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Sigma and Pi Notation

April 1, 2010

Sigma Notation

\sum is a capital letter from the Greek alphabet called “Sigma”… it corresponds to “S” in our alphabet (think of the starting sound of the word “sigma”). It is used in mathematics to represent the sum of a bunch of terms (think of the starting sound of the word “sum”: Sssigma = Sssum).

Sigma can be used all by itself to represent a generic sum… the general idea of a sum, of an unspecified number of unspecified terms:

\displaystyle\sum a_i~=~a_1+a_2+a_3+...

But this is not something that can be evaluated to produce a specific answer, as we have not been told how many terms to include in the sum, nor have we been told how to determine the value of each term.

A more typical use of Sigma notation will include an integer below the Sigma (the “starting term number”), and an integer above the Sigma (the “ending term number”). In the example below, the exact starting and ending numbers don’t matter much since we are being asked to add the same value, two, repeatedly. All that matters in this case is the difference between the starting and ending term numbers… that will determine how many twos we are being asked to add, one two for each term number.


Sigma notation is not usually worth the extra ink to describe simple sums such as the one above… multiplication could do that more simply.

Sigma notation is most useful when the “term number” can be used in some way to calculate each term. To facilitate this, a variable is usually listed below the Sigma with an equal sign between it an the starting term number. If this variable appears in the expression being summed, then the current term number should be substituted in for the variable:


Note that it is possible to have a variable below the Sigma, but never use it. In such cases, just as in the example that resulted in a bunch of twos above, the term being added never changes:


The “starting term number” need not be 1. It can be any value, including 0. For example:


That covers what you need to know to begin working with Sigma notation. However, since Sigma notation will usually have more complex expressions after the Sigma symbol, here are some further examples to give you a sense of what is possible:





Note that the previous example illustrates that, using the commutative property of addition, a sum of multiple terms can be broken up into multiple sums:


And lastly, this notation can be nested:

\displaystyle\sum_{i=1}^{2}\displaystyle\sum_{j=4}^{6}(3ij)\\*~\\*=\displaystyle\sum_{i=1}^{2}(3i\cdot4+3i\cdot5+3i\cdot6)\\*~\\*=(3\cdot1\cdot4+3\cdot1\cdot5+3\cdot1\cdot6)+ (3\cdot2\cdot4+3\cdot2\cdot5+3\cdot2\cdot6)

Pi Notation

\prod┬áis a capital letter from the Greek alphabet call “Pi”… it corresponds to “P” in our alphabet (think of the starting sound of the word “pi”). It is used in mathematics to represent the product of a bunch of terms (think of the starting sound of the word “product”: Pppi = Ppproduct). It is used in the same way as the Sigma notation described above, except that succeeding terms are multiplied instead of added:



\displaystyle\prod_{i=1}^{2}\displaystyle\prod_{j=4}^{6}(3ij)\\*~\\*=\displaystyle\prod_{i=1}^{2}((3i\cdot4)(3i\cdot5)(3i\cdot6))\\*~\\*=((3\cdot1\cdot4)(3\cdot1\cdot5)(3\cdot1\cdot6)) ((3\cdot2\cdot4)(3\cdot2\cdot5)(3\cdot2\cdot6))


Sigma and Pi notation are used in mathematics to indicate repeated addition or multiplication. Sigma notation provides a compact way to represent many sums, and is used extensively when working with Arithmetic or Geometric Series. Pi notation provides a compact way to represent many products.

To make use of them you will need a “closed form” expression (one that allows you to describe each term’s value using the term number) that describes all terms in the sum or product (just as you often do when working with sequences and series). Sigma and Pi notation save much paper and ink, as do other math notations, and allow fairly complex ideas to be described in a relatively compact notation.

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16 Comments leave one →
  1. Douglas maindo permalink
    February 3, 2012 7:46 am

    am very thankful 2 the information is very helpful to me

  2. February 17, 2012 3:49 pm

    I always see these equations on in technical papers but I never knew how to decode them. This was so helpful! It’s basically a for loop in scripting, makes so much sense. Also your blog is awesome, thank you for sharing!

    • February 17, 2012 4:25 pm

      Thank you! And yes, a little programming experience with loops makes Sigma and Pi Notation much easier to understand…

  3. Sundaram permalink
    September 3, 2012 8:23 am

    Very very useful. Thanks a lot – Sundaram

  4. Anonymous permalink
    November 10, 2012 9:18 am

    Thank you, this was very helpful. I was finding how to use Sigma notation, and finally found such a good one.

  5. Samama Fahim permalink
    July 11, 2013 7:20 am

    Indeed a very lucid exposition of Sigma and Pi notations! Thanks :)

  6. August 26, 2013 6:41 pm

    Very useful post. But what if the Pi notation is not in closed form, such as


    • August 26, 2013 9:33 pm

      If the index limit above the Pi symbol is a variable, as in the example you gave:
      then there are an indeterminate number of factors in the product until such time as “n” is specified.

      I suppose a problem could be posed this way if you are being asked to come up with an expression for such a product that does not involve Pi notation: is there some closed form expression involving “n” that represents this product?

      So, if n=3, then
      and if n=4, then
      and if you leave the final index as “n” becomes:
      =(1-\dfrac{1}{2})\cdot(1-\dfrac{1}{3})\cdot(1-\dfrac{1}{4})\cdot ... \cdot (1-\dfrac{1}{n})
      =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot ... \cdot\dfrac{n-1}{n}

      Is there some closed form expression that represents this product?

  7. May 14, 2014 6:23 am

    what is the relation between the two when they are logarithmic differentiated?

    • May 14, 2014 7:46 am

      It would help if you could provide an example of what you are asking about.

      If you need to differentiate a sum, I would not expect logarithmic differentiation to be very useful, as the laws of logarithms do not allow us to do anything with something like
      ln(y)=ln(\displaystyle\sum _i x^{ix})
      Differentiating this would turn the right side into the reciprocal of the original sum times its derivative = a mess.

      However, if you need to differentiate a product, logarithmic differentiation could make life simpler by converting a long succession of product rule applications into a sum of logs. Since
      we can rewrite the log of a product as a sum of logs:
      ln(y)=ln(\displaystyle\prod _i x^{ix})\\*\\ln(y)=\displaystyle\sum_i ln(x^{ix})
      which, in many cases, could simplify the differentiation process.

  8. Imad Ahmad permalink
    May 17, 2014 9:21 pm

    If sigma is for summation, and pi is for multiplication, are there any notations for division and subtraction? Just out of curiosity?

    • May 18, 2014 12:52 am

      Good question!

      Subtraction can be rewritten as the addition of a negative. So Sigma notation describes repeated subtraction when its argument is a negative quantity.

      Division can be rewritten as multiplication by the reciprocal. So Pi notation describes repeated division by the reciprocal of its argument.

      Therefore, additional notations are not needed to describe repeated subtraction or division… Which is quite convenient.

  9. john walter permalink
    July 31, 2014 4:10 am

    Sir, how about expressing thing one 1x2x3 + 2x3x4 + 3x4x5 + ….will it be a combination of
    sigma and pi? If you can illustrate it please.

    • July 31, 2014 7:26 am

      You are correct – this can be represented using a combination of Sigma and Pi notation:

      \displaystyle\sum_{i=1}^{N} \displaystyle\prod_{j=0}^{2}(i+j)

      In the above notation, i is the index variable for the Sum, and provides the starting number for each product. By having the Product index variable start at zero, the expression to generate each value is a bit simpler. If j went from one to three each time, the expression on the right would have to be (i + j – 1).


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