Question

For each of these generating functions, provide a closed formula for the sequence it determines.

a) \({(3x - 4)^3}\)

b) \({\left( {{x^3} + 1} \right)^3}\)

c) \(1/(1 - 5x)\)

d) \({x^3}/(1 + 3x)\)

e) \({x^2} + 3x + 7 + \left( {1/\left( {1 - {x^2}} \right)} \right)\)

f) \(\left( {{x^4}/\left( {1 - {x^4}} \right)} \right) - {x^3} - {x^2} - x - 1\)

g) \({x^2}/{(1 - x)^2}\)

h) \(2{e^{2x}}\)

Short answer

(a) The required result is\({a_0} = - 64,{a_1} = 144,{a_2} = - 108,{a_3} = 27,{a_n} = 0\)when\(n \ge 4\).

(b) The required result is\({a_0} = 1,{a_3} = 3,{a_6} = 3,{a_9} = 1,{a_n} = 0\)when\(n = 1,2,4,5,7,8\)and\(n \ge 10\).

(c) The required result is\({a_n} = {5^n},n = 0,1,2, \ldots \)

(d) The required result is\({a_0} = 0,{a_1} = 0,{a_2} = 0,{a_n} = {( - 3)^{n - 3}},n = 3,4,5, \ldots \)

(e) The required result is\({a_0} = 8,{a_1} = 3,{a_2} = 2,{a_n} = 1\)when\(n\)even and\(n > 2,{a_n} = 0\)otherwise.

(f) The required result is\({a_0} = - 1,{a_1} = - 1,{a_2} = - 1,{a_3} = - 1,{a_n} = 1\)when\(n\)is a multiple of 4 and\(n \ge 4,{a_n} = 0\)otherwise.

(g) The required result is\({a_0} = {a_1} = 0,{a_n} = n - 1\)when\(n \ge 2\).

(h) The required result is\({a_n} = \frac{{{2^{n + 1}}}}{{n!}},n = 0,1,2, \ldots \)

Step-by-step solution

Formula of generating function and binomial theorem

Generating function for the sequence\({a_0},{a_1}, \ldots ,{a_k}\)of real numbers is the infinite series

\(G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\)

Binomial theorem:

\({(x + y)^n} = \sum\limits_{j = 0}^n {\left( {\begin{array}{*{20}{l}}n\\j\end{array}} \right)} {x^{n - j}}{y^j}\)

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\({(3x - 4)^3}\)

Use binomial theorem:

\(\begin{array}{l}{(3x - 4)^3} = {(3x + ( - 4))^3}\\{(3x - 4)^3} = \left( {\begin{array}{*{20}{l}}3\\0\end{array}} \right){(3x)^{3 - 0}}{( - 4)^0} + \left( {\begin{array}{*{20}{l}}3\\1\end{array}} \right){(3x)^{3 - 1}}{( - 4)^1} + \left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){(3x)^{3 - 2}}{( - 4)^2} + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){(3x)^{3 - 3}}{( - 4)^3}\\{(3x - 4)^3} = 27{x^3} - 108{x^2} + 144x - 64\end{array}\)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\(\begin{array}{l}{a_0} = - 64\\{a_1} = 144\\{a_2} = - 108\end{array}\)

\(\begin{array}{l}{a_3} = 27\\{a_n} = 0{\rm{ when }}n \ge 4\end{array}\)

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\({\left( {{x^3} + 1} \right)^3}\)

Use the binomial theorem:

\(\begin{array}{l}{\left( {{x^3} + 1} \right)^3} = \left( {\begin{array}{*{20}{l}}3\\0\end{array}} \right){\left( {{x^3}} \right)^{3 - 0}}{(1)^0} + \left( {\begin{array}{*{20}{l}}3\\1\end{array}} \right){\left( {{x^3}} \right)^{3 - 1}}{(1)^1} + \left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){\left( {{x^3}} \right)^{3 - 2}}{(1)^2} + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){\left( {{x^3}} \right)^{3 - 3}}{(1)^3}\\{\left( {{x^3} + 1} \right)^3} = {x^9} + 3{x^6} + 3{x^3} + 1\end{array}\)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\({a_0} = 1\)

\({a_3} = 3\)

\({a_6} = 3\)

\({a_9} = 1\)

\({a_n} = 0\) when \(n = 1,2,4,5,7,8\) and \(n \ge 10\)

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\(1/(1 - 5x)\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\(\begin{array}{l}\frac{1}{{1 - 5x}} = \sum\limits_{k = 0}^{ + \infty } {{{(5x)}^k}} \\\frac{1}{{1 - 5x}} = \sum\limits_{k = 0}^{ + \infty } {{5^k}} {x^k}\end{array}\)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\(\begin{array}{l}{a_n} = {5^n}\\n = 0,1,2, \ldots \end{array}\)

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\({x^3}/(1 + 3x)\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\(\frac{{{x^3}}}{{1 + 3x}} = {x^3} \cdot \frac{1}{{1 - ( - 3x)}}\)

\(\frac{{{x^3}}}{{1 + 3x}} = \sum\limits_{m = 3}^{ + \infty } {{{( - 3)}^{m - 3}}{x^m}} \)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\(\begin{array}{l}{a_0} = 0\\{a_1} = 0\\{a_2} = 0\end{array}\)

\(\begin{array}{l}{a_n} = {( - 3)^{n - 3}}\,\\n = 3,4,5, \ldots \end{array}\)

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\({x^2} + 3x + 7 + \left( {1/\left( {1 - {x^2}} \right)} \right)\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\({x^2} + 3x + 7 + \frac{1}{{1 - {x^2}}} = {x^2} + 3x + 7 + \sum\limits_{k = 0}^{ + \infty } {{{\left( {{x^2}} \right)}^k}} \)

\({x^2} + 3x + 7 + \frac{1}{{1 - {x^2}}} = 8 + 3x + 2{x^2} + \sum\limits_{k = 2}^{ + \infty } {{x^{2k}}} \)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\({a_0} = 8\)

\({a_1} = 3\)

\({a_2} = 2\)

\({a_n} = 1\)when\(n\)even and\(n > 2\).

\({a_n} = 0\) otherwise.

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\(\left( {{x^4}/\left( {1 - {x^4}} \right)} \right) - {x^3} - {x^2} - x - 1\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\(\frac{{{x^4}}}{{1 - {x^4}}} - {x^3} - {x^2} - x - 1 = {x^4} \cdot \frac{1}{{1 - {x^4}}} - {x^3} - {x^2} - x - 1\)

\(\frac{{{x^4}}}{{1 - {x^4}}} - {x^3} - {x^2} - x - 1 = - 1 - x - {x^2} - {x^3} + \sum\limits_{m = 1}^{ + \infty } {{x^{4m}}} \)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\({a_0} = - 1\)

\({a_1} = - 1\)

\({a_2} = - 1\)

\({a_3} = - 1\)

\({a_n} = 1\)when\(n\)is a multiple of 4 and\(n \ge 4\).

\({a_n} = 0\) otherwise.

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\({x^2}/{(1 - x)^2}\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\(\frac{{{x^2}}}{{{{(1 - x)}^2}}} = {x^2} \cdot \frac{1}{{{{(1 - x)}^2}}}\)

\(\frac{{{x^2}}}{{{{(1 - x)}^2}}} = \sum\limits_{m = 2}^{ + \infty } {(m - 1)} {x^m}\)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\({a_0} = {a_1} = 0\)

\({a_n} = n - 1\) when\(n \ge 2\).

Use the definition of a generating function and binomial theorem to solve the sequence

For the sequence:

\(2{e^{2x}}\)

Use the fact:\(\sum\limits_{k = 0}^{ + \infty } {{x^k}} = \frac{1}{{1 - x}}\)

\(\begin{array}{l}2{e^{2x}} = 2\sum\limits_{k = 0}^{ + \infty } {\frac{{{{(2x)}^k}}}{{k!}}} \\2{e^{2x}} = 2\sum\limits_{k = 0}^{ + \infty } {\frac{{{2^k}}}{{k!}}} \cdot {x^k}\\2{e^{2x}} = \sum\limits_{k = 0}^{ + \infty } {\frac{{{2^{k + 1}}}}{{k!}}} \cdot {x^k}\end{array}\)

The generating function is of the form:

\(\begin{array}{l}G(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_k}{x^k} + \ldots \\G(x) = \sum\limits_{k = 0}^{ + \infty } {{a_k}} {x^k}\end{array}\)

\(\begin{array}{l}{a_n} = \frac{{{2^{n + 1}}}}{{n!}}\\n = 0,1,2, \ldots \end{array}\)