canceling and dividing by} \\ \text{$\tau^{n+2}$} \text{\, up and down)}}\\ %<<<< changed
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n+2}}{\frac{1}{\tau} - \frac{1}{\tau}\left(\frac{\sigma}{\tau}\right)^{n+1}}}} & \left(\text{Power properties}\right)\\
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n+2}}{\frac{1}{\tau}\left[ 1 - \left(\frac{\sigma}{\tau}\right)^{n+1}\right]}}} & \Block{}{\text{(Common Factor}\\\text{ Factorization)}}\\%<<<< changed
&&&\\
& = & \displaystyle{\lim_{n \to \infty}{\frac{1 - \left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)^{2}}{\frac{1}{\tau}\left[ 1 - \left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)\right]}}} & \left(\text{Power properties}\right)\\
&&&\\
& = & \frac{ \displaystyle{\lim_{n \to \infty}{1} - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\lim_{n \to \infty}\left(\frac{\sigma}{\tau}\right)^{2}}}}{\displaystyle{\lim_{n \to \infty}{\left(\frac{1}{\tau}\right)}\left[ \lim_{n \to \infty}{1} - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\lim_{n \to \infty}\left(\frac{\sigma}{\tau}\right)}\right]}} & \left(\text{Limit properties}\right)\\
&&&\\
& = & \frac{\displaystyle{1 - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}\left(\frac{\sigma}{\tau}\right)^{2}}}}{\displaystyle{\left(\frac{1}{\tau}\right)\left[1 - \lim_{n \to \infty}{\left(\frac{\sigma}{\tau}\right)^{n}}\left(\frac{\sigma}{\tau}\right)\right]}} & \left(\text{Solving constant limits}\right)\\
&&&\\
& = & \frac{\displaystyle{1 - 0\left(\frac{\sigma}{\tau}\right)^{2}}}{\displaystyle{\left(\frac{1}{\tau}\right)\left[1 - 0\left(\frac{\sigma}{\tau}\right)\right]}} & \Block{}{ \text{Using lemmas} \\ \text{ \textcolor{red}{\ref{L2}.(iv)} and \ref{L3}.)}}\\%<<<< changed
&&&\\
% \[\begin{array}{rclr}
& = & \displaystyle{\frac{\frac{1}{1}}{\frac{1}{\tau}}} & \left(\text{0 definition}\right)\\%<<<< changed
&&&\\
& = & \tau & \left(\text{Means and extremes law}\right)\\
&&& \\
% \end{array}\]
\end{NiceArray}\] % <<<<<<<< changed to here
As we wanted!
% \end{landscape}
\end{enumerate}
\end{exercise}
\end{document}