Homework 5 Sunday update

Documents/LaTeX/Homework 5.tex

@@ -39,7 +39,7 @@
 		\item \textbf{Section 3.3}
 		\begin{enumerate}
 			\item[2)] Let $x_1 > 1$ and $x_{n+1} := 2-1/x_n$ for $n \in \N$. Show that $(x_n)$ is bounded and monotone. Find the limit.
-			\\\\ The first five terms of this sequence are $x_1 \geq 2,x_2 \geq \frac{3}{2}, x_3 \geq \frac{4}{5}, x_4 \geq \frac{5}{4}, x_5 \geq \frac{6}{5}, \dots$. This sequence appears to be decreasing.
+			\\\\ The first five terms of this sequence are $x_1 \geq 2,x_2 \geq \frac{3}{2}, x_3 \geq \frac{4}{3}, x_4 \geq \frac{5}{4}, x_5 \geq \frac{6}{5}, \dots \approx x_1 \ge 2, x_2 \geq 1.5, x_3 \geq 1.3333, x_4 \geq 1.25, x_5 \geq 1.2, \dots$. This sequence appears to be decreasing.
 			\\\\Recall the Monotone Sequence Property:
 			\begin{theorem*}{Monotone Sequence Property}
 				A monotone sequence of real numbers is convergent if and only if it is bounded. Further,
@@ -151,7 +151,7 @@
 			\end{align*}
 			
 			\item[7)] Let $x_1 := a>0$ and $x_{n+1} := x_n+1/x_n$ for $n \in \N$. Determine whether $(x_n)$ converges or diverges.
-			\\\\The first 5 terms of this sequence are $x_1 \geq 1, x_2 \geq 2, x_3 \geq \frac{5}{2}, x_4 \geq \frac{29}{10}, x_5 \geq \frac{941}{290}, \dots$. This sequence appears to be increasing. We show this to be true as follows:
+			\\\\The first 5 terms of this sequence are $x_1 \geq 1, x_2 \geq 2, x_3 \geq \frac{5}{2}, x_4 \geq \frac{29}{10}, x_5 \geq \frac{941}{290}, \dots \approx x_1 \geq 1, x_2 \geq 2, x_3 \geq 2.5, x_4 \geq 2.9, x_5 \geq 3.244828, \dots$. This sequence appears to be increasing. We show this to be true as follows:
 			\begin{align*}
 				x_{n+1} \geq x_n &\iff x_n + \frac{1}{x_n} \geq x_n \\
 				&\iff x_n^2 + 1 \geq x_n^2 \\
@@ -191,7 +191,7 @@
 		\end{enumerate}
 		
 		\item $a_1 = 1,\ a_{n+1}=\frac{a_n^2+5}{2a_n}$
-		\\\\The first 5 terms of this sequence are $1, 3, \frac{7}{3}, \frac{47}{21}, \frac{2207}{987}, \dots$. This is a decreasing sequence.
+		\\\\The first 5 terms of this sequence are $1, 3, \frac{7}{3}, \frac{47}{21}, \frac{2207}{987}, \dots \approx 1, 3, 2.3333, 2.2381, 2.2361, \dots$. This is a decreasing sequence.
 		\\\\First, we must find the possible limits (fixed points) of the sequence. So,
 		\begin{align*}
 			a&=\frac{a^2+5}{2a} \\
@@ -242,50 +242,162 @@
 		\end{align*} 
 		
 		\item $a_1 = 5,\ a_{n+1}=\sqrt{4+a_n}$
-		\\\\The first 5 terms of this sequence are $5, \sqrt{5}, \sqrt{7}, \frac{\sqrt{57}}{3}, \frac{\sqrt{2751}}{21}, \dots$. This sequence is decreasing.
+		\\\\The first 5 terms of this sequence are 5, 3, $\sqrt{7}$, $ \frac{\sqrt{14}}{2} + \frac{\sqrt{2}}{2} $, $ \frac{\sqrt{2 \cdot (\sqrt{14}+\sqrt{2}+8)}}{2} $, $\dots$, $\approx$ 5, 3, 2.64575131106, 2.57793547457, 2.5647486182, $\dots$. This sequence is decreasing.
 		\\\\First, we must find the possible limits (fixed points) of the sequence. So,
 		\begin{align*}
 			a&=\sqrt{4+a} \\
-			a^2 &= 4 + a \\
+			\sqrt{4+a} &= a \\
-			a^2 -a - 4 &= 0 \\
+			4+a &= a^2 \\
+			-a^2+a+4 &= 0 \\
+			a^2-a-4 &= 0 \\
+			a^2-a&=4 \\
+			a^2-a+\frac{1}{4}&=4+\frac{1}{4} \\
+			a^2-a+\frac{1}{4}&=\frac{17}{4} \\
+			(a-\frac{1}{2})^2&=\frac{17}{4} \\
+			a-\frac{1}{2}&=\pm \frac{\sqrt{17}}{2}
 		\end{align*}
-		So we have that $a=\frac{1}{2} - \sqrt{17}$, or $a=\frac{1}{2}+\sqrt{17}$. Since we're given that $a_1=5$, we infer that $\frac{1}{2}+\sqrt{17}$ is a possible limit (fixed point) of $(a_n)$.
+		So we have that $a=\frac{1}{2} + \frac{\sqrt{17}}{2}$, or $a=\frac{1}{2}-\frac{\sqrt{17}}{2}$. We must now check these solutions for correctness; so, 
+		\begin{align*}
+			a \Rightarrow \frac{1}{2}-\frac{\sqrt{17}}{2} &=\frac{1}{2} \left(1-\sqrt{17}\right) \\
+			&\approx -1.56155 \\\\
+			\sqrt{a+4} &= \sqrt{\left(\frac{1}{2}-\frac{\sqrt{17}}{2}\right)+4} \\
+			&=\frac{\sqrt{9-\sqrt{17}}}{\sqrt{2}} \\
+			&\approx 1.56155
+		\end{align*}
+		Thus, this solution is incorrect. Now we must validate that $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is correct. So,
+		\begin{align*}
+			a \Rightarrow \frac{1}{2}+\frac{\sqrt{17}}{2} &= \frac{1}{2}\left(1+\sqrt{17}\right) \\
+			&\approx 2.56155 \\\\
+			\sqrt{a+4} &= \sqrt{\left(\frac{\sqrt{17}}{2}+\frac{1}{2}\right)+4} \\
+			&= \frac{\sqrt{9+\sqrt{17}}}{\sqrt{2}} \\
+			&\approx 2.56155
+		\end{align*}
+		Thus $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is a correct solution.
 		\\\\Now we want to show that $(a_n)$ is bounded below by $\frac{1}{2}+\sqrt{17}$.
 		\begin{proof}
-			We want to show that $a_n \geq \frac{1}{2}+\sqrt{17},\ \forall\ n \in \N$. We prove it by method of mathematical induction.
+			We want to show that $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$, by the definition of a lower bound. We prove this by method of mathematical induction.
-			\\\\\textbf{Basis Step:} $5 \geq \frac{1}{2}+\sqrt{17}$ yields that $a_1 \ge \frac{1}{2}+\sqrt{17}$
+			\\\\\textbf{Basis Step:} Since $5 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$, we have that $a_1 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$.
-			\\\\\textbf{Inductive Step:} Assume $a_n \geq \frac{1}{2}+\sqrt{17},\ \forall\ n \in \N$.
+			\\\\\textbf{Inductive Step:} Assume $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$.
-			\\\\\textbf{Show:} We want to show that $a_{n+1} \geq \frac{1}{2}+\sqrt{17},\ \forall\ n \in \N$.
+			\\\\\textbf{Show:} We now want to show that $ a_{n+1} \geq \frac{1}{2}+\frac{\sqrt{17}}{2}\ \forall\ n \in \N$. So,
-			\\So,
 			\begin{align*}
-				a_{n+1} \ge \sqrt{4+a_n}
+				a_{n+1} &= \sqrt{4+a_n}, &\text{by the definition of the sequence} \\
+				&\geq \sqrt{4+\left(\frac{1}{2}+\frac{\sqrt{17}}{2}\right)}, &\text{by the inductive hypothesis} \\
+				&\geq \sqrt{\frac{8}{2}+\frac{1}{2}+\frac{\sqrt{17}}{2}} \\
+				&\geq \sqrt{\frac{9+\sqrt{17}}{2}} \\
+				&\geq \sqrt{\frac{1}{2}\left(9+\sqrt{17}\right)} \\
+				&\geq \sqrt{\frac{1}{4}+\frac{\sqrt{17}}{2}+\frac{17}{4}}, &\text{by expressing } \frac{9+\sqrt{17}}{2} \text{ as a square} \\
+				&\geq \sqrt{\frac{1+2\sqrt{17}+17}{4}} \\
+				&\geq \sqrt{\frac{1+2\sqrt{17}+(\sqrt{17})^2}{4}} \\
+				&\geq \sqrt{\frac{(\sqrt{17}+1)^2}{4}} \\
+				&\geq \sqrt{\frac{1}{4}\left(1+\sqrt{17}\right)^2} \\
+				&\geq \frac{\sqrt{(1+\sqrt{17})^2}}{\sqrt{4}} \\
+				&\geq \frac{\sqrt{17}+1}{2} \\
+				&\geq \frac{1}{2} + \frac{\sqrt{17}}{2}
 			\end{align*}
-			
+			Thus we have that $ a_{n+1} \geq \frac{1}{2} + \frac{\sqrt{17}}{2}\ \forall\ n \in \N$.
 		\end{proof}
+		Now, we want to show that $ (a_n) $ is monotone decreasing; that is, we want to show that $(a_1 \geq a_2 \geq \dots \geq a_n)$.
+		\begin{proof}
+			We want to show that $(a_1 \geq a_2 \geq \dots \geq a_n),\ \forall\ n \in \N$. We prove this by method of mathematical induction.
+			\\\\\textbf{Basis Step:} Since $5 \geq 3$, we have that $a_1 \geq a_2$.
+			\\\\\textbf{Inductive Step:} Assume $a_n \geq a_{n+1}\ \forall\ n \in \N$.
+			\\\\\textbf{Show:} We want to show that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$. So,
+			\begin{align*}
+				a_{n+2} &= \sqrt{4+a_{n+1}} &\text{by the definition of the sequence} \\
+				&\leq \sqrt{4+a_n} &\text{by the inductive hypothesis} \\
+				&=a_{n+1}
+			\end{align*}
+			Thus we have that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$.
+		\end{proof}
+	Since $(a_n)$ is both bounded and monotone decreasing, by the \textit{Monotone Convergence Theorem}, we have that $(a_n)$ converges. Also by the \textit{Monotone Sequence Property}, we have that $(a_n)$ converges to the following:
+	\begin{align*}
+		\lim (a_n) &= \inf \{a_n: n \in \N\} \\
+		&= \frac{1}{2}+\frac{\sqrt{17}}{2} \approx 2.56155281281
+	\end{align*}
 	\end{enumerate}
 	
 	\item 
 	\begin{enumerate}
 		\item Show $a_n=\frac{3 \cdot 5 \cdot 7 \cdot \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$ converges to $A$ where $0 \leq A < 1/2$.
+		\\\\\textbf{TODO}
 		
 		\item Show $b_n = \frac{2 \cdot 4 \cdot 6 \dots (2n)}{3 \cdot 5 \cdot 7 \dots (2n+1)}$ converges to $B$ where $0 \leq B < 2/3$.
+		\\\\\textbf{TODO}
 	\end{enumerate}
 
 	\item \textbf{Section 3.4}
 	\begin{enumerate}
 		\item[1)] Give an example of an unbounded sequence that has a convergent subsequence.
+		\\\\Let $a_n:=(-1)^n$. This sequence diverges since it oscillates between 1 and $-1$, however if we define $b_n:=(a_{2n})$; that is, let $(b_n)$ be the sequence of all terms of $(a_n)$ such that $n$ is even. Thus, $(b_n)$ is a subsequence of $(a_n)$, but $(b_n)$ converges since $a_2=1, a_4=1, \dots, a_{2n}=1$. Thus while $(a_n)$ is a divergent sequence, the subsequence $(b_n)$ of $(a_n)$ converges for all $n \in \N$.\\
 		
 		\item[3)] Let $(f_n)$ be the Fibonacci sequence of Example 3.1.2(d), and let $x_n := f_{n+1}/f_n$. Given that $\lim (x_n) =L$ exists, determine the value of $L$.
+		\\\\We can rewrite $(x_n)$ as follows:
+		\begin{align*}
+			x_n &= \frac{f_{n+1}}{f_n} \\
+			&=\frac{f_n+f_{n-1}}{f_n} \\
+			&= 1+\frac{f_{n-1}}{f_n} \\
+			&= 1+\frac{\frac{1}{f_n}}{f_{n-1}} \\
+			&= 1+\frac{1}{x_{n-1}}
+		\end{align*}
+		Since we're given that $L=\lim (x_n)$ exists and since we just showed that it's equal to $\lim (x_{n-1})$, we get the following:
+		\begin{align*}
+			x_n &= \left. 1+\frac{1}{x_{n-1}}\ \ \ \ \ \right| \lim \\
+			\lim (x_n) &= 1+\frac{1}{\lim(x_{n-1})} \\
+			L &= \left. 1+\frac{1}{L}\ \ \ \ \ \right| \cdot L \\
+			L^2 &= L + 1 \\
+			L^2-L-1 &= 0 \\
+			L_{1,2} &= \frac{1 \pm \sqrt{1-4 \cdot 1 \cdot (-1)}}{2} \\
+			L_1 &= \frac{1-\sqrt{5}}{2} <0 \\
+			L_2 &= \frac{1+\sqrt{5}}{2}>0
+		\end{align*}
+		Now, since $f_n >0 \Rightarrow x_n >0 \Rightarrow L>0$, we can infer that the proper limit is
+		\[L=\frac{1+\sqrt{5}}{2}\]
+		\\
 		
 		\item[4a)] Show that the sequence $(1-(-1)^n+1/n)$ converges.
+		\\\\ Let $(x_n):=(1-(-1)^n+1/n)$. Let $(z_n)=(x_{2n})$, and $(w_n)=(x_{2n-1})$ be subsequence of $(x_n)$. Then $(z_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is even, and $(w_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is odd.
+		\\\\These subsequences yield the following:
+		\[z_n = x_{2n} = 1-(-1)^{2n}+\frac{1}{2n}=1-1+\frac{1}{2n}=\frac{1}{2n}\]
+		\[w_n=x_{2n-1}=1-(-1)^{2n-1}+\frac{1}{2n-1}=1+1+\frac{1}{2n-1}=2+\frac{1}{2n-1}\]
+		Now, if we take the limit of each sequence as $n \rightarrow \infty$ yields
+		\[\lim_{n \to \infty} (z_n) = 0 \neq 2 = \lim_{n \to \infty} (w_n)\]
+		Recall Theorem 3.4.5 \textit{Divergence Criteria}:
+		\begin{theorem*}
+			If a sequence $X=(x_n)$ of real numbers has either of the following properties, then $X$ is divergent.
+			\begin{enumerate}
+				\item $X$ has two convergent subsequences $X'=(x_{n_k})$ and $X''=(x_{r_k})$ whose limits are not equal.
+				
+				\item $X$ is unbounded
+			\end{enumerate}
+		\end{theorem*}
+		Thus by the \textit{Divergence Criteria}, we have that since $(z_n)$ and $(w_n)$ satisfy the first property of the \textit{Divergence Criteria}, we can conclude that the sequence $(x_n)$ is divergent. \\
 		
 		\item[16)] Give an example to show that Theorem 3.4.9 fails if the hypothesis that $X$ is a bounded sequences is dropped.
+		\\\\Recall \textit{Theorem 3.4.9}:
+		\begin{theorem*}
+			If $(x_n)$ is a bounded sequence of real numbers, then the following statements for a real number $x^*$ are equivalent:
+			\begin{enumerate}
+				\item $x^*= \lim \sup (x_n)$.
+				
+				\item If $\varepsilon > 0$, there are at most a finite number of $n \in \N$ such that $x^*+\varepsilon<x_n$, but an infinite number of $n \in \N$ such that $x^*-\varepsilon<x_n$.
+				
+				\item If $u_m=\sup \{x_n : n \geq m\}$, then $x^*=\inf \{u_m:m \in \N\}=\lim (u_m)$.
+				
+				\item If $S$ is the set of subsequential limits of $(x_n)$, then $x^*= \sup (S)$.\\
+			\end{enumerate}
+		\end{theorem*}
+		Consider the sequence $((-1)^nn)$. We note that any subsequence of this sequence is unbounded and thus this sequence has no convergent subsequence. Due to this, all of the conditions of \textit{Theorem 3.4.9} are satisfied vacuously, save the condition concerning boundedness. However, this sequence doesn't converge, but both oscillates and diverges towards $\infty$ and $-\infty$. Thus if the boundedness criterion of the theorem is dropped, this theorem fails.
 		
 		\item[18)] Show that if $(x_n)$ is a bounded sequence, then $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
+		\begin{proof}
+			Let $(x_n)$ be a bounded sequence. We want to show that $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
+			\\\\\textbf{TODO}
+		\end{proof}
 		
 		\item[19)] Show that if $(x_n)$ and $(y_n)$ are bounded sequences, then
 		\[\lim \sup (x_n + y_n) \leq \lim \sup (x_n) + \lim \sup (y_n).\]
 		Give an example in which the two sides are not equal.
+		\\\\\textbf{TODO}
 	\end{enumerate}
 	\item 
 	\begin{enumerate}
@@ -300,7 +412,7 @@
 		\item Compute the infimum, supremum, limit infimum, and limit supremum for $a_n = 3 - (-1)^n - (-1)^n/n$. 
 	\end{enumerate}
 
-	\item 
+	\item Prove or justify, if true. Provide a counterexample, if false.
 	\begin{enumerate}
 		\item If $a_n$ and $b_n$ are strictly increasing, then $a_n + b_n$ is strictly increasing.