Series representation for $(1+x)^{n-m}(1-x)^{m}$

Question

As the title suggests, for $n,m$ integers such that $0\leq m \leq n$, is there a series representation for

$$f(n,m)=(1+x)^{n-m}(1-x)^{m}$$

for each $m$? Been looking online to no avail...

Answer 1

First note that $$\left(\frac{d}{dx}\right)^k x^\alpha=p_k(\alpha)x^{\alpha-k}$$ where $$p_k(\alpha)=\prod_{j=1}^{k}(\alpha-j+1).$$ Hence for $|x|<1$, $$(1-x)^\alpha=\sum_{k\ge0}\frac{(-1)^kp_k(\alpha)}{k!}x^k$$ and $$(1+x)^\beta=\sum_{k\ge0}\frac{p_k(\beta)}{k!}x^k.$$ Hence $$(1+x)^\beta(1-x)^\alpha=\sum_{n\ge0}x^n\sum_{k=0}^{n}\frac{(-1)^kp_k(\alpha)p_{n-k}(\beta)}{k!(n-k)!},$$ which is $$(1+x)^\beta(1-x)^\alpha=\sum_{n\ge0}\frac{x^n}{n!}\sum_{k=0}^{n}(-1)^k{n\choose k}p_k(\alpha)p_{n-k}(\beta).\tag{1}$$ I leave the details of your specific case to you.

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As we may recognize, $$p_n(\alpha)=(-1)^n(-\alpha)_n=(-1)^n\frac{\Gamma(n-\alpha)}{\Gamma(-\alpha)},$$ so $(1)$ corresponds quite closely to the identity $$(a+b)_n=\sum_{k=0}^{n}{n\choose k}(a)_k(b)_{n-k}$$ which comes from the fact that $(1-x)^{-a}(1-x)^{-b}=(1-x)^{-(a+b)}$ and $$(1-x)^{-s}=\sum_{n\ge0}\frac{(s)_n}{n!}x^n=\,_1F_0(s;;x).$$

Answer 2

So I think you want to expand the expression? $$ (1+x)^{n-m}(1-x)^m = \sum_{i=0}^{n-m}\binom{n-m}{i} x^i \cdot \sum_{j=0}^{m}\binom{m}{j}(-x)^j = \sum_{k=0}^n\sum_{l=\max(0,k-n+m)}^{\min(k,m)} \binom{n-m}{k-l}\binom{m}{l} x^{k-l}(-x)^{l} $$ $$ = \sum_{k=0}^n x^k \sum_{l=\max(0,k-n+m)}^{\min(k,m)} (-1)^l\binom{n-m}{k-l}\binom{m}{l} $$ I'm not sure how to simplify the last sum with the binomial coefficients, but this is a power series expansion.

Answer 3

More than likely, too complex.

Using what has been given in previous answers $$f(n,m)=(1+x)^{n-m}(1-x)^{m}=\sum_{p=0}^\infty a_p\, x^p$$ where $$a_p=\binom{n-m}{p} \, _2F_1(-m,-p;n-m-p+1;-1)$$ where appears the gaussian hypergeometric function.