Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.\tan^{-1}x,x_0=0

Ayaana Buck 2021-09-16 Answered
Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.
\(\displaystyle{{\tan}^{{-{1}}}{x}},{x}_{{0}}={0}\)

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Expert Answer

faldduE
Answered 2021-09-17 Author has 14953 answers
We’ll answer the first question since the exact one wasn’t specified. Please submit a new question specifying the one you’d like answered.
We have \(\displaystyle{f{{\left({x}\right)}}}={{\tan}^{{-{1}}}{\left({x}\right)}},{x}_{{0}}={0}\)
The general form of a taylor expansion centered at a of an function
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{{f}^{{n}}{\left({a}\right)}}}}{{{n}!}}}{\left({x}-{a}\right)}^{{n}},\)
Here \(\displaystyle{f}^{{n}}\) represent the n th derivative of function.
We have a=0 then taylor expression reduce to
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{{f}^{{n}}{\left({a}\right)}}}}{{{n}!}}}{x}^{{n}}\)
Now,
\(\displaystyle{f{{\left({x}\right)}}}={{\tan}^{{-{1}}}{\left({x}\right)}}\)
\(\displaystyle{{f}^{{1}}{\left({x}\right)}}={\frac{{{1}}}{{{1}+{x}^{{2}}}}}\)
\(\displaystyle{{f}^{{1}}{\left({0}\right)}}={1}\)
\(\displaystyle{{f}^{{2}}{\left({x}\right)}}=-{\frac{{{2}{x}}}{{{\left({1}+{x}^{{2}}\right)}^{{2}}}}}\)
\(\displaystyle{{f}^{{2}}{\left({0}\right)}}={0}\)
\(\displaystyle{{f}^{{3}}{\left({x}\right)}}={\frac{{{2}{\left({1}+{x}^{{2}}\right)}^{{2}}-{2}{\left({1}+{x}^{{2}}\right)}{\left({2}{x}\right)}{\left({2}{x}\right)}}}{{{\left({1}+{x}^{{2}}\right)}^{{4}}}}}\)
\(\displaystyle=-{\frac{{{2}{\left({1}+{x}^{{2}}\right)}^{{2}}-{8}{x}^{{2}}{\left({1}+{x}^{{2}}\right)}}}{{{\left({1}+{x}^{{2}}\right)}^{{4}}}}}\)
\(\displaystyle{{f}^{{3}}{\left({0}\right)}}=-{2}\)
Therefore,
\(\displaystyle{{\tan}^{{-{1}}}{\left({x}\right)}}={\frac{{{{f}^{{0}}{\left({0}\right)}}}}{{{0}!}}}{x}^{{0}}+{\frac{{{{f}^{{1}}{\left({0}\right)}}}}{{{1}!}}}{x}^{{1}}+{\frac{{{{f}^{{2}}{\left({0}\right)}}}}{{{2}!}}}{x}^{{2}}+{\frac{{{{f}^{{3}}{\left({0}\right)}}}}{{{3}!}}}{x}^{{3}}+\ldots\)
\(\displaystyle={0}+{\frac{{{1}}}{{{1}!}}}{x}\pm+{\frac{{{\left(-{2}\right)}}}{{{3}!}}}{x}^{{3}}+\ldots\)
\(\displaystyle={x}-{\frac{{{x}^{{3}}}}{{{3}}}}+\ldots\)
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