Solved 2nd, 3rd and 4th exercise and started work on 1st of dis hw2

1 files changed, 149 insertions, 4 deletions

diff --git a/Year 1/Mathematical analysis/Homeworks/Homework2.tex b/Year 1/Mathematical analysis/Homeworks/Homework2.tex
index f400f35..c19a139 100644
--- a/Year 1/Mathematical analysis/Homeworks/Homework2.tex
+++ b/Year 1/Mathematical analysis/Homeworks/Homework2.tex
@@ -18,6 +18,10 @@
 \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*.}
 
@@ -46,7 +50,113 @@
 
 \paragraph{Решение:}
 
-[TODO]
+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} {{\; >0} \atop {}} \\
+	& \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} {{\; >0} \atop {}} \\
+	& \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} {{\; >0} \atop {}} \\
+	& \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} {{\; >0} \atop {}} \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}
+
+[TODO: FINISH]
 
 \paragraph{\colorbox{yellow}{Задача 2.}}
 
@@ -56,8 +166,16 @@
 \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*}
 
-[TODO]
+\emph{2)} $\frac{3}{2}$ и 
 
 \paragraph{\colorbox{yellow}{Задача 3.}}
 
@@ -68,7 +186,15 @@
 
 \paragraph{Решение:}
 
-[TODO]
+\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))} \stackrel{(\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.}}
 
@@ -79,7 +205,26 @@
 
 \paragraph{Решение:}
 
-[TODO]
+\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.}}