How can I find the coefficient of $x^{23}$ in $(x^3+x^4+x^5+x^6)^4$?

Question

How can I find the coefficient of $x^{23}$ with polynomial expansion?

Answer 1

The highest term is $x^{24}$, obtained by using $x^6$ in each factor. To get a term in $x^{23}$, three of the four factors must be $x^{6}$ and one $x^5$. There are $4$ choices of which of the four factors is $x^5$. Thus the coefficient is $4$.

Answer 2

You will need to multiply $x^6$ three times by $x^5$; this can be done in $\color{blue}{4}$ ways, \begin{eqnarray*} (x^3+x^4+x^5+\color{red}{x^6})(x^3+x^4+x^5+\color{red}{x^6})(x^3+x^4+x^5+\color{red}{x^6})(x^3+x^4+\color{red}{x^5}+x^6)=\cdots+\color{blue}{4}x^{23}+x^{24}. \end{eqnarray*}

Answer 3

Note that: $(a_1+a_2+...+a_m)^n=\sum \frac{n!}{k_1!k_2!...k_m!}a_1^{k_1}a_2^{k_2}...a_m^{k_m}$ with $k_1+k_2+...+k_m=n$ And: 23 = 6 + 6 + 6 + 5

Answer 4

Just to give a different approach, the coefficient of $x^{23}$ in $(x^3+x^4+x^5+x^6)^4$ is the same as the coefficient of $x$ in $f(x)=x^{24}\left({1\over x^3}+{1\over x^4}+{1\over x^5}+{1\over x^6}\right)^4=(x^3+x^2+x+1)^4$, which is $f'(0)=4$, since $f'(x)=4(x^3+x^2+x+1)^3(3x^2+2x+1)$.

The value (if any) in this approach is that you can get a few other coefficients as well, without doing any combinatorics. For example, the coefficient of $x^{22}$ (for the original expression) is ${1\over2}f''(0)$, where

$$f''(x)=12(x^3+x^2+x+1)^2(3x^2+2x+1)+4(x^3+x^2+x+1)^3(6x+2)$$

so that ${1\over2}f''(0)={1\over2}(12+8)=10$. Eventually, though, the taking of derivatives becomes cumbersome.

Answer 5

$$\begin{eqnarray*}[x^{23}](x^3+x^4+x^5+x^6)^4&=&[x^{23}]\,x^{12}(1+x+x^2+x^3)^4\\&=&[x^{11}](1+x+x^2+x^3)^4\\&=&[x^{11}]\frac{(1-x^4)^4}{(1-x)^4}\\&=&[x^{11}](1-4x^4+6x^8)\sum_{n\geq 0}\binom{n+3}{3}x^n\\&=&\binom{11+3}{3}-4\binom{7+3}{3}+6\binom{3+3}{3}=\color{red}{4}.\end{eqnarray*}$$

Answer 6

Just for fun the clever argument from @RobertIsreal algebraically encoded:

We obtain \begin{align*} \color{blue}{[x^{23}]\left(x^3+x^4+x^5+x^6\right)^4} &=[x^{-1}]\left(1+x^{-1}+x^{-2}+x^{-3}\right)^4\tag{1}\\ &=[x^{-1}]\left(1+x^{-1}\right)^4\tag{2}\\ &\color{blue}{=4}\tag{3} \end{align*}

Comment:

• In (1) we factor out $x^{24}$ and apply the shifting rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

• In (2) we observe that only constant terms and $x^{-1}$ provide a non-zero contribution.

• In (3) we note that there are $4$ different ways in (2) to select $x^{-1}$.