Question

Find the coefficient of \({x^{10}}\) in the power series of each of these functions.

a) \({\left( {1 + {x^5} + {x^{10}} + {x^{15}} + \cdots } \right)^3}\)

b) \({\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7} + \cdots } \right)^3}\)

c) \(\left( {{x^4} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7}} \right)(1 + x + \left. {{x^2} + {x^3} + {x^4} + \cdots } \right)\)

d) \(\left( {{x^2} + {x^4} + {x^6} + {x^8} + \cdots } \right)\left( {{x^3} + {x^6} + {x^9} + } \right. \cdots \left( {{x^4} + {x^8} + {x^{12}} + \cdots } \right)\)

e) \(\left( {1 + {x^2} + {x^4} + {x^6} + {x^8} + \cdots } \right)\left( {1 + {x^4} + {x^8} + {x^{12}} + } \right. \cdots )\left( {1 + {x^6} + {x^{12}} + {x^{18}} + \cdots } \right)\)

Short answer

The coefficient of \({x^{10}}\) in the power series of each of these functions is given below;

(a) 6

(b) 3

(c) 9

(d) 0

(e) 5

Step-by-step solution

a).Step 1: UseExtended Binomial Theorem:

The Extended Binomial Theorem: Let\(x\)be a real number with\(|x| < 1\)and let\(u\)be a real number. Then

\({(1 + x)^u} = \sum\limits_{k = 0}^\infty {\left( {\begin{array}{*{20}{l}}u\\k\end{array}} \right)} {x^k}\)

\(\begin{array}{c}{\left( {1 + {x^5} + {x^{10}} + {x^{15}} + \ldots } \right)^3} = {\left( {{x^{5(0)}} + {x^{5(1)}} + {x^{5(2)}} + {x^{5(3)}} + \ldots } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^{5k}}} } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{{\left( {{x^5}} \right)}^k}} } \right)^3}\\ = {\left( {\frac{1}{{1 - {x^5}}}} \right)^3}\\ = {\left( {1 + \left( { - {x^5}} \right)} \right)^{ - 3}}\\ = \sum\limits_{k = 0}^{ + \infty } {( - 3)} {\left( { - {x^5}} \right)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{5k}}\end{array}\)

The coefficient of \({x^{10}}\) is then the case\(k = 2\):

\(\begin{array}{c}{a_{5k}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right){( - 1)^2}\\ = \frac{{( - 3)( - 4)}}{{2!}}\\ = \frac{{12}}{2}\\ = 6\end{array}\)

Thus, the coefficient of \({x^{10}}\) is 6.

b).

UseExtended Binomial Theorem:

\(\begin{array}{c}{\left( {{x^3} + {x^4} + {x^5} + \ldots } \right)^3} = {\left( {{x^3}} \right)^3}{\left( {{x^0} + {x^1} + {x^2} + {x^3} + \ldots } \right)^3}\\ = {x^9} \cdot {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^k}} } \right)^3}\\ = {x^9} \cdot {\left( {\frac{1}{{1 - x}}} \right)^3}\\ = {x^9} \cdot {(1 + ( - x))^{ - 3}}\\ = {x^9} \cdot \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - x)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 9}}\end{array}\)

The coefficient of \({x^{10}}\) is then the case \(k = 1\) :

\(\begin{array}{c}{a_{k + 9}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right){( - 1)^1}\\ = - \frac{{( - 3)}}{{1!}}\\ = 3\end{array}\)

Thus, the coefficient of \({x^{10}}\) is 3

c).

UseExtended Binomial Theorem:

\(\begin{array}{c}\left( {{x^4} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + \ldots } \right)\\ = {x^4}\left( {1 + x + {x^2}} \right)\left( {{x^3}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + \ldots } \right)\\ = {x^7} \cdot \sum\limits_{k = 0}^2 {{x^k}} \cdot \sum\limits_{k = 0}^4 {{x^k}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^k}} \\ = {x^7} \cdot \sum\limits_{k = 0}^2 {{x^k}} \cdot \sum\limits_{k = 0}^4 {{x^k}} \cdot \frac{1}{{1 - x}}\\ = {x^7} \cdot \frac{{1 - {x^3}}}{{1 - x}} \cdot \frac{{1 - {x^5}}}{{1 - x}} \cdot \frac{1}{{1 - x}}\end{array}\)

By further simplification

\(\begin{array}{c} = \frac{{{x^7}\left( {1 - {x^3}} \right)\left( {1 - {x^5}} \right)}}{{{{(1 - x)}^3}}}\\ = \frac{{{x^7}\left( {1 - {x^3} - {x^5} + {x^8}} \right)}}{{{{(1 - x)}^3}}}\\ = \left( {{x^7} - {x^{10}} - {x^{12}} + {x^{15}}} \right) \cdot {(1 + ( - x))^{ - 3}}\\ = \left( {{x^7} - {x^{10}} - {x^{12}} + {x^{15}}} \right) \cdot \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - x)^k}\end{array}\)

By further simplification

\(\begin{array}{c} = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 7}} - \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 10}}\\ - \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 12}} + \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 15}}\end{array}\)

The coefficient of \({x^{10}}\) is then the case \(k = 3\) in the first summation and the case \(k = 0\) in the second summation:

\(\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\3\end{array}} \right){( - 1)^3} - \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right){( - 1)^0}\\ = - \frac{{( - 3)( - 4)( - 5)}}{{3!}} - 1\\ = 10 - 1\\ = 9\end{array}\)

d).

Use Extended Binomial Theorem:

\(\begin{array}{c}\left( {{x^2} + {x^4} + {x^6} + \ldots } \right)\left( {{x^3} + {x^6} + {x^9} + \ldots } \right)\left( {{x^4} + {x^8} + {x^{12}} + \ldots } \right)\\ = {x^2}\left( {1 + {x^2} + {x^4} + \ldots } \right)\left( {{x^3}} \right)\left( {1 + {x^3} + {x^6} + \ldots } \right)\left( {{x^4}} \right)\left( {1 + {x^4} + {x^8} + \ldots } \right)\\ = {x^9} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{2k}}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{3k}}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{4k}}} \end{array}\)

Since the first factor is\({x^9}\), we require a factor \(x\) in at least one of the sums. We then note that we cannot obtain the factor \(x\) from any of the sums (as the sums only provide the second, third, and fourth powers of \(x\) )

Thus \({x^{10}}\) is not present in the series and thus the coefficient of \({x^{10}}\) has to be zero.

e).

Use Extended Binomial Theorem:

\(\begin{array}{l}\left( {1 + {x^2} + {x^4} + \ldots } \right)\left( {1 + {x^4} + {x^8} + \ldots } \right)\left( {1 + {x^6} + {x^{12}} + \ldots } \right)\\ = \sum\limits_{k = 0}^{ + \infty } {{x^{2k}}} \cdot \sum\limits_{m = 0}^{ + \infty } {{x^{4m}}} \cdot \sum\limits_{n = 0}^{ + \infty } {{x^{6n}}} \end{array}\)

We then obtain \({x^{10}}\) if\(2k + 4m + 6n = 10\). Since\(k\), \(m\) and \(n\) are non-negative integers:

\(\begin{array}{*{20}{r}}{k = 5,m = 0,n = 0}\\{{\rm{ or }}k = 3,m = 1,n = 0}\\{{\rm{ or }}k = 1,m = 2,n = 0}\\{{\rm{ or }}k = 0,m = 1,n = 1}\\{{\rm{ or }}k = 2,m = 0,n = 1}\end{array}\)

Since the coefficient of each combination is 1 and since there are 5 combinations, the coefficient of \({x^{10}}\) is:

\(1 + 1 + 1 + 1 + 1 = 5\)