The sigma symbol in Math appears when we want to use sigma notation.
Sigma notation is used in Math usually when one wants to represent a situation where a number of terms are to be added up and summed.
So the notation can be helpful in writing long sums in much a much shorter and clearer way. Sometimes this notation can also be called summation notation.
The symbol used in these situations is the Greek letter sigma.
Say you want to sum up a finite list or sequence of n terms: a1 + a2 + a3 + …….. + an
We can let ai stand for a general term in the sequence.
Then using notation with sigma write:
\displaystyle \sum_{i=1}^n \space a_i
This means that we sum up the ai terms from 1, up to n.
For example n = 5:
\displaystyle \sum_{i=1}^5 \space a_i \space=\space a_1 + a_2 + a_3 + a_4 + a_5
The summation doesn’t always have to start at i = 1.
We could also have:
Examples
1.1
\displaystyle \sum_{b=1}^4 \space b \space= ?
Here we just sub in the numbers 1 to 4 where ever there is a b, as the function is only b, it’s the case that just the numbers themselves are summed up.
\displaystyle \sum_{b=1}^4 \space a_i \space=\space 1 + 2 + 3 + 4 \space=\space 10
1.2
\displaystyle \sum_{b=1}^4 \space 2b \space= ?
Here we input the numbers 1 to 4 where there is a b, but each number is multiplied by 2.
\displaystyle \sum_{b=1}^4 \space 2b \space=\space 2(1) + 2(2) + 2(3) + 2(4) \space=\space 2 + 4 + 6 + 8 \space=\space 20
1.3
\displaystyle \sum_{n=2}^5 \space 2^n \space= ?
This case requires each number to be used as a power/exponent of 2.
But we start with 2, instead of 1.
\displaystyle \sum_{n=2}^5 \space 2^n \space=\space 2^2 + 2^3 + 2^4 + 2^5 \space=\space 4 + 8 + 16 + 32 \space=\space 60
Sigma Symbol in Math,
Properties/Rules
Sigma symbol notation does have some properties or rules that are handy to remember at certain times.
A) \displaystyle \sum_{x=1}^n \space c\space.\space f(x) \space\space\space=\space\space\space c \sum_{x=1}^n \space f(x) where c is a constant.
Ex
\displaystyle \sum_{x=1}^3 \space 2(x^2) \space\space=\space\space 2(1^2) + 2(2^2) + 2(3^2) \space\space=\space\space 2 + 8 + 18 \space=\space 28
\displaystyle 2 \sum_{x=1}^3 \space (x^2) \space\space=\space\space 2(1^2 + 2^2 + 3^2) \space\space=\space\space 2(1 + 4 + 9) \space=\space 2(14) \space=\space 28
B) \displaystyle \sum_{x=1}^n \space [\space f(x) \space \pm \space g(x) \space] \space\space\space=\space\space\space \sum_{x=1}^n \space f(x) \space\space \pm \space\space \sum_{x=1}^n \space g(x)
Ex
\displaystyle \sum_{x=1}^2 \space 2x + x^2 \space\space=\space\space (2(1) + (1)^2) \space\space + \space\space (2(2) + (2)^2) \space\space = \space\space 3 + 8 \space=\space 11
=> \displaystyle \sum_{x=1}^2 \space 2x \space\space=\space\space 2(1) + 2(2) \space\space=\space\space 2 + 4 \space\space = \space\space 6
=> \displaystyle \sum_{x=1}^2 \space x^2 \space\space=\space\space (1)^2 + (2)^2 \space\space=\space\space 1 + 4 \space\space = \space\space 5
\displaystyle \sum_{x=1}^2 \space 2x \space\space+\space\space \sum_{x=1}^2 \space x^2 \space\space=\space\space 6 + 5 \space\space = \space\space 11
C) Where c is a constant. \displaystyle \sum_{x=1}^n \space c \space\space=\space\space c \times n
\displaystyle \sum_{x=1}^n \space 1 + 1 + 1 \space .......... \space n \space {\tt{times}}
\displaystyle \sum_{x=1}^4 \space 1 + 1 + 1 + 1 \space\space=\space\space 1 \times 4 \space=\space 4
\displaystyle \sum_{x=1}^3 \space 2 + 2 + 2 \space\space=\space\space 2 \times 3 \space=\space 6
Some Shortcuts
With sigma notation, there are some shortcuts that can be used with some specific sums.
Below are 3 of the most common.
A) \displaystyle \sum_{x=1}^n \space x \space=\space \frac{n(n+1)}{2}
The sum for \displaystyle \sum_{x=1}^n \space x is normally 1 + 2 + 3 + …… + n.
\displaystyle \sum_{x=1}^5 \space x \space=\space 1 + 2 + 3 + 4 + 5 \space=\space 15
But instead, for any such sum, the shortcut shown at A) can be used as opposed to the longer process of summing up.
Such as for the situation above summing up to 5.
B) \displaystyle \sum_{x=1}^n \space x^2 \space=\space \frac{n(n+1)(2n+1)}{6}
C) \displaystyle \sum_{x=1}^3 \space x^2 \space=\space ( \frac{n(n+1)}{2} )^2