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Sequence math homework help

  1. Jul 3, 2006 #1
    Hi... I am working on a problem...

    [tex]41.25 \sum_{n=0}^\24 \frac{n}{x^n}[/tex]

    (on the top of the Sigma, it should say 24, NOT 4)

    I am searching, but can't seem to find a way to reduce that.

    Computing that up to [tex]n=24[/itex] is pretty tedious...

    Anybody know if there is a simpler way to compute this?
     
    Last edited: Jul 3, 2006
  2. jcsd
  3. Jul 3, 2006 #2

    0rthodontist

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    You mean
    [tex]41.25 \sum_{n=0}^{24} \frac{n}{x^n}[/tex]

    I'm pretty sure this can be simplified by a generating function, step 1 is to get x only in positive powers by factoring out x-24.
     
  4. Jul 3, 2006 #3
    Yea that's what I meant...
     
  5. Jul 3, 2006 #4

    0rthodontist

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    The g.f. you are looking for generates 24, 23, 22, 21, ... 3, 2, 1, 0, 0, 0, ... (you can see that, right?)
    So step 1 is to find the g.f. for the sequence 0, -1, -2, -3, etc.
    Step 2 is to find the g.f. for the sequence 24, 24, 24 and then add it to the g.f. from step 1 to get 24, 23, ... 1, 0, -1, -2, -3, ...
    Step 3 is to find the g.f. that generates 25 0's and then generates 1, 2, 3, .... Then add it to the summed g.f. from step 2 (to get rid of the negative terms in that g.f.) and you have the function you want.
     
  6. Jul 3, 2006 #5
    Thank you Orthodontist... yea i can see your method it looks good!
    Thanks!
     
  7. Jul 3, 2006 #6

    Hurkyl

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    What's the derivative of the following function?

    [tex]
    f(x) = \sum_{n = 0}^{24} x^{-n}
    [/tex]
     
  8. Jul 3, 2006 #7
    Im sorry Latex is difficult to write in...

    but f'(x) = sum(n=1,24) -nx^(-n-1)

    Is that correct?
     
  9. Jul 3, 2006 #8

    Hurkyl

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    Yep. And that looks an awful lot like the sum you wanted. (And, to boot, you already know how to compute my sum!)
     
  10. Jul 3, 2006 #9
    Im sorry... I don't see how that makes it any easier?
     
  11. Jul 3, 2006 #10

    Hurkyl

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    Because you already know a simpler expression for [itex]
    f(x) = \sum_{n = 0}^{24} x^{-n}
    [/itex]
     
  12. Jul 3, 2006 #11
    O I am so confused right now....

    I was actually looking for a FAST way to compute [tex]41.25 \sum_{n=0}^{24} \frac{n}{x^n}[/tex]

    (it is 24 on top of the sigma, not 4)

    I'm confused because I don't know what taking the first derivative of [tex]f(x) = \sum_{n = 0}^{24} x^{-n}[/tex] had to do with anything
     
  13. Jul 3, 2006 #12

    shmoe

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    Take the derivative, then modify it to look like the sum you are after.

    You have a simple expression for f(x), it's a geometric series, so you can write it in a form without the summation. Follow the same steps (differentiate, etc.) and you will have the sum you are after in a a nice closed form (no summation)
     
  14. Jul 3, 2006 #13
    ohhh okay... now this is making a lot more sense...wow...
     
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