# Combinatorics-binomial expansion?

1. Jun 12, 2010

### pupeye11

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

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

$$\sum_{k>=0}x^k(\sum_{i=0}a_{i}b_{k-i})$$

I am guessing this is wrong?

2. Jun 12, 2010

### Tedjn

Maybe I'm wrong, but isn't this by definition? What is your definition of polynomial multiplication?

3. Jun 12, 2010

### pupeye11

to be honest i am not sure, i am quite lost in this class

4. Jun 12, 2010

### Tedjn

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...

5. Jun 12, 2010

### pupeye11

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.

6. Jun 12, 2010

### pupeye11

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?

7. Jun 12, 2010

### Tedjn

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.

8. Jun 12, 2010

### vela

Staff Emeritus
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

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

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

Then

$$A(x)B(x) = (a_0 + a_1 x + a_2 x^2 + a_3 x^3)(b_0 + b_1 x + b_2 x^2 + b_3 x^3)$$

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.

9. Jun 12, 2010

### pupeye11

alright, so if i expand A(x) i'll get something like

A(X) = $$a_{0}+a_{1}x+a_{2}x^2+...+a_{k}x^k?$$

then i do the same for B(X) and multiply them together?

10. Jun 12, 2010

### Tedjn

Yes.

11. Jun 12, 2010

### pupeye11

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 $$\frac{(x+y)^9}{1-x}$$ and I do not get how the answer to the previous part is supposed to help me...

12. Jun 12, 2010

### vela

Staff Emeritus
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.

13. Jun 12, 2010

### pupeye11

Well $$A(x)=\frac{1}{1-x}=1+x+x^2+x^3+...$$. So this coefficient from $$x^5$$ 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 $$a_{1}b_{4}x^5$$. Would that be the solution to this problem?

14. Jun 12, 2010

### Tedjn

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.

15. Jun 12, 2010

### pupeye11

Ok then, so I can use Newton's binomial Theorem to find that

$$\sum_{k=0} nCr(9,5) x^5y^4 = 126 x^5y^4$$. Then I would have to add that 1 to give me 127 correct?

16. Jun 12, 2010

### Tedjn

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.

17. Jun 12, 2010

### pupeye11

Alright I understand the B(X) part. The one would be added because when we change A(X) into the series the $$x^5$$ has a coefficient of one. Does it not?

18. Jun 12, 2010

### Tedjn

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.

19. Jun 12, 2010

### pupeye11

When I multiplied that out the pattern I saw is the same one from before, that $$a_{i}b_{k-i}$$... sorry I really am not getting this. I understand that 126 is part of it though...

20. Jun 12, 2010

### Tedjn

Yes, so what are b0 through b5 in B(x) = (x+y)9?