Exponential Generating Functions With Coefficients

Question

Given the exponential generating function, $B(x)=\sum_{n\geq 0}b_n\frac{x^n}{n!}=b_0+b_1\frac{x}{1!}+b_2\frac{x^2}{2!}+\cdots$, I want to find a formula for the generating function \begin{equation*} \tilde{B}(x)=b_1+b_0\frac{x}{1!}+b_3\frac{x^2}{2!}+b_2\frac{x^3}{3!}+b_5\frac{x^4}{4!}+b_4\frac{x^5}{5!}+\cdots \end{equation*} in terms of $B(x)$. Now to know if the formula is correct then if \begin{equation*}B(x)=e^x=\tilde{B}(x). \end{equation*} Likewise if \begin{equation*} B(x)=xe^x=\sum_{n\geq 0}n\frac{x^n}{n!}\end{equation*} then \begin{equation*} \tilde{B}(x)=xe^x+e^{-x}=1+0x+3\frac{x^2}{2!}+2\frac{x^3}{3!}+5\frac{x^4}{4!}+\cdots.\end{equation*} How would i go about swapping coefficients, would the formula involve a negative exponent like \begin{equation*}\frac{x^{-n}}{n!}?\end{equation*}

Answer 1

You can bisect a generating function into its even and odd parts with the formulae

$$ \sum_{n \text{ even}} b_n \frac{x^n}{n!} = \frac{B(x) + B(-x)}{2} $$

and

$$ \sum_{n \text{ odd}} b_n \frac{x^n}{n!} = \frac{B(x) - B(-x)}{2}. $$

You can shift the terms in an exponential generating function by integrating or differentiating.

If you put this together you get the formula

$$ \bar B(x)= \frac{C(x) - C(-x)}{2} + \frac{D(x) + D(-x)}{2} $$

where $C = \int B$ and $D = \frac{d}{dx} B$.

For $B(x) = xe^x$ we have $C(x) = (x - 1)e^x + 1$ and $D(x) = e^x + xe^x$. Therefore

$$ \bar B(x) = \frac{(x - 1)e^x - (-x - 1)e^{-x}}{2} + \frac{e^x + xe^x + e^{-x} - xe^{-x}}{2} = xe^x - e^{-x}. $$