227 lines
6.6 KiB
TeX
227 lines
6.6 KiB
TeX
\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 9}
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\usepackage{amsthm}
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\usepackage{mathabx}
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\begin{document}
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\ntnuTitle{}
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\break{}
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\begin{excs}
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\exc{}
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\begin{align*}
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p \rightarrow (q \vee r) &\equiv \neg p \vee (q \vee r) \\
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&\equiv \neg p \vee q \vee r \\
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&\equiv \neg (p \vee \neg q) \vee r \\
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&\equiv (p \vee \neg q) \rightarrow r \\
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\end{align*}
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\exc{}
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R does not define a partial orderering, because it is not transitive.
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$aRb$ and $bRc$, however $\neg aRc$
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{gather*}
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xyz + xy\overline{z}+\overline{x}y \\
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xy + \overline{x}y \\
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y
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\end{gather*}
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\subexc{}
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\begin{gather*}
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y + \overline{x}z + x\overline{y} \\
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y + \overline{x}z + x \\
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y + z + x \\
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x + y + z
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\end{gather*}
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\end{subexcs}
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\exc{}
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Step 1:
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\begin{align*}
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\sum^1_{n=1}\frac{1}{(2n-1)(2n+1)} &= \frac{1}{2\cdot1 + 1} \\
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\frac{1}{(2\cdot1-1)(2\cdot1+1)} &= \frac{1}{3} \\
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\frac{1}{(1)(3)} &= \frac{1}{3} \\
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\frac{1}{3} &= \frac{1}{3} \\
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\end{align*}
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Step 2:
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Assume
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\[ \sum^k_{n=1} \frac{1}{(2n-1)(2n+1)} = \frac{k}{2k+1} \]
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then
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\begin{align*}
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\sum^{k+1}_{n=1} \frac{1}{(2n-1)(2n+1)} &= \frac{k}{2k+1} \\[2ex]
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&= \frac{1}{(2\cdot1 - 1)(2\cdot1+1)} + \frac{1}{(2\cdot2 - 1)(2\cdot2+1)} + \ldots \\[2ex]
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&\qquad + \frac{1}{(2\cdot k - 1)(2\cdot k+1)} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex]
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&= \frac{k}{2k+1} + \frac{1}{(2\cdot(k+1) - 1)(2\cdot(k+1)+1)} \\[2ex]
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&= \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} \\[2ex]
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&= \frac{k(2k1)(2k+3) + (2k+1)}{(2k+1)^2(2k+3)} \\[2ex]
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&= \frac{k(2k+3) + 1}{(2k+1)(2k+3)} \\[2ex]
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&= \frac{2k^2+3k + 1}{(2k+1)(2k+3)} \\[2ex]
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&= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} \\[2ex]
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&= \frac{(k+1)}{(2k+3)} \\[2ex]
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&= \frac{k+1}{2(k+1)+1}
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\end{align*}
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\exc{}
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\textbf{Injective:}
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Suppose $a,b \in \R$
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\begin{align*}
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f(a) &= f(b) \\
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2a-3 &= 2b-3 \\
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2a &= 2b \\
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a &= b \\
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\end{align*}
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thus
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\[ f(a) = f(b) \Leftrightarrow a = b \]
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which means that $f$ is injective \\
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\textbf{Surjective:}
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Suppose $a \in \R$
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\begin{align*}
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a &= 2x-3 \\
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x &= \frac{a+3}{2} \\
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a &\in \R
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\end{align*}
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therefore $f$ is surjective \\
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\textbf{Inverse:}
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\begin{align*}
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y &= 2x-3 \\[2ex]
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x &= \frac{y+3}{2} \\[2ex]
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f^{-1}(y) &= \frac{y+3}{2}
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\end{align*}
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\exc{}
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\begin{gather*}
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(\overline{X \cap Y \cap Z}) \\
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\{ x \mid x \in \overline{X \cap Y \cap Z} \} \\
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\{ x \mid x \notin X \cap Y \cap Z \} \\
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\{ x \mid x \notin X \wedge x \notin Y \wedge x \notin Z \} \\
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\{ x \mid x \in \overline{X} \vee x \in \overline{Y} \vee x \in \overline{Z} \} \\
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\{ x \mid x \in \overline{X} \cup \overline{Y} \cup \overline{Z} \} \\
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\overline{X} \cup \overline{Y} \cup \overline{Z}
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\end{gather*}
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\exc{}
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Assuming 'dozen' is to be interpreted as 12
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\begin{subexcs}
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\subexc{}
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\[ \nPr{31}{12} = 67\ 596\ 957\ 267\ 840\ 000 \]
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\subexc{}
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\[ 31^{12} = 787\ 662\ 783\ 788\ 549\ 761 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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If we imagine a row of numbers going from 1 to 40, we can rephrase the question as how many ways we can split
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the numbers into $5$ chunks. Imagine a chunk as inserting $4$ delimiters like this:
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\[ 1\ 2\ 3\ |\ 4\ 5\ 6\ 7\ |\ 8\ 9\ 10\ |\ 11\ \ldots\ 39\ |\ 40 \]
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In this case, we split the amount of numbers so that $x_1 = 3, x_2 = 4, x_3 = 3, x_4 = 29, x_5 = 1$
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By doing $\nCr{n}{r}$ where n is the number of numbers and delimiters, and r is the number of delimiters,
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we will get all combinations of $x_1 + x_2 + x_3 + x_4 + x_5 = 40$A
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In order to make it $x_1 + x_2 + x_3 + x_4 + x_5 \leq 40$, we will add a fifth delimiter, indicating the last block of unused numbers.
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$x_1 + x_2 + x_3 + x_4 + x_5 < 40 \Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 \leq 39$
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This leaves us with $\nCr{39 + 5}{5} = 1086008$ different combinations.
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\subexc{}
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In this case, we modify the problem by adjusting the inequality of $x_i$ like the following
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\begin{align*}
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x_1 + x_2 + x_3 + x_4 + x_5 &< 40 \qquad &&x_i \geq -3 \\
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y_1 - 3 + y_2 - 3 + y_3 - 3 + y_4 - 3 + y_5 - 3 &< 40 &&(y_i - 3 = x) \\
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y_1 + y_2 + y_3 + y_4 + y_5 &< 40 + 5 \cdot 3 \\
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y_1 + y_2 + y_3 + y_4 + y_5 &< 55
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\end{align*}
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\[ (y_i - 3 = x) \Rightarrow (y_i - 3 \geq - 3) \Leftrightarrow (y_i \geq 0) \]
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With this information, we use the same way of solving as in \textbf{a)}
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\[\nCr{54 + 5}{5} = 5006386 \]
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\end{subexcs}
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\break{}
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\exc{}
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This is a Venn diagram of the set containing the elements that satisfy $\overline{c}_1$, $\overline{c}_2$ and $\overline{c}_3$.
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\includeDiagram[width=11cm, caption={$N(\overline{c}_2\overline{c}_3\overline{c}_4$)}]{diagrams/ex9_1.tex}
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The following diagrams show the terms of the RHS.
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\includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_2.tex}
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\includeDiagram[width=11cm, caption={$N(\overline{c_1}\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_3.tex}
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When we lay these two diagrams on top of each other, we can see that $N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$ is equal to $N(\overline{c}_2\overline{c}_3\overline{c}_4)$
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\includeDiagram[width=11cm, caption={$N(c_1\overline{c}_2\overline{c}_3\overline{c}_4) + N(\overline{c}_1\overline{c}_2\overline{c}_3\overline{c}_4)$}]{diagrams/ex9_4.tex}
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\exc{}
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let
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$c_1 = 2 \mid n $ \\
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$c_2 = 3 \mid n $ \\
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$c_3 = 5 \mid n $ \\
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$c_4 = 7 \mid n $
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\begin{align*}
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N(\overline{c}_1\overline{c}_2\overline{c}_3c_4) &= N(c_4) - \left( N(c_1c_7) + N(c_2c_7) + N(c_3c_7) \right) \\
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&\quad + \left( N(c_1c_2c_4) + N(c_1c_3c_4) + N(c_2c_3c_4) \right) \\
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&\quad - N(c_1c_2c_3c_4) \\[2ex]
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&= \left\lfloor \frac{2000}{7} \right\rfloor - \left( \left\lfloor \frac{2000}{2 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{5 \cdot 7} \right\rfloor \right) \\[2ex]
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&\quad + \left( \left\lfloor \frac{2000}{2 \cdot 3 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{2 \cdot 5 \cdot 7} \right\rfloor + \left\lfloor \frac{2000}{3 \cdot 5 \cdot 7} \right\rfloor \right) \\[2ex]
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&\quad - \left\lfloor \frac{2000}{2 \cdot 3 \cdot 5 \cdot 7} \right\rfloor \\[2ex]
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&= 76
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\end{align*}
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\end{excs}
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\end{document} |