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Find the coefficient of $x^6$ in the expansion of $(1-2x+3x^2-4x^3+5x^4)^4$.

My attempt:

Consider $(1-y)^{-2}=1+2y+3y^2+4y^3+\dots$

Then I substitute $y=-x$, get $(1+x)^{-2}= 1-2x+3x^2-4x^3+\dots$

The original polynomial can be written as $[(1+x)^{-2} + (6x^5-7x^6\dots)]^4$.

Now, I consider $(1+x)^{-2}$ as $A$, $(6x^5-7x^6......)$ as $B$, so I have $(A+B)^4= A^4 +4C_3\cdot A^3\cdot B+4C_2\cdot A^2\cdot B^2+4C_1AB^3+B^4$.

For coefficient of $x^6$, I can just consider the first two terms: $A^4 +4C_3\cdot A^3\cdot B$, since for the other two terms, $\min \deg(x)>6$. So I plug in back, I get: $(1+x)^{-8}+4(1+x)^{-6}\cdot(6x^5-7x^6)$.

I don't know how I can proceed next. Can I take $x=-y$ and use the negative binomial expansion, I'm not sure whether it is valid or not.

Dstarred
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user1228735
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2 Answers2

3

The mechanical way: we enumerate sets $\{a,b,c,d\}$ such that $a+b+c+d = 6$ and $a \ge b \ge c \ge d \in \{0, 1, 2, 3, 4\}$. These are: $$\{4,2,0,0\}, \{4,1,1,0\}, \{3,3,0,0\}, \{3,2,1,0\}, \{3,1,1,1\}, \{2,2,2,0\}, \{2,2,1,1\}.$$ These represent the distinct ways to pick exponents in the expansion that will add up to exactly $6$. The next step is to compute the product of the coefficients of these terms, and then compute the number of times these products appear in the expansion.

For $\{4,2,0,0\}$, this corresponds to the product $(5)(3)(1)(1) = 15$, and there are $\frac{4!}{1!1!2!} = 12$ ways to pick these powers in the expansion, for a contribution of $12(15) = 180$.

For $\{4,1,1,0\}$, this corresponds to the product $(5)(-2)(-2)(1) = 20$, and there are $\frac{4!}{1!2!1!} = 12$ ways to pick these powers in the expansion, for a contribution of $240$.

These and the remaining partitions are summarized as follows:

$$\begin{array}{c|c|c|c} \text{Partition} & \text{Coefficients} & \text{Combinations} & \text{Total} \\ \hline \{4,2,0,0\} & (5)(3)(1)(1) = 15 & \frac{4!}{1!1!2!} = 12 & 180 \\ \{4,1,1,0\} & (5)(-2)(-2)(1) = 20 & \frac{4!}{1!2!1!} = 12 & 240 \\ \{3,3,0,0\} & (-4)(-4)(1)(1) = 16 & \frac{4!}{2!2!} = 6 & 96 \\ \{3,2,1,0\} & (-4)(3)(-2)(1) = 24 & \frac{4!}{1!1!1!1!} = 24 & 576 \\ \{3,1,1,1\} & (-4)(-2)(-2)(-2) = 32 & \frac{4!}{1!3!} = 4 & 128 \\ \{2,2,2,0\} & (3)(3)(3)(1) = 27 & \frac{4!}{3!1!} = 4 & 108 \\ \{2,2,1,1\} & (3)(3)(2)(2) = 36 & \frac{4!}{2!2!} = 6 & 216 \\ \hline & & & 1544 \end{array}$$ So the coefficient of $x^6$ is $1544$.

heropup
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1

Here is another approach. We try to use the nice regular structure of the polynomial for a suitable transformation before we extract the coefficient of $x^6$. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series and want to find \begin{align*} [x^6]\left(1-2x+3x^2-4x^3+5x^4\right)^4\tag{1} \end{align*} We obtain \begin{align*} \color{blue}{P(x)}&\color{blue}{=1-2x+3x^2-4x^3+5x^4}\\ &=\frac{d}{dx}\left(x-x^2+x^3-x^4+x^5\right)\\ &=\frac{d}{dx}\left(x\frac{1+x^5}{1+x}\right)\\ &=\frac{1+x^5}{1+x}-x\left(1-2x+3x^3-4x^4\right)\tag{$(uv)^{\prime}=u^{\prime}v+uv^{\prime}$}\\ &=\frac{1+x^5}{1+x}-x\left(P(x)-5x^4\right)\tag{2}\\ \end{align*}

From (2) we get $P(x)$ in the form \begin{align*} \color{blue}{P(x)=\frac{1+x^5}{(1+x)^2}+\frac{5x^5}{1+x}} \end{align*} and use this representation to extract the coefficient of $x^6$ from $\left(P(x)\right)^4$.

We obtain by consequently skipping powers of $x$ which do not contribute to $[x^6]$: \begin{align*} \color{blue}{[x^6]\left(P(x)\right)^4}&=[x^6]\left(\frac{1+x^5}{(1+x)^2}+\frac{5x^5}{1+x}\right)^4\\ &=[x^6]\frac{\left(1+x^5\right)^4}{(1+x)^8}+20[x]\frac{\left(1+x^5\right)^3}{(1+x)^7}\tag{3}\\ &=[x^6]\frac{\left(1+4x^5\right)}{(1+x)^8}+20[x]\frac{1}{(1+x)^7}\tag{4}\\ &=[x^6](1+x)^{-8}+4[x](1+x)^{-8}+20[x](1+x)^{-7}\\ &=\binom{-8}{6}+4\binom{-8}{1}+20\binom{-7}{1}\tag{5}\\ &=\binom{13}{6}-4\binom{8}{1}-20\binom{7}{1}\tag{6}\\ &=1\,716-32-140\\ &\,\,\color{blue}{=1\,544} \end{align*} and get the same result as calculated by @heropup.

Comment:

  • In (3) we apply the binomial theorem to $(a(x)+b(x))^4$ and take only $a^4(x)+4a^3(x)b(x)$ since other terms do not contribute to $x^6$. We also apply the rule $[x^{m-n}]A(x)=[x^m]x^nA(x)$.

  • In (4) we expand the numerators, again applying the binomial theorem and skip terms which do not contribute to $[x^6]$.

  • In (5) we use the binomial series expansion.

  • In (6) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

Markus Scheuer
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