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\({\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x}}\).

\({\rm{f(x) = x}}{{\rm{e}}^{{\rm{2x}}}}\).

Short answer

The series and radius of converges are\(\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{n!}}}}} {\rm{,}}\;\;\;{\rm{R = }}\infty \).

Step-by-step solution

Finding series.

Known that\({\rm{f(x) = x}}{{\rm{e}}^{{\rm{2x}}}}\).

Here to find the series \({e^{2x}}\)by using \({{\rm{e}}^{\rm{x}}}\) it will help to get the series.

\(\begin{array}{c}{{\rm{e}}^{\rm{x}}}{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{x}}^{\rm{n}}}}}{{{\rm{n!}}}}} \\{{\rm{e}}^{{\rm{2x}}}}{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{{\rm{(2x)}}}^{\rm{n}}}}}{{{\rm{n!}}}}} \\{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{\rm{n}}}}}{{{\rm{n!}}}}} \\{\rm{x}}{{\rm{e}}^{{\rm{2x}}}}{\rm{ = x}}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{\rm{n}}}}}{{{\rm{n!}}}}} \\{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{n!}}}}} \end{array}\)

To find the radius of convergence.

As per the ratio test,

\(\begin{array}{c}\left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\left| {\frac{{{{\rm{2}}^{{\rm{n + 1}}}}{{\rm{x}}^{{\rm{(n + 1) + 1}}}}}}{{{\rm{(n + 1)!}}}}{\rm{ \times }}\frac{{{\rm{n!}}}}{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{{\rm{n + 1}}}}}}} \right|\\{\rm{ = }}\left| {\frac{{{\rm{2x}}}}{{{\rm{n + 1}}}}} \right|\\\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{\rm{2x}}}}{{{\rm{n + 1}}}}} \right|{\rm{ = 0}}\end{array}\)

For everyone, the series will converge \({\rm{x}}\).so the radius of convergence is

\({\rm{R = }}\infty \)

Therefore, the series and radius of converges for the given equation are\(\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{\rm{x}}^{{\rm{n + 1}}}}}}{{{\rm{n!}}}}} {\rm{,}}\;\;\;{\rm{R = }}\infty \).