\documentclass {article}
\usepackage{fontspec}
\usepackage{amsmath}
\usepackage{sectsty}

\sectionfont{\normalfont\fontsize{9}{16}}
\setmainfont[Ligatures=TeX]{FreeSans}
\begin {document}
	\section*{Доказать, что $E_k = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$}
	\begin{math}
		\displaystyle
		A = \int_1^2\vec{F}d\vec{S} = \int_{t_1}^{t_2}(\vec{F}\vec{v})dt = \int_{t_1}^{t_2}\frac{d\vec{p}}{dt}\vec{v}dt =
		\int_{p_1}^{p_2}\vec{v}d\vec{p} = (\vec{p}\cdot\vec{v}) - \int_{v_1}^{v_2}(\vec{p}d\vec{v}) = \newline
		= (\vec{p}\cdot\vec{v}) - \int_{v_1}^{v_2}\frac{m\vec{v}d\vec{v}}{\sqrt{1 - \frac{v^2}{c^2}}} = 
		(\vec{p}\cdot\vec{v}) - \frac{1}{2}\int_{v_1}^{v_2}\frac{md\vec{v^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = 
		(\zeta = 1 - \frac{v^2}{c^2}) = \newline 
		(\vec{p}\cdot\vec{v}) + \frac{mc^2}{2}\int_{\zeta_1}^{\zeta_2}\zeta^{-\frac{1}{2}}d\zeta = 
		(\vec{p}\cdot\vec{v} + mc^2\zeta)\vert_{\zeta_1}^{\zeta_2} = 
		\left.\left(\frac{mv^2}{\sqrt{1 - \frac{v^2}{c^2}}} + mc^2\sqrt{1 - \frac{v^2}{c^2}}\right)\right|_{v_1}^{v_2} = \newline
		\left.\left[\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}\right]\right|_{v_1}^{v_2} = E_{k_2} - E_{k_1} \implies
		E_k = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} + K_0
	\end{math}
	\section*{Найдём $K_0$}
	При $v\rightarrow0$:\newline
	\begin {math} \displaystyle
		\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} + K_0 = \frac{mv^2}{2} = \star \newline
			\frac{d}{dv}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = -\frac{1}{2c^2}\frac{(-1) 2v^1}{\sqrt{1 - \frac{v^2}{c^2}}^3} \newline
			\frac{d^2}{dv^2}\frac{1}{\sqrt{1 - \frac{v^2}{2c^2}}} = \frac{1}{2c^2}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}^3} + \frac{1}{2c^4}\frac{3v^2}{\sqrt{1 - \frac{v^2}{c^2}}^5}, \newline
	\end {math}
	Как видно, при ${v\rightarrow 0}$, $E^{(n)}(v)\rightarrow C_n$, где $C_n$ - конечное число, которое при умножении на $v^n$ даёт $0$. \newline 
	\begin{math} \displaystyle
			% \frac{d^{(n)}}{dv^{(n)}}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 
			% (-1)^n\frac{(2n+1)!!}{2^n\cdot c^{2n}}\fra2c{(-1)^n\cdot v^n}{\sqrt{\left(1 - \frac{v^2}{c^2}\right)^{2n+1}}}  = 
			% \frac{(2n+1)!!}{2^n\cdot c^{2n}}\frac{v^n}{\sqrt{\left(1 - \frac{v^2}{c^2}\right)^{2n+1}}} \newline
			E(v) = \frac{E(v=0)v^0}{0!} + \frac{E'(v=0)v^1}{1!} + \frac{E''(v=0)v^2}{2!} + \dots \newline
		\star = mc^2\left(1 + \frac{v^2}{2c^2} + \frac{v^2}{4c^2} + \frac{3v^4}{4c^4} + \dots\right) + K_0 = \frac{mv^2}{2} \newline
		mc^2 + \frac{mv^2}{2} + \frac{mv^2}{4} + \frac{3mv^4}{4c^2} + \dots + K_0 = \frac{mv^2}{2} \newline
		mc^2 + 0 + 0 + \dots + K_0 = 0 \implies K_0 = -mc^2 , \newline
		E_k = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2
	\end{math}
\end{document}
