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Sigma notation

Sigma notation is often used to write complicated sums in a concise and compact way. One of the most common forms is

For instance, if we wanted to represent the sum of all integers between 1 and 100 we would write:

here we are taking j as summation index, and we are adding f⁢(j)=j where j runs over all the integers starting at 1 and ending at 100.

Now suppose we wanted to represent the sum 42+52+62+⋯+202. The straightforward way to convert it to sigma notation is

However, it must be noted that such representation is not unique. For instance, here is another way to express the same summation:

If we wanted to sum all positive even numbersMathworldPlanetmath not greater than 50 we would write

since if we substitute k=1,2,3,…,25 into f⁢(k)=2⁢k we get 2,4,6,8,…,50. Now, if we wanted to write all odd numbersMathworldPlanetmath not greater than 50 we would write

∑k=125(2⁢k-1) or ∑k=024(2⁢k+1).

It must be noted that, although the running variable usually takes integer values, the summation function needs not, and it can lie on any algebraic structurePlanetmathPlanetmath where a sum is defined. So, we can write ∑j=110j for representing 1+2+⋯+10 even though the summing terms aren’t integers. If we wanted to sum all the fifth roots of unityMathworldPlanetmath (complex numbersMathworldPlanetmathPlanetmath) we could write ∑k=04e2⁢π⁢i⁢k/5.

There are several variations to the notation. Often the starting and ending values for the running parameter are ommited if the set of values it can take is not relevant or is understood from context. So on some contexts one could see the sum

written as ∑k12k where the summation is understood from context to be done over all positive integers.

Also notice another variation in the preceding example: the upper limit is ∞, which means the sum doesn’t stop after some terms. In other words, the previous example represents the sum

where there are an infiniteMathworldPlanetmathPlanetmath number of summands.

Another variation is to give further specifications for the allowed values of the running parameter. So, if we wanted to sum if we wanted to sum the reciprocals of all prime numbersMathworldPlanetmath we could simply write

where we also use the previous variation where omitting the omiting starting and stopping values indicates summing over all possible values the context allows. Now, if we wanted to sum all prime numbers between 1 and 50 we have choices:

∑p=1p⁢prime50p,∑p⁢primep≤50p.

Evaluating sums

We have also a “changing limits” property, which we used to give two different expressions for 42+52+62+⋯+202:

∑k=abf⁢(k)=∑k=a+rb+rf⁢(k-r) (4)

where r is any integer.

As example, supose we wanted to sum all the values for x2+2⁢x+1 when x runs from 3 to 7. That is, we wish to evaluate ∑j=37j2+2⁢j+1. By using the mentioneds properties we have

∑(j2+2⁢j+1) =∑j2+∑2⁢j+∑1
=∑j2+2⁢∑j+∑1

where context indicates that in all previous sums, j runs from 3 to 7. The problem now is reduced to find formulas for the sums in the last expression.

Notice also that, since (x2+2⁢x+1) is the same as (x+1)2 we could have also done the following changes:

∑j=37(j2+2⁢j+1) =∑j=37(j+1)2
=∑j=48j2.

and again we are left with the task of evaluating a simpler summation.

Useful formulas

We now give formulas for evaluating many common summations, which can be combined using the mentioned properties to evaluate a wide range of sums.

Constants. The simplest case is when the summation terms do not involve the running variable. Two examples are:

which respectively represent the sums 2+2+2+2 and x2+x2+x2+x2. It’s obvious that if the summand does not depend on the running variable, all terms will be the same, and thus the sum will be the productPlanetmathPlanetmath of any summand by the nuber of summands. In other words,

Remark: If a summand does not depend on the summation index, we say it is constant (with respect the summation). So in the previous example x2 was “constant” since it didn’t depend on the running variable k, and thus

Small powers.Suppose we wanted to calculate 1+2+3+⋯+98+99+100. In other words, we want to calculate ∑k=1100k. We can use the formula

which is credited to Gauss. Applying such formula we find that the sought answer is 100⁢(101)/2=5050. Notice that we can use the formula to evaluate sums of consecutive integers not necessarily at 1. For instance, if we wanted50+51+52+⋯+77 we could proceed as following:

∑k=5077k=50+51+52+⋯+77 =(1+2+⋯+77)-(1+2+⋯+49)
=∑k=177k-∑k=149k
=77⋅782-49⋅502
=3003-225=1778

Similar formulas exist for evaluating sums of small powers of consecutive integers:

∑k=1nk2 =n⁢(n+1)⁢(2⁢n+1)6
∑k=1nk3 =n2⁢(n+1)24
∑k=1nk4 =n⁢(n+1)⁢(2⁢n+1)⁢(3⁢n2+3⁢n-1)30
∑k=1nk5 =n2⁢(n+1)2⁢(2⁢n2+2⁢n-1)12
∑k=1nk6 =n⁢(n+1)⁢(2⁢n+1)⁢(3⁢n4+6⁢n3-3⁢n+1)42
∑k=1nk7 =n2⁢(n+1)2⁢(3⁢n4+6⁢n3-n2-4⁢n+2)24
∑k=1nk8 =n⁢(n+1)⁢(2⁢n+1)⁢(5⁢n6+15⁢n5+5⁢n4-15⁢n3-n2+9⁢n-3)90
∑k=1nk9 =n2⁢(n+1)2⁢(n2+n-1)⁢(2⁢n4+4⁢n3-n2-3⁢n+3)20
∑k=1nk10 =n⁢(n+1)⁢(2⁢n+1)⁢(n2+n-1)⁢(3⁢n6+9⁢n5+2⁢n4-11⁢n3+3⁢n2+10⁢n-5)66

The corresponding sum a+(a+d)+(a+2⁢d)+⋯+(a+n⁢d) can be written with sigma notation as∑k=0n(a+k⁢d). We can use all formulas we have so far to calculate it:

∑k=0n(a+k⁢d) =∑k=0na+∑k=0nk⁢d
=(n+1)⁢a+d⁢∑k=0nk
=(n+1)⁢a+d⁢n⁢(n+1)2=(n+1)⁢(d⁢n+2⁢a)2

If a0,a1,a2,…,an represent the terms of the progression, then we can also rewrite the last result as

∑k=0n(a+k⁢d)=(n+1)⁢(d⁢n+2⁢a)2=(n+1)⁢(a0+an)2

A particular case of arithmetic sequence is summing consecutive odd (or consecutive even ) numbers, since the common difference is always 2.

∑k=1n2⁢k=2⁢∑k=1n=n⁢(n+1) ⁢even numbers
∑k=1n(2⁢k-1)=2⁢∑k=1nk-∑k=1n1=n⁢(n+1)-n=n2 ⁢odd numbers

The corresponding sum a+a⁢r+a⁢r2+a⁢r3+⋯+a⁢rn can be reduced to calculating 1+r+r2+⋯+rn by factoring a out of the summation. The corresponding formula is

∑k=0nrk=1+r+r2+⋯+rn=rn+1-1r-1

A particularly nice formula is obtained when r=2:

So, in the old story about a chess board where the board is filled doubling the number of seeds on the previous position, the answer would be

∑k=0632k=264-1=18446744073709551615.

Other formulas.Many other formulas can be found to evaluate sums. Here is a small miscellanea of remarkable formulas.

∑k=1nk⁢(k+1)=n⁢(n+1)⁢(n+2)3
∑k=1nk⁢(k+1)⁢(k+2)=n⁢(n+1)⁢(n+2)⁢(n+3)4
∑k=1nk⁢(k+1)⁢(k+2)⁢(k+3)=n⁢(n+1)⁢(n+2)⁢(n+3)⁢(n+4)5
∑k=0n(nk)=2n
∑k=0n(-1)k⁢(nk)=0
∑k=0nk⁢(nk)=n⁢2n-1
∑k=0nk2⁢(nk)=n⁢(n+1)⁢2n-2
∑k=0nk3⁢(nk)=n2⁢(n+3)⁢2n-3
∑k=0(nk)2=(2⁢nn)

If fn denotes the n-th Fibonacci numberMathworldPlanetmath, then

∑k=0nfk=fn+2-1
∑k=0nfk2=fn⁢fn+1
∑k=0⌊n/2⌋(n-k-1k)=fn

Notes

The sigma notation was introduced by the French mathematician Joseph Fourier in 1820 [1].

References