Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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]


    [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
  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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook