Coefficient
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Steven G
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How many ways can you get x^8 from (1+x^2-x^3)^9?
Note that every term in the expansion will be using exactly one term from each of the nine factors.
You can multiply 1*1*1*1*1*1x^2*(-x^3)(-x^3) = 1x^8. How many ways can you do that?
You can multiply 1*1*1*1*1*x^2*x^2*x^2*x^2 = 1x^8. How many ways can you do that?
Any other ways to get x^8?
Note that every term in the expansion will be using exactly one term from each of the nine factors.
You can multiply 1*1*1*1*1*1x^2*(-x^3)(-x^3) = 1x^8. How many ways can you do that?
You can multiply 1*1*1*1*1*x^2*x^2*x^2*x^2 = 1x^8. How many ways can you do that?
Any other ways to get x^8?
Here is one way to think about it.
[math]\text {Let } p = 1 + x^2 \text { and } q = x^3.\\ \therefore (1 + x^2 - x^3)^9 = (p - q)^9 = \sum_{j=0}^9 \dbinom {9}{j} * (-1)^j * p^{(9-j)} * q^j =\\ \sum_{j=0}^9 \dbinom{9}{j} * (-1)^j * (1 + x^2)^{(n-j)}x^{3j}.[/math]
Follow that? Now consider the term resulting when j = 3.
[math]\dbinom{9}{3} * (-1)^3 * (1 + x^2)^6 * (x^3)^3 = - x^9 * \text {stuff.}[/math]
How many instances of [imath]x^8[/imath] do you expect to find in that term?
Is it not clear that all you need to think about are the first three terms?
[math]\text {Let } p = 1 + x^2 \text { and } q = x^3.\\ \therefore (1 + x^2 - x^3)^9 = (p - q)^9 = \sum_{j=0}^9 \dbinom {9}{j} * (-1)^j * p^{(9-j)} * q^j =\\ \sum_{j=0}^9 \dbinom{9}{j} * (-1)^j * (1 + x^2)^{(n-j)}x^{3j}.[/math]
Follow that? Now consider the term resulting when j = 3.
[math]\dbinom{9}{3} * (-1)^3 * (1 + x^2)^6 * (x^3)^3 = - x^9 * \text {stuff.}[/math]
How many instances of [imath]x^8[/imath] do you expect to find in that term?
Is it not clear that all you need to think about are the first three terms?
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Why 3?
Not sure I understand...
I asked you to think. Whatever the expansion of [imath]\dbinom{9}{3} (-1)^3 * (1 + x^2)^6[/imath] may be, when you multiply it by [imath](x^3)^3 = x^9[/imath], how many terms involving [imath]x^8[/imath] are you going to get?
It should quickly be obvious that if j = 3, every term in the expansion will be a multiple of a power of x of AT LEAST 9. If j = 4, every term in the expansion will be a multiple of a power of x of AT LEAST 12.
The first thing to do on some problem that is unfamiliar is to think rather than calculate.
I am afraid I don't get it.
BigBeachBanana
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JeffM's approach is perfectly fine, but I'll present using the multinomial theorem, which is the generalized version of the binomial expansion.
Multinomial Theorem:
[math](a_1+a_2+\dots+a_k)^n=\sum_{b_1+b_2+\dots+b_k=n} {n \choose b_1,b_2,\dots,b_k}\prod_{j=1}^{k}a_j^{b_j}[/math]
Define the variables:
\(\displaystyle a_1=1\,,a_2=x^2\,,a_3=-x^3\,, n=9\)
Apply the multinominal theorem:
[math](1+x^2-x^3)^9= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}[/math]This means a general term for the expansion of [imath](1+x^2-x^3)^9[/imath] will be in the form of
[math]{9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}}_{\text{How to get }x^8?}[/math]
Question:
Now, for what value of [imath]b_1,b_2,b_3[/imath] will give you [imath]x^8?[/imath] Remeber that [imath]b_1+b_2+b_3=9[/imath].
Multinomial Theorem:
[math](a_1+a_2+\dots+a_k)^n=\sum_{b_1+b_2+\dots+b_k=n} {n \choose b_1,b_2,\dots,b_k}\prod_{j=1}^{k}a_j^{b_j}[/math]
Define the variables:
\(\displaystyle a_1=1\,,a_2=x^2\,,a_3=-x^3\,, n=9\)
Apply the multinominal theorem:
[math](1+x^2-x^3)^9= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}[/math]This means a general term for the expansion of [imath](1+x^2-x^3)^9[/imath] will be in the form of
[math]{9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}}_{\text{How to get }x^8?}[/math]
Question:
Now, for what value of [imath]b_1,b_2,b_3[/imath] will give you [imath]x^8?[/imath] Remeber that [imath]b_1+b_2+b_3=9[/imath].
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Steven G
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When you multiply out (1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3) you pick one term from each factor to multiply.
For example, if you were going to multiply out (2x^3+3x^2+x)(3x^2+4x+2) how would you get x^3 terms. The answer is you'll get x^3 terms by multiplying the 2x^3 in the first bracket by the 2 in the 2nd bracket, the 3x^2 in the 1st bracket by the 4x in the 2nd bracket and the x in the 1st bracket by the 3x^2 in the 2nd bracket. Now tally up how many x^3 you have all together.
Steven G
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Using my results above, the answer would be 9!/(6!2!) + 9!/(5!4!).
Two comments. This is the easiest way to do such a problem and more importantly students should now that when multiply out polynomials you are taking every combination of one term from each polynomial to multiply together.
Two comments. This is the easiest way to do such a problem and more importantly students should now that when multiply out polynomials you are taking every combination of one term from each polynomial to multiply together.
BigBeachBanana
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Since Steve provided the answer, using my method:
[math]{9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}[/math]Notice we achieve [imath]x^8[/imath], by letting [imath]b_1=5\,,b_2=4\,,b_3=0[/imath]
[math]{9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8[/math]Or by letting [imath]b_1=6\,,b_2=1\,,b_3=2[/imath]
[math]{9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8[/math]Therefore, the answer is
[math]\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378[/math]
I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.
BigBeachBanana
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Are you referring to the n and k in the binomial coefficient [imath]{n \choose k}?[/imath]
BigBeachBanana
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The binomial coefficient is a special case of the multinomial coefficient. As their names suggest, the prefix "bi-" deals with 2 groups, while the multinomial can deal with multiple groups, i.e. [imath]\ge 2[/imath].
First, observe the binomial coefficient:
[math]{n \choose k}=\frac{n!}{k!(n-k)!}[/math]Notice that in the binomial coefficient denominator, you have 2 groups: [imath]k![/imath] and [imath](n-k)![/imath] It means that you're partitioning n objects into 2 groups.
Group 1: k objects, which means whatever is left belongs in Group 2, i.e. (n-k). Of course, the sum of the groups must always add up to n objects because that's what you started with. Let's look at the binomial coefficient from the multinomial perspective. Instead of letting group 2 be (n-k) objects, simply call it [imath]k_2[/imath]. So we have:
[math]{n \choose k_1,k_2}=\frac{n!}{k_1!k_2!}=\underbrace{\frac{n!}{k_1!(n-k_1)!}}_\text{This is what you're familiar with}[/math]
Now, what if you want to partition n objects into more than 2 groups. In your case, you want to partition into 3 groups, let's call those groups [imath]k_1,k_2,k_3[/imath]. So we have
[math]{n \choose k_1,k_2,k_3}=\frac{n!}{k_1!k_2!k_3!}[/math]But remember, the sum of the objects in the groups must add up to n, so we can rewrite them as:
[math]{n \choose k_1,k_2,k_3}=\frac{n!}{k_1!k_2!k_3!}=\frac{n!}{k_1!k_2!(n-k_1-k_2)!}[/math]
In short, the [imath]b_i[/imath] is just my variable convention. You use k or r, whatever you're comfortable with, as long as you understand the concept.
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Steven G
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BBB, Please see post #2 where I stated the method which you used. Not a big deal. You can have the credit!
BigBeachBanana
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The early bird gets the worm, but the second mouse gets the cheese
When I read your post the first time, I didn't see where you were going with that, but now that I've looked again, I understand.
Others have tried to explain the multinomial theorem to you. I avoid memorizing things that I seldom use. So here is an alternative way.
[math]y = (1 + x^2 - x^3)^9 = (p - q)^9, \text { where } p = (x^0 + x^1) \text { and } q = x^3.[/math]
This is just a basic substitution of variables that will let us use the binomial theorem.
When we expand the binomial, we get
[math]y = (p - q)^9 = \sum_{j=0}^9 \dbinom{9}{j} (-1)^j p^{(9-j)} q^j.[/math]
Binomial theorem. No questions there I presume. But
[math]q = x^3 \implies q^3 = x^9, \ q^4 = x^{12}, \ q^5 = x^{15}, \text { etc.}[/math]
We are interested in the coefficient of [imath]x^8[/imath] in the expansion of y. But if j > 2, the terms will be multiples of powers of x of 9 or greater and so are irrelevant to powers of 8. The only terms that are relevant are when j = 0, j = 1, and j = 2. Let’s consider them one by one.
[math]j = 0 \implies \dbinom{9}{j} * (-1)^j p^{(9-j)}q^j = p^9 = (1 + x^2)^9 =\\ \sum_{k=0}^9 \dbinom{9}{k} 1^{(9-k)} (x^2)^k.[/math]
The only way we can get [imath]x^8[/imath] when j = 0 is when k = 4.
[math]\dbinom{9}{4} = \dfrac{9!}{4! * 5!} = \dfrac{9 * 8 * 7 * 6}{4 * 3 * 2} = 2 * 63 = 126.[/math]
Let’s think about j = 1.
[math]j = 1 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = -9 p^8q = -9x^3(1 + x^2)^8.[/math]
Is it not obvious that multiplying an odd power of x by even powers of x will never generate [imath]x^8[/imath]?
j = 1 is irrelevant.
Let’s think about j = 2.
[math]j = 2 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = 36x^6(1 + x^2)^7.[/math]
The only way to get [imath]x^8[/imath] out of that is with the term containing [imath]x^2[/imath].
[math](1 + x^2)^9 = \sum_{k=0}^7 \dbinom{7}{k}1^{(7-k)}(x^2)^k.[/math]
We need to look at k = 1. So the coefficient we want is 36 * 7 = 252.
Putting this altogether we get 252 + 126 = 378.
That is EXACTLY the number BBB got. His way is admittedly much more efficient. My way involves avoids memorizing another formula and is more easily explicable.
Steven G
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Jeff,
The way I did it was basically counting the number of ways that you get an x^8 term (or any other term) by multiplying nine terms together (they happened to be the same nine factors but that doesn't change anything). In multiplication of polynomials (or even non-polynomials) you choose one term from each factor and multiply them together. This method does not require you to know the multinomial theorem. I got the answer without even thinking of the multinomial theorem.
There are, for example, 9!/(6!2!) ways to get 1*1*1*1*1*1x^2*(-x^3)(-x^3) (= 1x^8). Is this clear?