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Difficult series

  1. Jul 15, 2004 #1
    Good morning ,
    I have to calculate these series , Help me

    [tex]\displaystyle\sum_{k=1}^{i-1}C_{i}^{k}(tk)^{k-2}k(i-k)^{i-k+1}[/tex]
    Thanks
     
  2. jcsd
  3. Jul 15, 2004 #2
    Are you sure it is [tex]C_{i}^{k}[/tex]? When k is smaller than i, I think [tex]C_{i}^{k}[/tex] is 0?
     
  4. Jul 15, 2004 #3
    YA I'm sure
    so???????????
     
  5. Jul 15, 2004 #4

    Zurtex

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    I don't really understand the maths involved here, but I think the point was if [itex]C_{i}^{k} = 0[/itex] then your sum is 0 + 0 + 0 + ... + 0.
     
  6. Jul 15, 2004 #5
    u write it is [itex]C_{k}^{i} [/itex]
     
  7. Jul 15, 2004 #6
    So how calculate these serieS?
     
  8. Jul 19, 2004 #7
    Please help me how we calculate these SERIES??
     
  9. Jul 19, 2004 #8
    Maybe you should try to use Maple for this particular problem? Some guys told me that Maple can handle this kind of symbolic computations.
     
  10. Jul 19, 2004 #9
    Good morning
    I'm must proof the calculation step of this series and not the result from maple!!!!
     
  11. Jul 19, 2004 #10
    Now, your series contains a variable t. Surely if the sum makes sense, it should sum to a polynomial. So, the problem is what do you want to prove? The problem makes sense if the proposition is like "prove that the series sums to a polynomial in t with coefficients of the form..." or like "prove that the series sums to the nth derivative of ... function".

    So what exactly is the proposition?
     
  12. Jul 19, 2004 #11
    I need the steps calculation of these series!!
     
  13. Jul 19, 2004 #12
    sorry then....maybe I can't help, as I am unclear about what you want to prove. The expression is a polynomial in t. It may be "simplified" into any form that you may deem suitable.
     
  14. Jul 19, 2004 #13
    Can u calculate me this series?
     
  15. Jul 21, 2004 #14
    So Wong can u help me?
     
  16. Jul 21, 2004 #15
    Sorry, maybe I can't help....
     
  17. Jul 22, 2004 #16

    Zurtex

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    Have you tried expanding it out and trying the first few terms? I'm just doing that now.

    But it does look complex, I get the 1st 2 terms as being [itex](i/t)(i - 1)^i[/itex] and [itex]i(i-1)(i-2)^{i-1}[/itex]

    Perhaps looking at this wrong, it may be more easy for trying different values of i.
     
  18. Jul 22, 2004 #17
    Write out, on paper, the first 5 terms, the kth term and the last two terms. See if you can find a pattern. For example, the first term MIGHT pair with the last term to equal zero, the second might pair with the second to last, etc.

    If that doesn't work, try induction on k, see how far you get.

    At least have a go!

    EDIT: Zurtex ... I didn't notice your post, I'm just pretty much repeating what you said.
     
    Last edited: Jul 22, 2004
  19. Jul 22, 2004 #18

    Zurtex

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    Edit: It's all wrong, sorry.
     
    Last edited: Jul 23, 2004
  20. Jul 22, 2004 #19

    Zurtex

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    Edit: Sorry I tried reducing the seris in a way that was just plain wrong. However it does occur to me that you can take out [itex]i!/t^2[/itex] as a common factor.
     
    Last edited: Jul 23, 2004
  21. Jul 26, 2004 #20
    Zurtex
    THe [itex]i[/itex] is not a complex it is the index!!!!!!!!!!!!!!!!!!!!
     
  22. Jul 26, 2004 #21

    Zurtex

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    I know, however it is just a constant and thus is common to the whole sum, I'm trying to help you please do not talk to me like i'm an idoit. I've been working on this a bit more and I don't see how it can be reduced.

    Whatever value of i you choose you get a polynomial in t, where the coefficient of each power of t is dependent on both on the power t and the value of i. Such that the final form is:

    [tex]\sum_{k=1}^{i-1} a_{k,i}t^{k-2}[/tex]

    Well the sum you have is:

    [tex]\sum_{k=1}^{i-1} \frac{i!k^k(i-k)^{i+1}}{(i-k)!k!k(i-k)^k}t^{k-2}[/tex]

    This is of the same form :uhh:. But I would like to point out you can take [itex]i!/t^2[/itex] out of it like so:

    [tex]\frac{i!}{t^2}\sum_{k=1}^{i-1} \frac{k^k(i-k)^{i+1}}{(i-k)!k!k(i-k)^k}t^k[/tex]
     
    Last edited: Jul 26, 2004
  23. Jul 27, 2004 #22
    Excuse Me Zurtex But How we can continue ?
     
  24. Jul 27, 2004 #23

    Zurtex

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    Sorry I don't understand your question, but choose a value for i and work it out.
     
  25. Jul 27, 2004 #24
    Zurtex , I need the exact value of my series for every i
    So what should i do?
     
  26. Jul 27, 2004 #25

    Zurtex

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    Well I wrote a little program and this is what I got:

    For i = 2: [tex]2t^{-1}[/tex]

    For i = 3: [tex]24t^{-1}+6[/tex]

    For i = 4: [tex]324t^{-1}+96+36t[/tex]

    For i = 5: [tex]5120t^{-1}+1620+720t+320t^2[/tex]

    For i = 6: [tex]93750t^{-1}+30720+14580t+7680t^2+3750t^3[/tex]

    For i = 7: [tex]1959552t^{-1}+656250+322560t+181440t^2+105000t^3+54432t^4[/tex]

    I can give you up to 12 if you want, if you can spot a pattern and prove it inductively good luck.

    Edit: I'm not 100% confident in my calculations, I'll check them over later.
     
    Last edited: Jul 27, 2004
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