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Oh wow, a math question came up at work! Finite series

  1. Apr 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Being professionals now we've all forgotten our math skills and I'm trying to impress everyone. P=sum from k=0 to n of (x)^(m-k)*(1-x)


    Sorry for the hurried lack of latex, it's x^(m-k), and that term is multiplied by (1-x)


    2. Relevant equations
    Uh-oh


    3. The attempt at a solution

    Unfortunately this isn't really homework and I don't even know where to start, so I guess it'd be too much to ask to just do it! If someone could just get me started methinks hazy memories could kick in
     
    Last edited: Apr 1, 2008
  2. jcsd
  3. Apr 1, 2008 #2

    tiny-tim

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    Hi blochwave! :smile:

    Well, most of it is constant, so it's (x^m)(1-x)∑x^(-k),

    (or is it (x^-m)(1-x)∑x^k ?)

    so all you have to sum is ∑x^(-k). :smile:

    (or ∑x^k)
     
  4. Apr 1, 2008 #3

    Hootenanny

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    I assume you mean,

    [tex]P = \sum_{k=0}^{n} x^{n-k}(1-x) = (1-x)\sum_{k=0}^{n}\frac{x^n}{x^k}[/tex]

    [tex] = (1-x)\left(x^n+x^{n-1} + x^{n-2} + \ldots + x^2 + x +1\right)[/tex]

    [tex] = \left(x^n+x^{n-1} + x^{n-2} + \ldots + x^2 + x +1\right) - \left(x^{n+1}+x^{n} + x^{n-1} + \ldots + x^3 + x^2 +x\right)[/tex]

    Notice that all the terms where the exponent is between n and 1 inclusive cancel leaving,

    [tex]P = 1-x^{n+1}[/tex]

    Edit: Ooops, I thought the m was an n, never mind. See TT's post.
     
  5. Apr 1, 2008 #4
    You guys rock so hard

    EDIT: Unfortunately I don't

    A)It WAS x^(m-k), m is a constant distinct from n, sorry

    it was (1-x)^k, to make it I believe more difficult
     
    Last edited: Apr 1, 2008
  6. Apr 1, 2008 #5

    Hootenanny

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    So the series is,

    [tex]P = \sum_{k=0}^{n}x^{m-k}\left(1-x\right)^k[/tex]

    Correct?
     
  7. Apr 1, 2008 #6
    Yes, I wrote it down this time to avoid further embarrassment >_>
     
  8. Apr 1, 2008 #7

    Hootenanny

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    The contribution from the x(m-k) term will be similar to the series detailed in post #3. For the contribution of the parenthesised term, one may consider using the binomial theorem.

    Edit: I'm curious as to your line of work, in what context did the series arise?
     
    Last edited: Apr 1, 2008
  9. Apr 1, 2008 #8
    I don't even know what he's doing it for, it's something to do with probabilities like I said

    Check this though: I did the ratio convergence test to make sure he didn't ask a stupid question, forgetting that it only works if the series is geometric

    Well if you divide any subsequent terms you get x^-1(1-x), which I realized is r

    so the series can be written as x^m[r]^k, if a=x^m, r is that thing above(I checked this for at least the first few terms), you can just write down the solution to a geometric series using those terms and BAM

    Did I do good? He's an aerospace engineer and I have a degree in physics, we work for an engineering consulting type place. I felt obliged to solve this after I didn't know his other question of which was more efficient, a turbofan or turboprop :(

    Edit: Well duh, it's a finite series, I shouldn't have even been doing a convergence test, BUT in the process I found out I could rewrite it as a geometric series, and right is right. I hope
     
    Last edited: Apr 1, 2008
  10. Apr 1, 2008 #9

    Hootenanny

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    You know what, that's real nice blochwave, I didn't spot that. Looks good to me :approve:
     
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