New Notes Blog: `f(x)=2/x n=3,c=1` Find the n'th Taylor Polynomial centered at c

Thursday, 9 April 2015

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}{n}={3},{c}={1}$ Find the n'th Taylor Polynomial centered at c

Taylor series is an example of infinite series derived from the expansion of f(x) 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:

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

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

To determine the Taylor polynomial of degree $\displaystyle{n}={3}$ from the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}$ centered at $\displaystyle{x}={1}$ , we may apply the definition of Taylor series.

To determine the list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ up to $\displaystyle{n}={3}$ , we may apply Law of Exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ and Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ .

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}{\quad\text{or}\quad}{2}{{x}}^{{-{1}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}\frac{{2}}{{x}}$

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

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

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

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

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

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

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

$\displaystyle={4}{{x}}^{{-{3}}}{\quad\text{or}\quad}\frac{{4}}{{{x}}^{{3}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{4}{{x}}^{{-{3}}}$

$\displaystyle={4}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{-{3}}}$

$\displaystyle={4}\cdot{\left(-{3}{{x}}^{{-{3}-{1}}}\right)}$

$\displaystyle=-{12}{{x}}^{{-{4}}}{\quad\text{or}\quad}-\frac{{12}}{{{x}}^{{4}}}$

Plug-in $\displaystyle{x}={1}$ , we get:

$\displaystyle{f{{\left({2}\right)}}}=\frac{{2}}{{1}}={2}$

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

$\displaystyle{{f}}^{{2}}{\left({2}\right)}=\frac{{4}}{{{1}}^{{3}}}={4}$

$\displaystyle{{f}}^{{3}}{\left({2}\right)}=-\frac{{12}}{{{1}}^{{4}}}=-{12}$

Applying the formula for Taylor series, we get:

$\displaystyle{\sum_{{{n}={0}}}^{{3}}}\frac{{{{f}}^{{n}}{\left({1}\right)}}}{{{n}!}}{{\left({x}-{1}\right)}}^{{n}}$

$\displaystyle={f{{\left({1}\right)}}}+{f{'}}{\left({1}\right)}{\left({x}-{1}\right)}+\frac{{{{f}}^{{2}}{\left({1}\right)}}}{{{2}!}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{{{f}}^{{3}}{\left({1}\right)}}}{{{3}!}}{{\left({x}-{1}\right)}}^{{3}}$

$\displaystyle={2}+{\left(-{2}\right)}{\left({x}-{1}\right)}+\frac{{4}}{{{2}!}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{-{12}}}{{{3}!}}{{\left({x}-{1}\right)}}^{{3}}$

$\displaystyle={2}-{2}{\left({x}-{1}\right)}+\frac{{4}}{{2}}{{\left({x}-{1}\right)}}^{{2}}-\frac{{12}}{{6}}{{\left({x}-{1}\right)}}^{{3}}$

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

The Taylor polynomial of degree $\displaystyle{n}={3}$ for the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}$ centered at $\displaystyle{x}={1}$ will be:

$\displaystyle{P}_{{3}}{\left({x}\right)}={2}-{2}{\left({x}-{1}\right)}+{2}{{\left({x}-{1}\right)}}^{{2}}-{2}{{\left({x}-{1}\right)}}^{{3}}$

New Notes Blog

Thursday, 9 April 2015

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}{n}={3},{c}={1}$ 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, 9 April 2015

Find the n'th Taylor Polynomial centered at c

No comments:

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?

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{x}}{n}={3},{c}={1}$ 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?