# 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

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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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,
$$\displaystyle{{\tan}^{{-{1}}}{\left({x}\right)}}={\frac{{{{f}^{{0}}{\left({0}\right)}}}}{{{0}!}}}{x}^{{0}}+{\frac{{{{f}^{{1}}{\left({0}\right)}}}}{{{1}!}}}{x}^{{1}}+{\frac{{{{f}^{{2}}{\left({0}\right)}}}}{{{2}!}}}{x}^{{2}}+{\frac{{{{f}^{{3}}{\left({0}\right)}}}}{{{3}!}}}{x}^{{3}}+\ldots$$
$$\displaystyle={0}+{\frac{{{1}}}{{{1}!}}}{x}\pm+{\frac{{{\left(-{2}\right)}}}{{{3}!}}}{x}^{{3}}+\ldots$$
$$\displaystyle={x}-{\frac{{{x}^{{3}}}}{{{3}}}}+\ldots$$