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Homework Help: Combinatorics-binomial expansion?

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A(x) = [tex]\sum_{k>=0}a_{k}x^k[/tex] and B(x) = [tex]\sum_{k>=0}b_{k}x^k[/tex] show that:
    A(X)B(X) = [tex]\sum_{k>=0}(\sum_{i=0}a_{i}b_{k-i})x^k[/tex]

    The second sum sign in the answer should be from i=0 to k.


    3. The attempt at a solution

    I factored out like terms and then multiplied them together and got

    [tex]\sum_{k>=0}x^k(a_{k}b_{k})[/tex] then if i={0,1,2,...,k} we would get

    [tex]\sum_{k>=0}x^k(\sum_{i=0}a_{i}b_{k-i})[/tex]

    I am guessing this is wrong?
     
  2. jcsd
  3. Jun 12, 2010 #2
    Maybe I'm wrong, but isn't this by definition? What is your definition of polynomial multiplication?
     
  4. Jun 12, 2010 #3
    to be honest i am not sure, i am quite lost in this class
     
  5. Jun 12, 2010 #4
    Which class, and from where did this question come? It's possible all you're supposed to say is that a term with xk clearly appears iff you multiply a term with xi by a term with xk-i, which gives you a coefficient of akbk-i. Combining like terms gives you the sum...
     
  6. Jun 12, 2010 #5
    It's an Intro to Combinatorics class and its from some homework that covers Binomial coefficients, the binomial theorem, and newtons binomial theorem. That is the sections I believe the answer should come from out of all the choices I have. If you don't agree with that I can tell you the other sections we have covered for this homework.
     
  7. Jun 12, 2010 #6
    I also don't get how your idea would work. I start with the two x^k therefor i never multiplied a x^i by an x^{k-i} so the coefficient wouldn't come about that way?
     
  8. Jun 12, 2010 #7
    When you multiply two polynomials, you need to combine their individual terms in every way possible, so you will need to multiply the aixi by bk-ixk-i sometime. It's possible for ai or bk-i to be 0.
     
  9. Jun 12, 2010 #8

    vela

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    It's often helpful to expand the summations to see what you're actually doing and where you're going wrong in your calculations. For example, suppose you had

    [tex]A(x) = \sum_{k=0}^3 a_k x^k = a_0 + a_1 x + a_2 x^2 + a_3 x^3[/tex]

    [tex]B(x) = \sum_{k=0}^3 b_k x^k = b_0 + b_1 x + b_2 x^2 + b_3 x^3[/tex]

    Then

    [tex]A(x)B(x) = (a_0 + a_1 x + a_2 x^2 + a_3 x^3)(b_0 + b_1 x + b_2 x^2 + b_3 x^3)[/tex]

    From this, you can see your first step of factoring like terms doesn't work in that there will be no terms that look like akbkxk except for k=0. Also, if you multiply it out, you should see a pattern for the coefficients in the final series.
     
  10. Jun 12, 2010 #9
    alright, so if i expand A(x) i'll get something like

    A(X) = [tex]a_{0}+a_{1}x+a_{2}x^2+...+a_{k}x^k?[/tex]

    then i do the same for B(X) and multiply them together?
     
  11. Jun 12, 2010 #10
    Yes.
     
  12. Jun 12, 2010 #11
    Ok, I got that answer then. The second part of the question says to use this result to find the coefficient of x^5 in the expansion of [tex]\frac{(x+y)^9}{1-x}[/tex] and I do not get how the answer to the previous part is supposed to help me...
     
  13. Jun 12, 2010 #12

    vela

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    Use A(x)=1/(1-x) and B(x)=(x+y)9. It looks like the book expects you to know how to write A(x) as a series.
     
  14. Jun 12, 2010 #13
    Well [tex]A(x)=\frac{1}{1-x}=1+x+x^2+x^3+...[/tex]. So this coefficient from [tex]x^5[/tex] is just going to be 1 but I don't know how to get the one from B(x)? Unless its not looking for the specific number which I would think that the coefficient would then be written as [tex]a_{1}b_{4}x^5[/tex]. Would that be the solution to this problem?
     
  15. Jun 12, 2010 #14
    Remember you are have a sum, of which a1b4 is certainly one part, but not all. Likely the book is looking for an expression involving y; you can find the coefficients for B(x) using binomial coefficients.
     
  16. Jun 12, 2010 #15
    Ok then, so I can use Newton's binomial Theorem to find that

    [tex]\sum_{k=0} nCr(9,5) x^5y^4 = 126 x^5y^4[/tex]. Then I would have to add that 1 to give me 127 correct?
     
  17. Jun 12, 2010 #16
    The summation sign you have there should not be there. Indeed, in (x+y)9, the coefficient in front of the x5 term is 9C5y4 (note the y in the expression). However, why would you add 1? Take a look again at the formula from the first part.
     
  18. Jun 12, 2010 #17
    Alright I understand the B(X) part. The one would be added because when we change A(X) into the series the [tex]x^5[/tex] has a coefficient of one. Does it not?
     
  19. Jun 12, 2010 #18
    I suggest, along the lines of vela's advice, to multiply out

    (a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5)(b0 + b1x + b2x2 + b3x3 + b4x4 + b5x5).​

    See what pattern you can spot. It should help you get a feel for what the formula for multiplication is really saying.
     
  20. Jun 12, 2010 #19
    When I multiplied that out the pattern I saw is the same one from before, that [tex]a_{i}b_{k-i}[/tex]... sorry I really am not getting this. I understand that 126 is part of it though...
     
  21. Jun 12, 2010 #20
    Yes, so what are b0 through b5 in B(x) = (x+y)9?
     
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