Step By Step Calculus » 8.1 - Summation and Product

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Synopsis
When we want to work with the sum or product of a long list of numbers, it is convenient to introduce efficient notation.
  • An example of summation notation is: \sum\limits_{i=1}^{4}i\sum\limits_{i=1}^{4}i, which translates into english as “add up the natural number ii, starting with i=1i=1 and stopping with i=4i=4, in other words 1+2+3+41+2+3+4. The general form of summation notation is:
    \displaystyle \sum\limits_{i=m}^{n}a_i = a_{m} + a_{m+1} + a_{m+2} + \dots + a_{n}, \quad \mathrm{where} \; i,\; m,\; n\in \mathbb{Z}, .
    \displaystyle \sum\limits_{i=m}^{n}a_i = a_{m} + a_{m+1} + a_{m+2} + \dots + a_{n}, \quad \mathrm{where} \; i,\; m,\; n\in \mathbb{Z}, .
  • Product notation is very similar, written as \prod\limits_{i=m}^{n}a_i=a_m \times a_{m+1}\times \cdots\times a_{n-1} \times a_n \, . \prod\limits_{i=m}^{n}a_i=a_m \times a_{m+1}\times \cdots\times a_{n-1} \times a_n \, .
  • A frequently-used case of product notation is \prod\limits_{i=1}^{n}i\prod\limits_{i=1}^{n}i. It is written simply as n!n! and called \mathbf{n}\mathbf{n} factorial. Remember that 0!=10!=1.
  • One of the basic summation formulae is \displaystyle\sum_{i=m}^ni=\frac{(m+n)(n+1-m)}{2}\displaystyle\sum_{i=m}^ni=\frac{(m+n)(n+1-m)}{2}.
The following are the basic properties of summations and products.
$ \newcommand{\T}{\rule{0pt}{4.6ex}} \newcommand{\B}{\rule[-3.8ex]{0pt}{0pt}} \begin{array}{|l|l|}\hline \T \B \textrm{Partial Sums}& \displaystyle \sum_{i=m}^n a_i=\sum_{i=m}^ka_i+\sum_{i=k+1}^na_i\quad\textrm{for any $m\leq k<n$}\\ \hline \T \B \textrm{Partial Products}& \displaystyle \prod_{i=m}^n a_i=\left(\prod_{i=m}^ka_i\right)\left(\prod_{i=k+1}^na_i\right)\quad\textrm{for any $m\leq k<n$}\\ \hline \T \B\textrm{Index Change}& \displaystyle \sum_{i=m}^n a_i=\sum_{j=0}^{n-m}a_{j+m}\quad\textrm{and}\quad \prod_{i=m}^na_i=\prod_{j=0}^{n-m}a_{j+m}\quad\textrm{for any }m\leq n\\ \hline \T \B \textrm{Linearity}& \displaystyle \sum_{i=m}^n (ca_i+db_i)=c\cdot\sum_{i=m}^na_i+d\cdot\sum_{i=m}^nb_i\quad\textrm{for any }m\leq n\\ \hline \T \B \textrm{Reversibility}\index{Properties of sum and product!Reversibility}& \displaystyle \sum_{i=m}^n a_i=\sum_{i=m}^n a_{n+m-i}\quad\textrm{and}\quad \prod_{i=m}^na_i =\prod_{i=m}^n a_{n+m-i}\\ \hline \end{array} $