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Sigma Notation Help?

  1. Mar 27, 2011 #1
    Express this in sigma notation?
    3^3 - 3^4 + 3^5 - ... - 3^100

    Evaluate these two sigmas?
    n
    ∑ (i-2)^2
    i =1

    n
    ∑ (4-i^2)
    i =1

    I don't really understand sigma notation so I'm really interested in the process and explanation of how to do it. Any help would be greatly appreciated!
     
  2. jcsd
  3. Mar 28, 2011 #2

    Stephen Tashi

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    Science Advisor

    Did your course materials cover how to add up the squares of consecutive integers?
    Do they give a formula for
    [tex] \sum_{i=1}^n i^2 [/tex]
     
  4. Mar 28, 2011 #3
    yes.

    n
    ∑ (i^2) = n(n+1)/2
    i =1

    How do I apply this? :S
     
  5. Mar 28, 2011 #4

    Stephen Tashi

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    For example

    [tex] \sum_{i=1}^n 3i^2 + 2i + 1 [/tex]

    The summation can be distributed to obtain

    [tex] = 3 \sum_{i=1}^n i^2 + 2 \sum_{i=1}^n i + \sum_{i=1}^n 1 [/tex]

    [tex] = 3 ( n (n+1)/2) + 2 \sum_{i=1}^n i + n [/tex]

    The materials probably give the the formula for [tex] \sum_{i=1}^n i [/tex] and you can use it also.

    In the problems you asked about, you need to multiply out the expressions like [tex] (i-2)^2 [/tex] before you distribute the summation.
     
  6. Mar 28, 2011 #5
    Alright, I think I understand what you're saying. What about the first question with expressing 3^3 - 3^4 + 3^5 - ... - 3^100 as sigma notation? Is there a formula that could be used to acquire the expression for the sigma notation?
     
  7. Mar 28, 2011 #6

    Stephen Tashi

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    Science Advisor

    No, there isn't a mechanical procedure for determining the formula (the function of i) that would be used in the sigma notation to get that series. You have to do it by trial and error. The terms alternate, so it could involve a negative number raised to a power. You can throw in a factor of [tex] (-1)^i [/tex] or [tex] (-1)^{i+1} [/tex] in the formula if it is needed to make the signs come out correctly. The things that are increasing by 1 in each term should tell you that the formula involves [tex]i [/tex] as an exponent. It looks like the first exponent starts as 3 not as 1. If your materials want you to start all summations with [tex] i = 1 [/tex] then use the expression [tex] i+2 [/tex] in the exponent to make the first term have an exponent of 3.
     
  8. Mar 28, 2011 #7
    This is wrong. n(n+1)/2 is the sum of the first n integers.

    To write the first in sigma notation, just look at the pattern. you've got a coefficient that is oscliating between -1 and 1, you have some odd numbers in the denominator. Just try to find a pattern.
     
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