A calculus of the absurd
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Chapter 3 Sequences and series
3.1 Sigma notation
One way of writing long sums (commonly used at GCSE), is to use an elipsis (the \(...\) symbol). For example, if we were to write the sum of all the positive integers from \(1\) to \(21\) we could write it out in full as
\(\seteqnumber{0}{3.}{0}\)
\begin{align}
& 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 \\ \notag & + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21
\end{align}
The other way we could write it is as
\(\seteqnumber{0}{3.}{1}\)
\begin{equation}
1 + 2 + 3 + ... + 21
\end{equation}
Here we’ve used \(...\) to stand for all the terms between 2 and 21. This notation works, but there’s another way we could write this sum; using sigma notation! In this case, we could write this as
\(\seteqnumber{0}{3.}{2}\)
\begin{equation}
\sum _{i=1}^{21} i
\end{equation}
Sigma notation is not as bad as it looks! All this means (when read aloud) is “the sum of all the values of \(i\) where \(i\) starts at \(1\) and ends at \(21\) (inclusive11 i.e. we include \(1\) and \(21\))”.
In general there are three main parts to sigma notation - the place where we start counting from, the place where we finish counting and the "variable of indexation". In a more general case, we would have something of the form
\(\seteqnumber{0}{3.}{3}\)
\begin{equation}
\sum _{i=0}^{n} \left [ \text {some expression depending on $i$} \right ]
\end{equation}
This would mean that we start at \(i=0\) and find the value of whatever the expression depending on \(i\) is. Say, for the sake of example, it happened to be22 This is just an example - the expression could be
anything! \(3i^2\). In this case, we would have \(3 \cdot 0 ^2 = 0\). We would then add this to the value of the expression at \(1\) (\(3 \cdot 1^2\)), at \(2\) (\(3 \cdot 2^2\)), at \(3\) (\(3 \cdot 3^2\)), and so on (all the way to \(n\) - \(3 \cdot n^2\)).