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Sigma (or
) Notation
In many branches of Mathematics, especially applied Mathematics and/or Statistics, it routinely occurs that one wants to talk about sums of large numbers of measurements of some quantity. For example, one might one to find the average value of a set of readings, or one might wish to know how a set of readings deviates from its average.
Instead of writing out an expression containing thousands, or millions, or billions, of summands, scientists have developed a shorthand nota- tion:
Instead of 10 + 20 + 30 + 40 + · · · + 5270 + 5280 we write
i= 1
10 i,
In general, instead of
am + am+ 1 + am+ 2 + am+ 3 + · · · + an− 3 + an− 2 + an− 1 + an, we write
i∑=n
i=m
ai or, even more briefly,
∑^ n
i=m
ai;
instead of
xm + xm+ 1 + xm+ 2 + xm+ 3 + · · · + xn− 3 + xn− 2 + xn− 1 + xn,
we write
j ∑=n
j=m
xj or
∑^ n
j=m
xj.
The notation
k∑=n
k=m
f (k) consists of four components:
first , there is the Σ which tells us that a sum is to be taken over a range of values,
second , there is the k = m under the Σ which tells us that the summa- tion index is k and that k starts out at the lower summation limit m ,
third , there is the k = m , or just m over the Σ which tells us that k stops at the upper summation limit n ,
fourth , there is the f (k) which gives us a formula for the summands in terms of the summation index k.
It is required that the summation limits be integers and that the summa- tion index increases by 1 as it runs through all integer values between the lower and upper summation limits.
The number of summands is n − m + 1 , and the average value of the summands is ∑ n i=m f (i) n − m + 1
f (m) + f (m + 1 ) + f (m + 2 ) + · · · + f (n − 2 ) + f (n − 1 ) + f (n) n − m + 1
Such averages are often denoted using the Greek letter μ (pronounced “mew”).
The lower case Greek letter σ is pronounced “sigma”. The upper case Greek letter Σ is pronounced exactly the same, but it can, and often is, read as “the sum of... as i runs from m to n”.
Examples
∑^10
i= 1
i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
∑^10
i= 1
i^2 = 1 2 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 =
∑^10
i= 1
S n^2 =
∑^ n i= 1
i^2 = 12 + 22 + 32 + · · · + (n − 1 )^2 + n^2 = n(n + 1 )( 2 n + 1 ) 6
S n^3 =
∑^ n
i= 1
i^3 = 13 + 23 + 33 + · · · + (n − 1 )^3 + n^3 =
(n(n + 1 ) 2
Also, it is useful to know the basic formula for a
Geometric Progression :
∑^ n
i= 0
x i^ = x^0 + x^1 + x^2 + x^3 + · · · + x n−^2 + x n−^1 + x n^ = x n+^1 − 1 x − 1
or ∑^ n
i= 0
x i^ = 1 + x + x^2 + x^3 + · · · + x n−^2 + x n−^1 + x n^ = x n+^1 − 1 x − 1
Properties
∑^ n
i=m
(ai + bi ) =
∑^ n
i=m
ai +
∑^ n
i=m
bi
∑n
i=m
cai = c
∑n i=m
ai
∑n
i=m
ai =
∑^ p i=m
ai +
∑^ n i=p
ai if m ≤ p ≤ n
Note that the ranges of summation in first two equations are identical. If they are different much extra care must be taken.
Examples:
Sums are usually evaluated by reducing them to one or more of the above forms by algebraic manipulation:
∑^6
i= 0
2 i^ = 1 + 2 + 4 + 8 + 16 + 32 + 64 =
26 +^1 − 1
∑^30
i= 0
5 i+^2 =
∑^30
i= 1
5 i 52 = 25
∑^30
i= 1
5 i^ = 25
530 +^1 − 1
∑^40
i= 0
4 −i+^2 =
∑^40
i= 1
4 −i 42 = 16
∑^40
i= 1
)i = 16
4
4
4 −^41 − 1
−^34
( 1 − 4 −^41 )
∑^40
i= 20
i^2 =
∑^40
i= 1
i^2 −
∑^19
i= 1
i^2 =
= 20 ( 41 )( 27 ) − 19 ( 10 )( 13 ) = 10 [ 2 ( 41 )( 27 ) −
19 ( 13 )] =
10 [ 2214 − 247 ] = 10 ( 1967 ) = 19 , 670
Telescoping Sums
It often happens that sums can be collapsed nicely, for example:
∑^ n
k= 1
(k + 1 )m^ − km^ =
( 2 m− 1 m^ )+( 3 m− 2 m^ )+( 4 m− 3 m^ )+· · ·+(n m−(n− 1 )m^ )+((n+ 1 ) m−n m^ ) =
(n + 1 )m^ − 1
In general, we have a sum of the form
∑^ n
i= 1
(ai+ 1 −a 1 ) = (a 2 −a 1 )+(a 3 −a 2 )+· · ·+(an− 1 −an− 2 )+(an −an− 1 )+
(an+ 1 − an ) = an+ 1 − a 1
Another possibility is that although the two sums have the same number of terms, the ranges are offset. For example, we might wish to evaluate
i∑=n
i= 1
ai +
i=∑n+ 2
i= 3
ai.
Note that both sums have the same number of terms.
We have to adjust one of the sums so that its summation range equals
that of the other. We will choose to work on
n∑+ 2
i= 3
ai and we will introduce
a new summation index k = i − 2, so that k = 1 when i = 3, and k = n when i = n + 2.
We of course can solve for i in terms of k: i = k + 2, so we replace every
occurrence of i in
n∑+ 2
i= 3
ai with k + 2:
i=∑n+ 2
i= 3
ai =
k+ (^2) ∑=n+ 2
k+ 2 = 3
ak+ 2 =
k∑=n
k= 1
ak+ 2.
Now it doesn’t matter what symbol we use for our summation index, so we write: k∑=n
k= 1
ak+ 2 =
i∑=n
i= 1
ai+ 2.
Using this in our original expression, we get i∑=n
i= 1
ai +
i=∑n+ 2
i= 3
ai =
i∑=n
i= 1
ai +
i∑=n
i= 1
ai+ 2 =
i∑=n
i= 1
(ai + ai+ 2 ).
