Notes: `f(x)=1/x ,c=1` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Friday, 8 April 2016

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

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:

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

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

To apply the definition of Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{x}}$ centered at $\displaystyle{c}={1}$ , we list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ using the Power rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ and basic differentiation property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{c}\cdot{f{{\left({x}\right)}}}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}$ .

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

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}\frac{{1}}{{x}}$

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

$\displaystyle=-{1}\cdot{{x}}^{{-{1}-{1}}}$

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

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

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

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

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

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

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

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

$\displaystyle=-{6}{{x}}^{{-{4}}}{\quad\text{or}\quad}-\frac{{6}}{{{x}}^{{4}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}-{6}{{x}}^{{-{4}}}$

$\displaystyle=-{6}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{-{4}}}$

$\displaystyle=-{6}\cdot{\left(-{4}{{x}}^{{-{4}-{1}}}\right)}$

$\displaystyle={24}{{x}}^{{-{5}}}{\quad\text{or}\quad}\frac{{24}}{{{x}}^{{5}}}$

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

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

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

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

$\displaystyle{{f}}^{{3}}{\left({1}\right)}=-\frac{{6}}{{{1}}^{{4}}}=-{6}$

$\displaystyle{{f}}^{{4}}{\left({1}\right)}=\frac{{24}}{{{1}}^{{5}}}={24}$

Plug-in the values on the formula for Taylor series, we get:

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

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

$\displaystyle={1}+{\left(-{1}\right)}\cdot{\left({x}-{1}\right)}+\frac{{2}}{{{2}!}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{-{6}}}{{{3}!}}{{\left({x}-{1}\right)}}^{{3}}+\frac{{24}}{{{4}!}}{{\left({x}-{1}\right)}}^{{4}}+\ldots$

$\displaystyle={1}-{\left({x}-{1}\right)}+\frac{{2}}{{2}}{{\left({x}-{1}\right)}}^{{2}}-\frac{{6}}{{6}}{{\left({x}-{1}\right)}}^{{3}}+\frac{{24}}{{24}}{{\left({x}-{1}\right)}}^{{4}}+\ldots$

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

The Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{x}}$ centered at $\displaystyle{c}={1}$ will be:

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

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

Notes

Friday, 8 April 2016

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

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Friday, 8 April 2016

Use the definition of Taylor series to find the Taylor series, centered at c for the function.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

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

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?