Question

Find the Maclaurin series\({\rm{f}}\)and its radius of convergence. You may use either the direct method (definition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for\({{\rm{e}}^{\rm{x}}}\),\({\rm{sinx}}\)and\({\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x}}\).

\({\rm{f(x) = ln(4 - x)}}\).

Short answer

The Series and radius of converges is \({\rm{ln4 - }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{x}}^{\rm{n}}}}}{{{\rm{n}}{{\rm{4}}^{\rm{n}}}}}} {\rm{,}}\;\;\;{\rm{R = 4}}\).

Step-by-step solution

Finding series.

Known that\({\rm{f(x) = ln(4 - x)}}\).

Here to find the series for \({{\rm{f}}^{{\rm{ - 1}}}}\)it will help to get the series.

\(\begin{array}{c}{{\rm{f}}'}{\rm{(x) = }}\frac{{{\rm{ - 1}}}}{{{\rm{4 - x}}}}\\{{\rm{f}}'}{\rm{(x) = - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{1 - }}\frac{{\rm{x}}}{{\rm{4}}}}}\\{{\rm{f}}'}{\rm{(x) = - }}\frac{{\rm{1}}}{{\rm{4}}}\sum\limits_{{\rm{n = 0}}}\infty {{{\left( {\frac{{\rm{x}}}{{\rm{4}}}} \right)}{\rm{n}}}} \end{array}\)

Now integrate,

\(\int {\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}\sum\limits_{{\rm{n = 0}}}^\infty {{{\left( {\frac{{\rm{x}}}{{\rm{4}}}} \right)}^{\rm{n}}}} } \right)} {\rm{dx = C - }}\frac{{\rm{1}}}{{\rm{4}}}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}{{\rm{4}}^{\rm{n}}}}}} \)

Here \({\rm{x = 0}}\)means to find \({\rm{C}}\).

\(\begin{array}{c}{\rm{ln(4 - x) = C - }}\frac{{\rm{1}}}{{\rm{4}}}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}{{\rm{4}}^{\rm{n}}}}}} \\{\rm{ln4 = C}}\\{\rm{ln(4 - x) = ln4 - }}\frac{{\rm{1}}}{{\rm{4}}}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}{{\rm{4}}^{\rm{n}}}}}} \\{\rm{ = ln4 - }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}{{\rm{4}}^{{\rm{n + 1}}}}}}} \\{\rm{ln(4 - x) = ln4 - }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{x}}^{\rm{n}}}}}{{{\rm{n}}{{\rm{4}}^{\rm{n}}}}}} \end{array}\)

To find the radius of convergence.

As per the ratio test

\(\begin{array}{c}\left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\left| {\frac{{{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)}}{{\rm{4}}^{{\rm{n + 1}}}}}}{\rm{ \times }}\frac{{{\rm{n}}{{\rm{4}}^{\rm{n}}}}}{{{{\rm{x}}^{\rm{n}}}}}} \right|\\{\rm{ = }}\left| {\frac{{{\rm{nx}}}}{{{\rm{(n + 1)4}}}}} \right|\\{\rm{ = }}\left| {\frac{{{\rm{nx}}}}{{{\rm{4n + 4}}}}} \right|\\\left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\left| {\frac{{\rm{x}}}{{{\rm{4 + }}\frac{{\rm{4}}}{{\rm{n}}}}}} \right|\end{array}\)

When the series reaches a point of convergence,\(\left| {\frac{{\rm{x}}}{{\rm{4}}}} \right|{\rm{ < 1}}\)or \({\rm{|x| < 4}}\)As a result, the convergence radius is \({\rm{R = 4}}\)

Therefore, the series and radius of converges is \({\rm{ln4 - }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{x}}^{\rm{n}}}}}{{{\rm{n}}{{\rm{4}}^{\rm{n}}}}}} {\rm{,}}\;\;\;{\rm{R = 4}}\).