General form of the coefficients of the polynomial $p(z)=\binom{q+z}{n}+\binom{q-z}{n}$

Question

In my mathematical wanderings I found this polynomial in $z$ $$ p(z)=\binom{q+z}{n}+\binom{q-z}{n} $$ Now , I'm wondering what the general terms for the coefficients are. First of all its easy to see that $p(z)=p(-z)$. The parameter $q$ is a rational number and $n \in \mathbb{N}_{0}$ may vary in the interval $0\leq n \leq \lfloor q \rfloor$. If we write this polynomial as $p(z)=a_{n}z^{2n}+\cdots+a_{0}$, the degree only jumps for $n$ even. The general term for $a_{0}$ is $$ a_{0}=2\binom{q}{n} $$ My question: what is the general form of the remaining terms?

Thanks.

Answer 1

We use falling factorials $z^{\underline{n}}=z(z-1)\cdots (z-n+1)$ and the Stirling numbers of the first kind $\left[n\atop k\right]$ with the identity \begin{align*} z^{\underline{n}}=\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]z^k \end{align*}

We obtain \begin{align*} p(z)&=\binom{q+z}{n}+\binom{q-z}{n}\\ &=\frac{1}{n!}(q+z)^{\underline{n}}+\frac{1}{n!}(q-z)^{\underline{n}}\\ &=\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right](q+z)^k +\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right](q-z)^k\\ &=\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]\left\{(q+z)^k+(q-z)^{k}\right\}\tag{1} \end{align*}

We use the coefficient of operator $[z^j]$ to denote the coefficient of $z^j$ in a series.

Since $p$ is an even polynomial we consider $[z^{2j}]$ for $0\leq 2j\leq n$ and obtain from (1): \begin{align*} \color{blue}{[z^{2j}]p(z)}&=[z^{2j}]\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]\left\{(q+z)^k+(q-z)^{k}\right\}\\ &=\frac{1}{n!}\sum_{k=2j}^n(-1)^{n-k}\left[n\atop k\right][z^{2j}]\left\{\sum_{l=0}^k\binom{k}{l}z^lq^{k-l}+\sum_{l=0}^k\binom{k}{l}(-z)^lq^{k-l}\right\}\tag{2}\\ &\color{blue}{=\frac{2}{n!}\sum_{k=2j}^n(-1)^{n-k}\left[n\atop k\right]\binom{k}{2j}q^{k-2j}}\tag{3} \end{align*}

We get in accordance with OP's calculation \begin{align*} a_0=[z^0]p(z)=\frac{2}{n!}\sum_{k=0}^n(-1)^{n-k}\left[n\atop k\right]q^{k} =\frac{2}{n!}q^{\underline{n}}=2\binom{q}{n}. \end{align*}

Comment:

• In (2) we use the linearity of the coefficient of operator and apply the binomial theorem twice.

• In (3) we select the coefficient of $z^{2j}$ accordingly.

Answer 2

$$ a_i=\frac{1}{i!}\frac{\partial^i(p(z))}{\partial z^i} \Big|_{z=0}. $$ Formulas for derivation of binomial coefficients you may find here