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Algabreic Manipulation of Sigma Notation

  1. Feb 17, 2007 #1


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    Even though this question deals mostly with arithmetic and geometic series, this notation is used in linear algebra and differential geomety quite a bit so I will inquire of this matter here.

    What are the rules for algabreically dealing with sigma notation. When you change the value of an index from for example [itex]i=0[/itex] becomes [itex]i=1[/itex] how do you adjust the rest of the problem taking that transformation into account. Another question I have is how do you deal with injecting more into and taking quantities out of the sum, (passing through the sigma), and how that effects the rest of the sum. My final big concern is what conditions need to be met to combine / pull apart sums.

    Of course any more rules or comments would be greatly appreciated.

    Here is an example:
    According to http://en.wikipedia.org/wiki/Evaluating_sums [Broken] for the derivation of a general rule for an geometric series they have this proof:

    [tex] S = \sum_{i=0}^{n} ar^{i} [/tex]
    [tex] S-rS = S(1-r) = \sum_{i=0}^{n} ar^{i} - \sum_{i=0}^{n} ar^{i+1} =
    a (\sum_{i=0} ^{n} r^{i} - \sum_{i=1}^{i+1} r^{i}) = a(1-r^{n+1}) [/tex]

    I get confused when [itex] \sum_{i=0}^{n} ar^{i+1} [/itex] ends up as [itex] \sum_{i=1}^{i+1} r^{i} [/itex]. How does [itex]n[/itex] become [itex] i+1 [/itex] and [itex] i=0 [/itex] becomes [itex] i=1 [/itex].

    I become futher confused on how the sums dissappear into the answer, how the [itex]i[/itex] becomes an [itex]n[/itex].
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 17, 2007 #2


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    (The following applies to finite sums. Some things will change for infinite sums)

    Remember that a sum really is just adding together every element in a given sequence, so all the properties of addition apply: in particular, the distributive, commutative, and associative rules.

    As for indexing, when you're adding a bunch of numbers together, it doesn't matter in what order you do it, nor how you label the numbers: as long as the multiset of summands remains unchanged, the sum remains unchanged.

    For example, the (n+1) summands in the expression [itex]\sum_{i=0}^n r^{i}[/itex] are exactly the same as the (n+1) summands in the expression [itex]\sum_{i=0}^{n} r^{n - i}[/itex], so the sums must be equal.

    (p.s. you've made a typo in your proof of the geometric series formula)
  4. Feb 17, 2007 #3


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    thx that really helps: breaking it down back into addition. Ill remember that for the future
  5. Feb 17, 2007 #4
    I have made a small edit to that wiki page now. That should clear things up a bit.

    -- AI
  6. Feb 17, 2007 #5


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    How would things change if the series were an infinite series?
  7. Feb 17, 2007 #6


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    When an infinite sum is absolutely convergent, it tends to (always?) behave nicely. But otherwise, the order you do the addition matters: you can get different values by summing terms in different orders.

    One example of this is if I'm adding infinitely many 1's and infinitely many (-1)'s.

    I could alternate one of each, giving the sum

    [tex]\sum_{n = 0}^{+\infty} (1 + (-1)) = \sum_{n = 0}^{+\infty} 0 = 0[/tex]

    Or I could alternate between two 1's followed by a (-1):

    [tex]\sum_{n = 0}^{+\infty} (1 + 1 + (-1)) = \sum_{n = 0}^{+\infty} 1 = +\infty[/tex]

    Or, I could add up the 1's, then add up the (-1)'s, and find it doesn't exist:
    [tex]\sum_{n = 0}^{+\infty} 1 + \sum_{n = 0}^{+\infty} (-1) = +\infty + -\infty = \mathrm{D.N.E.}[/tex]

    One can get this effect simply by reordering a summation too, without doing the regrouping I did above: see the Riemann series theorem.
  8. Feb 17, 2007 #7


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    I had an inkling it was incorrect, i just didnt have the confidence to change it
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