empty sumEmptySum2009-01-01 12:12:252009-01-01 13:15:35Topicsummation% this is the default PlanetMath preamble. as your knowledge
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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}
\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:
$$(-n)a\!+\!na \,=\, (-n\!+\!n)a = 0a$$
This empty sum equals the additive identity 0 of the ring, since the multiple $(-n)a$ is defined to be
$$\underbrace{(-a)\!+\!(-a)\!+\ldots+\!(-a)}_{n\; \mathrm{copies}}$$
\item In using the \PMlinkname{sigma notation}{Summing}
\begin{align}
\sum_{i=m}^nf(i)
\end{align}
one sometimes sees a case
\begin{align}
\sum_{i=m}^{m-1}f(i).
\end{align}
It must be an empty sum, because in
\begin{align}
\sum_{i=m}^mf(i)
\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.
\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
\begin{align}
\sum_{i=m}^nf(i)+\sum_{i=n+1}^kf(i) \;=\; \sum_{i=m}^kf(i)
\end{align}
would be true also in such a cases, then one has to define
$$\sum_{i=m}^nf(i) \;=\; -\sum_{i=n+1}^{m-1}f(i) \qquad\qquad(n < m\!-\!1),$$
because by (4) one could calculate
$$0 \,=\, -\sum_{i=n+1}^{m-1}f(i)+\sum_{i=n+1}^{m-1}f(i) \,=\,\sum_{i=m}^nf(i)+\sum_{i=n+1}^{m-1}f(i)
\,=\, \sum_{i=m}^{m-1}f(i).$$