Coefficient

[Loki123]

Find the coefficient that goes with x^8 in development form of (1+x^2-x^3)^9.
I have no idea how to determine where we get just x^8...
Here is my work although it's pretty useless.

 

[Steven G]

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?

 

[JeffM]

Here is one way to think about it.

$$\text {Let } p = 1 + x^2 \text { and } q = x^3.\\ \therefore (1 + x^2 - x^3)^9 = (p - q)^9 = \sum_{j=0}^9 \dbinom {9}{j} * (-1)^j * p^{(9-j)} * q^j =\\ \sum_{j=0}^9 \dbinom{9}{j} * (-1)^j * (1 + x^2)^{(n-j)}x^{3j}.$$
Follow that? Now consider the term resulting when j = 3.

$$\dbinom{9}{3} * (-1)^3 * (1 + x^2)^6 * (x^3)^3 = - x^9 * \text {stuff.}$$
How many instances of $x^8$ do you expect to find in that term?

Is it not clear that all you need to think about are the first three terms?

 

[Loki123]

Follow that? Now consider the term resulting when j = 3.

Why 3?

 

[Loki123]

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?

Not sure I understand...

 

[JeffM]

Why 3?

I asked you to think. Whatever the expansion of $\dbinom{9}{3} (-1)^3 * (1 + x^2)^6$ may be, when you multiply it by $(x^3)^3 = x^9$, how many terms involving $x^8$ 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.

 

[Loki123]

I asked you to think. Whatever the expansion of $\dbinom{9}{3} (-1)^3 * (1 + x^2)^6$ may be, when you multiply it by $(x^3)^3 = x^9$, how many terms involving $x^8$ 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]

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:
$$(a_1+a_2+\dots+a_k)^n=\sum_{b_1+b_2+\dots+b_k=n} {n \choose b_1,b_2,\dots,b_k}\prod_{j=1}^{k}a_j^{b_j}$$
Define the variables:
$\displaystyle a_1=1\,,a_2=x^2\,,a_3=-x^3\,, n=9$

Apply the multinominal theorem:
$$(1+x^2-x^3)^9= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}= \sum_{b_1+b_2+b_3=9}{9\choose b_1,b_2,b_3}(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}$$This means a general term for the expansion of $(1+x^2-x^3)^9$ will be in the form of
$${9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(x^3)^{b_3}(-1)^{b_3}}_{\text{How to get }x^8?}$$
Question:
Now, for what value of $b_1,b_2,b_3$ will give you $x^8?$ Remeber that $b_1+b_2+b_3=9$.

 

[Steven G]

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?

Not sure I understand...

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.

 

[pka]

The expansion in this link may help check your work.
Can you explain why the terms that have a negative coefficient, $-Cx^n$, are those where the exponent, $n$, is an odd multiple of three?

 

[Steven G]

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.

 

[BigBeachBanana]

Using my results above, the answer would be 9!/(6!2!) + 9!/(5!4!).

Since Steve provided the answer, using my method:
$${9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}$$Notice we achieve $x^8$, by letting $b_1=5\,,b_2=4\,,b_3=0$
$${9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8$$Or by letting $b_1=6\,,b_2=1\,,b_3=2$
$${9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8$$Therefore, the answer is
$$\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378$$

 

[Loki123]

Since Steve provided the answer, using my method:
$${9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}$$Notice we achieve $x^8$, by letting $b_1=5\,,b_2=4\,,b_3=0$
$${9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8$$Or by letting $b_1=6\,,b_2=1\,,b_3=2$
$${9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8$$Therefore, the answer is
$$\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378$$

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]

I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.

Are you referring to the n and k in the binomial coefficient ${n \choose k}?$

 

[Loki123]

Are you referring to the n and k in the binomial coefficient ${n \choose k}?$

Yes

 

[BigBeachBanana]

I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.

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. $\ge 2$.
First, observe the binomial coefficient:
$${n \choose k}=\frac{n!}{k!(n-k)!}$$Notice that in the binomial coefficient denominator, you have 2 groups: $k!$ and $(n-k)!$ 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 $k_2$. So we have:
$${n \choose k_1,k_2}=\frac{n!}{k_1!k_2!}=\underbrace{\frac{n!}{k_1!(n-k_1)!}}_\text{This is what you're familiar with}$$

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 $k_1,k_2,k_3$. So we have
$${n \choose k_1,k_2,k_3}=\frac{n!}{k_1!k_2!k_3!}$$But remember, the sum of the objects in the groups must add up to n, so we can rewrite them as:
$${n \choose k_1,k_2,k_3}=\frac{n!}{k_1!k_2!k_3!}=\frac{n!}{k_1!k_2!(n-k_1-k_2)!}$$
In short, the $b_i$ is just my variable convention. You use k or r, whatever you're comfortable with, as long as you understand the concept.

 

[Steven G]

Since Steve provided the answer, using my method:
$${9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}$$Notice we achieve $x^8$, by letting $b_1=5\,,b_2=4\,,b_3=0$
$${9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8$$Or by letting $b_1=6\,,b_2=1\,,b_3=2$
$${9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8$$Therefore, the answer is
$$\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378$$

BBB, Please see post #2 where I stated the method which you used. Not a big deal. You can have the credit!

 

[BigBeachBanana]

BBB, Please see post #2 where I stated the method which you used. Not a big deal. You can have the credit!

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.

 

[JeffM]

I am afraid I don't get it.

Others have tried to explain the multinomial theorem to you. I avoid memorizing things that I seldom use. So here is an alternative way.

$$y = (1 + x^2 - x^3)^9 = (p - q)^9, \text { where } p = (x^0 + x^1) \text { and } q = x^3.$$
This is just a basic substitution of variables that will let us use the binomial theorem.

When we expand the binomial, we get

$$y = (p - q)^9 = \sum_{j=0}^9 \dbinom{9}{j} (-1)^j p^{(9-j)} q^j.$$
Binomial theorem. No questions there I presume. But

$$q = x^3 \implies q^3 = x^9, \ q^4 = x^{12}, \ q^5 = x^{15}, \text { etc.}$$
We are interested in the coefficient of $x^8$ 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.

$$j = 0 \implies \dbinom{9}{j} * (-1)^j p^{(9-j)}q^j = p^9 = (1 + x^2)^9 =\\ \sum_{k=0}^9 \dbinom{9}{k} 1^{(9-k)} (x^2)^k.$$
The only way we can get $x^8$ when j = 0 is when k = 4.

$$\dbinom{9}{4} = \dfrac{9!}{4! * 5!} = \dfrac{9 * 8 * 7 * 6}{4 * 3 * 2} = 2 * 63 = 126.$$
Let’s think about j = 1.

$$j = 1 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = -9 p^8q = -9x^3(1 + x^2)^8.$$
Is it not obvious that multiplying an odd power of x by even powers of x will never generate $x^8$?

j = 1 is irrelevant.

Let’s think about j = 2.

$$j = 2 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = 36x^6(1 + x^2)^7.$$
The only way to get $x^8$ out of that is with the term containing $x^2$.

$$(1 + x^2)^9 = \sum_{k=0}^7 \dbinom{7}{k}1^{(7-k)}(x^2)^k.$$
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]

Others have tried to explain the multinomial theorem to you. I avoid memorizing things that I seldom use. So here is an alternative way.

$$y = (1 + x^2 - x^3)^9 = (p - q)^9, \text { where } p = (x^0 + x^1) \text { and } q = x^3.$$
This is just a basic substitution of variables that will let us use the binomial theorem.

When we expand the binomial, we get

$$y = (p - q)^9 = \sum_{j=0}^9 \dbinom{9}{j} (-1)^j p^{(9-j)} q^j.$$
Binomial theorem. No questions there I presume. But

$$q = x^3 \implies q^3 = x^9, \ q^4 = x^{12}, \ q^5 = x^{15}, \text { etc.}$$
We are interested in the coefficient of $x^8$ 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.

$$j = 0 \implies \dbinom{9}{j} * (-1)^j p^{(9-j)}q^j = p^9 = (1 + x^2)^9 =\\ \sum_{k=0}^9 \dbinom{9}{k} 1^{(9-k)} (x^2)^k.$$
The only way we can get $x^8$ when j = 0 is when k = 4.

$$\dbinom{9}{4} = \dfrac{9!}{4! * 5!} = \dfrac{9 * 8 * 7 * 6}{4 * 3 * 2} = 2 * 63 = 126.$$
Let’s think about j = 1.

$$j = 1 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = -9 p^8q = -9x^3(1 + x^2)^8.$$
Is it not obvious that multiplying an odd power of x by even powers of x will never generate $x^8$?

j = 1 is irrelevant.

Let’s think about j = 2.

$$j = 2 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = 36x^6(1 + x^2)^7.$$
The only way to get $x^8$ out of that is with the term containing $x^2$.

$$(1 + x^2)^9 = \sum_{k=0}^7 \dbinom{7}{k}1^{(7-k)}(x^2)^k.$$
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.

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?