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'example of jump discontinuity'
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| Title of object: |
example of jump discontinuity |
| Canonical Name: |
ExampleOfJumpDiscontinuity |
| Type: |
Example |
| Created on: |
2006-11-18 11:01:27 |
| Modified on: |
2006-11-20 00:06:38 |
| Classification: |
msc:54C05, msc:26A15 |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
The \PMlinkname{elementary}{ElementaryFunction} real function
$$f\colon\,x \mapsto \frac{1}{1+e^\frac{1}{x}}$$
has a jump discontinuity at the origin, since
$$\lim_{x\to 0-}f(x) = 1\quad \mathrm{and}\quad \lim_{x\to 0+}f(x) =0.$$
Indeed,
\begin{itemize}
\item if\, $x \to 0-$,\, then\, $\displaystyle \frac{1}{x} \to -\infty$,\;
$\displaystyle e^\frac{1}{x} \to 0$,\;
$\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 1$;
\item if\, $x \to 0+$,\, then\, $\displaystyle \frac{1}{x} \to \infty$,\;
$\displaystyle e^\frac{1}{x} \to \infty$,\;
$\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 0$.
\end{itemize}
Obviously such a function cannot be represented by Taylor series about $x=0$, but for $x\geq0$ it may be given by the series
\begin{align*}
\frac{1}{1+e^{\frac{1}{x}}}=\sum_{k=1}^\infty(-1)^{k-1}e^{-\frac{k}{x}},
\end{align*}
and so
\begin{align*}
\lim_{x\to0+}\sum_{k=1}^\infty(-1)^{k-1}e^{-\frac{k}{x}}=0.
\end{align*}
However for $x\leq0$ if we consider the partial sum of that series and by applying the ratio test, we see that the test resultant function $e^{-1/x}$ grows without bounds as $x\to0-$. Therefore, the function in question cannot be represented by the series above indicated in that case. |
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