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Polynomial expansion

  1. Mar 23, 2006 #1
    Hello
    I'm trying to do some research on polynomial expansion however my math isn't that good to do high level calculations, proofs and so on. A while ago my friend came up with a formula for finding number of terms in any expansion of a polynomial to the y power. For example for
    (a+b+c+d)^4 = 35 terms
    Formula: n-y-1Cy for n = number of terms in paranthases.
    Can anyone help me, give some clues on how to proof this formula or show me a proof?
    Thanks!
     
  2. jcsd
  3. Mar 23, 2006 #2

    arildno

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    What is y? What is C? Are you sure you haven't omitted any parantheses here?
     
  4. Mar 23, 2006 #3
    Basically, what you're doing in your example to get the 35 terms is picking 4 objects from a collection of 4, allowing repetition. In general, you have [itex] (a_1 + a_2 + \dots + a_n)^y [/itex]. Are you saying that the formula is [itex] \binom{n-y-1}{y} [/itex]?
     
  5. Mar 23, 2006 #4
    Sorry for bad explanation and wrong formula after all,
    for any polynomial (with number of terms n) to the yth power we may find number of terms after expansion by
    n+y-1 C y
    C means 'choose' for example aCb = a!
    -----------
    (a-b)! b!

    for polynomial (a+b+c+d)^4 where number of terms is 4 (a, b, c, d) we may find number of terms after expansion by 4+3C4 which is 7C4 = 35
    Thanks for all the help.
     
  6. Mar 23, 2006 #5
    nCr is the same as [tex] \binom{n}{r} [/tex] Just different notation.
     
  7. Mar 24, 2006 #6
    Clearly, each term in the expansion of [tex](a_{1} + a_{2} + \dots + a_{n})^y [/tex] will have degree y.
    Let an arbitary term be [tex]a_{1}^{p}a_{2}^{q}a_{3}^r\dots[/tex] then according to the condition p+q+r....=y.
    Thus, find the number of integer solutions of the above equation. thats your required number of terms.
     
  8. Mar 24, 2006 #7
    This is kind of interesting in that it might imply a quick solution to a question such as "how many pairs of integers satisfy a+b=100" and beyond, no?
     
  9. Mar 26, 2006 #8
    well, there is a quick solution to your question hypermonkey2, though i do not know the proof...

    the number of solutions to [tex] a_1x_1 + a_2x_2 + \dots + a_nx_n = p \ where \ b_i \leq x_i \leq c_i \mbox{for}\ 1 \leq i \leq n [/tex] is given by the coefficient of [tex] t^n [/tex] in the expression

    [tex] \prod ( (t^{a_i})^{b_i} + (t^{a_i})^{b_i + 1} + (t^{a_i})^{b_i + 2} + \dots + (t^{a_i})^{c_i})\where \ 1 \leq i \leq n [/tex]

    this however also includes solutions so that two [tex] x_i [/tex] may be equal. for distinct solutions introduce dummy variables so that condition of distinctness is removed. introduction of a dummy variable can be extended to number of solutions of the equation [tex] a_1x_1 + a_2x_2 + \dots + a_nx_n \leq p [/tex].
     
    Last edited: Mar 27, 2006
  10. Mar 29, 2006 #9
    After spending some time trying to figure out what this is about, I think I see what the problem is. Let p = the power and n=number of linear terms, C = the combinations, then the number of terms in

    [tex](\sum n)^p=(p+n-1) C (p).[/tex]

    For p=1 it is trivial, and easy enough for p=2; I'll go ahead and show it for all n for p=3.

    If n=1, then by the formula we should have 3C3=1, which is correct, and represents (a)^3=a^3, which is one term.

    Now if it is true for n terms, which I represent as u and we add one more term, call it b, then we have by the binominal formula:

    [tex](u+b)^3=u^3+3u^2b+3ub^2+b^3[/tex]

    In this case b^3 adds only 1 term. 3ub^2 adds n more terms. 3u^2b adds (n+1)C2. (This by how the squares work) and u^3 by the induction hypothses adds (n+2)C3.

    So we have (n+2)(n+1)n/6+(n+1)n/2+n+1 =(n+1)/6{n^2+2n+3n+6}=
    (n+1)/6{(n+2)(n+3)}=(n+3)C3. Or the induction is complete for p=3.
     
    Last edited: Mar 29, 2006
  11. Mar 7, 2010 #10
    C actually means combination, not choose. Choose is only a simpler representative word for combination to help people to understand better.
     
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