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

\({\rm{f(x) = 1}}{{\rm{0}}^{\rm{x}}}\).

Short answer

The series and radius of converges are\(\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{{\rm{(ln10)}}}^{\rm{n}}}}}{{{\rm{n!}}}}} {{\rm{x}}^{\rm{n}}},{\rm{R = }}\infty \).

Step-by-step solution

Finding series.

Known that\({\rm{f(x) = 1}}{{\rm{0}}^{\rm{x}}}\).

Use the direct method\({\rm{a = 0}}\) for the Maclaurin series.

Here Find some derivatives\({{\rm{f}}^{{\rm{(n)}}}}{\rm{(x)}}\)to evaluate \({{\rm{f}}^{{\rm{(n)}}}}{\rm{(a)}}\).

\(\begin{array}{c}{\rm{f(x) = 1}}{{\rm{0}}^{\rm{x}}}\\{{\rm{f}}^{\rm{'}}}{\rm{(x) = 1}}{{\rm{0}}^{\rm{x}}}{\rm{ln10}}\\{{\rm{f}}^{{\rm{''}}}}{\rm{(x) = 1}}{{\rm{0}}^{\rm{x}}}{{\rm{(ln10)}}^{\rm{2}}}\\{{\rm{f}}^{{\rm{'''}}}}{\rm{(x) = 1}}{{\rm{0}}^{\rm{x}}}{{\rm{(ln10)}}^{\rm{3}}}\\ \vdots \\{{\rm{f}}^{{\rm{(n)}}}}{\rm{(x) = 1}}{{\rm{0}}^{\rm{x}}}{{\rm{(ln10)}}^{\rm{n}}}\end{array}\) \(\begin{array}{c}{\rm{f(0) = 1}}\\{{\rm{f}}^{\rm{'}}}{\rm{(0) = ln10}}\\{{\rm{f}}^{{\rm{'''}}}}{\rm{(0) = (ln10}}{{\rm{)}}^{\rm{2}}}\\{{\rm{f}}^{{\rm{'''}}}}{\rm{(0) = (ln10}}{{\rm{)}}^{\rm{3}}}\\ \vdots \\{{\rm{f}}^{{\rm{(n)}}}}{\rm{(0) = (ln10}}{{\rm{)}}^{\rm{n}}}\end{array}\)

Arrange this series,

\(\begin{array}{c}{\rm{f(x) = f(a) + }}\frac{{{\rm{f'(a)}}}}{{{\rm{1!}}}}{\rm{(x - a) + }}\frac{{{{\rm{f}}^{{\rm{''}}}}{\rm{(a)}}}}{{{\rm{2!}}}}{{\rm{(x - a)}}^{\rm{2}}}{\rm{ + }}\frac{{{{\rm{f}}^{{\rm{'''}}}}{\rm{(a)}}}}{{{\rm{3!}}}}{{\rm{(x - a)}}^{\rm{3}}}{\rm{ + L}}\\{\rm{1}}{{\rm{0}}^{\rm{x}}}{\rm{ = 1 + x ln10 + }}\frac{{{{{\rm{(ln10)}}}^{\rm{2}}}}}{{{\rm{2!}}}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}\frac{{{{{\rm{(ln10)}}}^{\rm{3}}}}}{{{\rm{3!}}}}{{\rm{x}}^{\rm{3}}}{\rm{ + L}}\\{\rm{f(x) = }}\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{{\rm{(ln10)}}}^{\rm{n}}}}}{{{\rm{n!}}}}} {{\rm{x}}^{\rm{n}}}\end{array}\)

To find the radius of convergence.

As per the ratio test,

\(\begin{array}{c}\left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\left| {\frac{{{{{\rm{(ln10)}}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)!}}}}{{\rm{x}}^{{\rm{n + 1}}}}{\rm{ \times }}\frac{{{\rm{n!}}}}{{{{{\rm{(ln10)}}}^{\rm{n}}}{{\rm{x}}^{\rm{n}}}}}} \right|\\\left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\left| {\frac{{{\rm{xln10}}}}{{{\rm{n + 1}}}}} \right|\end{array}\)

\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{\rm{xln10}}}}{{{\rm{n + 1}}}}} \right|{\rm{ = 0}}\)

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\(\sum\limits_{{\rm{n = 0}}}^\infty {\frac{{{{{\rm{(ln10)}}}^{\rm{n}}}}}{{{\rm{n!}}}}} {{\rm{x}}^{\rm{n}}},{\rm{R = }}\infty \).