Finding exponential generating function of sequence $\left\{0,1,2a,3a^2,4a^3,\ldots\right\}$

Question

I need to find exponential generating function for the sequence $\left\{0,1,2a,3a^2,4a^3,\ldots\right\}$.
My attempt:
I know that $e^x$ generates $\{1,1,1,1,\ldots\}$, so $\,e^{ax}$ must generate $\{1,a,a^2,a^3,a^4,\ldots\}$. Then I tried taking derivative of $\,e^{ax}$, which is $\,ae^{ax}$ so that I would have the sequence $\left\{1,2a,3a^2,4a^3,\ldots\right\}$. If I multiply the sequence by $\,x\,,\,$ that should give me the sequence $\left\{0,1,2a,3a^2,4a^3,\ldots\right\}$.
Is my approach correct?

Answer 1

As @TrevorGunn already pointed out, derivation of $e^{ax}$ with respect to $x$ to get $ae^{ax}$ leads from \begin{align*} \left(1,a,a^2,a^3,\ldots\right)\quad\to\quad\left(a,a^2,a^3,a^4,\ldots\right) \end{align*} which is not that helpful to obtain the exponential generating function (egf) of the sequence \begin{align*} \left(0,1,2a,3a^2,4a^3,\ldots\right)\tag{1} \end{align*}

Here is a slightly different approach. In order to reach the wanted egf from (1) it is sufficient to find a closed expression of the $n$-th term of the sequence. It's not too hard to see \begin{align*} \left(0,1,2a,3a^2,4a^3,\ldots\right)=\color{blue}{\left(na^{n-1}\right)_{n\geq 0}}\tag{2} \end{align*} From (2) we can immediately derive the wanted egf, namely \begin{align*} \color{blue}{\sum_{n=0}^{\infty}na^{n-1}\frac{x^n}{n!}}\tag{3} \end{align*} The representation (3) is given in a form, where we see it's an egf and the job is done.

A simplification of (3) as for instance \begin{align*} \color{blue}{\sum_{n=0}^{\infty}na^{n-1}\frac{x^n}{n!}} &=\sum_{n=1}^{\infty}na^{n-1}\frac{x^n}{n!}\\ &=x\sum_{n=1}^{\infty}a^{n-1}\frac{x^{n-1}}{(n-1)!}\\ &=x\sum_{n=0}^{\infty}a^n\frac{x^n}{n!}\\ &\,\,\color{blue}{=xe^{ax}} \end{align*} is just a nice add-on.

Answer 2

If the derivative is $ae^{ax}$ that is just multiplying everything by $a$ so $\{a, a^2, a^3, \dots\}$. For exponential generating functions, derivatives shift the sequence—the derivative of $x^n / n!$ is $x^{n - 1}/(n - 1)!$ can you see how this relates to what I just said about derivatives shifting the coefficients?.

Next, if you multiply

$$ x \cdot \sum a_n \frac{x^n}{n!} = \sum a_n \frac{x^{n+1}}{n!},$$

it isn't a simple shift because for this to be an EGF it needs to be in terms of $x^{n+1} / (n + 1)!$. We need to rewrite it as

$$\sum a_n \frac{x^{n+1}}{n!} \cdot \color{blue}{\frac{n + 1}{n + 1}} = \sum a_n(n + 1)\frac{x^{n+1}}{(n + 1)!}$$

Can you use this to correct your working out?