Exponential generating function - what happens when there is a leftover term?

Question

Find the coefficient on $x^2/2!$ in the following generating function: $$xe^{3x}-x^2$$

I got this far: $$x\sum_{n=0}^{\infty}\frac{3^nx^n}{n!}-x^2=\sum_{n=0}^{\infty}\frac{3^nx^{n+1}}{n!}-x^2$$

So, there are two terms here. To find the coefficient of $x^2/2!$, I can see that for the first sigma term, $n = 1$. So we have $\frac{3^1x^2}{1!}=\frac{6x^2}{2!}$. Thus the coefficient for the first term is 6. (is that right?) But what about the coefficient for the term $-x^2$? Is it just $-1$? I guess that $-x^2$ can be rewritten as $-\frac{2x^2}{2!}$, so maybe the coefficient should be $-2$? I would appreciate any clarification on this. Thanks.

Answer 1

It’s probably simplest first to find the coefficient of $x^2$. The $x^2$ term in the summation

$$\sum_{n\ge 0}\frac{3^nx^{n+1}}{n!}$$

is the $n=1$ term, $\dfrac{3^1x^{1+1}}{1!}=3x^2\;.$ The $x^2$ term in $xe^{3x}-x^2$ is therefore $3x^2-x^2=2x^2$.

To finish the problem, you need to express $2x^2$ as a multiple of $\dfrac{x^2}{2!}$:

$$2x^2=4\cdot\frac{x^2}{2!}\;,$$

so the desired coefficient is $4$. Your approach will also work: $x^2=\dfrac{2x^2}{2!}$, so the $x^2$ term is

$$\frac{6x^2}{2!}-\frac{2x^2}{2!}=4\cdot\frac{x^2}{2!}\;.$$

Answer 2

Pikov and Dropov, the limousine company of the Tappet Brothers will happily provide the solution. Pick off the $x^2$ terms from the two sums. Get the $x^2$ term for the sum. Now adjust for $x^2/2$.