| Version current |
Version 1 |
| The {\em empty sum} is such a borderline case of sum where the number of the addends is zero, i.e. the set of the addends is an empty set. |
The {\em empty sum} is such a borderline case of sum where the number of the addends is zero, i.e. the set of the addends is an empty set. |
|
|
| \begin{itemize} |
\begin{itemize} |
|
|
| \item One may think that the zeroth multiple $0a$ of a ring element $a$ is the empty sum; it can spring up by adding in the ring two multiples whose integer coefficients are opposite numbers: |
\item One may think that the zeroth multiple $0a$ of a ring element $a$ is the empty sum; it can spring up by adding in the ring two multiples whose integer coefficients are opposite numbers: |
| $$(-n)a\!+\!na \,=\, (-n\!+\!n)a = 0a$$ |
$$(-n)a\!+\!na \,=\, (-n\!+\!n)a = 0a$$ |
| This empty sum equals the additive identity 0 of the ring, since the multiple $(-n)a$ is defined to be |
This empty sum equals the additive identity 0 of the ring, since the multiple $(-n)a$ is defined to be |
| $$\underbrace{(-a)\!+\!(-a)\!+\ldots+\!(-a)}_{n\; \mathrm{copies}}$$ |
$$\underbrace{(-a)\!+\!(-a)\!+\ldots+\!(-a)}_{n\; \mathrm{copies}}$$ |
|
|
| \item In using the \PMlinkname{sigma notation}{Summing} |
\item In using the \PMlinkname{sigma notation}{Summing} |
| \begin{align} |
\begin{align} |
| \sum_{i=m}^nf(i) |
\sum_{i=m}^nf(i) |
| \end{align} |
\end{align} |
| one sometimes sees a case |
one sometimes sees a case |
| \begin{align} |
\begin{align} |
|
\sum_{i=m}^{m-1}f(i).
|
\sum_{i=m}^{m-1}f(i) |
| \end{align} |
\end{align} |
| It must be an empty sum, because in |
It must be an empty sum, because in |
| \begin{align} |
\begin{align} |
| \sum_{i=m}^mf(i) |
\sum_{i=m}^mf(i) |
| \end{align} |
\end{align} |
| the number of addends is clearly one and therefore in (2) the number is zero.\, Thus the value of (2) may be defined to be 0. |
the number of addends is clearly one and therefore in (2) the number is zero.\, Thus the value of (2) may be defined to be 0. |
| \end{itemize} |
\end{itemize} |
|
|
|
|
|
|
| \textbf{Note.}\, The sum (1) is not defined when $n$ is less than $m\!-\!1$, but if one would want that the usual rule |
\textbf{Note.}\, The sum (1) is not defined when $n$ is less than $m\!-\!1$, but if one would want that the usual rule |
| \begin{align} |
\begin{align} |
| \sum_{i=m}^nf(i)+\sum_{i=n+1}^kf(i) \;=\; \sum_{i=m}^kf(i) |
\sum_{i=m}^nf(i)+\sum_{i=n+1}^kf(i) \;=\; \sum_{i=m}^kf(i) |
| \end{align} |
\end{align} |
| would be true also in such a cases, then one has to define |
would be true also in such a cases, then one has to define |
| $$\sum_{i=m}^nf(i) \;=\; -\sum_{i=n+1}^{m-1}f(i) \qquad\qquad(n < m\!-\!1),$$ |
$$\sum_{i=m}^nf(i) \;=\; -\sum_{i=n+1}^{m-1}f(i) \qquad\qquad(n < m\!-\!1),$$ |
| because by (4) one could calculate |
because by (4) one could calculate |
| $$0 \,=\, -\sum_{i=n+1}^{m-1}f(i)+\sum_{i=n+1}^{m-1}f(i) \,=\,\sum_{i=m}^nf(i)+\sum_{i=n+1}^{m-1}f(i) |
$$0 \,=\, -\sum_{i=n+1}^{m-1}f(i)+\sum_{i=n+1}^{m-1}f(i) \,=\,\sum_{i=m}^nf(i)+\sum_{i=n+1}^{m-1}f(i) |
| \,=\, \sum_{i=m}^{m-1}f(i).$$ |
\,=\, \sum_{i=m}^{m-1}f(i).$$ |
|
|
