path: root / Year 1 / Mathematical analysis / Homeworks / Homework2.tex

% Document MUST be built with XeLaTeX!
\documentclass[11pt,a4paper]{article}

\usepackage[margin=1.5cm]{geometry}

% Поддръжка на български език
% Thanks to https://tex.stackexchange.com/a/136926
\usepackage{fontspec}
\defaultfontfeatures{Ligatures=TeX}
\setmainfont{CMU Serif}
\setsansfont{CMU Sans Serif}

\usepackage[bulgarian,english]{babel}

\usepackage{amsmath}            % Математически символи
\usepackage{extarrows}
\usepackage{amsfonts}           % Още математически символи
\usepackage{cancel}             % Зачертаване
\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
\newcommand{\answerbox}[1]{\colorbox{yellow}{$\displaystyle #1$}}

\usepackage{tikz}
\newcommand{\lipoint}[3]{\draw (#1, 3pt) -- (#1, -3pt) node[below] {#2} node[above] {#3}}
\newcommand{\parch}[2]{\node[above] at (#1, 0) {#2} ; \draw (#1 + 0.5, 0) arc (0:180:0.5)}

\usepackage{enumitem}
\setlist[enumerate]{label=\arabic*.}

\usepackage[pdfauthor={Камен Димитров Младенов, КН Зимен семестър, 2021},
		    pdftitle={Домашна работа 2 по ДИС 1. Компютърни науки, 1 курс, 2 поток},
		    pdfsubject={},
		    pdfkeywords={},
		    colorlinks=true,
		    urlcolor=blue]{hyperref}

\title{\vspace{-1.5cm}Домашна работа 2 по ДИС 1. Компютърни науки, 1 курс, 2 поток}
\author{Камен Димитров Младенов, Група 6}
\date{20 Ноември 2021}

\begin{document}

\maketitle

\paragraph{\colorbox{yellow}{Задача 1.}}

Дадена е редицата
\begin{align*}
	a_1 = \lambda, \; a_{n+1} = \frac{2a_n^2 - a_n + 6}{a_n - 6}
\end{align*}
Да се изследва за сходимост спрямо стойностите на параметъра $\lambda \neq 6$.

\paragraph{Решение:}

Разглеждаме за монотонност:
\begin{align*}
	& a_{n+1} - a_n = \frac{2a_n^2 - a_n + 6}{a_n - 6} - a_n = \frac{2a_n^2 - a_n + 6 -a_n^2 + 6a_n}{a_n - 6} = \frac{a_n^2 + 5a_n + 6}{a_n - 6} = \frac{(a_n+3)(a_n+2)}{a_n - 6} \\
	& \implies \text{Корените за } a_n \text{ са -3, -2 и 6}
\end{align*}
\hfill
\begin{tikzpicture}
	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
	\parch{3.5}{+} ;
	\lipoint{3}{6}{} ;
	\parch{2.5}{-} ;
	\lipoint{2}{-2}{} ;
	\parch{1.5}{+};
	\lipoint{1}{-3}{} ;
	\parch{0.5}{-} ;
\end{tikzpicture}

Ако съществува $l = \lim\limits_{n \rightarrow \infty} a_n$, то
\begin{align*}
	l = \frac{2l^2 - l + 6}{l - 6} \qquad l^2 - 6l = 2l^2 - l + 6 \qquad 0 = l^2 + 5l + 6 \implies l_1 = -2 \;,\; l_2 = -3
\end{align*}

Разглеждаме положението на $a_{n+1}$ спрямо -3, -2 и 6

\noindent
\begin{align*}
	& a_{n+1} - (-3) = \frac{2a_n^2 - a_n + 6}{a_n - 6} + 3 = \frac{2a_n^2 - a_n + 6 + 3a_n - 18}{a_n - 6} = \frac{2a_n^2 + 2a_n - 12}{a_n - 6} \\
	& \implies \text{Корените за } a_n \text{ са -3, 2 и 6}
\end{align*}
\hfill
\begin{tikzpicture}
	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
	\parch{3.5}{+} ;
	\lipoint{3}{6}{} ;
	\parch{2.5}{-} ;
	\lipoint{2}{2}{} ;
	\parch{1.5}{+};
	\lipoint{1}{-3}{} ;
	\parch{0.5}{-} ;
\end{tikzpicture}

\noindent
\begin{align*}
	& a_{n+1} - (-2) = \frac{2a_n^2 - a_n + 6}{a_n - 6} + 2 = \frac{2a_n^2 - a_n + 6 + 2a_n - 12}{a_n - 6} = \frac{2a_n^2 + a_n - 6}{a_n - 6} \\
	& \implies \text{Корените за } a_n \text{ са -2, $\frac{3}{2}$ и 6}
\end{align*}
\hfill
\begin{tikzpicture}
	\draw[-latex] (-0.25, 0) -- (4.25, 0) ;
	\parch{3.5}{+} ;
	\lipoint{3}{6}{} ;
	\parch{2.5}{-} ;
	\lipoint{2}{$\frac{3}{2}$}{} ;
	\parch{1.5}{+};
	\lipoint{1}{-2}{} ;
	\parch{0.5}{-} ;
\end{tikzpicture}

\noindent
\begin{align*}
	& a_{n+1} - 6 = \frac{2a_n^2 - a_n + 6}{a_n - 6} - 6 = \frac{2a_n^2 - a_n + 6 - 6a_n + 36}{a_n - 6} = \frac{2a_n^2 - 7a_n + 42}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
	& \implies \text{Единствения корен за } a_n \text{ е 6}
\end{align*}
\hfill
\begin{tikzpicture}
	\draw[-latex] (-0.25, 0) -- (2.25, 0) ;
	\parch{1.5}{+};
	\lipoint{1}{6}{} ;
	\parch{0.5}{-} ;
\end{tikzpicture}

Разглеждаме положението на $a_{n+1}$ спрямо $\frac{3}{2}$ и 2

\noindent
\begin{align*}
	& a_{n+1} - \frac{3}{2} = \frac{2a_n^2 - a_n + 6}{a_n - 6} - \frac{3}{2} = \frac{2a_n^2 - a_n + 6 - \frac{3}{2}a_n + 9}{a_n - 6} = \frac{2a_n^2 - \frac{5}{2}a_n + 15}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
	& \implies \text{Единствения корен за } a_n \text{ е 6}
\end{align*}

\noindent
\begin{align*}
	& a_{n+1} - 2 = \frac{2a_n^2 - a_n + 6}{a_n - 6} - 2 = \frac{2a_n^2 - a_n + 6 - 2a_n + 12}{a_n - 6} = \frac{2a_n^2 - 3a_n + 18}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \\
	& \implies \text{Единствения корен за } a_n \text{ е 6}
\end{align*}

Разглеждаме положението на $a_{n+1}$

\noindent
\begin{align*}
	& a_{n+1} = \frac{2a_n^2 - a_n + 6}{a_n - 6} \genfrac..{0pt}{0}{\; >0}{} \implies \text{Единствения корен за } a_n \text{ е 6}
\end{align*}

\begin{center}
	\begin{tabular}{c|ccccccccccc}
		$a_n$ & $-\infty$ & -3 & & -2 & & $\frac{3}{2}$ & & 2 & & 6 & $+\infty$ \\
		\hline
		$a_{n+1} - a_n$ & - & \vdots & + & \vdots & - & \vdots & - & \vdots & - & \vdots & + \\
		$a_{n+1} - (-3)$ & - & \vdots & + & \vdots & + & \vdots & + & \vdots & - & \vdots & + \\
		$a_{n+1} - (-2)$ & - & \vdots & - & \vdots & + & \vdots & - & \vdots & - & \vdots & + \\
		$a_{n+1} - 6$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & \vdots & + \\
		$a_{n+1} - \frac{3}{2}$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & \vdots & + \\
		$a_{n+1} - 2$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & \vdots & + \\
		$a_{n+1}$ & - & \vdots & - & \vdots & - & \vdots & - & \vdots & - & \vdots & + \\
	\end{tabular}
\end{center}

\textbf{I. Случай} $\lambda < -3$ \quad Знаем, че $a_{n+1} - a_n$ е винаги намаляваща $\implies \lim\limits_{n \rightarrow \infty} a_n = -\infty$

\textbf{II. Случай} $\lambda \in (-3, -2)$

\textbf{III. Случай} $\lambda \in (-2, \frac{3}{2})$

\textbf{IV. Случай} $\lambda \in (\frac{3}{2}, 2)$

\textbf{V. Случай} $\lambda \in (2, 6)$ \quad Знаем, че $a_{n+1} - a_n$ е винаги намаляваща $\implies \lim\limits_{n \rightarrow \infty} a_n = -\infty$

\textbf{VI. Случай} $\lambda > 6$ \quad Знаем, че $a_{n+1} - a_n$ е винаги растяща $\implies \lim\limits_{n \rightarrow \infty} a_n = +\infty$


\paragraph{\colorbox{yellow}{Задача 2.}}

Да се пресметне границата
\begin{align*}
	\lim_{x \rightarrow \infty} \frac{\sqrt{2x^2 + 1} - \sqrt[3]{4x^2 - 1}}{\sqrt[4]{x^4 + 7} - \sqrt[5]{x^4 - 3}}
\end{align*}

\paragraph{Решение:}
\begin{align*}
	\lim_{x \rightarrow \infty} \frac{\sqrt{2x^2 + 1} - \sqrt[3]{4x^2 - 1}}{\sqrt[4]{x^4 + 7} - \sqrt[5]{x^4 - 3}} = \lim_{x \rightarrow \infty} \frac{\cancel{x} \sqrt{2 + \frac{1}{x^2}} - \cancel{x} \sqrt[3]{\frac{4}{x} - \frac{1}{x^3}}}{\cancel{x} \sqrt[4]{1 + \frac{7}{x^4}} - \cancel{x} \sqrt[5]{\frac{1}{x} - \frac{3}{x^4}}} = L
\end{align*}
\begin{align*}
	& \sqrt{2} \xrightarrow[x \rightarrow \infty]{} \sqrt{2} \;,\; \frac{1}{x^2} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{4}{x} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{1}{x^3} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \\
	& \sqrt[4]{1} \xrightarrow[x \rightarrow \infty]{} \sqrt[4]{1} \;,\; \frac{7}{x^4} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{1}{x} \xrightarrow[x \rightarrow \infty]{} 0 \;,\; \frac{3}{x^4} \xrightarrow[x \rightarrow \infty]{} 0
	\qquad 	\implies L = \frac{\sqrt{2}}{\sqrt[4]{1}} = \answerbox{\sqrt{2}}
\end{align*}

\paragraph{\colorbox{yellow}{Задача 3.}}

Да се пресметне границата
\begin{align*}
	\lim_{x \rightarrow 0} \frac{\sqrt[3]{5x + 1} - \sqrt[4]{7x + 1}}{\arctg(\ln(x+1))}
\end{align*}

\paragraph{Решение:}

\begin{align*}
	\arctg(\ln(x+1)) \sim \ln(x+1) \quad x \rightarrow 0 \qquad \sqrt[3]{5x + 1} = \underbrace{(1 + 5x)^{\frac{1}{3}} - 1}_{\sim \frac{5x}{3} \quad x \rightarrow 0} + 1 \qquad \sqrt[4]{7x + 1} = \underbrace{(1 + 7x)^{\frac{1}{4}} - 1}_{\sim \frac{7x}{x} \quad x \rightarrow 0} + 1
\end{align*}
\begin{align*}
	\implies \lim_{x \rightarrow 0} \frac{\sqrt[3]{5x + 1} - \sqrt[4]{7x + 1}}{\arctg(\ln(x+1))} \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{\frac{5x}{3} + 1 - \left( \frac{7x}{4} + 1 \right)}{\ln(x+1)} = \lim_{x \rightarrow 0} \frac{\frac{20x - 21x}{12}}{\ln(x+1)} = \lim_{x \rightarrow 0} - \frac{x}{12 \ln(x+1)} = L
\end{align*}
\begin{align*}
	\frac{x}{\ln(x+1)} \xrightarrow[x \rightarrow 0]{} 1 \implies L = \answerbox{- \frac{1}{12}}
\end{align*}

\paragraph{\colorbox{yellow}{Задача 4.}}

Да се пресметне $f^{(n)}(x)$, където
\begin{align*}
	f(x) = \frac{x^3}{\sqrt{1 - 7x}}
\end{align*}

\paragraph{Решение:}

\begin{align*}
	g(x) = \frac{1}{\sqrt{1 - 7x}} = (1 - 7x)^{\frac{1}{2}} \qquad h(x) = x^3 \qquad\qquad \implies f(x) = g(x) \cdot h(x)
\end{align*}
\begin{align*}
	g^{(n)}(x) = \left( (1 - 7x)^{\frac{1}{2}} \right)^{(n)} & = \frac{1}{2} \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} - 2 \right) \cdots \left( \frac{1}{2} - n + 1 \right) \cdot x^{\frac{1}{2} - n} \cdot 1^n = \\
	& = \frac{1(-1)(-3) \cdots (1-2n+2)}{2^n} \cdot x^{\frac{1}{2} - n}
\end{align*}
\begin{align*}
	h^{(0)}(x) = h(x) = x^3 \quad h'(x) = 3x^2 \quad h''(x) = 6x \quad h'''(x) = 6 \quad h^{IV}(x) = 0 \quad \implies h^{(n)}(x) = 0 \quad n \geq 4
\end{align*}
\begin{align*}
	f^{(n)} & = \sum_{k = 0}^n g^{(n-k)} h^{(k)} = \binom{n}{0}g^{(n)}h^{(0)} + \binom{n}{1}g^{(n-1)}h^{(1)} + \binom{n}{2}g^{(n-2)}h^{(2)} + \binom{n}{3}g^{(n-3)}h^{(3)} + \\
	& \quad + \underbrace{\binom{n}{4}g^{(n-4)}h^{(4)} + \cdots + \binom{n}{0}g^{(0)}h^{(n)}}_{0} = \\
	& = 1 \cdot \frac{(-1)(-3) \cdots (3-2n)}{2^n} \; x^{\frac{1}{2} - n} \cdot x^3 + n \cdot \frac{(-1)(-3) \cdots (5-2n)}{2^n} \; x^{\frac{3}{2} - n} \cdot 3x^2 + \\
	& \quad + \frac{n!}{2(n-2)!} \cdot \frac{(-1)(-3) \cdots (7-2n)}{2^n} \; x^{\frac{5}{2} - n} \cdot 6x + \frac{n!}{6(n-3)!} \cdot \frac{(-1)(-3) \cdots (9-2n)}{2^n} \; x^{\frac{7}{2} - n} \cdot 6 = \\
	& = \frac{(-1)(-3) \cdots (3-2n)}{2^n} \; x^{\frac{7}{2} - n} + 3n \cdot \frac{(-1)(-3) \cdots (5-2n)}{2^n} \; x^{\frac{7}{2} - n} + \\
	& \quad + 3n(n-1) \cdot \frac{(-1)(-3) \cdots (7-2n)}{2^n} \; x^{\frac{7}{2} - n} + n(n-1)(n-2) \cdot \frac{(-1)(-3) \cdots (9-2n)}{2^n} \; x^{\frac{7}{2} - n} = \\
	& = \answerbox{x^{\frac{7}{2} - n} \; \frac{(-1)(-3) \cdots (9-2n)}{2^n} \left( (8-2n)\cdots (3-2n) + 3n \cdot (8-2n) \cdots (5-2n) + \right.} \\
	& \;\,\quad \answerbox{\left. + 3n^{\underline{2}} \cdot (8-2n)(7-2n) + n^{\underline{3}} \right)}
\end{align*}

\paragraph{\colorbox{yellow}{Задача 5.}}

Да се пресметне границата
\begin{align*}
	\lim_{x \rightarrow 0} \left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}}
\end{align*}

\paragraph{Решение:}

\begin{align*}
	\lim_{x \rightarrow 0} \left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}} = \lim_{x \rightarrow 0} e^{\ln\left( \frac{\sin x}{\arcsin x} \right) ^ {\frac{1}{x^2}}} = \lim_{x \rightarrow 0} e^{\frac{1}{x^2} \cdot \ln\left( \frac{\sin x}{\arcsin x} \right)} = e^l
\end{align*}
\begin{align*}
	l & = \lim_{x \rightarrow 0} \frac{1}{x^2} \ln\left( \frac{\sin x}{\arcsin x} \right) \qquad \ln\left( \frac{\sin x}{\arcsin x} - 1 + 1 \right) \sim \frac{\sin x}{\arcsin x} - 1 \quad x \rightarrow 0 \\
	& \implies l \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x}{\arcsin x} - 1 \right) = \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x - \arcsin x}{\arcsin x} \right) \qquad \arcsin x \sim x \quad x \rightarrow 0 \\
	& \implies l \xlongequal{(\sim)} \lim_{x \rightarrow 0} \frac{1}{x^2} \left(  \frac{\sin x - \arcsin x}{x} \right) = \lim_{x \rightarrow 0} \frac{\sin x - \arcsin x}{x^3} \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{\cos x - \frac{1}{\sqrt{1-x^2}}}{3x^2} \\
	& \qquad \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-\frac{1}{2}} \quad \left( (1-x^2)^{-\frac{1}{2}} \right)' = -\frac{1}{2} (1-x^2)^{-\frac{3}{2}} \cdot \left( (1-x^2) \right)' = x(1-x^2)^{-\frac{3}{2}} \\
	& \implies l \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{-\sin x - x(1-x^2)^{-\frac{3}{2}}}{6x} \\
	& \qquad \left( x(1-x^2)^{-\frac{3}{2}} \right)' = \left( \frac{x}{(1-x^2)^{\frac{3}{2}}} \right)' = \frac{(1-x^2)^{\frac{3}{2}} + \frac{3}{2}(1-x^2)^{\frac{1}{2}} 2x \cdot x}{(1-x^2)^3} = \frac{(1-x^2)\sqrt{1-x^2} + 3x^2\sqrt{1-x^2}}{(1-x^2)^3} = \\
	& \qquad = \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}} \\
	& \implies l \xlongequal{\text{Лопитал } \left[ \frac{0}{0} \right]} \lim_{x \rightarrow 0} \frac{-\cos x - \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}}}{6} \qquad \cos x \xrightarrow[x \rightarrow 0]{} 1 \qquad \frac{2x^2+1}{(1-x^2)^{\frac{5}{2}}} \xrightarrow[x \rightarrow 0]{} 1 \quad \implies l = -\frac{1}{3}
\end{align*}

Следователно, търсената граница е: $e^l = e^{-\frac{1}{3}} = \answerbox{\frac{1}{\sqrt[3]{e}}}$

\end{document}