# Use the definition of Taylor series to find the Taylor

Use the definition of Taylor series to find the Taylor series, centered at c for the function. $$\displaystyle{f{{\left({x}\right)}}}={\ln{{x}}},{c}={1}$$

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Sact1989

The Taylor series of $$\displaystyle{\ln{{x}}}$$ around c=1. We know from the definition of Taylor series of the function f
$$\displaystyle{f{{\left({x}\right)}}}={f{{\left({x}\right)}}}{\mid}_{{{x}={c}}}+{\frac{{{f}'{\left({x}\right)}{\mid}_{{{x}={c}}}}}{{{1}!}}}{\left({x}-{c}\right)}+{\frac{{{f}{''}{\left({x}\right)}{\mid}_{{{x}={c}}}}}{{{2}!}}}{\left({x}-{c}\right)}^{{2}}+{\frac{{{f}{'''}{\left({x}\right)}{\mid}_{{{x}={c}}}}}{{{3}!}}}{\left({x}-{c}\right)}^{{3}}+\ldots$$
Now the values of differential functions at given point
$$\displaystyle{f{{\left({1}\right)}}}={\ln{{\left({1}\right)}}}={0},$$ $$f'(1)=\frac{1}{x}|_{x=1}=1$$
$$\displaystyle{f}{''}{\left({1}\right)}=-{\frac{{{1}}}{{{x}^{{2}}}}}_{{{x}={1}}}=-{1},\ {f}{'''}{\left({1}\right)}={\frac{{{2}}}{{{x}^{{3}}}}}{\left|_{\left\lbrace{x}={1}\right\rbrace}={2}{f}{''''}{\left({1}\right)}=-{\frac{{{6}}}{{{x}^{{4}}}}}\right|}_{{{x}={1}}}=-{6}$$
$$\displaystyle{{f}^{{n}}{\left({1}\right)}}={\left(-{1}\right)}^{{{n}+{1}}}{\frac{{{\left({n}-{1}\right)}!}}{{{x}^{{n}}}}}{\mid}_{{{x}={1}}}={\left(-{1}\right)}^{{n}}{\left({\left({n}-{1}\right)}!\right)}$$
Thus the Taylor series expression become
$$\displaystyle{f{{\left({x}\right)}}}={0}+{\left({x}-{1}\right)}-{\frac{{{1}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}+{2}!{\frac{{{1}}}{{{3}!}}}{\left({x}-{1}\right)}^{{3}}-{3}!{\frac{{{1}}}{{{4}!}}}+{\left(-{1}\right)}^{{{n}+{1}}}{\left({n}-{1}\right)}!{\frac{{{1}}}{{{n}!}}}{\left({x}-{1}\right)}^{{n}}$$
Or we can write as
$$\displaystyle{f{{\left({x}\right)}}}={\left({1}-{x}\right)}-{\frac{{{1}}}{{{2}}}}{\left({x}-{1}\right)}^{{2}}+{\frac{{{1}}}{{{3}}}}{\left({x}-{1}\right)}^{{3}}-{\frac{{{1}}}{{{4}}}}{\left({x}-{1}\right)}^{{4}}+{\left(-{1}\right)}^{{{n}+{1}}}{\left({n}-{1}\right)}!{\frac{{{1}}}{{{n}!}}}{\left({x}-{1}\right)}^{{n}}-\ldots$$