MA0301/exercise4/main.tex

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\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\author{Øystein Tveit}
\title{MA0301 Exercise 4}
\usepackage{amsthm}
\begin{document}
\ntnuTitle{}
\break{}
\begin{excs}
\exc{}
\begin{subexcs}
\subexc{}
\begin{gather*}
\overline{xy} + \overline{x}\ \overline{y} \\
\overline{1 \cdot 0} + (\overline{1}\cdot \overline{0}) \\
\overline{0} + (0 \cdot 1) \\
1 + 0 \\
1
\end{gather*}
\subexc{}
\begin{gather*}
w + \overline{x}y \\
1 + (\overline{1} \cdot 0) \\
1 + 0 \\
1
\end{gather*}
\subexc{}
\begin{gather*}
wx + \overline{y} + yz \\
(1 \cdot 1) + \overline{0} + (0 \cdot 0) \\
1 + 1 + 0 \\
1
\end{gather*}
\subexc{}
\begin{gather*}
(wx + y\overline{z}) + w\overline{y} + \overline{(w + y)(\overline{x} + y)} \\
((1 \cdot 1) + (0 \cdot \overline{0})) + (1 \cdot \overline{0}) + \overline{(1 + 0)(\overline{1} + 0)} \\
(1 + 0) + (1 \cdot 1) + \overline{(1)(0 + 0)} \\
1 + 1 + \overline{(1)(0)} \\
1 + 1 + \overline{0} \\
1 + 1 + 1 \\
1
\end{gather*}
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
\begin{gather*}
xy + (x + y)\overline{z} + y \\
(xy + y) + \overline{z}x + \overline{z}y \\
y + \overline{z}x + \overline{z}y \\
\overline{z}x + (y + \overline{z}y) \\
\overline{z}x + y
\end{gather*}
\subexc{}
\begin{gather*}
x + y + \overline{(\overline{x} + y + z)} \\
x + y + \overline{\overline{x}}\ \overline{y}\ \overline{z} \\
x + y + x\overline{y}\ \overline{z} \\
(x + x\overline{y}\ \overline{z}) + y \\
x + y
\end{gather*}
\subexc{}
\begin{gather*}
yz + wx + z +[wz(xy + wz)] \\
(yz + z) + wx + (xywz + wz) \\
(z + xywz + wz) + wx \\
z + wx
\end{gather*}
\end{subexcs}
\exc{}{}
Base case
\begin{align*}
\sum^{1}_{i=0}i^2 &= \frac{1 \cdot (1 + 1)(2 \cdot 1 + 1)}{6} \\[2ex]
1^2 &=\frac{2 \cdot 3}{6} \\[2ex]
1 &=\frac{6}{6} \\[2ex]
1 &= 1
\end{align*}
Assume:
\[ \sum^{k}_{i=0}i^2 = \frac{k (k + 1)(2k + 1)}{6} \]
\begin{align*}
\sum^{k+1}_{i=0}i^2 &= 0^2 + 1^2 + 2^2 + \ldots + k^2 + (k + 1)^2 \\[2ex]
&= \frac{k (k + 1)(2k + 1)}{6} + (k + 1)^2 \\[2ex]
&= \frac{k (k + 1)(2k + 1) + 6(k + 1)^2}{6} \\[2ex]
&= \frac{ (k + 1)(k (2k + 1) + 6(k + 1))}{6} \\[2ex]
&= \frac{ (k + 1)(2k^2 + k + 6k + 6)}{6} \\[2ex]
&= \frac{ (k + 1)(2k^2 + 7k + 6)}{6} \\[2ex]
&= \frac{ (k + 1)(k + 2)(2k + 3)}{6} \\[2ex]
&= \frac{(k + 1) ((k + 1) + 1)(2(k + 1) + 1)}{6}
\end{align*}
\qed
\exc{}
\begin{subexcs}
\subexc{}
\begin{align*}
S(0) &= 2^{-0} = 1 \\
S(1) &= 2^{-0} + 2^{-1} = 1.5 \\
S(2) &= 2^{-0} + 2^{-1} + 2^{-2} = 1.75 \\
S(3) &= 2^{-0} + 2^{-1} + 2^{-2} + 2^{-3} = 1.875
\end{align*}
\subexc{}
Based on the results from a, I conjecture that
\[ S(n) = 2 - 2^{-n} \]
\subexc{}
Base case
\begin{align*}
\sum^{0}_{i=0}2^{-i} &= 2-2^{-0} \\
2^{-0} &= 2-1 \\
1 &= 1
\end{align*}
Assume:
\[ \sum^{n}_{i=0}2^{-i} = 2-2^{-n} \]
\begin{align*}
\sum^{n+1}_{i=0}2^{-i} &= 2^{-0} + 2^{-1} + 2^{-2} + \ldots + 2^{-n} + 2^{-(n+1)} \\
&= 2-2^{-n} + 2^{-(n+1)} \\
&= 2-2^{-n} + 2^{-n-1} \\
&= 2-2^{-n} + 2^{-n}2^{-1} \\
&= 2-2^{-n}(1-2^{-1}) \\
&= 2-2^{-n}(\frac{2}{2}-\frac{1}{2}) \\
&= 2-2^{-n}(\frac{1}{2}) \\
&= 2-2^{-n}(2^{-1}) \\
&= 2-2^{-n-1} \\
&= 2-2^{-(n+1)}
\end{align*}
\qed
\subexc{}
\begin{align*}
S(n) &> \epsilon \\
2-2^{-n} &> \epsilon \\
2^{-n} &> \epsilon - 2 \\
-n &> \log_2(\epsilon - 2) \\
n &< -\log_2(\epsilon - 2) \\
\end{align*}
Assuming $S(n)$ never can reach n,
for $S(n)$ to be within $\epsilon$ of $2$, n has to be less than $-\log_2(\epsilon - 2)$
\end{subexcs}
\exc{}
Base case
\begin{align*}
\sum^{1}_{i=1}2^{i-1} \cdot i &= 2^n \cdot (n-1) + 1 \\
2^{1-1} \cdot 1 &= 2^1 \cdot (1-1) + 1 \\
2^0 \cdot 1 &= 2 \cdot 0 + 1 \\
1 \cdot 1 &= 1 \\
1 &= 1
\end{align*}
Assume:
\[ \sum^{n}_{i=1}2^{i-1} \cdot i = 2^n \cdot (n-1) + 1 \]
\begin{align*}
\sum^{n+1}_{i=1}2^{i-1} \cdot i &= (2^{1-1} \cdot 1) + (2^{2-1} \cdot 2) + \ldots
+ (2^{n-1} \cdot n) + (2^{(n+1)-1} \cdot (n+1)) \\
&= 2^n \cdot (n-1) + 1 + (2^{(n+1)-1} \cdot (n+1)) \\
&= 2^n \cdot (n-1) + 1 + 2^{n} \cdot (n+1) \\
&= (2^n \cdot n - 2^n) + 1 + (2^{n} \cdot n + 2^{n}) \\
&= 2^n \cdot n - 2^n + 1 + 2^{n} \cdot n + 2^{n} \\
&= 2(2^n \cdot n) - 2^n + 2^{n}+ 1 \\
&= (4^n \cdot n) + 1 \\
&= (2^{n+1} \cdot ((n+1)-1)) + 1
\end{align*}
\qed
\end{excs}
\end{document}