From cdb6b286b395a4c91dc99079baa00788bf4b2441 Mon Sep 17 00:00:00 2001 From: ceciliachan1979 <75919064+ceciliachan1979@users.noreply.github.com> Date: Mon, 5 Feb 2024 20:26:53 -0800 Subject: [PATCH 1/4] Midterm --- .github/workflows/blank.yml | 2 + Exams/Rutgers-Analysis/midterm.tex | 142 +++++++++++++++++++++++++++++ 2 files changed, 144 insertions(+) create mode 100644 Exams/Rutgers-Analysis/midterm.tex diff --git a/.github/workflows/blank.yml b/.github/workflows/blank.yml index a643a38..877604e 100644 --- a/.github/workflows/blank.yml +++ b/.github/workflows/blank.yml @@ -41,6 +41,7 @@ jobs: Exams/HKALE/1999/HKALE-1999.tex Exams/HKCEE/2001/HKCEE-2001.tex Exams/HKDSE/2020/HKDSE-2020.tex + Exams/Rutgers-Analysis/midterm.tex MIT/Solutions/MIT.tex Misc/Bernoulli/Bernoulli.tex Misc/Cosine/Cosine.tex @@ -76,6 +77,7 @@ jobs: HKALE-1999.pdf HKCEE-2001.pdf HKDSE-2020.pdf + midterm.pdf MIT.pdf Bernoulli.pdf Cosine.pdf diff --git a/Exams/Rutgers-Analysis/midterm.tex b/Exams/Rutgers-Analysis/midterm.tex new file mode 100644 index 0000000..255f188 --- /dev/null +++ b/Exams/Rutgers-Analysis/midterm.tex @@ -0,0 +1,142 @@ +\documentclass{article} +\usepackage{amssymb} +\usepackage{nth} +\usepackage[utf8]{inputenc} +\usepackage{graphicx} +\usepackage[final]{pdfpages} + + +\title{Number Theory} +\author{Cecilia Chan} +\date{February 2024} + +\begin{document} +\section*{Problem 1} +\subsection*{Part 1} +True, because otherwise the real number would be the disjoint union of two countable sets. +\subsection*{Part 2} +Yes, because it is the union of countably many countable sets. +\subsection*{Part 3} +It depends on the space. If the space is complete (e.g. $ \mathbb{R}^N $), then the sequence is convergent (that's the definition of a space being complete). +\subsection*{Part 4} +False, consider the sequence $ \{1, -1, 1, \ldots \} $, it is bounded but not convergent. +\subsection*{Part 5} +True, to satisfy the epsilon challenge, just pick the maximum $ N $ for both subsequences. +\subsection*{Part 6} +True, because if that's not true, $ t - \epsilon $ is a smaller upper bound. +\subsection*{Part 7} +True, to satisfy the epsilon challenge, just pick the $ N $ corresponding to $ \frac{1}{L} \epsilon $ for $ f(x) $ where $ -L < g(x) < L$. +\subsection*{Part 8} +True, because otherwise at least one of the greatest lower bounds is actually not greatest. +\subsection*{Part 9} +True, the least upper bound exists and it is the limit. +\subsection*{Part 10} +False, 0 is also an accumulation point. + +\section*{Problem 2} +Let $ N = \lceil \frac{2}{\epsilon} \rceil $, then for $ n > N $, we have +\begin{eqnarray*} + n & > & N \\ + n & > & \frac{2}{\epsilon} \\ + \epsilon & > & \frac{2}{N} \\ + \frac{2}{n} & < & \epsilon \\ + \frac{2}{n} - 3 + 3 & < & \epsilon \\ + \frac{2}{n} - \frac{3n}{n} + 3 & < & \epsilon \\ + -(\frac{3n - 2}{n} - 3) & < & \epsilon +\end{eqnarray*} + +We also have $ \frac{3n - 2}{n} = 3 - \frac{2}{n} < 3 $, section + +\begin{eqnarray*} + \frac{3n - 2}{n} &=& 3 - \frac{2}{n} < 3 \\ + \frac{3n - 2}{3} - 3 &<& 0 \\ + &<& \epsilon +\end{eqnarray*} + +Together with have $ |\frac{3n - 2}{3} - 3| < \epsilon $ when $ n > \lceil \frac{2}{\epsilon} \rceil $, so we conclude + +\begin{eqnarray} + \lim_{n \to \infty} \frac{3n - 2}{n} = 3 +\end{eqnarray} + +\section*{Problem 3} +Let $ \delta = \min (1, \frac{\epsilon}{8}) $, note that $ 0 < \delta < 1 $, so $ 0 < \delta^2 < \delta $. + +Whenever $ | x - 1 | < \delta $, we have + +\begin{eqnarray*} + -\delta &< x - 1 &< \delta \\ + 2 -\delta &< x + 1 &< 2 + \delta \\ + (2 -\delta)^2 &< (x + 1)^2 &< (2 + \delta)^2 \\ + 4 - 4\delta + \delta^2 &< x^2 + 2x + 1 &< 4 + 4\delta + \delta^2 \\ + -4\delta + \delta^2 &< (x^2 + 2x + 1) - 4 &< 4\delta + \delta^2 \\ + -4\delta &< (x^2 + 2x + 1) - 4 &< 5\delta \\ + -\frac{\epsilon}{2} &< (x^2 + 2x + 1) - 4 &< \frac{5\epsilon}{8} \\ + -\epsilon &< (x^2 + 2x + 1) - 4 &< \epsilon +\end{eqnarray*} + +So we conclude that $ \lim_{x \to 1} x^2 + 2x + 1 = 4 $. + +\section*{Problem 4} + +We claim that the limit is $ L = \frac{1 + \sqrt{17}}{2} $ + +Note that when $ L $ is designed to be the larger root of $ x^2 - x - 4 $, so when $ x > L $, $ x^2 - x - 4 > 0 $, and so we have: + +\begin{eqnarray*} +x^2 - x - 4 > 0 +x^2 > x + 4 +x > \sqrt{4 + x} +\end{eqnarray*} + +Since $ a_1 = 4 > L $, the iteration will be decreasing. It is obviously bounded below by 0, so a limit exists. + +Suppose the limit exists, then + +\begin{eqnarray*} + lim_{n \to \infty} a_n \\ + &=& lim_{n \to \infty} \sqrt{4 + a_{n-1}} \\ + &=& \sqrt{4 + lim_{n \to \infty} a_{n-1}} \\ +\end{eqnarray*} + +So it must be a root of $ x = \sqrt{4 + x} $, and apparently it must be > 0, so it must be $ L $. + +\section*{Problem 5} +\subsection*{Part a} +The use of L'Hopital's rule is justified by the $ \frac{\infty}{\infty} $ form of the limit. We have: +\begin{eqnarray*} + & & \lim_{n \to \infty}\frac{3n^3 - n + 100}{n^3 - n + 2} \\ + &=& \lim_{n \to \infty}\frac{9n^2 - 1}{3n^2 - 1} \\ + &=& \lim_{n \to \infty}\frac{18n}{6n} \\ + &=& 3 +\end{eqnarray*} + +\subsection*{Part b} +The use of L'Hopital's rule is justified by the $ \frac{0}{0} $ form of the limit. We have: +\begin{eqnarray*} + & & \lim_{x \to 0}\frac{\sqrt{x + 4} - 2}{x} \\ + &=& \lim_{x \to 0}\frac{\frac{1}{2}(x + 4)^{-\frac{1}{2}}}{1} \\ + &=& \frac{1}{4} +\end{eqnarray*} + +\section*{Problem 6} +$ \lim_{x_0 \to 0} f(x) = 2 $ implies for any $ \epsilon > 0 $, there exists $ \delta > 0 $ such that $ 0 < |x - x_0| < \delta $ implies $ |f(x) - 2| < \epsilon $. In particular, for $ \epsilon = 1 $, there exists $ \delta > 0 $ such that $ 0 < |x - x_0| < \delta $ implies $ |f(x) - 2| < 1 $. + +Note that $ 0 < | x - x_0| < \delta $ is the same as $ x \in (x_0 - \delta, x_0 + \delta) $, and $ |f(x) - 2| < 1 $ is the same as $ f(x) \in (1, 3) $, in particular, $ f(x) > 1 $. + +The fact that $ x_0 $ is an accumulation point implies some $ x $ will exist in the given set. It is not strictly necessary because otherwise the statement would simply be vacuously true. + +\section*{Problem 7} +The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence. The sequence $ \{a_n\} $ is bounded, so it has a convergent subsequence $ \{a_{n_k}\} $. + +We can prove that by showing a monotone subsequence exists, because then the sequence of monotone and bounded, so it must converge. + +Assuming the sequence has a monotonic increasing suffix, then we can take that suffix. + +Otherwise it must have a maximum, so we can take that point. The rest of the sequence is not monotonically increasing either, so we can build the sequence inductively. + +By definition, this sequence is monotonically decreasing. + +So we know there exists a monotonic subsequence, and it is also bounded, so it must converge. + +\end{document} From 356a165ad9206c76e9dc66f08efab2867ef2fc07 Mon Sep 17 00:00:00 2001 From: ceciliachan1979 <75919064+ceciliachan1979@users.noreply.github.com> Date: Tue, 6 Feb 2024 06:31:59 -0800 Subject: [PATCH 2/4] Some updates --- .github/workflows/blank.yml | 4 ++-- .../{midterm.tex => midterm.cecilia.tex} | 24 +++++++++++-------- 2 files changed, 16 insertions(+), 12 deletions(-) rename Exams/Rutgers-Analysis/{midterm.tex => midterm.cecilia.tex} (89%) diff --git a/.github/workflows/blank.yml b/.github/workflows/blank.yml index 877604e..f1fbc60 100644 --- a/.github/workflows/blank.yml +++ b/.github/workflows/blank.yml @@ -41,7 +41,7 @@ jobs: Exams/HKALE/1999/HKALE-1999.tex Exams/HKCEE/2001/HKCEE-2001.tex Exams/HKDSE/2020/HKDSE-2020.tex - Exams/Rutgers-Analysis/midterm.tex + Exams/Rutgers-Analysis/midterm.cecilia.tex MIT/Solutions/MIT.tex Misc/Bernoulli/Bernoulli.tex Misc/Cosine/Cosine.tex @@ -77,7 +77,7 @@ jobs: HKALE-1999.pdf HKCEE-2001.pdf HKDSE-2020.pdf - midterm.pdf + midterm.cecilia.pdf MIT.pdf Bernoulli.pdf Cosine.pdf diff --git a/Exams/Rutgers-Analysis/midterm.tex b/Exams/Rutgers-Analysis/midterm.cecilia.tex similarity index 89% rename from Exams/Rutgers-Analysis/midterm.tex rename to Exams/Rutgers-Analysis/midterm.cecilia.tex index 255f188..edb9e82 100644 --- a/Exams/Rutgers-Analysis/midterm.tex +++ b/Exams/Rutgers-Analysis/midterm.cecilia.tex @@ -4,13 +4,17 @@ \usepackage[utf8]{inputenc} \usepackage{graphicx} \usepackage[final]{pdfpages} +\usepackage{hyperref} - -\title{Number Theory} +\title{Midterm Exam, Introduction to Real Analysis I} \author{Cecilia Chan} \date{February 2024} \begin{document} +\section*{Course Information} +The course homepage is \href{https://math.rutgers.edu/academics/undergraduate/courses/955-01-640-311-introduction-to-real-analysis-i}{Introduction to Real Analysis I}. + +The midterm pdf link is \href{https://math.rutgers.edu/images/test-311-10-2016.pdf}{here}. \section*{Problem 1} \subsection*{Part 1} True, because otherwise the real number would be the disjoint union of two countable sets. @@ -38,7 +42,7 @@ \section*{Problem 2} \begin{eqnarray*} n & > & N \\ n & > & \frac{2}{\epsilon} \\ - \epsilon & > & \frac{2}{N} \\ + \epsilon & > & \frac{2}{n} \\ \frac{2}{n} & < & \epsilon \\ \frac{2}{n} - 3 + 3 & < & \epsilon \\ \frac{2}{n} - \frac{3n}{n} + 3 & < & \epsilon \\ @@ -84,9 +88,9 @@ \section*{Problem 4} Note that when $ L $ is designed to be the larger root of $ x^2 - x - 4 $, so when $ x > L $, $ x^2 - x - 4 > 0 $, and so we have: \begin{eqnarray*} -x^2 - x - 4 > 0 -x^2 > x + 4 -x > \sqrt{4 + x} + x^2 - x - 4 &> 0 \\ + x^2 &> x + 4 \\ + x &> \sqrt{4 + x} \end{eqnarray*} Since $ a_1 = 4 > L $, the iteration will be decreasing. It is obviously bounded below by 0, so a limit exists. @@ -94,12 +98,12 @@ \section*{Problem 4} Suppose the limit exists, then \begin{eqnarray*} - lim_{n \to \infty} a_n \\ - &=& lim_{n \to \infty} \sqrt{4 + a_{n-1}} \\ - &=& \sqrt{4 + lim_{n \to \infty} a_{n-1}} \\ + & & \lim_{n \to \infty} a_n \\ + &=& \lim_{n \to \infty} \sqrt{4 + a_{n-1}} \\ + &=& \sqrt{4 + \lim_{n \to \infty} a_{n-1}} \\ \end{eqnarray*} -So it must be a root of $ x = \sqrt{4 + x} $, and apparently it must be > 0, so it must be $ L $. +So it must be a root of $ x = \sqrt{4 + x} $, and apparently it must be greater than 0, so it must be $ L $. \section*{Problem 5} \subsection*{Part a} From 364ab30052fc26381fc6f148f14d8d47bad2321f Mon Sep 17 00:00:00 2001 From: ceciliachan1979 <75919064+ceciliachan1979@users.noreply.github.com> Date: Sat, 10 Feb 2024 07:32:51 -0800 Subject: [PATCH 3/4] Improved problem 3 --- Exams/Rutgers-Analysis/midterm.cecilia.tex | 41 +++++++++++++++------- 1 file changed, 28 insertions(+), 13 deletions(-) diff --git a/Exams/Rutgers-Analysis/midterm.cecilia.tex b/Exams/Rutgers-Analysis/midterm.cecilia.tex index edb9e82..973d679 100644 --- a/Exams/Rutgers-Analysis/midterm.cecilia.tex +++ b/Exams/Rutgers-Analysis/midterm.cecilia.tex @@ -1,4 +1,5 @@ \documentclass{article} +\usepackage{amsmath} \usepackage{amssymb} \usepackage{nth} \usepackage[utf8]{inputenc} @@ -64,22 +65,36 @@ \section*{Problem 2} \end{eqnarray} \section*{Problem 3} -Let $ \delta = \min (1, \frac{\epsilon}{8}) $, note that $ 0 < \delta < 1 $, so $ 0 < \delta^2 < \delta $. +Let $ \delta = \min (\sqrt{\frac{\epsilon}{2}}, \frac{\epsilon}{8}) $. -Whenever $ | x - 1 | < \delta $, we have +On one hand, we have: -\begin{eqnarray*} - -\delta &< x - 1 &< \delta \\ - 2 -\delta &< x + 1 &< 2 + \delta \\ - (2 -\delta)^2 &< (x + 1)^2 &< (2 + \delta)^2 \\ - 4 - 4\delta + \delta^2 &< x^2 + 2x + 1 &< 4 + 4\delta + \delta^2 \\ - -4\delta + \delta^2 &< (x^2 + 2x + 1) - 4 &< 4\delta + \delta^2 \\ - -4\delta &< (x^2 + 2x + 1) - 4 &< 5\delta \\ - -\frac{\epsilon}{2} &< (x^2 + 2x + 1) - 4 &< \frac{5\epsilon}{8} \\ - -\epsilon &< (x^2 + 2x + 1) - 4 &< \epsilon -\end{eqnarray*} +\begin{align*} + |x - 1| &< \delta \\ + |x - 1| &< \sqrt{\frac{\epsilon}{2}} \\ + |(x-1)^2| &< \frac{\epsilon}{2} &\text{square is okay because LHS $\ge$ 0} \\ +\end{align*} + +On the other hand, we have: + +\begin{align*} + |x - 1| &< \delta \\ + |x - 1| &< \frac{\epsilon}{8} \\ + |4(x-1)| &< \frac{\epsilon}{2} \\ +\end{align*} + +Therefore, by the triangle inequality, we have: + +\begin{align*} + & |x^2 + 2x + 1 - 4 | \\ + =& |x^2 - 2x + 1 + 4x - 4| \\ + =& |(x-1)^2 + 4(x-1)| \\ + <& |(x-1)^2| + |4(x-1)| \\ + <& \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ + =& \epsilon +\end{align*} -So we conclude that $ \lim_{x \to 1} x^2 + 2x + 1 = 4 $. +Therefore we proved that $ \lim\limits_{x \to 1} x^2 + 2x + 1 = 4 $ by the definition of the limit, it is possible to find a $ \delta $ such that $ |x - 1| < \delta $ implies $ |x^2 + 2x + 1 - 4| < \epsilon $. \section*{Problem 4} From 5dec0574bd016dc46379579d70ec77ae628b3b22 Mon Sep 17 00:00:00 2001 From: ceciliachan1979 <75919064+ceciliachan1979@users.noreply.github.com> Date: Sun, 11 Feb 2024 10:33:41 -0800 Subject: [PATCH 4/4] Some updates --- Exams/Rutgers-Analysis/midterm.cecilia.tex | 39 ++++++++++++---------- 1 file changed, 21 insertions(+), 18 deletions(-) diff --git a/Exams/Rutgers-Analysis/midterm.cecilia.tex b/Exams/Rutgers-Analysis/midterm.cecilia.tex index 973d679..58c97fd 100644 --- a/Exams/Rutgers-Analysis/midterm.cecilia.tex +++ b/Exams/Rutgers-Analysis/midterm.cecilia.tex @@ -12,11 +12,12 @@ \date{February 2024} \begin{document} +\maketitle \section*{Course Information} The course homepage is \href{https://math.rutgers.edu/academics/undergraduate/courses/955-01-640-311-introduction-to-real-analysis-i}{Introduction to Real Analysis I}. The midterm pdf link is \href{https://math.rutgers.edu/images/test-311-10-2016.pdf}{here}. -\section*{Problem 1} +\section*{Problem 1 (ok)} \subsection*{Part 1} True, because otherwise the real number would be the disjoint union of two countable sets. \subsection*{Part 2} @@ -102,25 +103,27 @@ \section*{Problem 4} Note that when $ L $ is designed to be the larger root of $ x^2 - x - 4 $, so when $ x > L $, $ x^2 - x - 4 > 0 $, and so we have: -\begin{eqnarray*} - x^2 - x - 4 &> 0 \\ +\begin{align*} + x^2 - x - 4 &> 0 & \text{ By the geometry of the graph} \\ x^2 &> x + 4 \\ - x &> \sqrt{4 + x} -\end{eqnarray*} + x &> \sqrt{4 + x} & \text{ Because $ x > 0 $ and square root is increasing } \\ +\end{align*} Since $ a_1 = 4 > L $, the iteration will be decreasing. It is obviously bounded below by 0, so a limit exists. -Suppose the limit exists, then +Next, we have: -\begin{eqnarray*} - & & \lim_{n \to \infty} a_n \\ - &=& \lim_{n \to \infty} \sqrt{4 + a_{n-1}} \\ - &=& \sqrt{4 + \lim_{n \to \infty} a_{n-1}} \\ -\end{eqnarray*} +\begin{align*} + & \lim_{n \to \infty} a_n \\ + =& \lim_{n \to \infty} \sqrt{4 + a_{n-1}} \\ + =& \sqrt{\lim_{n \to \infty} 4 + a_{n-1}} & \text{ by the continuity of the square root function} \\ + =& \sqrt{4 + \lim_{n \to \infty} a_{n-1}} \\ + =& \sqrt{4 + \lim_{n \to \infty} a_{n}} \\ +\end{align*} So it must be a root of $ x = \sqrt{4 + x} $, and apparently it must be greater than 0, so it must be $ L $. -\section*{Problem 5} +\section*{Problem 5 (ok)} \subsection*{Part a} The use of L'Hopital's rule is justified by the $ \frac{\infty}{\infty} $ form of the limit. We have: \begin{eqnarray*} @@ -138,7 +141,7 @@ \subsection*{Part b} &=& \frac{1}{4} \end{eqnarray*} -\section*{Problem 6} +\section*{Problem 6 (ok)} $ \lim_{x_0 \to 0} f(x) = 2 $ implies for any $ \epsilon > 0 $, there exists $ \delta > 0 $ such that $ 0 < |x - x_0| < \delta $ implies $ |f(x) - 2| < \epsilon $. In particular, for $ \epsilon = 1 $, there exists $ \delta > 0 $ such that $ 0 < |x - x_0| < \delta $ implies $ |f(x) - 2| < 1 $. Note that $ 0 < | x - x_0| < \delta $ is the same as $ x \in (x_0 - \delta, x_0 + \delta) $, and $ |f(x) - 2| < 1 $ is the same as $ f(x) \in (1, 3) $, in particular, $ f(x) > 1 $. @@ -146,16 +149,16 @@ \section*{Problem 6} The fact that $ x_0 $ is an accumulation point implies some $ x $ will exist in the given set. It is not strictly necessary because otherwise the statement would simply be vacuously true. \section*{Problem 7} -The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence. The sequence $ \{a_n\} $ is bounded, so it has a convergent subsequence $ \{a_{n_k}\} $. +The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence; in other words, if a sequence $ \{ a_n \} $ is bounded, then it has a convergent subsequence $ \{ a_{n_k} \} $. We can prove that by showing a monotone subsequence exists, because then the sequence of monotone and bounded, so it must converge. -Assuming the sequence has a monotonic increasing suffix, then we can take that suffix. +Assuming the sequence has a monotonic increasing subsequence, then we can take that subsequence. -Otherwise it must have a maximum, so we can take that point. The rest of the sequence is not monotonically increasing either, so we can build the sequence inductively. +Otherwise it must have a maximum, so we can take that point. The rest of the sequence is not does not have a monotonic increasing subsequence either, so we can define the subsequence inductively by keep taking the maximum of the rest. -By definition, this sequence is monotonically decreasing. +By definition, this subsequence is monotonically decreasing. -So we know there exists a monotonic subsequence, and it is also bounded, so it must converge. +So we know there exists a monotonic subsequence, and it is also bounded, so it must converges. \end{document}