How to expand product of $n$ factors.

Question

I have a product say \begin{equation} F(a,n,x) = \prod _{j=0}^{n}(1-{a}^{n-2\,j}x) \end{equation} I want to expand and hope to have general terms of the coefficients. I did for $n= 2,3,4,5,6,7,8...$ I see it will be different for $n$ even or $n$ odd. We have \begin{equation*} F(a,2,x)= 1-{x}^{3}+ \left( {a}^{2}+1+{a}^{-2} \right) {x}^{2}+ \left( -{a}^{2}- 1-{a}^{-2} \right) x \end{equation*} \begin{equation} F(a,3,x)= 1+{x}^{4}+ \left( -{a}^{3}-a-{a}^{-3}-{a}^{-1} \right) {x}^{3}+ \left( {a}^{4}+2+{a}^{-2}+{a}^{2}+{a}^{-4} \right) {x}^{2}+ \left( -{ a}^{3}-a-{a}^{-3}-{a}^{-1} \right) x \end{equation} The coefficients of $a$ gets more interesting as $n$ grows. I am interested in the coefficient of $a.$ Does anyone know how to expand product of $n$ factors.

Answer 1

When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.

We obtain \begin{align*} F(a,n,x)&=\prod_{j=0}^n\left(1-a^{n-2j}x\right)\tag{1}\\ &=\sum_{S\subseteq \{0,1,\ldots,n\}}(-x)^{|S|}a^{n|S|}\prod_{j\in S}a^{-2j}\tag{2}\\ &=1+\sum_{k=1}^{n+1}\sum_{{S\subseteq \{0,1,\ldots,n\}}\atop{|S|=k}}(-x)^{|S|}a^{n|S|}\prod_{j\in S}a^{-2j}\tag{3}\\ &=1+\sum_{k=1}^{n+1}(-x)^ka^{nk}\sum_{0\leq j_1<\cdots<j_k\leq n}a^{-2(j_1+\cdots+j_k)}\tag{4}\\ &=1+\sum_{k=1}^{n+1}(-x)^ka^{nk}\sum_{l=k(k-1)/2}^{k(2n-k+1)/2}\sum_{{0\leq j_1<\cdots<j_k\leq n}\atop{j_1+\cdots +j_k=l}}a^{-2(j_1+\cdots +j_k)}\tag{5}\\ &=1+\sum_{k=1}^{n+1}(-1)^k\sum_{l=k(k-1)/2}^{k(2n-k+1)/2}\left(\sum_{{0\leq j_1<\cdots<j_k\leq n}\atop{j_1+\cdots +j_k=l}}1\right)a^{nk-2l}x^k\tag{6}\\ &=1+\sum_{k=1}^{n+1}(-1)^k\sum_{l=0}^{k(n-k+1)} \left(\sum_{{0\leq j_1<\cdots<j_k\leq n}\atop{j_1+\cdots +j_k=l+k(k-1)/2}}1\right)a^{k(n-k+1)-2l}x^k\tag{7}\\ &=1+\sum_{k=1}^{n+1}\color{blue}{(-1)^k}\sum_{l=0}^{k(n-k+1)} \left(\color{blue}{\sum_{{0\leq q_1\leq \cdots\leq q_k\leq n}\atop{q_1+\cdots +q_k=l}}1}\right)a^{k(n-k+1)-2l}x^k\tag{8}\\ \end{align*} The coefficients of $a$ in $P_{n,k}(a)$ are given in (8) as the blue marked sum times $(-1)^k$.

We conclude the absolute value of the coefficient gives the number of integer partitions of $l$ into $k$ parts with largest part at most $n$.

Comment:

• In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $S\subseteq \{0,1,\ldots,n\}$.

• In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=\emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.

• In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S=\{j_1,j_2,\ldots,j_k\}$ for each specific choice.

• In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+\cdots+j_k$ of the $k$-tupels.

  We observe the smallest sum comes from the $k$-tupel $(0,1,2,\ldots,k-1)$ which gives $$\sum_{j=0}^k j=k(k-1)/2$$

  while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,\ldots,n)$ which gives \begin{align*}\sum_{j=n-k+1}^n j&=\sum_{j=1}^n j-\sum_{j=1}^{n-k} j=\frac{n(n+1)}{2}-\frac{(n-k)(n-k+1)}{2}\\ &=\frac{k(2n-k+1)}{2}. \end{align*}

• In (6) we factor out $a^{-2(j_1+\cdots+j_k)}=a^{-2l}$.

• In (7) we shift the index of $l$ to start from $0$.

• In (8) we finally change the index variables $j_t=q_t+t-1, 1\leq t\leq k$. We have \begin{align*} j_1+j_2+\cdots+j_k&=l+\frac{k(k-1)}{2}\\ q_1+(q_2+1)+\cdots+(q_k+k-1)&=l+\frac{k(k-1)}{2}\\ q_1+q_2+\cdots+q_k&=l \end{align*}

  This way we get an inequality chain containing $\leq$ symbols only which admits a nice interpretation via partitions.

Example $F(a,3,x)$:

We evaluate the expression (8) for the case $n=3$. We obtain \begin{align*} \color{blue}{F(a,3,x)}&= 1+\sum_{k=1}^4(-1)^k\sum_{l=0}^{k(4-k)}\left(\sum_{{0\leq q_1\leq \cdots\leq q_k\leq 3}\atop{q_1+\cdots +q_k=l}} 1\right)a^{k(4-k)-2l}x^k\\ &=1-\sum_{l=0}^3\left(\sum_{{0\leq q_1\leq 3}\atop{q_1=l}}1\right)a^{3-2l}x +\sum_{l=0}^4\left(\sum_{{0\leq q_1\leq q_2\leq 3}\atop{q_1+q_2=l}} 1\right)a^{4-2l}x^2\\ &\qquad -\sum_{l=0}^3\left(\sum_{{0\leq q_1\leq q_2\leq q_3\leq 3}\atop{q_1+q_2+q_3=l}} 1\right)a^{3-2l}x^3 -\sum_{l=0}^0\left(\sum_{{0\leq q_1\leq q_2\leq q_3\leq q_4\leq 3}\atop{q_1+q_2+q_3+q_4=l}} 1\right)a^{-2l}x^4\\ &\,\,\color{blue}{=1-\left(a^3+a+a^{-1}+a^{-3}\right)x+\left(a^4+a^2+2+a^{-2}+a^{-4}\right)x^2}\\ &\qquad\,\,\color{blue}{-\left(a^3+a+a^{-1}+a^{-3}\right)x^3+x^4} \end{align*} in accordance with OPs calculation.