Consider the provided function,

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\) with \(\displaystyle{a}={4}\), approximate \(\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}\)

Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.

The Taylor series as a form of \(\displaystyle{\sum_{{{k}={0}}}^{\infty}}{c}_{{k}}{\left({x}-{a}\right)}^{{k}}\), where \(\displaystyle{c}_{{k}}={\frac{{{{f}^{{k}}{\left({a}\right)}}}}{{{k}!}}},{k}={0},{1},{2},\ldots\)

We evaluate the derivative of the function at point \(\displaystyle{a}={4}\)

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\Rightarrow{f{{\left({4}\right)}}}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{f}'{\left({x}\right)}=-{\frac{{{1}}}{{{2}{x}^{{{\frac{{{3}}}{{{2}}}}}}}}}\Rightarrow{f}'{\left({4}\right)}=-{\frac{{{1}}}{{{16}}}}\)

\(\displaystyle{f}{''}{\left({x}\right)}={\frac{{{3}}}{{{4}{x}^{{{\frac{{{5}}}{{{2}}}}}}}}}\Rightarrow{f}{''}{\left({4}\right)}={\frac{{{3}}}{{{128}}}}\)

\(\displaystyle{f}{'''}{\left({x}\right)}=-{\frac{{{15}}}{{{8}{x}^{{{\frac{{{7}}}{{{2}}}}}}}}}\Rightarrow{f}{'''}{\left({4}\right)}=-{\frac{{{15}}}{{{1024}}}}\)

Hence, the first four term of the series is shown below.

\(\displaystyle{\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({x}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({x}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({x}-{4}\right)}^{{3}}\)

So, \(\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}={\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{3}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}-{\frac{{{1}-{4}\sqrt{{{3}}}}}{{{16}\sqrt{{{3}}}}}}+{\frac{{{49}\sqrt{{{3}}}-{24}}}{{{128}\sqrt{{{3}}}}}}-{\frac{{{5}{\left({145}-{204}\sqrt{{{3}}}\right)}}}{{{1024}\sqrt{{{3}}}}}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}+{\frac{{{1668}-{327}\sqrt{{{3}}}}}{{{1024}}}}\)

Hence.

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\) with \(\displaystyle{a}={4}\), approximate \(\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}\)

Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.

The Taylor series as a form of \(\displaystyle{\sum_{{{k}={0}}}^{\infty}}{c}_{{k}}{\left({x}-{a}\right)}^{{k}}\), where \(\displaystyle{c}_{{k}}={\frac{{{{f}^{{k}}{\left({a}\right)}}}}{{{k}!}}},{k}={0},{1},{2},\ldots\)

We evaluate the derivative of the function at point \(\displaystyle{a}={4}\)

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\Rightarrow{f{{\left({4}\right)}}}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{f}'{\left({x}\right)}=-{\frac{{{1}}}{{{2}{x}^{{{\frac{{{3}}}{{{2}}}}}}}}}\Rightarrow{f}'{\left({4}\right)}=-{\frac{{{1}}}{{{16}}}}\)

\(\displaystyle{f}{''}{\left({x}\right)}={\frac{{{3}}}{{{4}{x}^{{{\frac{{{5}}}{{{2}}}}}}}}}\Rightarrow{f}{''}{\left({4}\right)}={\frac{{{3}}}{{{128}}}}\)

\(\displaystyle{f}{'''}{\left({x}\right)}=-{\frac{{{15}}}{{{8}{x}^{{{\frac{{{7}}}{{{2}}}}}}}}}\Rightarrow{f}{'''}{\left({4}\right)}=-{\frac{{{15}}}{{{1024}}}}\)

Hence, the first four term of the series is shown below.

\(\displaystyle{\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({x}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({x}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({x}-{4}\right)}^{{3}}\)

So, \(\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}={\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{3}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}-{\frac{{{1}-{4}\sqrt{{{3}}}}}{{{16}\sqrt{{{3}}}}}}+{\frac{{{49}\sqrt{{{3}}}-{24}}}{{{128}\sqrt{{{3}}}}}}-{\frac{{{5}{\left({145}-{204}\sqrt{{{3}}}\right)}}}{{{1024}\sqrt{{{3}}}}}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}+{\frac{{{1668}-{327}\sqrt{{{3}}}}}{{{1024}}}}\)

Hence.