| Version 5 |
Version 4 |
| The \PMlinkname{elementary}{ElementaryFunction} real function |
The \PMlinkname{elementary}{ElementaryFunction} real function |
| $$f\colon\,x \mapsto \frac{1}{1+e^\frac{1}{x}}$$ |
$$f\colon\,x \mapsto \frac{1}{1+e^\frac{1}{x}}$$ |
| has a jump discontinuity at the origin, since |
has a jump discontinuity at the origin, since |
| $$\lim_{x\to 0-}f(x) = 1\quad \mathrm{and}\quad \lim_{x\to 0+}f(x) =0.$$ |
$$\lim_{x\to 0-}f(x) = 1\quad \mathrm{and}\quad \lim_{x\to 0+}f(x) =0.$$ |
| Indeed, |
Indeed, |
| \begin{itemize} |
\begin{itemize} |
| \item if\, $x \to 0-$,\, then\, $\displaystyle \frac{1}{x} \to -\infty$,\; |
\item if\, $x \to 0-$,\, then\, $\displaystyle \frac{1}{x} \to -\infty$,\; |
| $\displaystyle e^\frac{1}{x} \to 0$,\; |
$\displaystyle e^\frac{1}{x} \to 0$,\; |
| $\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 1$; |
$\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 1$; |
| \item if\, $x \to 0+$,\, then\, $\displaystyle \frac{1}{x} \to \infty$,\; |
\item if\, $x \to 0+$,\, then\, $\displaystyle \frac{1}{x} \to \infty$,\; |
| $\displaystyle e^\frac{1}{x} \to \infty$,\; |
$\displaystyle e^\frac{1}{x} \to \infty$,\; |
| $\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 0$. |
$\displaystyle \frac{1}{1+e^\frac{1}{x}} \to 0$. |
| \end{itemize} |
\end{itemize} |
|
These results can be seen also from the series expansions of the function got by performing the divisions: For\, $x < 0$\, we obtain the converging alternating series
|
Obviously such a function cannot be represented by Taylor series about\, $x=0$,\, but for\, $x > 0$\, the by the series
|
| \begin{align*} |
\begin{align*} |
|
1:(1+e^{\frac{1}{x}}) = \sum_{k=0}^\infty(-1)^ke^{\frac{k}{x}},
|
\frac{1}{1+e^{\frac{1}{x}}}=\sum_{k=0}^\infty(-1)^ke^{-\frac{k}{x}},
|
| \end{align*} |
\end{align*} |
|
and for\, $x > 0$\,
|
and so
|
| \begin{align*} |
\begin{align*} |
|
1:(e^{\frac{1}{x}}+1) = \sum_{k=1}^\infty(-1)^{k+1}e^{-\frac{k}{x}}=0.
|
\lim_{x\to 0+}\sum_{k=0}^\infty(-1)^ke^{-\frac{k}{x}}=0.
|
| \end{align*} |
\end{align*} |
| |
However for\, $x < 0$\, 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\, $x \to 0-$.\, Therefore the function in question cannot be represented by the series above indicated, in that case.
|
