Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [

illusiia 2021-10-19 Answered
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that \(\displaystyle{R}{n}{\left({x}\right)}→{0}\).] Also find the associated radius of convergence. \(\displaystyle{f{{\left({x}\right)}}}={2}^{{x}}\)

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Expert Answer

Alix Ortiz
Answered 2021-10-20 Author has 19822 answers
Find a few derivatives, and calculate their values at a=0.
\(\displaystyle{f{{\left({x}\right)}}}={2}^{{x}}\ {f{{\left({0}\right)}}}={1}\)
\(\displaystyle{f}'{\left({x}\right)}={\left({\ln{{2}}}\right)}{2}^{{x}}\ {f}'{\left({0}\right)}={\ln{{2}}}\)
\(\displaystyle{f}{''}{\left({x}\right)}={\left({\ln{{2}}}\right)}^{{2}}{2}^{{x}}\ {f}{''}{\left({0}\right)}={\left({\ln{{2}}}\right)}^{{2}}\)
\(\displaystyle{f}{'''}{\left({x}\right)}={\left({\ln{{2}}}\right)}^{{3}}{2}^{{x}}\ {f}{'''}{\left({0}\right)}={\left({\ln{{2}}}\right)}^{{3}}\)
\(\displaystyle{{f}^{{{\left({4}\right)}}}{\left({x}\right)}}={\left({\ln{{2}}}\right)}^{{4}}{2}^{{x}}\ {{f}^{{{\left({4}\right)}}}{\left({0}\right)}}={\left({\ln{{2}}}\right)}^{{4}}\)
Plug everything into the Maclaurin general form
\(\displaystyle{f{{\left({x}\right)}}}={f{{\left({0}\right)}}}+{\frac{{{f}'{\left({0}\right)}}}{{{1}!}}}{x}+{\frac{{{f}{''}{\left({0}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{f}{'''}{\left({0}\right)}}}{{{3}!}}}{x}^{{3}}+\ldots\)
\(\displaystyle{f{{\left({x}\right)}}}={1}+{\frac{{{\ln{{2}}}}}{{{1}!}}}{x}+{\frac{{{\left({\ln{{2}}}\right)}^{{2}}}}{{{2}!}}}{x}^{{2}}+{\frac{{{\left({\ln{{2}}}\right)}^{{3}}}}{{{3}!}}}{x}^{{3}}+{\frac{{{\left({\ln{{2}}}\right)}^{{4}}}}{{{4}!}}}{x}^{{4}}+\ldots\)
Find the pattern of the numbers to erite in summation form
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{\left({\ln{{2}}}\right)}^{{n}}{x}^{{n}}}}{{{n}!}}}\)
Use the Ratio Test
\(\displaystyle{a}_{{n}}={\frac{{{\left({\ln{{2}}}\right)}^{{n}}{x}^{{n}}}}{{{n}!}}}\)
\(\displaystyle{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{n}}}}}\right|}={\left|{\frac{{{\left({\ln{{2}}}\right)}^{{{n}+{1}}}{x}^{{{n}+{1}}}}}{{{\left({n}+{1}\right)}!}}}\cdot{\frac{{{n}!}}{{{\left({\ln{{2}}}\right)}^{{n}}{x}^{{n}}}}}\right|}={\left|{\frac{{{\left({\ln{{2}}}\right)}{x}}}{{{n}+{1}}}}\right|}\)
\(\displaystyle\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({\ln{{2}}}\right)}{x}}}{{{n}+{1}}}}\right|}={0}\)
The limit is
\(\displaystyle{R}=\infty\)
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