Sigma (Summation) Notation
As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter \(\Sigma ,\) sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write
We could probably skip writing a couple of terms and write
which is better, but still cumbersome. With sigma notation, we write this sum as
which is much more compact.
Typically, sigma notation is presented in the form
where \(a_{i}\) describes the terms to be added, and the i is called the index. Each term is evaluated, then we sum all the values, beginning with the value when \(i = 1\) and ending with the value when \(i = n .\) For example, an expression like \(\sum_{i = 2}^{7} s_{i}\) is interpreted as \(s_{2} + s_{3} + s_{4} + s_{5} + s_{6} + s_{7} .\) Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use i, j, k, m, and n for indices.
Let’s try a couple of examples of using sigma notation.
Example 5.1
Using Sigma Notation
- Write in sigma notation and evaluate the sum of terms \(3^{i}\) for \(i = 1 , 2 , 3 , 4 , 5 .\)
- Write the sum in sigma notation:
\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} .\]
Solution
- Write
\[\begin{aligned} \sum_{i = 1}^{5} 3^{i} & = 3 + 3^{2} + 3^{3} + 3^{4} + 3^{5} \\ & = 363. \end{aligned}\] - The denominator of each term is a perfect square. Using sigma notation, this sum can be written as \(\sum_{i = 1}^{5} \frac{1}{i^{2}} .\)
Checkpoint 5.1
Write in sigma notation and evaluate the sum of terms 2i for \(i = 3 , 4 , 5 , 6 .\)
The properties associated with the summation process are given in the following rule.
Rule: Properties of Sigma Notation
Let \(a_{1} , a_{2} ,\ldots, a_{n}\) and \(b_{1} , b_{2} ,\ldots, b_{n}\) represent two sequences of terms and let c be a constant. The following properties hold for all positive integers n and for integers m, with \(1 \leq m \leq n .\)
- \[\sum_{i = 1}^{n} c = n c\]
- \[\sum_{i = 1}^{n} c a_{i} = c \sum_{i = 1}^{n} a_{i}\]
- \[\sum_{i = 1}^{n} \left(\right. a_{i} + b_{i} \left.\right) = \sum_{i = 1}^{n} a_{i} + \sum_{i = 1}^{n} b_{i}\]
- \[\sum_{i = 1}^{n} \left(\right. a_{i} - b_{i} \left.\right) = \sum_{i = 1}^{n} a_{i} - \sum_{i = 1}^{n} b_{i}\]
- \[\sum_{i = 1}^{n} a_{i} = \sum_{i = 1}^{m} a_{i} + \sum_{i = m + 1}^{n} a_{i}\]
Proof
We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.
2. We have
3. We have
□
A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers, and we use them in the next set of examples.
Rule: Sums and Powers of Integers
- The sum of n integers is given by
\[\sum_{i = 1}^{n} i = 1 + 2 + ⋯ + n = \frac{n \left(\right. n + 1 \left.\right)}{2} .\] - The sum of consecutive integers squared is given by
\[\sum_{i = 1}^{n} i^{2} = 1^{2} + 2^{2} + ⋯ + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6} .\] - The sum of consecutive integers cubed is given by
\[\sum_{i = 1}^{n} i^{3} = 1^{3} + 2^{3} + ⋯ + n^{3} = \frac{n^{2} \left(\right. n + 1 \left.\right)^{2}}{4} .\]
Example 5.2
Evaluation Using Sigma Notation
Write using sigma notation and evaluate:
- The sum of the terms \(\left(\right. i - 3 \left.\right)^{2}\) for \(i = 1 , 2 ,\ldots, 200 .\)
- The sum of the terms \(\left(\right. i^{3} - i^{2} \left.\right)\) for \(i = 1 , 2 , 3 , 4 , 5 , 6 .\)
Solution
- Multiplying out \(\left(\right. i - 3 \left.\right)^{2} ,\) we can break the expression into three terms.
\[\begin{aligned} \sum_{i = 1}^{200} \left(\right. i - 3 \left.\right)^{2} & = \sum_{i = 1}^{200} \left(\right. i^{2} - 6 i + 9 \left.\right) \\ \\ \\ & = \sum_{i = 1}^{200} i^{2} - \sum_{i = 1}^{200} 6 i + \sum_{i = 1}^{200} 9 \\ & = \sum_{i = 1}^{200} i^{2} - 6 \sum_{i = 1}^{200} i + \sum_{i = 1}^{200} 9 \\ & = \frac{200 \left(\right. 200 + 1 \left.\right) \left(\right. 400 + 1 \left.\right)}{6} - 6 \left[\right. \frac{200 \left(\right. 200 + 1 \left.\right)}{2} \left]\right. + 9 \left(\right. 200 \left.\right) \\ & = 2,686,700 - 120,600 + 1800 \\ & = 2,567,900 \end{aligned}\] - Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.
\[\begin{aligned} \sum_{i = 1}^{6} \left(\right. i^{3} - i^{2} \left.\right) & = \sum_{i = 1}^{6} i^{3} - \sum_{i = 1}^{6} i^{2} \\ \\ \\ \\ & = \frac{6^{2} \left(\right. 6 + 1 \left.\right)^{2}}{4} - \frac{6 \left(\right. 6 + 1 \left.\right) \left(\right. 2 \left(\right. 6 \left.\right) + 1 \left.\right)}{6} \\ & = \frac{1764}{4} - \frac{546}{6} \\ & = 350 \end{aligned}\]
Checkpoint 5.2
Find the sum of the values of \(4 + 3 i\) for \(i = 1 , 2 ,\ldots, 100 .\)
Example 5.3
Finding the Sum of the Function Values
Find the sum of the values of \(f \left(\right. x \left.\right) = x^{3}\) over the integers \(1 , 2 , 3 ,\ldots, 10 .\)
Solution
Using the formula, we have
Checkpoint 5.3
Evaluate the sum indicated by the notation \(\sum_{k = 1}^{20} \left(\right. 2 k + 1 \left.\right) .\)
This lesson is part of:
Integration