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Polynomial Expansion of a Constant and Variable

  1. Sep 17, 2009 #1
    Is it possible to find a formula to expand this polynomial: [tex](n+1)(n+2)\ldots(n+x)[/tex] where [tex]n,x\in\textbf{N}[/tex]. In other words, is it possible to deduce a formula [tex]F[/tex] such that

    [tex]\displaystyle\prod_{k=1}^x{(n+k)}=\displaystyle\sum_{LB}^{UB}F[/tex]

    Where LB and UB are the respective lower and upper bounds. I'm assuming it has something to do with binomial coefficients since the obvious

    [tex]{(a+b)}^n=\displaystyle\sum_{k=0}^{n}{\binom{n}{k}a^{n-k}b^{k}}[/tex]
     
  2. jcsd
  3. Sep 17, 2009 #2
    Sorry no such formula - at least nothing simple like what you're looking for.
     
  4. Sep 18, 2009 #3
    there appears to be something like

    [tex]=\displaystyle\sum_{k=0}^{n}A_{k,n}n^{n-k}[/tex]

    [tex]A(i,n)[/tex] seems to be a polynominal in n to the 2*i degree with the initial term which begins with 0 repeated i times, then with i!. The remainder of the terms follow the formula
    [tex]A_{i,n} = A_{i-1,n-1}*n + A_{i-1,n}[/tex]. I don't know if there is a simple way to denote the polynominals. But the one to the fourth degree begins 0,0,2,11,35.... I say this seems to be the case because I am not positive.
     
  5. Sep 19, 2009 #4

    CRGreathouse

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    I'll use x as the symbolic variable rather than n, if you don't mind; that's more natural for me.

    Here's what I have for the first 10:
    1: x + 1
    2: x^2 + 3x + 2
    3: x^3 + 6x^2 + 11x + 6
    4: x^4 + 10x^3 + 35x^2 + 50x + 24
    5: x^5 + 15x^4 + 85x^3 + 225x^2 + 274x + 120
    6: x^6 + 21x^5 + 175x^4 + 735x^3 + 1624x^2 + 1764x + 720
    7: x^7 + 28x^6 + 322x^5 + 1960x^4 + 6769x^3 + 13132x^2 + 13068x + 5040
    8: x^8 + 36x^7 + 546x^6 + 4536x^5 + 22449x^4 + 67284x^3 + 118124x^2 + 109584x + 40320
    9: x^9 + 45x^8 + 870x^7 + 9450x^6 + 63273x^5 + 269325x^4 + 723680x^3 + 1172700x^2 + 1026576x + 362880
    10: x^10 + 55x^9 + 1320x^8 + 18150x^7 + 157773x^6 + 902055x^5 + 3416930x^4 + 8409500x^3 + 12753576x^2 + 10628640x + 3628800

    Looks like Stirling numbers.
     
  6. Sep 19, 2009 #5
    Yes that is the idea,but the polynominals to the 2i th power that I noted (corresponding with each column of numbers) appear to start from the row corresponding to -2 with off 1, (0, repeated i +1 times , then i fractorial follow by the completing terms. These polynominals appear to follow a form of Wilson's theorem since [tex]A_{i,j} \mod p [/tex] for j = -2 mod P always equals 1, it appears.
     
    Last edited: Sep 19, 2009
  7. Sep 21, 2009 #6
    I did some more investigation on this, while the second column of numbers are listed in the Online Encyclopedia of Integer Sequences as the Stirling numbers of the first kind, the number sequences in the higher number columns are not listed. Here are the results.

    formula as a function of y = x-2 where n and x are set forth in the original post

    [tex]\displaystyle\prod_{k=1}^x{(n+k)}=\displaystyle\sum_{m=0}^{n}A_{ij}n^m[/tex]

    0 [tex]A_{ij} = \Sum_{t=0}^{2i+1}B_{jt}\Binom{y}{t} [/tex]where [b_{jt} is as follows

    Column j : t = 0,1,2,3,...

    0 : 1
    1 : 1, -1, 1
    2 : 1, -1, 1, -1, 3
    3 : 1, -1, 1, -1, 1, 5, 15
    4 : 1, -1, 1, -1, 1,-1, 25, 105, 105
    5 : 1, -1, 1, -1, 1, -1, 1, 119, 805,1575, 945

    If the above is stacked as an isosceles triangle the righ half is
    i\h = 0,1,2,3,4....
    1 : 1
    2:-1, 3
    3 : 1, 5, 15
    4:-1, 25, 105, 105
    5 :1, 119,805,1575,945

    [tex] B_{ih} = (h+i)B_{i-1,h-1} +(h+i-1)B_{i-1,h}[/tex]
     
    Last edited: Sep 21, 2009
  8. Oct 1, 2009 #7
    I kinda have one, it was proven by someone else, and I never got to see that proof. What they showed is the following (their notation):

    [tex]\prod_{i=1}^{n} (x - a_i ) = x^n + k_{n-1}x^{n-1} + k_{n-2}x^{n-2} + ... k_1 x + (-1)^n a_1 a_2 ... a_n[/tex]

    where

    [tex]k_r=(-1)^{n-r} \bigg[\sum_{\displaystyle{c_1<c_2<...<c_{n-r}}} c_1 c_2 \cdot\cdot\cdot c_{n-r} \bigg][/tex]

    where [tex]c_{i}'s \ \epsilon \{ a_1 , a_2 , ... , a_n \}[/tex] and [tex]0<r<n[/tex]


    The idea is this:
    If you expand the product by distribution, you will notice that each term in the expansion (without simplification) will be a product of n factors. Also, you will notice that the original product has n factors, and from combinatronics that the terms (in the unsimplified expansion) are actually formed by multiplying one of the terms (x or a) in each factor (of the original product).

    Not sure if that was clear, so this is a small illustration:

    The two most obvious terms are [tex]x^n[/tex] (we got that by multiplying the x's in each factor together) and [tex]a_1 a_2 ... a_n[/tex] (we got that by multiplying all the a's together, instead of the x's).

    Another term would be [tex]a_5 a_8 a_{13} x^{n-3}[/tex]. This was taken by multiplying the x's from all the factors except for the fifth, eighth, and thirteenth factors.

    So, we know how to get each term in the "super-expanded" form. We will simplify this by combining similar terms.

    For [tex]x^{n-1}[/tex], the coefficient will be [tex]a_1 + a_2 + ... + a_n[/tex]
    For [tex]x^{n-2}[/tex], the coefficient will be [tex]a_1 a_2 + a_1 a_3 + ... + a_2 a_3 + a_2 a_4 + ... + a_{n-1} a_n[/tex]

    and if you follow the pattern, that's basically how you get the above formula for k
     
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