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Math Help - Sequence

  1. #1
    MHF Contributor Quick's Avatar
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    Sequence

    Interesting, my teacher taught the class sequences and how to solve them, such as tables with the terms (n) and the value (f(n))

    She showed us something I though was strange:

    Code:
    
      n | 1 |  2 |  3 |  4  |  5   
    f(n)| 3 | 17 | 55 | 129 | 251
    Then she said find the difference in the values:

    Code:
    
      n | 1 |  2 |  3 |  4  |  5   
    f(n)| 3 | 17 | 55 | 129 | 251 
            14   38   74   122
    Do it again, until you have a common difference:

    Code:
    
      n | 1 |  2 |  3 |  4  |  5   
    f(n)| 3 | 17 | 55 | 129 | 251 
            14   38   74   122
               24   36   48
                  12   12
    Then count the number of times you took the difference:


    Code:
    
      n | 1 |  2 |  3 |  4  |  5   
    f(n)| 3 | 17 | 55 | 129 | 251 
            14   38   74   122    :
               24   36   48       : 3
                  12   12         :
    Then she said that this means the sequence is a cubed equation (because it is three rows down).

    I looked over it and it works (I think) for all exponents.

    I then discovered that if the equation was in the form: f(n)=an^x+b
    That the last term (the common difference) would be: x!\times a

    How would someone prove this?

    BTW, the equation for the tables was: f(n)=2n^3+1
    Last edited by Quick; October 25th 2006 at 05:01 PM.
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  2. #2
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    Quote Originally Posted by Quick View Post
    How would someone prove this?
    I realized a lot of math fans played around with these sequences and realized things like this, including me and you. (But I proved it )*). I really do not want to post a prove of this I happen to know it will get messy with latex. But basically it is all induction arguments (remember in Quicklopedia). All was able to show several properties, including the fact that the last term is n! but I did not prove it directly (it really got complicated) rather with a different approach.


    *)Note I mention myself first, can you guess why?
    ---------
    The fist idea is done like this. You want to show a polynomial of degree n, terminates after n differences (or derivatives like I call them). You show it works for linear polynomials and quaratics. Then you show that a polynomial of degree k+1 always reduces to k polynomial and you use induction on that.

    The converse of the theorem I can prove to you, would you like to see? If you are given "n" elements (finite number) then you can express it as a polynomial (unique) up to order "n-1".

    ~~~
    This is not for you. Has anyone ever considered developing this theory for the fields \mathbb{Z}_{p^n} it might lead to some beautiful results.
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