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Homework Help: Closed Form Summation

  1. Sep 28, 2010 #1

    How were they able to derive this?
  2. jcsd
  3. Sep 28, 2010 #2


    Staff: Mentor

    The sequence in this sum is a geometric sequence.
  4. Sep 28, 2010 #3


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    Denote the sum by [tex]s_p[/tex]. Can you find a linear relationship between [tex]s_{p+1}[/tex] and [tex]s_p[/tex]? Now compute [tex]s_{p+1} - s_p[/tex] explicitly and use the relation to solve for [tex]s_p[/tex].
  5. Sep 28, 2010 #4
    So S_p = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5+ … + + 1/2^p
    What would (1/2)*S_p look like?
    What about S_p - (1/2)*S_p?
    Now note: S_p - (1/2)*S_p = S_p(1 – ½) = (½)S_p So ask yourself what 2*(½)S_p = S_p looks like, and you should get your desired result
  6. Sep 28, 2010 #5
    Sorry the last post confused me. I mean is there a known formula for something like this?
  7. Sep 28, 2010 #6
    There is a known formula if this thing starts at 0. The question is how do I make it start at 0 rather than 1. I knew there was a way but I forgot.
  8. Sep 28, 2010 #7


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    [tex]s_p = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{p-1}} + \frac{1}{2^p}[/tex]

    [tex]\frac{1}{2}s_p = \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^p} + \frac{1}{2^{p+1}}[/tex]

    [tex]s_{p+1} = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{p-1}} + \frac{1}{2^p}+\frac{1}{2^{p+1}}[/tex]

    What relations between these expressions can you write down?
  9. Sep 28, 2010 #8
    None, I really dont see what you did from the 2nd step to the third step.
  10. Sep 28, 2010 #9
    I think subtracting them is easier to see.

    Sp = 1/2 + 1/22 + 1/23 + 1/24 + 1/25 + … + 1/2p

    ½Sp = 1/22 + 1/23 + 1/24 + 1/25 + … + 1/2p + 1/2p+1

    if we take Sp - ½Sp, which terms would cancel out? Which would be left over? I color coded it to help you see it.
    Last edited: Sep 29, 2010
  11. Sep 28, 2010 #10


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    The 3rd step is just the expansion of the definition

    [tex]s_{p} = \sum_{k=1}^p \frac{1}{2^k}[/tex]

    applied to p+1.
  12. Sep 28, 2010 #11
    Why would we subtract 1/2sp from sp? Can we just make a quick generalization? There was a trick to it that I forgot. If the summation starts at 1 and you want it to start at 0, then just take out the first term, or something like that? But now I still dont get the original question. Any other way to explain-you guys are really complicating this thing.
  13. Sep 28, 2010 #12
    Because it nearly gives us our desired result. Look at how I color coded it. Sp - ½Sp = ½ - 1/2p+1
  14. Sep 28, 2010 #13


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    We're telling you how to find the sum without referring to another result. If you know the answer for

    [tex]S_p = \sum_{k=0}^p \frac{1}{2^p},[/tex]

    then explain where you got it and we'll tell you how to use it here.
  15. Sep 28, 2010 #14
    Ok forget about this question. I'm just not seeing it. There should be an equation in the book. But now for the other question, changing the index of the summation, how can I do that? If any summation starts at i=2, and I want to make it start at 0, how would I do that?
  16. Sep 28, 2010 #15
    Don't give up on this problem yet, look at my color coded post (#9), do you see a pattern between the 1/2sp and sp?

    Consider this example for changing indexes, actually write out the terms

  17. Sep 29, 2010 #16


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    The only difference between [itex]\sum_{n=0}^\infty r^n[/itex] and [itex]\sum_{n=1}^\infty r^n[/itex] is that the first is missing the first term, [itex]r^0= 1[/itex]. To find [itex]\sum_{n=1}^\infty r^n[/itex], just find [itex]\sum_{n=0}^\infty r^n[/itex] and subtract 1.

    For example, to find [itex]]\displaytype\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n[/itex], first find [itex]\displaytype\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n= \frac{1}{1- \frac{1}{3}}= \frac{3}{2}[/itex].

    Now subtract 1: [itex]\frac{3}{2}- 1= \frac{1}{2}[/itex]

    In your original problem, with a finite sum, to find [itex]\displaytype\sum_{k=1}^p\frac{1}{2}^p[/itex], start from
    [tex]\displaytype\sum_{k=0}^p \frac{1}{2^p}= \frac{1- \left(\frac{1}{2}\right)^{p+1}}{1- \frac{1}{2}[/tex]
    [tex]= 2(1- \frac{1}{2^{p+1}})= 2- \frac{1}{2^p}[/tex]

    Now subtract 1:
    [tex]\left(2- \frac{1}{2^p}\right)- 1= 1- \frac{1}{2^p}[/tex]
    Last edited by a moderator: Sep 29, 2010
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