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Finite Sum - Modified Geometric Series

  1. Feb 26, 2008 #1

    cepheid

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    Does anyone know how to evaluate

    [tex] S_n = \sum_{i=0}^{n-1} i2^i [/tex]​

    I tried the following. Let r = 2, and figure out the terms in

    [tex] S_n - rS_n [/tex]​

    Unlike with a regular geometric series, this does not make all but two of the terms disappear. But it does make all but one of the terms turn into a simple power of 2 (once you collect like powers of 2). In other words, it turns into something plus a regular geometric series. For my final answer, solving for S_n, I got:

    [tex] S_n = (n-2)2^n + 2 [/tex]​

    but I have reason to believe this is incorrect. Can anybody help me out?
     
  2. jcsd
  3. Feb 26, 2008 #2

    CRGreathouse

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    I think you're right. Check yourself against http://www.research.att.com/~njas/sequences/A036799 [Broken] if you like (but watch the offset!).
     
    Last edited by a moderator: May 3, 2017
  4. Feb 26, 2008 #3

    mathman

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    Using elemtary calculus I got (n-1)2n+2
     
  5. Feb 26, 2008 #4
    Perhaps, it is unnecessary but one way is to use the properties of the z transform. The sum could be equal to the z transform with z=1. I’ll check later. There are two time domain operations performed on the infinite geometric series. They are differentiation and multiplication by a rec function. These operations have equivalents in the z domain.
     
  6. Feb 26, 2008 #5

    cepheid

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    The online encyclopedia of integer sequences? I had no idea that it existed. My answer looks consistent with theirs.

    Let n = 4, and use their formula (we should get 98)

    [tex] 2+2^5(4-1) = 2+32(3) = 2+ 96 = 98 [/tex]

    Let n-1 = 4 and use MY formula (we should get 98)

    [tex] 2 + 2^n(n-2) = 2 + 2^5(3) = 98 [/tex]

    Edit: That confirms my answer. Thanks for the link CRGreathouse!
     
    Last edited by a moderator: May 3, 2017
  7. Feb 26, 2008 #6

    cepheid

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    Exactly what method did you use?
     
  8. Feb 26, 2008 #7

    cepheid

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    Right, from what I can remember, the z-transform of a sequence a(n) is a function of a complex variable f(z) such that {a(n)} are just the coefficients (in reverse order) of the Laurent series expansion of f(z) (about zero?)

    [tex] f(z) = \sum_{n=-\infty}^{\infty} a(n)z^{-n} [/tex]

    And you were thinking of the property that:

    [tex] \frac{d}{dz}f(z) = - \sum_{n=-\infty}^{\infty} na(n)z^{-n-1} = - z^{-1}\sum_{n=-\infty}^{\infty} na(n)z^{-n} = -z^{-1}\mathcal{Z}\{na(n)\} [/tex]

    Or in other words

    [tex] \mathcal{Z}\{na(n)\} = -z\frac{d}{dz}\mathcal{Z}\{a(n)\} [/tex]

    And so my series would be:

    [tex] -(z)\frac{d}{dz}\mathcal{Z}\{1\} [/tex]

    evaluated at

    [tex] z = 2^{-1} [/tex]

    EDIT: But I just realized that these are INFINITE series, and I'm dealing with a FINITE series, so I don't think that any of this is relevant. AARRGH!
     
    Last edited: Feb 26, 2008
  9. Feb 26, 2008 #8

    Hurkyl

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    Any finite series can be extended into an equal infinite series....
     
  10. Feb 26, 2008 #9

    cepheid

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    Right, of course. I was being stupid. Of course you can take the z transform of a finite sequence. I've done it many times. Formally, you let a(i) = 1 for 0 <= i <= n-1, and a(i) = 0 everywhere else.

    Then the sum you evaluate gives you:

    [tex] \mathcal{Z}\{a(i)\} = \frac{1-z^{-n}}{1-z^{-1}} [/tex]

    Then you differentiate this wrt z and then you multiply that by -z and then you plug in z = 1/2. I did that and got the same answer as in my original post.

    So the answer has been verified by two separate methods.
     
  11. Feb 27, 2008 #10

    mathman

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    f(x)=sum(0,n-1)xk=(xn-1)/(x-1)

    f'(x)=sum(0,n-1)kxk-1={(x-1)nxn-1-(xn-1)}/(x-1)2

    Note in the above sum, the k=0 term is 0.

    Desired sum=2f'(2)=(n-2)2n+2

    (sorry for my mistake!!)
     
  12. Feb 22, 2009 #11
    This is the sum of a Arithmetic-Geometric Series:

    S_n=n2^n-2^(n+1)+2
     
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