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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: |
2006-12-14 21:25:19 |
| Classification: |
msc:54C05, msc:26A15 |
Revision comment (for changes between this and next version):
| fix the plot; it was mirrored over y axis before |
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}
These results can be seen also from the series \PMlinkescapetext{expansions} of the function gotten by performing the divisions:\, for\, $x < 0$\, we obtain the \PMlinkname{converging}{Converge} \PMlinkname{alternating series}{LeibnizEstimateForAlternatingSeries}
\begin{align*}
1:(1+e^{\frac{1}{x}}) = \sum_{k=0}^\infty(-1)^ke^{\frac{k}{x}}
= 1-e^{\frac{1}{x}}+e^{\frac{2}{x}}-e^{\frac{3}{x}}+-\ldots
\end{align*}
and for\, $x > 0$\, the series
\begin{align*}
1:(e^{\frac{1}{x}}+1) = \sum_{k=1}^\infty(-1)^{k+1}e^{-\frac{k}{x}}
= e^{-\frac{1}{x}}-e^{-\frac{2}{x}}+e^{-\frac{3}{x}}-+\ldots
\end{align*}
\textbf{Note.}\, The derivative of the function may be written as
$$f'(x) = \frac{1}{x^2(e^{-\frac{1}{x}}+1)(1+e^\frac{1}{x})},$$
and thus we have the one-sided limits\, $\displaystyle \lim_{x\to 0\pm}f'(x) = 0$.
\begin{figure}[!tb]
\begin{center}
\includegraphics{e1xjump.eps}
\end{center}
\caption{Graph of the function $f$ with jump discontinuity}
\end{figure}
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