New Notes Blog: `f(x)=root(3)(x),n=3,c=8` Find the n'th Taylor Polynomial centered at c

Thursday, 22 January 2015

$\displaystyle{f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}},{n}={3},{c}={8}$ Find the n'th Taylor Polynomial centered at c

Taylor series is an example of infinite series derived from the expansion of $\displaystyle{f{{\left({x}\right)}}}$ about a single point. It is represented by infinite sum of $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ centered at $\displaystyle{x}={c}$ .The general formula for Taylor series is:

$\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({c}\right)}}}{{{n}!}}{{\left({x}-{c}\right)}}^{{n}}$

$\displaystyle{f{{\left({x}\right)}}}={f{{\left({c}\right)}}}+{f{'}}{\left({c}\right)}{\left({x}-{c}\right)}+\frac{{{f{'}}{\left({c}\right)}}}{{{2}!}}{{\left({x}-{c}\right)}}^{{2}}+\frac{{{f{'}}{\left({c}\right)}}}{{{3}!}}{{\left({x}-{c}\right)}}^{{3}}+\frac{{{f{'}}{\left({c}\right)}}}{{{4}!}}{{\left({x}-{c}\right)}}^{{4}}+\ldots$

To evaluate the given function $\displaystyle{f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}}$ , we may express it in terms of fractional exponent using the radical property: $\displaystyle{\sqrt[{{n}}]{{{x}}}}={{x}}^{{\frac{{1}}{{n}}}}$ . The function becomes:

$\displaystyle{f{{\left({x}\right)}}}={{\left({x}\right)}}^{{\frac{{1}}{{3}}}}$ .

Apply the definition of the Taylor series by listing the $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ up to $\displaystyle{n}={3}$ .

We determine each derivative using Power Rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ .

$\displaystyle{f{{\left({x}\right)}}}={{\left({x}\right)}}^{{\frac{{1}}{{3}}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{3}}\cdot{{x}}^{{\frac{{1}}{{3}}-{1}}}$

$\displaystyle=\frac{{1}}{{3}}{{x}}^{{-\frac{{2}}{{3}}}}{\quad\text{or}\quad}\frac{{1}}{{{3}{{x}}^{{\frac{{2}}{{3}}}}}}$

$\displaystyle{{f}}^{{2}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{3}}{{x}}^{{-\frac{{2}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{3}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{x}}^{{-\frac{{2}}{{3}}}}\right)}$

$\displaystyle=\frac{{1}}{{3}}\cdot{\left(-\frac{{2}}{{3}}{{x}}^{{-\frac{{2}}{{3}}-{1}}}\right)}$

$\displaystyle=-\frac{{2}}{{9}}{{x}}^{{-\frac{{5}}{{3}}}}{\quad\text{or}\quad}-\frac{{2}}{{{9}{{x}}^{{\frac{{5}}{{3}}}}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-\frac{{2}}{{9}}{{x}}^{{-\frac{{5}}{{3}}}}\right)}$

$\displaystyle=-\frac{{2}}{{9}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{x}}^{{-\frac{{5}}{{3}}}}\right)}$

$\displaystyle=-\frac{{2}}{{9}}\cdot{\left(-\frac{{5}}{{3}}{{x}}^{{-\frac{{5}}{{3}}-{1}}}\right)}$

$\displaystyle=\frac{{10}}{{27}}{{x}}^{{-\frac{{8}}{{3}}}}{\quad\text{or}\quad}\frac{{10}}{{{27}{{x}}^{{\frac{{8}}{{3}}}}}}$

Plug-in x=8, we get:

$\displaystyle{f{{\left({8}\right)}}}={{\left({8}\right)}}^{{\frac{{1}}{{3}}}}$

$\displaystyle={2}$

$\displaystyle{f{'}}{\left({8}\right)}=\frac{{1}}{{{3}\cdot{{8}}^{{\frac{{2}}{{3}}}}}}$

$\displaystyle=\frac{{1}}{{{3}\cdot{4}}}$

$\displaystyle=\frac{{1}}{{12}}$

$\displaystyle{{f}}^{{2}}{\left({8}\right)}=-\frac{{2}}{{{9}\cdot{{8}}^{{\frac{{5}}{{3}}}}}}$

$\displaystyle=-\frac{{2}}{{{9}\cdot{32}}}$

$\displaystyle=-\frac{{2}}{{288}}$

$\displaystyle=-\frac{{1}}{{144}}$

$\displaystyle{{f}}^{{3}}{\left({8}\right)}=\frac{{10}}{{{27}\cdot{{8}}^{{\frac{{8}}{{3}}}}}}$

$\displaystyle=\frac{{10}}{{{27}\cdot{256}}}$

$\displaystyle=\frac{{10}}{{6912}}$

$\displaystyle=\frac{{5}}{{3456}}$

Applying the formula for Taylor series centered at $\displaystyle{c}={8}$ , we get:

$\displaystyle{\sum_{{{n}={0}}}^{{3}}}\frac{{{{f}}^{{n}}{\left({8}\right)}}}{{{n}!}}{{\left({x}-{8}\right)}}^{{n}}$

$\displaystyle={f{{\left({8}\right)}}}+{f{'}}{\left({8}\right)}{\left({x}-{8}\right)}+\frac{{{f{'}}{\left({8}\right)}}}{{{2}!}}{{\left({x}-{8}\right)}}^{{2}}+\frac{{{f{'}}{\left({8}\right)}}}{{{3}!}}{{\left({x}-{8}\right)}}^{{3}}$

$\displaystyle={2}+{\left(\frac{{1}}{{12}}\right)}{\left({x}-{8}\right)}+\frac{{-\frac{{1}}{{144}}}}{{{2}!}}{{\left({x}-{8}\right)}}^{{2}}+\frac{{\frac{{5}}{{3456}}}}{{{3}!}}{{\left({x}-{8}\right)}}^{{3}}$

$\displaystyle={2}+\frac{{1}}{{12}}{\left({x}-{8}\right)}-\frac{{1}}{{{144}\cdot{2}}}{{\left({x}-{8}\right)}}^{{2}}+\frac{{5}}{{{3456}\cdot{6}}}{{\left({x}-{8}\right)}}^{{3}}$

$\displaystyle={2}+\frac{{1}}{{12}}{\left({x}-{8}\right)}-\frac{{1}}{{288}}{{\left({x}-{8}\right)}}^{{2}}+\frac{{5}}{{20736}}{{\left({x}-{8}\right)}}^{{3}}$

The Taylor polynomial of degree $\displaystyle{n}={3}$ for the given function $\displaystyle{f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}}$ centered at $\displaystyle{c}={8}$ will be:

$\displaystyle{P}{\left({x}\right)}={2}+\frac{{1}}{{12}}{\left({x}-{8}\right)}-\frac{{1}}{{288}}{{\left({x}-{8}\right)}}^{{2}}+\frac{{5}}{{20736}}{{\left({x}-{8}\right)}}^{{3}}$

New Notes Blog

Thursday, 22 January 2015

$\displaystyle{f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}},{n}={3},{c}={8}$ Find the n'th Taylor Polynomial centered at c

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Thursday, 22 January 2015

Find the n'th Taylor Polynomial centered at c

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{f{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}},{n}={3},{c}={8}$ Find the n'th Taylor Polynomial centered at c

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?