Notes: `f(x) = 1/(1-x) ,c=2` 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 $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{1}-{x}}}$ centered at c=2, we list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ using the Power rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{u}}^{{n}}={n}\cdot{{u}}^{{{n}-{1}}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ and basic differentiation property: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{c}\cdot{f{{\left({x}\right)}}}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{f{{\left({x}\right)}}}$ .

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

Let $\displaystyle{u}={1}-{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=-{1}$ .

The derivative of f(x) will be:

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

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

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

Then, we list the derivatives of $\displaystyle{f{{\left({x}\right)}}}$ as:

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

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

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

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

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

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

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

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

$\displaystyle={6}{{\left({1}-{x}\right)}}^{{-{4}}}{\quad\text{or}\quad}\frac{{6}}{{{\left({1}-{x}\right)}}^{{4}}}$

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

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

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

$\displaystyle={24}{{\left({1}-{x}\right)}}^{{-{5}}}{\quad\text{or}\quad}\frac{{24}}{{{\left({1}-{x}\right)}}^{{5}}}$

Plug-in $\displaystyle{x}={2}$ , we get:

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

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

$\displaystyle=-{1}$

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

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

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

$\displaystyle={1}$

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

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

$\displaystyle=\frac{{2}}{{-{1}}}$

$\displaystyle=-{2}$

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

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

$\displaystyle=\frac{{6}}{{1}}$

$\displaystyle={6}$

$\displaystyle{{f}}^{{4}}{\left({2}\right)}=\frac{{24}}{{{\left({1}-{2}\right)}}^{{5}}}$

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

$\displaystyle=\frac{{24}}{{-{1}}}$

$\displaystyle=-{24}$

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

$\displaystyle\frac{{1}}{{{1}-{x}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({2}\right)}}}{{{n}!}}{{\left({x}-{2}\right)}}^{{n}}$

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

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

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

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

The Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{1}-{x}}}$ centered at $\displaystyle{c}={2}$ will be:

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

$\displaystyle\frac{{1}}{{{1}-{x}}}={\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{{n}+{1}}}{{\left({x}-{2}\right)}}^{{n}}$

Notes

Wednesday, 14 May 2014

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{1}-{x}}},{c}={2}$ Use the definition of Taylor series to find the Taylor series, centered at c for the function.

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

Wednesday, 14 May 2014

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

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

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