Skip to content

Commit 99bc3d6

Browse files
wiobberwiobber
committed
add folder review (from main@mooculus)
1 parent 48cf164 commit 99bc3d6

File tree

75 files changed

+5152
-0
lines changed

Some content is hidden

Large Commits have some content hidden by default. Use the searchbox below for content that may be hidden.

75 files changed

+5152
-0
lines changed

‎review/exerciseTemplate.tex‎

Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
\documentclass{ximera}
2+
3+
\input{../preamble.tex}
4+
%\input{../../preamble.tex}
5+
6+
\author{}
7+
\license{Creative Commons 3.0 By-NC}
8+
\outcome{}
9+
\begin{document}
10+
11+
\begin{exercise}
12+
13+
14+
15+
\end{exercise}
16+
\end{document}

‎review/refreshConicSections.tex‎

Lines changed: 86 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,86 @@
1+
%\documentclass{ximera}
2+
%
3+
%\input{../preamble.tex}
4+
%
5+
%\title[Refresh:]{Conic sections and their algebraic descriptions}
6+
%
7+
%\begin{document}
8+
%\begin{abstract}
9+
% Conic sections are examples of famous curves in mathematics.
10+
%\end{abstract}
11+
%\maketitle
12+
%
13+
%\section{Conics Sections}
14+
%The \dfn{conic sections} are circles, ellipses, parabolas, and
15+
%hyperbolas. In this section you'll explore the origin of the name
16+
%``conic sections.''
17+
%
18+
%The interactive figure shows a cone, formed by rotating a line in
19+
%space around some fixed axis, and a plane. Use the sliders for
20+
%$\alpha$ and $\beta$ to produce each of the four conic sections:
21+
%
22+
%If the embedded interactive below does not display correctly, click \link[here]{https://www.geogebra.org/m/xHXwN8wK}
23+
%
24+
%\geogebra{xHXwN8wK}{1366}{480}
25+
%
26+
%\begin{problem}
27+
% Use the figure to help you select the true statements:
28+
% \begin{selectAll}
29+
% \choice{Any plane perpendicular to the cone's axis produces a circle.}
30+
% \choice[correct]{Any plane parallel to the cone's axis and doesn't pass through the vertex produces an hyperbola.}
31+
% \choice[correct]{A plane skew to the cone that doesn't pass through the vertex produces an ellipse.}
32+
% \choice[correct]{A plane can intersect a cone in one point.}
33+
% \end{selectAll}
34+
%\end{problem}
35+
%
36+
%\section{Algebraic description of conic sections}
37+
%The general algebraic description of all the conic sections in cartesian coordinates is given by
38+
%\[
39+
% Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,
40+
%\]
41+
%for constants $A$, $B$, $C$, $D$, $E$, and $F$.
42+
%For most of our cases, the conic sections will be symmetric with respect to either the $x$- or $y$-axes.
43+
%
44+
%Therefore we can take $B = 0$ in the above equation and further
45+
%simplifications are possible.
46+
%
47+
%You can find the standard forms of circles, ellipses, and hyperbolas online. For instance, check \link[here]{https://www.varsitytutors.com/hotmath/hotmath_help/topics/conic-sections-and-standard-forms-of-equations}
48+
%
49+
%\begin{problem}
50+
% What is the standard from of an ellipse centered at the origin with
51+
% vertices at $(\pm a,0)$ and foci at $(\pm c ,0)$?
52+
% \begin{explanation}
53+
% The standard form of an ellipse is
54+
% \[
55+
% \frac{x^2}{a^2} + \frac{y^2}{b^2}=1\qquad c^2 = a^2 - b^2
56+
% \]
57+
% \end{explanation}
58+
%\end{problem}
59+
%
60+
%
61+
%
62+
%\begin{problem}
63+
% What is the standard from of a hyperbola centered at the origin with
64+
% vertices at $(0,\pm a)$ and foci at $(0,\pm c)$?
65+
% \begin{explanation}
66+
% The standard form of an ellipse is
67+
% \[
68+
% \frac{y^2}{a^2} - \frac{x^2}{b^2}=1\qquad c^2 = a^2 + b^2
69+
% \]
70+
% \end{explanation}
71+
%\end{problem}
72+
%
73+
%
74+
%\begin{problem}
75+
% Consider
76+
% \[
77+
% y^2 = -36x
78+
% \]
79+
% This is the formula for a
80+
% \wordChoice{\choice{ellipse}\choice{hyperbola}\choice[correct]{parabola}}.
81+
% Its focus is at $\left(-9,0\right)$ and its directrix is the line $x
82+
% = \answer{9}$.
83+
%\end{problem}
84+
%
85+
%
86+
%\end{document}

‎review/refreshFactorials/exerciseList.tex‎

Lines changed: 13 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
\documentclass{xourse}
2+
3+
\input{../../preamble.tex}
4+
5+
6+
\begin{document}
7+
\part{Start Here}
8+
9+
\practice{introduction}
10+
\practice{exponents}
11+
\practice{factorials}
12+
\practice{puttingItTogether}
13+
\end{document}

‎review/refreshFactorials/exponents.tex‎

Lines changed: 149 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,149 @@
1+
\documentclass{ximera}
2+
3+
\input{../../preamble.tex}
4+
5+
\title[Refresh:]{Exponentials}
6+
\author{Jim Talamo}
7+
\begin{document}
8+
\begin{abstract}
9+
Remember our facts about exponentials.
10+
\end{abstract}
11+
\maketitle
12+
13+
14+
\begin{exercise}
15+
Ratios of exponential terms simplify nicely. In order to simplify expressions with exponentials, it is helpful to remember the laws of exponents.
16+
17+
\underline{Laws of Exponents}
18+
\begin{itemize}
19+
\item $a^{n+m} = a^n \cdot a^m$
20+
\item $a^{n-m} = \frac{a^n}{a^m}$
21+
\item $\left(a^n\right)^m = a^{n \cdot m}$
22+
\item $(a \cdot b)^n = a^n \cdot b^n$
23+
\item $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
24+
\end{itemize}
25+
26+
Let's see these in action.
27+
28+
\begin{example}
29+
Simplify $\frac{9^n\cdot 2^{n+3}}{2^{n+1} \cdot 3^{2n}}$.
30+
31+
\begin{explanation}
32+
We use the laws of exponents above.
33+
34+
\[
35+
\frac{9^n\cdot 2^{n+3}}{2^{n+1} \cdot 3^{2n}} = \frac{9^n \cdot 2^n \cdot 2^3}{2^n \cdot 2^1 \cdot \left(3^2\right)^n}= \frac{\cancel{9^n} \cdot \cancel{2^n} \cdot 2^3}{\cancel{2^n} \cdot 2^1 \cdot \cancel{9^n} } =4
36+
\]
37+
\end{explanation}
38+
\end{example}
39+
40+
\begin{example}
41+
Suppose that $a_k = 5\left(\frac{1}{2}\right)^{2k}$. Simplify $\frac{a_{k+1}}{a_k}$.
42+
43+
\begin{explanation}
44+
We write out terms. Notice that
45+
46+
\[
47+
a_{k+1} = 5\left(\frac{1}{2}\right)^{2(k+1)} = 5\left(\frac{1}{2}\right)^{2k+2}.
48+
\]
49+
50+
Thus, we can write out the desired ratio and simplify.
51+
52+
\[
53+
\frac{a_{k+1}}{a_k} = \frac{5\left(\frac{1}{2}\right)^{2k+2}}{5\left(\frac{1}{2}\right)^{2k}}= \frac{\cancel{5\left(\frac{1}{2}\right)^{2k}}\cdot \left(\frac{1}{2}\right)^2}{\cancel{5\left(\frac{1}{2}\right)^{2k}}} = \frac{1}{4}
54+
\]
55+
\end{explanation}
56+
\end{example}
57+
58+
Now, try some examples.
59+
60+
\begin{problem}
61+
Suppose $a_k=2^k$
62+
and $b_k=3^{2k}$. Simplify the following.
63+
\[
64+
\frac{a_{k+1}}{a_k} = \answer{2}
65+
\]
66+
\[
67+
\frac{b_{k+1}}{b_k}=\answer{9}
68+
\]
69+
\end{problem}
70+
71+
72+
\begin{problem}
73+
Suppose $a_k$ is a sequence whose $k$th term is given by:
74+
\[
75+
a_k=4^{k^2}
76+
\]
77+
Note: The $k^2$ is in the exponent. Simplify
78+
\[
79+
\frac{a_{k+1}}{a_k} = \answer{4^{2k+1}}
80+
\]
81+
\end{problem}
82+
%
83+
%\section{Factorials}
84+
%
85+
%\begin{problem}
86+
% Given an integer $n$, the notation ``$n!$'' is defined as follows:
87+
% \[
88+
% n! = n(n-1)(n-2)\dots (3)(2)(1)
89+
% \]
90+
% For instance, $3!=3\cdot2\cdot 1 = 6$. Compute:
91+
% \begin{align*}
92+
% 4! &= \answer{24}\\
93+
% 6! &= \answer{720}\\
94+
% \frac{6!}{4!} &= \answer{30}
95+
% \end{align*}
96+
% \begin{problem}
97+
% How did you compute $\frac{6!}{4!}$ in the last problem? You could
98+
% certainly find $6!$ and $4!$ separately, then divide, but there is
99+
% a nicer way to do this:
100+
% \begin{align*}
101+
% \frac{6!}{4!} &= \frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}\\
102+
% &=\frac{6\cdot5\cdot\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}{\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}\\
103+
% &= 6\cdot 5.
104+
% \end{align*}
105+
% Ratios of factorials are always easiest to compute by canceling
106+
% like terms! Compute the following, simplify your final answers.
107+
% \begin{align*}
108+
% \frac{5!}{4!} &= \answer{5}\\
109+
% \frac{6!}{3!\cdot 3!} &= \answer{20}\\
110+
% \frac{k!}{(k+2)!} &=\answer{\frac{1}{(k+2)(k+1)}}
111+
% \end{align*}
112+
% \end{problem}
113+
%\end{problem}
114+
%
115+
%\begin{problem}
116+
% Often, factorials arise in sequences, and it is important to
117+
% understand how to write out various terms. Suppose $a_k$ is a
118+
% sequence whose $k$th term is given by:
119+
% \[
120+
% a_k = (2k+1)!
121+
% \]
122+
% \begin{enumerate}
123+
% \item $a_2 = \answer{120}$
124+
% \item Find an expression for $a_{k+1}$. Express your answer in the form $(ck+d)!$.
125+
% \[
126+
% a_{k+1} = \left(\answer{2} \cdot k + \answer{3}\right)!
127+
% \]
128+
% \item Calculate and simplify:
129+
% \[
130+
% \frac{a_{k+1}}{a_k} = \answer{(2k+3)(2k+2)}
131+
% \]
132+
% \end{enumerate}
133+
%\end{problem}
134+
%
135+
%\begin{problem}
136+
% Let $a_k$ be a sequence whose $k$th term is given by:
137+
% \[
138+
% a_k = \frac{(2k)!}{(k!)^2}
139+
% \]
140+
% Calculate and simplify:
141+
% \begin{align*}
142+
% \frac{a_{k+1}}{a_k} &= \answer{\frac{4k+2}{k+1}}\\
143+
% \lim_{k\to\infty} \frac{a_{k+1}}{a_k} &= \answer{4}
144+
% \end{align*}
145+
%\end{problem}
146+
147+
\end{exercise}
148+
149+
\end{document}

‎review/refreshFactorials/factorials.tex‎

Lines changed: 113 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,113 @@
1+
\documentclass{ximera}
2+
3+
\input{../../preamble.tex}
4+
5+
\title[Refresh:]{Factorials}
6+
\author{Jim Talamo}
7+
\begin{document}
8+
\begin{abstract}
9+
Remember our facts about exponentials.
10+
\end{abstract}
11+
\maketitle
12+
13+
14+
\begin{exercise}
15+
Ratios of factorials simplify nicely as well. To begin, we review some properties of factorials.
16+
17+
\section{Factorials}
18+
19+
\begin{problem}
20+
Given an integer $n$, the notation ``$n!$'' is defined as follows:
21+
\[
22+
n! = n(n-1)(n-2)\dots (3)(2)(1)
23+
\]
24+
For instance, $3!=3\cdot2\cdot 1 = 6$.
25+
26+
Compute the following.
27+
\begin{align*}
28+
4! &= \answer{24}\\
29+
6! &= \answer{720}\\
30+
\frac{6!}{4!} &= \answer{30}
31+
\end{align*}
32+
33+
\begin{problem}
34+
How did you compute $\frac{6!}{4!}$ in the last problem? You could
35+
certainly find $6!$ and $4!$ separately, then divide, but there is
36+
a nicer way to do this.
37+
\begin{align*}
38+
\frac{6!}{4!} &= \frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}\\
39+
&=\frac{6\cdot5\cdot\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}{\not{4}\cdot\not{3}\cdot\not{2}\cdot\not{1}}\\
40+
&= 6\cdot 5.
41+
\end{align*}
42+
43+
Note that we could make this even more notationally efficient by noting that $6! = 6 \cdot 5 \cdot 4!$ so we can write
44+
45+
\[
46+
\frac{6!}{4!} = \frac{6 \cdot 5 \cdot \cancel{4!}}{\cancel{4!}} =30
47+
\]
48+
Ratios of factorials are always easiest to compute by canceling
49+
like terms! Compute the following, simplify your final answers.
50+
\begin{align*}
51+
\frac{5!}{4!} &= \answer{5}\\
52+
\frac{6!}{3!\cdot 3!} &= \answer{20}\\
53+
\frac{k!}{(k+2)!} &=\answer{\frac{1}{(k+2)(k+1)}}
54+
\end{align*}
55+
56+
57+
58+
59+
\end{problem}
60+
\end{problem}
61+
62+
\begin{problem}
63+
Often, factorials arise in sequences, and it is important to
64+
understand how to write out various terms. Suppose $a_k$ is a
65+
sequence whose $k$th term is given by:
66+
\[
67+
a_k = (2k+1)!
68+
\]
69+
\begin{enumerate}
70+
\item $a_2 = \answer{120}$
71+
\item Find an expression for $a_{k+1}$. Express your answer in the form $(ck+d)!$.
72+
\[
73+
a_{k+1} = \left(\answer{2} \cdot k + \answer{3}\right)!
74+
\]
75+
\item Calculate and simplify:
76+
\[
77+
\frac{a_{k+1}}{a_k} = \answer{(2k+3)(2k+2)}
78+
\]
79+
\end{enumerate}
80+
\end{problem}
81+
82+
\begin{problem}
83+
Let $a_k$ be a sequence whose $k$-th term is given by
84+
\[
85+
a_k = \frac{(2k)!}{(k!)^2}.
86+
\]
87+
Calculate and simplify.
88+
\begin{align*}
89+
\frac{a_{k+1}}{a_k} &= \answer{\frac{4k+2}{k+1}}\\
90+
\lim_{k\to\infty} \frac{a_{k+1}}{a_k} &= \answer{4}
91+
\end{align*}
92+
93+
\begin{hint}
94+
Note that $a_{k+1} = \frac{(2(k+1))!}{\big[(k+1)!\big]^2} = \frac{(2k+2)!}{(k+1)! \cdot (k+1)!}$. Thus,
95+
96+
\[
97+
\frac{a_{k+1}}{a_k} = \frac{(2k+2)!}{(k+1)! \cdot (k+1)!} \cdot \frac{k! \cdot k!}{(2k)!}.
98+
\]
99+
100+
Rearranging to collect like terms gives
101+
102+
\begin{align*}
103+
\frac{a_{k+1}}{a_k} &= \frac{(2k+2)!}{(2k)!} \cdot \frac{k!}{(k+1)!} \cdot \frac{k!}{(k+1)!} \\
104+
&= \frac{(2k+2)(2k+1)(2k)!}{(2k)!} \cdot \frac{k!}{(k+1) \cdot k!} \cdot \frac{k!}{(k+1) \cdot k!}.\\
105+
\end{align*}
106+
107+
Now simplify.
108+
\end{hint}
109+
\end{problem}
110+
111+
\end{exercise}
112+
113+
\end{document}

0 commit comments

Comments
 (0)