bash test2

testforbash

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sudo apt install texlive
echo "test"

\documentclass[12pt, a4paper]{article}
% Allow the usage of UTF-8 characters
\usepackage[utf8]{inputenc}
% Allow the usage of graphics (.png, .jpg)
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{kotex}
\title{proof that any $n^{th}$ order derivatives of $f(x)$ can be expressed via limits of $f(x)$}
\author{Eugene Kang}
\date{}
\usepackage{amssymb}
\usepackage{cancel}
\begin{document}
\maketitle
Let function $f(x) $ $n$ times differentiable function , and it's $n^{th}$ derivative is also continuous ,
$$\lim_{h \to 0}{\frac{f(x+2h)-f(x+h)}{h}}=\lim_{h\to 0}{f'(x+h)}$$ is clear,
than$$\displaystyle {\lim_{h \to 0+}{\frac{f'(x+h)-f'(x)}{h}}=
\lim_{h \to 0+}\frac{{\frac{f(x+2h)-f(x+h)}{h}} - {\frac{f(x+h)-f(x)}{h}}}{h}}$$
\\thus we can see,
$$\displaystyle {\lim_{h \to 0+}{f''(x)}=
\lim_{h \to 0+}{\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}}}$$
\\using this equation, we can see that
$$\displaystyle {\lim_{h \to 0}{f'''(x)}={
\lim_{h\to0}{\frac{f''(x+h)-f''(x)}{h}}}=
\lim_{h \to 0}{\frac{f(x+3h)-3f(x+2h)+3f(x+2h)-f(x)}{h^3}}}$$
these can be also shown with L'Hôpital's rule.\\\\we now expect a pattern,  for $f''(x)$ , each $f$, have coefficients as same as those of   $(x-1)^2$ , same goes for $f'''(x)$ and $(x-1)^3$ .
\newline \newline now I will prove that this pattern , i.e. $$ \lim_{h\to0+} \sum_{k=0}^{n} \frac{_nC_k (-1)^{n-k}f(x+kh)}{h^n}= \frac{d^n{f}}{d{x}^n}$$ holds for every $n^{th}$ order derivatives.
\\\\
lets define a function $g_{n}(x)$ , \\
$g_0 (x)=f(x)$, $g_1(x)=f(x+h)-f(x)$, $g_2(x)=\left(\begin{array}{cc}f(x+2h)-f(x+h)\\ \end{array}\right)  - \left(\begin{array}{cc}f(x+h)-f(x)\\ \end{array}\right) $ $\cdots$
\\\\\newline we'll also define a  function  $\triangle(x)$,which represents x-axis transformation. \\\\$\triangle(f(x))=f(x+h)$ , $\triangle(x) = x +h $ ,\newline for constant $a$ , $\triangle(ax) = a\times \triangle(x) = a(x+h)$\newline\newline
\\for constant $a$ , define $(\triangle + a )(f(x)) = \triangle(f(x)) +a f(x)$ for natural number $n$ and constant $a$ , we'll define $${(\triangle + a) }^n (f(x)) = \overbrace {{(\triangle +a )}\circ {(\triangle  +a) }\circ \cdots }^{\text{{n times}}} f(x)$$, also for constant $a,b$ : let $$ (\triangle +a )\circ(\triangle + b)\circ(f(x))=(\triangle + a)\circ(\triangle(f(x))+ bf(x))  = \triangle ^2 f(x) + ( a + b ) \triangle (f(x) )+ ab f(x)$$
than,by definition,
$$(\triangle ^2 + ( a + b ) \triangle + ab)(x) =  (\triangle +a)(\triangle+ b )(x) =(\triangle +a )\circ((\triangle + b)\circ(x) )$$ \newline if for a   natural number $n$ and constant $a$ , \\$(\triangle +a)^n (f) = \sum_{k=0}^{n} _nC_k (a)^{n-k}\triangle^{k} f$ \\is true,than:\newline$$(\triangle +a)^{n+1} (f) =(\triangle +a)\sum_{k=0}^{n} _nC_k (a)^{n-k}\triangle^{k} f$$
is true,   by setting $p = k+1$ we get\\
 $$=(\sum_{p=1}^{n+1} _nC_{p-1} (a)^{n+1-p}\triangle^{p} f )+ (\sum_{k=0}^{n} _nC_{k} (a)^{n+1-k}\triangle^{k} f)$$\\

 $$=(\sum_{p=1}^{n} _nC_{p-1} (a)^{n+1-p}\triangle^{p} f )+
(\sum_{k=1}^{n} _nC_{k} (a)^{n+1-k}\triangle^{k} f)+ (a\triangle^{n+1}f)+(a^{n+1}\triangle^0 f)$$
$$=(\sum_{k=1}^{n} _{n+1}C_{k} (a)^{n+1-k}\triangle^{k}f)+ (a\triangle^{n+1}f)+(a^{n+1}\triangle^0 f)$$\\
$$=\sum_{k=0}^{n+1} _{n+1}C_{k} (a)^{n+1-k}\triangle^{k}f$$is true, and when $n = 2 $,
$(\triangle +a)^2 (f) = \sum_{k=0}^{2} _2C_k (a)^{2-k}\triangle^{k} f$.\\Thus for any natural number  $N$ and constant $a$ $$(\triangle +a)^N (f) = \sum_{k=0}^{N} _NC_k (a)^{N-k}\triangle^{k}f $$ holds.\\\\now for $g_{n}(x)$ , we can say that $$g_1 (x)=(\triangle-1)(g_0(x))~, ~g_2 (x) =(\triangle-1)(g_1(x))~, \cdots ~g_{n+1}(x)=(\triangle-1)(g_n(x))$$
$$g_{n}(x)=(\triangle-1)(g_{n-1}(x)) =(\triangle-1)^{2}(g_{n-2}(x)) = (\triangle-1)^{3}(g_{n-3}(x)) = \cdots (\triangle-1)^{n}(g_{0}(x)) $$ since $f(x)=g_{0}(x)$, and we can say that $\frac{d^{n}f}{dx^n} = \lim\limits_{h \to 0} {\frac{g_n (x)}{h^n}} $  Thus it is proven that $$\lim_ {h \to 0}\frac{g_n}{h^n} = \lim_{h\to 0}\frac{d^n{f}}{d{x}^n} = \lim_{h \to 0}{\frac{(\triangle -1)^n (f(x))}{h^n}}= \lim_{h\to0} \sum_{k=0}^{n} \frac{_nC_k (-1)^{n-k}\triangle^{k}f(x)}{h^n} $$
  $$ \therefore \lim_{h\to0+} \sum_{k=0}^{n} \frac{_nC_k (-1)^{n-k}f(x+kh)}{h^n}= \frac{d^n{f}}{d{x}^n}$$\newline\newline \large{therefore, any $nth$ order derivatives of $f(x)$ can be expressed via limits of $f(x)$.}
\end{document}

$f(x0)$