diff --git a/Documents/LaTeX/Homework 5.pdf b/Documents/LaTeX/Homework 5.pdf index d005e1d..07b3899 100644 Binary files a/Documents/LaTeX/Homework 5.pdf and b/Documents/LaTeX/Homework 5.pdf differ diff --git a/Documents/LaTeX/Homework 5.tex b/Documents/LaTeX/Homework 5.tex index b20fd78..da71423 100644 --- a/Documents/LaTeX/Homework 5.tex +++ b/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 \\ - a^2 -a - 4 &= 0 \\ + \sqrt{4+a} &= a \\ + 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. - \\\\\textbf{Basis Step:} $5 \geq \frac{1}{2}+\sqrt{17}$ yields that $a_1 \ge \frac{1}{2}+\sqrt{17}$ - \\\\\textbf{Inductive Step:} Assume $a_n \geq \frac{1}{2}+\sqrt{17},\ \forall\ n \in \N$. - \\\\\textbf{Show:} We want to show that $a_{n+1} \geq \frac{1}{2}+\sqrt{17},\ \forall\ n \in \N$. - \\So, + 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:} 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 $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$. + \\\\\textbf{Show:} We now want to show that $ a_{n+1} \geq \frac{1}{2}+\frac{\sqrt{17}}{2}\ \forall\ n \in \N$. So, \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