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%%% Copyright (c) 2004 by Oliver Schneider                                  %%%
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%%%   Some lemmas and helpful information about Math II (ERM)               %%%
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\special{pdf: docinfo << /Author   (Oliver Schneider [mathII AT assarbad DOT net])
                         /Title    (Math II - oral exam preparation)
                         /Subject  (Mathematics, Series, Sequences, Limits)
                         /Keywords (BTU Cottbus, Assarbad, Excersises, Solutions)
                         /Creator  (LaTeX2e) >>}
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\begin{document}
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%% ---------------------------------------------------------------------------
  \section{Mathematics II}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
  Occasionally the $\lim$ will be omitted throughout the script, but
  the fact that it is a limit will be clear by the comments.
  \smallskip

%% ---------------------------------------------------------------------------

  \subsection{General Assumptions}
  \begin{itemize}
  \item For equations which contain infinity or zero, the conclusion
  wether the result is zero, infinity or undefined is often hard
  to make:
  \begin{enumerate}
  \item $\infty \pm a      = \infty \qquad \forall a \in \mathds{R}$
  \item $\infty +   \infty = \infty$
  \item $\infty -   \infty   \quad \mbox{is undefined}$
  \item $  a \cdot \infty  =  \left\{ \begin{array}{rcl}
      \infty & \mbox{for} & a > 0 \\
      - \infty & \mbox{for} & a < 0 \\
      \mbox{undefined} & \mbox{for} & a = 0 \\
    \end{array} \right. $
  \item $\frac{\infty}{a} = \left\{ \begin{array}{rcl}
    \infty & \mbox{for} & a > 0 \\
    - \infty & \mbox{for} & a < 0 \\
  \end{array}\right. $
  \item $\frac{a}{\infty} = 0 \qquad \forall a \in \mathds{R}$
  \item $\infty \cdot \infty = \infty$
  \item $\frac{\infty}{\infty} \quad\mbox{is undefined!}$
  \item $\frac{a}{0} = \left\{ \begin{array}{rcl}
    \infty & \mbox{for} & a > 0 \\
    -\infty & \mbox{for} & a < 0  \\
    undefined & \mbox{for} & a = 0 \\
  \end{array} \right.$
  \end{enumerate}
  \item $f \in C(I)$ means "all functions which are continuous on
  the interval $I$.
  \item $f \in C^n(I)$ means "all functions which are $n$ times continuously
  on differentiable on the interval $I$.
  \item $a^x = e^{x \cdot \ln a}$
  \item $e = e^1 = exp(1)$
  \item $\ln e = 1 \qquad \leadsto \qquad e^x = e^{x \cdot \ln e} = e^{x \cdot 1}$
  \item The power function follows these rules:
  \begin{eqnarray*}
  a^0 & = & 1 \\
  a^1 & = & a \\
  a^{x+y} & = & a^x \cdot a^y, \quad x,y \in \mathds{R} \\
  (a \cdot b)^{x} & = & a^x \cdot b^x, \quad x \in \mathds{R} \\
  a^{-x} & = & \frac{1}{a^x}, \quad x \in \mathds{R} \\
  \log(a^x) & = & x \log (a), \quad x \in \mathds{R} \\
  (a^x)^y & = & a^{x \cdot y} \ne a^{x^y}, \quad x,y \in \mathds{R} \\
  \end{eqnarray*}
  \item Some rules for trigonometric functions:
  \begin{eqnarray*}
  \sin^2 x + \cos^2 x & = & 1 \\
  \sin(x \pm y) & = & \sin(x)\cos(y) \pm \cos(x)\sin(y) \\
  \cos(x \pm y) & = & \cos(x)\cos(y) \mp \sin(x)\sin(y) \\
  \cos(2x) & = & \cos^2 (x) - \sin^2 (x) = 2 \cos^2 (x) - 1 \\
  \sin(2x) & = & 2 \sin(x)\cos(x) \\
  \tan x &=& \frac{\sin x}{\cos x} \\
  \cot x &=& \frac{\cos x}{\sin x} \\
  \tan x \cdot \cot x &=& 1 \\
  1 + \tan^2 x &=& \frac{1}{\cos^2 x} \\
  1 + \cot^2 x &=& \frac{1}{\sin^2 x}
  \end{eqnarray*}
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Special Limits}
  \begin{itemize}
  \item Some special limits of sequences:
  \begin{eqnarray*}
   \lim_{n \to \infty} \frac{1}{n} &=& 0 \\
   \lim_{n \to \infty} \sqrt[n]{n} &=& 1 \\
   \lim_{n \to \infty} \sqrt[n]{n!} &=& \infty\\
   \lim_{n \to \infty} \sqrt[n]{x} &=& 1, \quad x > 0\\
   \lim_{n \to \infty} \frac{a^n}{n!} &=& 0 \\
   \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n &=& e^x \qquad \Leftrightarrow \qquad e = \lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \left( \frac{1 + n}{n}\right)^n \\
   \lim_{n \to \infty} k^n &=& \left\{ \begin{array}{rcl}
    0 & \mbox{for} & \left| k \right| < 1 \\
    1 & \mbox{for} & k=1 \\
  \end{array}\right. \mbox{The sequence (in the last example) diverges for} \left| k \right| > 1 \\
  \end{eqnarray*}
  \item The \textsc{Euler} transcendent number:
  \begin{displaymath}
  e^x = \lim_{n \to \infty}   \left( 1+ \frac{x}{n}\right)^n =
  \sum_{k = 0}^{\infty} \frac{x^k}{k!},\quad x \in \mathds{R}
  \end{displaymath}
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Sequences}
  \begin{itemize}

  \item A sequence is called \textbf{arithmetic sequence} if:
  \begin{displaymath}
  a_{n+1} - a_n = d = \mbox{const.} \qquad \mbox{for} \quad \forall n \in
  \mathds{N} \quad\longrightarrow\quad a_n = a_1 + (n-1)d \qquad n \in
  \mathds{N}
  \end{displaymath}
  that is, if the difference between two elements of the sequence is
  an argument dependent on the first element and the product of $n$
  and a constant - hence the difference between two elements is
  constant.

  \item A sequence is called \textbf{geometric sequence} if:
  \begin{displaymath}
  \frac{a_{n+1}}{a_n} = q = \mbox{const.} \qquad \mbox{for} \quad \forall n \in
  \mathds{N} \quad\longrightarrow\quad a_n = a_1q^{n-1} \qquad n \in
  \mathds{N}
  \end{displaymath}
  that is, if the difference between two elements of the sequence
  is an argument determined by a quotient depending on $n$.

  \item We say that a sequence $\langle x_n \rangle _{n \in
  \mathds{N}}$ converges to $a \in \mathds{R}$ as $n$ tends to
  $\infty$ if, for each $\epsilon > 0$, there is a natural number
  $N := N(\epsilon)$, such that one has
  \begin{displaymath}
    \left| x_a - a \right| < \epsilon, \qquad n \ge \mathds{N}
  \end{displaymath}
  In this case we write $\lim\limits_{n \to \infty} x_n = a$
  \item A sequence is called \textbf{null sequence} if it
  converges to zero.
  \item If a sequence $\langle x_n \rangle_{n \in \mathds{N}}$
  converges to $a \in \mathds{R}$ then also each of its
  subsequences converges to $a$.
  \item A sequence $\langle x_n \rangle_{n \in \mathds{N}}$
  diverges or is divergent if it is not convergent. In this case
  we also say that the limit $\lim\limits_{n \to \infty} x_n$
  does not exist.
  \item For $\lim_{n \to \infty} x_n = a$ and $\lim_{n \to \infty} y_n =
  b$ it holds:
  \begin{enumerate}
    \item $
      \lim\limits_{n \to \infty} \left( x_n \pm y_n \right)
          = a \pm b
          $
    \item $
      \lim\limits_{n \to \infty} \left( x_n \cdot y_n \right)
      = a \cdot b \\
          $
    \item $
      \lim\limits_{n \to \infty} \left( \alpha x_n \right)
      = \alpha \cdot \lim_{n \to \infty} x_n  = \alpha \cdot a, \qquad \alpha \in \mathds{R}\\
          $
    \item $
      \lim\limits_{n \to \infty} \left( \alpha  x_n \right)
      = \alpha \cdot \lim_{n \to \infty} x_n = \alpha \cdot a, \qquad \alpha \in \mathds{R}\\
          $
    \end{enumerate}
  \item A sequence $\langle a_n \rangle_{n \in \mathds{N}}$ is being called \textbf{limited} for
  $\forall n \in \mathds{N}$, if:
  \begin{displaymath}
  \left| a_n \right| \le k \qquad \mbox{for}\quad k \in \mathds{R}
  \end{displaymath}

  \item A sequence $\langle a_n \rangle_{n \in \mathds{N}}$ is called \textbf{strictly monotonic}
  if for all $\forall n \in \mathds{N}$ holds:
  \begin{eqnarray*}
  a_{n+1} & > & a_n \qquad \mbox{(strictly monotonic increasing)}\\
  a_{n+1} & \ge & a_n \qquad \mbox{(monotonic non-decreasing)}\\
  a_{n+1} & < & a_n \qquad \mbox{(strictly monotonic decreasing)}\\
  a_{n+1} & \le & a_n \qquad \mbox{(monotonic non-increasing)}\\
  \end{eqnarray*}

  \item If one has to calculate the limit of a sequence, the limit is
  the quotient of the highest powers of the enumerator n, for the
  form:
  \begin{displaymath}
  a_n  = \frac{b \cdot n^w + c \cdot n^x}{d \cdot n^y + e \cdot n^z}
  \end{displaymath}
  Simply divide the whole term by the highest power and cancel
  everything tending to zero. The result will usually be a simple
  expression.

  \item If one has a sequence similar to this:
  \begin{eqnarray*}
  \lim_{n \to \infty} a_n & = & \left( \frac{n}{n + x}\right)^n
  \to\quad
  \mbox{add \& subtract x:} \quad \left( \frac{n + x - x}{n +
  x}\right)^n = \left( \frac{n + x}{n + x} - \frac{x}{n +
  x}\right)^n \\
  & = & \left( 1 - \frac{x}{n + x} \right)^n \quad \mbox{which
  yields:} \quad
  \left( 1 \pm \frac{x}{n + x} \right)^n = e^{\pm x} \\
  \end{eqnarray*}

  \item If one has something similar to (without denominator):
  \begin{displaymath}
  a_n = \sqrt{x} - \sqrt{y} = \left( \sqrt{x} -
  \sqrt{y} \right) \cdot 1 = \left( \sqrt{x} - \sqrt{y} \right)
  \cdot \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}}
  \end{displaymath}
  Divide or multiply with $\sqrt{x} + \sqrt{y}$ to transform the
  term into the form $(a-b)\cdot(a+b) = a^2+b^2$ \\ \\
  \textbf{Example} assuming $\lim_{n \to \infty} a_n$:
  \begin{eqnarray*}
    a_n &=& n\sqrt{n^4 + n} - n\sqrt{n^4 - n} \quad \longrightarrow \quad \mbox{substitute:} \quad a = \sqrt{n^4 + n} \quad\mbox{\&}\quad b = \sqrt{n^4 - n}\\
    n \cdot (a-b) &=& n \cdot (a-b) \cdot 1 = n \cdot (a-b) \cdot \frac{a+b}{a+b} = n \cdot \frac{(a-b) \cdot (a+b)}{a+b} =  n \cdot \frac{a^2+b^2}{a+b}\\
     &=& n \cdot \left( \frac{(n^4 + n) - (n^4 - n)}{\sqrt{n^4 + n} - \sqrt{n^4 - n}}
     \right) = \frac{2n^2}{\sqrt{n^4 + n} - \sqrt{n^4 - n}} \\
     &=& \frac{\frac{2n^2}{n^2}}{\sqrt{\frac{n^4}{n^4} + \frac{n}{n^4}} - \sqrt{\frac{n^4}{n^4} -
     \frac{n}{n^4}}}= \frac{2}{\sqrt{1 + \frac{1}{n^3}} - \sqrt{1 -
     \frac{1}{n^3}}} = \frac{1}{\sqrt{1}} = 1
  \end{eqnarray*}

  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Series}
  For a series $S_k = \sum\limits_{k=1}^{\infty}{a_k}$
  one should first examine the sequence $a_k$!


  \begin{itemize}
  \item Similarly to arithmetic and geometric sequences we also have
  \textbf{harmonic} and \textbf{geometric series}:
    \begin{itemize}
      \item For the \textbf{geometric} series it is even possible to
      determine the limit:
      \begin{displaymath}
      \sum_{k=1}^{\infty} x^k = \left\{\begin{array}{r}
      \frac{1}{1-x}, \quad \left|x\right| < 1 \\
      \mbox{diverges for} \left|x\right| \ge 1
      \end{array}\right.
      \end{displaymath}
      Example:
      \begin{displaymath}
        \sum_{k = 0}^{\infty} 9 \cdot \left( \frac{1}{10}\right)^k =
        9 \cdot \sum_{k = 0}^{\infty} \left( \frac{1}{10}\right)^k =
        9 \cdot \frac{1}{1-\frac{1}{10}} = 9 \cdot
        \frac{1}{\frac{9}{10}} = 10 \cdot \frac{9}{9} = 10
      \end{displaymath}

      \item For the \textbf{harmonic} series:
      \begin{displaymath}
      \sum_{k=1}^{\infty} \frac{1}{k} = + \infty
      \end{displaymath}

      \item There also exists the special case of the \textbf{alternating harmonic} series:
      \begin{displaymath}
      \sum_{k=1}^{\infty} (-1)^k \frac{1}{k}, \quad \mbox{where the \textsc{Leibniz} test is employed to determine a possible convergence.}
      \end{displaymath}

      \item There are also the so called \textbf{power series} (the geometric series is a special case of it):
      \begin{displaymath}
      \sum_{k=0}^{\infty} g_k (x), \quad x \in I
      \end{displaymath}
      where the $g_k$'s are functions defined on an interval $I$.

    \end{itemize}

  \item A series may converge, if the limit of the respective
  sequence tends to zero:
  \begin{displaymath}
    \sum_{k = 1}^{\infty} a_k \qquad \mbox{converges} \qquad
    \Rightarrow
    \qquad \lim_{k \to \infty} a_k = 0
  \end{displaymath}
  \textbf{Proof:}
  \begin{displaymath}
    s_k := \sum_{k=1}^{\infty} a_k \Rightarrow \left\{
    \begin{array}{rcl}
       \lim\limits_{k \to \infty} s_k & = & s \\
       \lim\limits_{k \to \infty} s_{k-1} & = & s
    \end{array}
    \right.
  \end{displaymath}
  Hence: $a_k = s_k - s_{k-1} \quad \leadsto \quad s - s = 0 \qquad (n \to
  \infty)$\\
  \textbf{This also implies} that, if $\lim\limits_{k \to \infty} a_k \ne
  0$ or $\lim\limits_{k \to \infty} a_k$ does not exist
  $\Rightarrow\quad\sum\limits_{k = 1}^{\infty} a_k$
  \textbf{diverges}!

  \item \textbf{Definition}: $\sum\limits_{k=1}{\infty} a_k$ is
  \textbf{absolutely convergent} if $\sum\limits_{k=1}{\infty}
  \left| a_k \right|$. Absolute convergence implies normal
  convergence!

  \item \textbf{Rearrangement theorem}: If a series is
  absolutely convergent, then each series formed by rearrangement
  of its terms converges to the same limit.

  \item The \textbf{radius of convergence} is determined by the
  following rule for $\sum\limits_{k=a}^{b} c_k$:
  \begin{displaymath}
    R = \frac{c_k}{c_{k+1}}
  \end{displaymath}

  \item \textit{\textbf{Convergence criteria}} for series of the
  type $\sum\limits_{k=1}^{\infty} a_k$ ...
  \begin{enumerate}
  \item \textbf{Comparison test}: Let some $k_0 \in \mathds{N}$
  exist such that $\left| a_k \right| \le b_k, \quad k \ge k_0$.
  Then one has
  \begin{enumerate}
  \item $\sum\limits_{k=1}^{\infty} b_k\quad$ converges $\qquad \leadsto \qquad \sum\limits_{k=1}^{\infty}
  a_k\quad$ converges absolutely.
  \item $\sum\limits_{k=1}^{\infty} \left| a_k \right| = 0 \qquad \leadsto \qquad \sum\limits_{k=1}^{\infty} b_k = 0$
  \end{enumerate}
  \begin{displaymath}
  \end{displaymath}
  \item \textbf{Ratio test}: If $a_k \ne 0$ is true for all $k \ge
  k_0$ with some $k_0 \in \mathds{N}$ and if the sequence
  $\frac{a_{k+1}}{a_k}$ converges properly or improperly then one
  has:
  \begin{displaymath}
  \lim_{k \to \infty} \frac{a_{k+1}}{a_k} \left\{ \begin{array}{rl}
    < 1 & \mbox{The series converges} \\
    > 1 & \mbox{The series diverges} \\
    =1  & \mbox{Undefined} \\
  \end{array} \right.
  \end{displaymath}
  \textbf{Example} assuming ($a_k = \frac{x^k}{k!} ;\qquad a_{k+1} = \frac{x^{k+1}}{(k+1)!}$):
  \begin{displaymath}
  \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \frac{x^{k+1} \cdot k!}{(k+1)! \cdot
  x^k} = \lim_{k \to \infty} \frac{x \cdot x^k \cdot k!}{k! \cdot (k+1) \cdot x^k} =
  \lim_{k \to \infty} \frac{x}{k+1} < 1 \quad \longrightarrow \quad \mbox{series converges}
  \end{displaymath}
  \item \textbf{Root test}: Let the sequence $\left( \sqrt[k]{\left| a_k \right|} \right)_{k \ge
  1}$ converge properly or improperly, then the following holds
  true:
  \begin{displaymath}
    \lim_{k \to \infty} \sqrt[k]{ \left| a_k \right|} \left\{%
    \begin{array}{ll}
        < 1, & \mbox{The series absolutely converges} \\
        > 1, & \mbox{The series diverges} \\
        = 1, & \mbox{Undefined} \\
    \end{array}%
    \right.
  \end{displaymath}
  \textbf{Example} assuming ($a_k = \frac{x^k}{k!}$):
  \begin{displaymath}
  \sqrt[k]{\left| a_k \right|} =
  \sqrt[k]{\frac{x^k}{k!}} = \frac{\sqrt[k]{x^k}}{\sqrt[k]{k!}} =
  \frac{x}{\sqrt[k]{k!}} = 0 < 1 \quad \longrightarrow \quad \mbox{series converges}
  \end{displaymath}
  \item \textbf{Leibniz test}: Assume $c_k \ge c_{k+1} \ge 0, \quad k \in
  \mathds{N}$ and $\lim\limits_{k \to \infty} c_k = 0$ then the
  alternating series $\sum\limits_{k=1}^{\infty} (-1)^k c_k$
  converges.
  \end{enumerate}
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Polynomials}
  \begin{itemize}
  \item Polynomials should first be checked for a zero. Afterwards
  one divides the polynomial $P(x)$ through the term $(x-x_1)$
  where $x_1$ is the first found zero of the polynomial. \\
  If the polynomial is of second degree after this, the following
  approach is used:
  \begin{displaymath}
    ax^2 + bx + c \qquad \leadsto \qquad \frac{-b \pm \sqrt{b^2 - 4 ac}}{2a}
  \end{displaymath}
  or the special case where the quadratic argument has no
  coefficient:
  \begin{displaymath}
    x^2 + px + q \qquad \leadsto \qquad \frac{-p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2 - q}
  \end{displaymath}
  \item If the sum of the coefficients is zero, the first zero of
  the polynomial is $x_1 = 1$.\\
  For an "ordered" polynomial $P(x) = a_k x^n + a_{k-1} x^{k-1} + ... + a_1 x + a_0$ one
  has $a_k$ as the first "odd" coefficient and  $a_{k+1}$ as the
  first "even" coefficient. If there are coefficients which are
  zero, they should be written to not mix up the order of odd and
  even elements.
  \item If the sum of the "odd" and the "even" coefficients is
  equal, another zero of the polynomial is at $x_2 = -1$.
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Limits of Functions}
  \begin{itemize}
  \item For limits of functions there hold similar relations true
  as for limits of sequences:
  \begin{enumerate}
    \item $
      \lim\limits_{x \to x_0} \left( f(x) \pm \pm g(x) \right) =
      \lim\limits_{x \to x_0} f(x) \pm \lim\limits_{x \to x_0} g(x) =
          a \pm b
          $
    \item $
      \lim\limits_{x \to x_0} \left( f(x) \cdot \pm g(x) \right) =
      \lim\limits_{x \to x_0} f(x) \cdot \lim\limits_{x \to x_0} g(x) =
          a \cdot b
          $
    \item $
      \lim\limits_{x \to x_0} \frac{f(x)}{g(x)} =
      \frac{\lim\limits_{x \to x_0} f(x)}{\lim\limits_{x \to x_0} g(x)}=
          \frac{a}{b}
          $
    \end{enumerate}
  \item If one has a function of the form:
  \begin{displaymath}
    \lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{0}{0} \quad \mbox{or} \quad \lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{\infty}{\infty}
  \end{displaymath}
  the rule of \textsc{L'Hospital} says that:
  \begin{displaymath}
    \lim_{x \to x_0} \frac{f(x)}{g(x)} = \lim_{x \to x_0} \frac{f'(x)}{g'(x)}
  \end{displaymath}
  Hence if one faces such an undetermined form, derive the two
  functions and check the limit. In case the first (second, ...)
  derivative does not yield a suitable solution, just derive once
  more. Cannot be used with forms like $\frac{a}{0}$ or $\frac{a}{\infty}$
  because they are determined as shown in the section \textbf{General
  assumptions}.
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{The Derivative}
  \begin{itemize}
  \item The definition of the derivative is as follows:
  \begin{displaymath}
    \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta
    x} \qquad \leadsto \qquad \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) -
    f(x)}{\Delta x} = f'(x)
  \end{displaymath}
  \textbf{Example} $f(x) = 3x^2 + 7x - 8$:
  \begin{eqnarray*}
    f'(x) &=& \lim_{\Delta x \to \infty} \frac{\left( 3(x+ \Delta x)^2 + 7(x + \Delta x) - 8 \right) - \left( 3x^2 + 7x - 8 \right)}{\Delta x} \\
          &=& \lim_{\Delta x \to \infty} \frac{\left( 3(x^2 + 2 x \Delta x + \Delta x^2) + 7(x + \Delta x) - 8 \right) - \left( 3x^2 + 7x - 8 \right)}{\Delta x} \\
          &=& \lim_{\Delta x \to \infty} \frac{3x^2 + 6 x \Delta x + 3\Delta x^2 + 7x + 7\Delta x - 8 - 3x^2 - 7x + 8}{\Delta x} \\
          &=& \lim_{\Delta x \to \infty} \frac{6 x \Delta x + 3\Delta x^2 + 7\Delta x}{\Delta x} = \lim_{\Delta x \to \infty} \frac{\Delta x \left(6 x + 3\Delta x + 7 \right)}{\Delta x} \\
          &=& \lim_{\Delta x \to \infty} \left(6 x + 3\Delta x + 7 \right) = 6x + 7
  \end{eqnarray*}

  \item Some basic derivatives (for valid ranges of definition):\\
  \begin{tabular}{|c|c|c|}
    \hline
    % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
    $f(x)$ & $f'(x)$ & $f''(x)$ \\ \hline \hline
    $a = const.$ & $0$ & $0$ \\ \hline
    $x^n$ & $nx^{n-1}$ & $n(n-1)x^{n-2}$ \\ \hline
    $\sin x$ & $\cos x$ & $-\sin x$ \\ \hline
    $\cos x$ & $-\sin x$ & $-\cos x$ \\ \hline
    $\tan x$ & $\frac{1}{\cos^2 x} = 1 + \tan^2 x$ & $2\tan x(1+\tan^2 x)$ \\ \hline
    $\arcsin x$ & $\frac{1}{\sqrt{1-x^2}}$ & $\frac{x}{(1-x^2)\sqrt{1-x^2}}$ \\ \hline
    $\arccos x$ & $-\frac{1}{\sqrt{1-x^2}}$ & $-\frac{x}{(1-x^2)\sqrt{1-x^2}}$ \\ \hline
    $\arctan x$ & $\frac{1}{1+x^2}$ & $\frac{-2x}{(1+x^2)^2}$ \\ \hline
    $e^x$ & $e^x$ & $e^x$ \\ \hline
    $e^{c \cdot x}$ & $c \cdot e^{c \cdot x}$ & $c^2 \cdot e^{c \cdot x}$ \\ \hline
    $\log_a x$ & $\frac{1}{x \cdot \ln a}$ & $\frac{-1}{x^2 \cdot \ln a}$ \\ \hline
    $\ln x$ & $\frac{1}{x}$ & $-\frac{1}{x^2}$ \\ \hline
  \end{tabular}
  \item \textbf{Mean value theorem of differentiation}: Let $f \in
  C^1[a,b]$, then for each pair of points $x,y \in \left[ a,b
  \right]$with $x < y$ some point $\xi \in (x,y)$ exists such that
  \begin{displaymath}
    f(y) - f(x) = f'(\xi)(x-y)
  \end{displaymath}
  \end{itemize}

%% ---------------------------------------------------------------------------

  \hrule\smallskip
  \subsection{Taylor Polynomial}
  \begin{displaymath}
    P_n(x) = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + ... +
    \frac{f^(n)(x_0)}{n!}(x-x_0)^n + R_n(x)
  \end{displaymath}

%% ---------------------------------------------------------------------------
  \newpage
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\end{document}
