Write out he first four terms of the Maclaurin series of f(x) if f(0)=2,\ f'(

Clifland 2021-11-06 Answered
Write out he first four terms of the Maclaurin series of f(x) if \(\displaystyle{f{{\left({0}\right)}}}={2},\ {f}'{\left({0}\right)}={3},\ {f}{''}{\left({0}\right)}={4},\ {f}{'''}{\left({0}\right)}={12}\)

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Expert Answer

Bentley Leach
Answered 2021-11-07 Author has 27528 answers

Maclaurin series:
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{f}^{{{\left({n}\right)}{\left({0}\right)}}}}}{{{n}!}}}{x}^{{n}}={f{{\left({0}\right)}}}+{f}'{\left({0}\right)}{x}+{\frac{{{f}{''}{\left({0}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{f}{'''}{\left({0}\right)}}}{{{3}!}}}{x}^{{3}}+{\frac{{{{f}^{{{\left({4}\right)}}}{\left({0}\right)}}}}{{{4}!}}}{x}^{{4}}+\ldots\)
Here given \(f(0)=2,f'(0)=3,f''(0)=4\) and we need to find out the first 4 terms of the Maclaurin Series
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{{f}^{{{\left({n}\right)}}}{\left({0}\right)}}}}{{{n}!}}}{x}^{{n}}={f{{\left({0}\right)}}}+{f}'{\left({0}\right)}{x}+{\frac{{{f}{''}{\left({0}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{f}{'''}{\left({0}\right)}}}{{{3}!}}}{x}^{{3}}+\ldots\)
We need to find 4 terms
\(\displaystyle{f{{\left({x}\right)}}}={2}+{3}{x}+{\frac{{{4}}}{{{2}}}}{x}^{{2}}+{\frac{{{12}}}{{{6}}}}{x}^{{3}}\)
\(\displaystyle{f{{\left({x}\right)}}}={2}+{3}{x}+{2}{x}^{{2}}+{2}{x}^{{3}}\)
Result: \(f(x)=2+3x+2x^2+2x^3\)

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