Question

Find the sequence with each of these functions as its exponential generating function$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{-}\mathbf{3}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$.

Short answer

$a_k=3^k-3·2^k$.

Step-by-step solution

Given data

The given function is $f\left(x\right)=e^{3x}-3e^{2x}$

Concept used of generating function

The ordinary generating function of a sequence ${\mathit{a}}_{\mathbf{n}}$ is

$\mathit{G}\mathbf{(}{\mathbf{a}}_{\mathbf{n}}\mathbf{;}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}\mathbf{.}$

Solve the function

Using $\sum _{k=0}^{+\infty }\frac{a_k}{k!}{x}^k=e^x$, we have

$e^{3x}-3e^{2x}$=$\sum _{k=0}^{+\infty }\frac{{\left(3x\right)}^k}{k!}-3\sum _{k=0}^{+\infty }\frac{{\left(2x\right)}^k}{k!}$

$=\sum _{k=0}^{+\infty }{3}^k\frac{x^k}{k!}-3\sum _{k=0}^{+\infty }{2}^k\frac{x^k}{k!}$

$=\sum _{k=0}^{+\infty }\left(3^k-3·2^k\right)\frac{x^k}{k!}$

The sequence $\left\{a_k\right\}$are then the coefficients of $\frac{x^k}{k!}$in the above sum: $a_k=3^k-3·2^k$.