Results 1 to 4 of 4

Math Help - Proving a theorem by Polya

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    New York
    Posts
    6
    Thanks
    1

    Proving a theorem by Polya

    I've been trying to figure out where to start with this problem and am quite lost. Any help would be appreciated.

    Theorem:
    Let Ck(x)= [x(x-1)(x-2)...(x-(k-1))] / k!
    Then any polynomial p(x) with the property that p(n) is in Z, for all n in Z
    can be written as an integer linear combination
    p(x) = (summation from 1 to k) MiCi(x)
    with each Mi in Z, i = 1,...,k

    Any help or place to start would be greatly appreciated. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2013
    From
    BC, Canada
    Posts
    95
    Thanks
    15

    Re: Proving a theorem by Polya

    Seems that there may be something missing. If x = 0, then \sum_{i=1}^k M_i C_i(0) = 0 since C_k(0) = 0, \forall k. But, what if p(0)\ne 0? Is the theorem (which I tried to find online with no luck) perhaps this?:

    p(x) = M_0 + \sum_{i=1}^k M_i C_i(x)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2013
    From
    New York
    Posts
    6
    Thanks
    1

    Re: Proving a theorem by Polya

    Ah, yes I forgot to say that we define C_0(x) = 1.
    So I think they are actually equivalent, as C_0 (x) = 1, so the first term in the sequence would just be M_0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2013
    From
    BC, Canada
    Posts
    95
    Thanks
    15

    Re: Proving a theorem by Polya

    Quote Originally Posted by Magnechu View Post
    Ah, yes I forgot to say that we define C_0(x) = 1.
    So I think they are actually equivalent, as C_0 (x) = 1, so the first term in the sequence would just be M_0.
    I've been trying to think about the series given above. One thing is bothering me that I don't now how to tackle. It seems that the C_k(x) vanish beyond a certain term l where x>l, which is great if you want to show that certain M_i exist. You can construct the M_i in this way by starting at p(0), then p(1) (which requires p(0), computed previously), and p(3) similarly. However, when one considers the negative integers the C_k(x) do not vanish, and also, the M_i's would be completely determined by the procedure I mentioned above (for positive integers). I'm not sure how one reconciles this to also include negative integers. I can't help you beyond this (sorry!).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using Polya strategy and the sub-goal or others
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: August 15th 2011, 05:45 PM
  2. Replies: 0
    Last Post: November 13th 2009, 05:41 AM
  3. Polya's Urn (Discrete Martingale)
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 29th 2009, 12:22 PM
  4. Polya Problems
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 17th 2007, 06:38 AM
  5. polya strikes out
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 29th 2007, 06:39 PM

Search Tags


/mathhelpforum @mathhelpforum