Equality of Two Series Expansions

Question

I want to show the function \begin{equation} f(x)=x^{-m} e^x\frac{d^n}{dx^n}(x^{m+n}e^{-x}) \end{equation} is equal to the infinite series \begin{equation} f(x) = \frac{(m+n)!}{m!} \sum_{p=0}^n \frac{(-n)_p}{(m+1)_p p!}x^p. \end{equation} where $(a)_p = a(a+1)\dots(a+p-1)$. I can make it easier by rewriting the second version as \begin{align} f(x) &= \frac{(m+n)!}{m!} \sum_{p=0}^n \frac{(-n)_p}{(m+1)_p p!}x^p \\ &= \frac{(m+n)!}{m!} \sum_{p=0}^n \frac{(-n)(1-n)\dots(p-1-n)}{(m+1)(m+2)\dots(m+p) p!}x^p \\ &= \frac{(m+n)!}{m!} \sum_{p=0}^n \frac{n(n-1)\dots(n-p+1)}{(m+1)(m+2)\dots(m+p) p!}(-1)^p x^p \\ &= (m+n)! \sum_{p=0}^n \frac{n(n-1)\dots(n-p+1)}{p! p!}(-1)^p x^p \\ &= (m+n)! \sum_{p=0}^n \frac{n!}{p!(n-p)! p!}(-1)^p x^p. \end{align} But when I expand the first I get \begin{align} f(x) &= x^{-m} e^x\frac{d^n}{dx^n}(x^{m+n}e^{-x}) \\ &= x^{-m} e^x \sum_{p=0}^n {n \choose p} \frac{d^{n-p}}{dx^{n-p}}(x^{m+n})(\frac{d^p}{dx^p}e^{-x}) \\ &= x^{-m} e^x \sum_{p=0}^n \frac{n!}{p!(n-p)!} (m+n)(m+n-1)\dots (m+n-(n-p+1)) x^{m+n-n+p}(-1)^pe^{-x} \\ &= \sum_{p=0}^n \frac{n!}{p!(n-p)!} (m+n)(m+n-1)\dots (m+p-1)) x^p (-1)^p \end{align} i get stuck and when I expand a different way \begin{align} f(x) &= x^{-m} e^x\frac{d^n}{dx^n}(x^{m+n}e^{-x}) \\ &= x^{-m} e^x \sum_{p=0}^n {n \choose p} \frac{d^p}{dx^p}(x^{m+n})(\frac{d^{n-p}}{dx^{n-p}}e^{-x}) \\ &= x^{-m} e^x \sum_{p=0}^n \frac{n!}{p!(n-p)!} (m+n)(m+n-1)\dots(m+n-p+1) x^{m+n-p}(-1)^{n-p} e^{-x} \\ &= \sum_{p=0}^n \frac{n!}{p!(n-p)!} \frac{(m+n)!}{(m+n-p)!} x^{p}(-1)^{n-p} \\ &= (m+n)! \sum_{p=0}^n \frac{n!}{p!(n-p)!} \frac{1}{(m+n-p)!} (-1)^{n-p} x^{p} \\ &= (m+n)! \sum_{p=0}^n \frac{n(n-1) \dots (n - p + 1) (n-p)!}{p!(n-p)!} \frac{1}{(m+n-p)!} (-1)^{n-p} x^{p} \\ &= (m+n)! \sum_{p=0}^n \frac{(-n)(1-n) \dots (p-1-n)}{p!} \frac{1}{(m+n-p)!} (-1)^{n} x^{p} \end{align} I get stuck again. Help?

Answer 1

When expanding the sum you use $m! (m+1) \cdots (m+p) = p!$ instead of $m! (m+1) \cdots (m+p) = (m+p)!$ at one point. With this adjustment you will find the correct result $$ f(x) = (m+n)! \sum \limits_{p=0}^n \frac{n!}{p!(n-p)!(m+p)!} (-1)^p x^p \, . $$

In the evaluation of the Rodrigues formula you should replace every $i$ by a $p$. In your first try there is just a small mistake: you should have $$ \frac{\mathrm{d}^{n-p}}{\mathrm{d}x^{n-p}} x^{m+n} = (m+n) \cdots (m+n - (n-p) \color{red}{+1}) x^{m+p} = \frac{(m+n)!}{(m+p)!} x^{m+p} \, . $$ Then the final results agree and you are done.

As for your second try, I do not know what $$ \frac{\mathrm{d}^p}{\mathrm{d} x^{n-p}}$$ is supposed to mean, but you can simply take the first approach anyway.