Question

If $\mathit{G}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is the generating function for the sequence$\left\{a_k\right\}$, what is the generating function for each of these sequences?

a)$\mathbf{2}{\mathit{a}}_{\mathbf{0}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{3}}\mathbf{,}\mathbf{\dots }$

b) $\mathbf{0}\mathbf{,}{\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{,}\mathbf{\dots }$(Assuming that terms follow the pattern of all but the first term)

c) $\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{,}\mathbf{\dots }$(Assuming that terms follow the pattern of all but the first four terms)

d)${\mathit{a}}_{\mathbf{2}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{,}{\mathit{a}}_{\mathbf{4}}\mathbf{,}\mathbf{\dots }$

e) ${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{3}{\mathit{a}}_{\mathbf{3}}\mathbf{,}\mathbf{4}{\mathit{a}}_{\mathbf{4}}\mathbf{,}\mathbf{\dots }$[Hint: Calculus required here.]

f)${\mathit{a}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{0}}{\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{a}}_{\mathbf{0}}{\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{0}}{\mathit{a}}_{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{2}{\mathit{a}}_{\mathbf{0}}{\mathit{a}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{3}}\mathbf{+}{\mathit{a}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{,}\mathbf{\dots }$

Short answer

The generating function for each of the given sequences is below:

(a) 2G(x)

(b) x G(x)

(c) $x^2\left(G\left(x\right)-a_0-a_{1}x\right)$

(d)$\frac{1}{x^2}\left(G\left(x\right)-a_0-a_{1}x\right)$

(e)$G^{\text{'}}\left(x\right)$

(f)$G^2\left(x\right)=G\left(x\right)·G\left(x\right)$

Step-by-step solution

Use Generating Function:

Generating function for the sequence${\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}\mathbf{,}\mathbf{\dots }$of real numbers is the infinite series

$\mathit{G}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}$

The given sequence is:

$2a_0,2a_1,2a_2,2a_3,\dots$

By generating function:

$$\begin{gathered} G_a\left(x\right)=2a_0+2a_{1}x+2a_2{x}^2+\dots \\ =2\left(a_0+a_{1}x+a_2{x}^2+\dots \right) \\ =2\sum _{k=0}^{+\infty }{a}_k{x}^k \\ =2G\left(x\right) \end{gathered}$$

Use Generating Function:

G(x) is the generating function for the sequence $\left\{a_k\right\}$

$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

The given sequence is:

$0,a_0,a_1,a_2,a_3,\dots$

By generating function:

$$\begin{gathered} G_a\left(x\right)=0+a_{0}x+a_1{x}^2+a_2{x}^3+\dots \\ =a_{0}x+a_1{x}^2+a_2{x}^3+\dots . \\ =x\left(a_0+a_{1}x+a_2{x}^2+\dots \right) \\ =x\sum _{k=0}^{+\infty }{a}_k{x}^k \\ =xG\left(x\right) \end{gathered}$$

Use Generating Function:

G(x) is the generating function for the sequence $\left\{a_k\right\}$$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

The given sequence is:

$0,0,0,a_2,a_3,\dots$

By generating function:

$$\begin{gathered} G_a\left(x\right)=0+0x+0x^2+0x^3+a_2{x}^4+a_3{x}^5+a_4{x}^6+\dots \\ =a_2{x}^4+a_3{x}^5+a_4{x}^6+\dots \\ =-a_0{x}^2-a_1{x}^3+a_0{x}^2+a_1{x}^3+a_2{x}^4+a_3{x}^5+a_4{x}^6+\dots \\ =-a_0{x}^2-a_1{x}^3+x^2\left(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\dots \right) \end{gathered}$$

By further simplification:

$$\begin{gathered} G_a\left(x\right)=-a_0{x}^2-a_1{x}^3+x^2\sum _{k=0}^{+\infty }{a}_k{x}^k \\ =-a_0{x}^2-a_1{x}^3+x^{2}G\left(x\right) \\ =x^{2}G\left(x\right)-a_0{x}^2-a_1{x}^3 \\ =x^2\left(G\left(x\right)-a_0-a_{1}x\right) \end{gathered}$$

Use Generating Function:

G(x) is the generating function for the sequence $\left\{a_k\right\}$$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

The given sequence is:

$a_2,a_3,a_4\dots$

By generating function:

$$\begin{gathered} G_a\left(x\right)=a_2+a_{3}x+a_4{x}^2+\dots . \\ =\frac{1}{x^2}\left(a_2{x}^2+a_3{x}^3+a_4{x}^4+\dots .\right) \\ =\frac{1}{x^2}\left(-a_0-a_{1}x+a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\dots .\right) \\ =\frac{1}{x^2}\left(\sum _{k=0}^{+\infty }{a}_k{x}^k-a_0-a_{1}x\right) \\ =\frac{1}{x^2}\left(G\left(x\right)-a_0-a_{1}x\right) \end{gathered}$$

Use Generating Function:

G(x) is the generating function for the sequence $\left\{a_k\right\}$$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

The given sequence is:

$a_1,2a_2,3a_3,4a_4\dots$

Take the derivative with respect to x :

$$\begin{gathered} G^{\text{'}}\left(x\right)=a_1+2a_{2}x+3a_3{x}^2+\dots .+k{a}_k{x}^{k-1}+\dots \\ =\sum _{k=0}^{+\infty }k{a}_k{x}^{k-1} \end{gathered}$$

We then note that $G^{\text{'}}\left(x\right)$is the generating function of the given sequence.

Use the Theorem of addition and multiplication of generating functions.

If $f\left(x\right)=\sum _{k=0}^{+\infty }{a}_k{x}^k$any then $g\left(x\right)=\sum _{k=0}^{+\infty }{b}_k{x}^k$,

$f\left(x\right)·g\left(x\right)=\sum _{k=0}^{+\infty }\left(\sum _{j=0}^k{a}_j{b}_{k-j}\right)x^k$

The given sequence is:

$a_0^2,2a_0{a}_1,a_1^2+2a_0{a}_2,2a_0{a}_3+2a_1{a}_2,2a_0{a}_4+2a_1{a}_3+a_2^2,\dots$

By the above theorem:

$$\begin{gathered} G^2\left(x\right)=G\left(x\right)·G\left(x\right) \\ =\sum _{k=0}^{+\infty }\left(\sum _{j=0}^k{a}_j{a}_{k-j}\right)x^k \\ =a_0^2+2a_0{a}_{1}x+\left(a_1^2+2a_0{a}_2\right)x^2+\left(2a_0{a}_3+2a_1{a}_2\right)x^3+\left(2a_0{a}_4+2a_1{a}_3+a_2^2\right)x^4+\dots \end{gathered}$$

Thus$G^2\left(x\right)=G\left(x\right)·G\left(x\right)$ has the required sequence of coefficients.