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Advanced summation (sigma)

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    xObic.png
    2. Relevant equations
    3. The attempt at a solution
    Obviously I don't need a solution because it's right there. What I need to understand is what happened after the third equation sign and more importantly, how would I learn to solve these kinds of problems on my own. I looked at Wikipedia and even though it was helpful, I still don't understand this specific exercise for example. I would be grateful if someone could point me at a resource on this topic.
     
  2. jcsd
  3. Sep 11, 2011 #2

    gb7nash

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    Homework Helper

    So you're wondering how to go from:

    [tex]\frac{1}{2}(\frac{1}{1}-\frac{1}{3}) + \frac{1}{2}(\frac{1}{3}-\frac{3}{5}) + ... + \frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})[/tex]

    to:

    [tex]\frac{1}{2}(1-\frac{1}{2n+1})[/tex]

    Factor a 1/2 out of all the terms and then add the stuff inside. Something interesting happens. For example:

    (1/2)(a+b) + (1/2)(c+d) = (1/2)(a+b+c+d)
     
  4. Sep 11, 2011 #3
    Ok so everything but 1/1 and -1/(2n+1) cancel each other out. Now I get this exercise but what about the bigger question? Are there any tricks to learn here (about the sigma) or is it just about using your knowledge of algebra?
     
  5. Sep 11, 2011 #4

    gb7nash

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    Homework Helper

    It depends on the problem, but a lot of times you can use mathematical induction to prove summation results. There's an example of it on wiki:

    http://en.wikipedia.org/wiki/Mathematical_induction

    This assumes that you know what you're trying to prove ahead of time (like in the case of your example). However, if you want to know a general way of finding the sum of n terms, there's no standard way, as far as I know. The best you can usually do is some trick like the one I showed you, or see if you're looking at a geometric sum. Otherwise, memorize these:

    http://en.wikipedia.org/wiki/Summation
     
  6. Sep 11, 2011 #5
    Partial fractions:

    [tex]\frac{1}{(2x-1)(2x+1)} = \frac{1/2}{2x-1} - \frac{1/2}{2x+1}[/tex]
     
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