Notes: `f(x)=ln(x^2+1), c=0` 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:

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

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

To determine the Taylor series for the function $\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({{x}}^{{2}}+{1}\right)}}}$ centered at $\displaystyle{c}={0}$ , we may list the $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

$\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({{x}}^{{2}}+{1}\right)}}}$

Applying derivative formula for logarithmic function: d $\displaystyle/{\left({\left.{d}{x}\right.}\right)}{\ln{{\left({u}\right)}}}=\frac{{1}}{{u}}\cdot\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}$ .

Let $\displaystyle{u}={{x}}^{{2}}+{1}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={2}{x}$

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

$\displaystyle=\frac{{1}}{{{{x}}^{{2}}+{1}}}\cdot{2}{x}$

$\displaystyle=\frac{{{2}{x}}}{{{{x}}^{{2}}+{1}}}$

Applying Quotient rule for differentiation: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{u}}{{v}}\right)}=\frac{{{u}'\cdot{v}-{u}\cdot{v}'}}{{{v}}^{{2}}}$ .

Let $\displaystyle{u}={2}{x}$ then $\displaystyle{u}'={2}$

$\displaystyle{v}={{x}}^{{2}}+{1}$ then $\displaystyle{v}'={2}{x}$ and $\displaystyle{{v}}^{{2}}={{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$

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

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

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

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

Let $\displaystyle{u}={2}-{2}{{x}}^{{2}}$ then $\displaystyle{u}'=-{4}{x}$

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

then $\displaystyle{{v}}^{{2}}={{\left({{\left({{x}}^{{2}}+{1}\right)}}^{{2}}\right)}}^{{2}}={{\left({{x}}^{{2}}+{1}\right)}}^{{4}}$

and $\displaystyle{v}'={2}\cdot{{\left({{x}}^{{2}}+{1}\right)}}^{{{2}-{1}}}\cdot{2}{x}={4}{x}{\left({{x}}^{{2}}+{1}\right)}$

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

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

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

$\displaystyle=\frac{{{\left(-{4}{{x}}^{{3}}-{4}{x}\right)}-{\left({8}{x}-{8}{{x}}^{{3}}\right)}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{3}}}$

$\displaystyle=\frac{{-{4}{{x}}^{{3}}-{4}{x}-{8}{x}+{8}{{x}}^{{3}}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{3}}}$

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

Let $\displaystyle{u}={\left({4}{{x}}^{{3}}-{12}{x}\right)}$ then $\displaystyle{u}'={12}{{x}}^{{2}}-{12}$

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

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

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

$\displaystyle={{\left({{x}}^{{2}}+{1}\right)}}^{{6}}$

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

$\displaystyle={6}{x}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}$

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

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

$\displaystyle=\frac{{{\left({12}{{x}}^{{4}}-{12}\right)}-{\left({24}{{x}}^{{4}}-{72}{{x}}^{{2}}\right)}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

$\displaystyle=\frac{{{12}{{x}}^{{4}}-{12}-{24}{{x}}^{{4}}+{72}{{x}}^{{2}}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

$\displaystyle=\frac{{-{12}{{x}}^{{4}}+{72}{{x}}^{{2}}-{12}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{4}}}$

$\displaystyle{{f}}^{{5}}{\left({x}\right)}=\frac{{-{480}{{x}}^{{3}}+{28}{{x}}^{{5}}+{240}{x}}}{{{\left({x}+{1}\right)}}^{{5}}}$

$\displaystyle{{f}}^{{6}}{\left({x}\right)}=\frac{{-{240}{{x}}^{{6}}+{3600}{{x}}^{{4}}-{3600}{{x}}^{{2}}+{240}}}{{{\left({x}+{1}\right)}}^{{6}}}$

Plug-in $\displaystyle{x}={0}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

$\displaystyle{f{{\left({0}\right)}}}={\ln{{\left({{0}}^{{2}}+{1}\right)}}}$

$\displaystyle={\ln{{\left({1}\right)}}}$

$\displaystyle={0}$

$\displaystyle{f{'}}{\left({0}\right)}=\frac{{{2}\cdot{0}}}{{{{0}}^{{2}}+{1}}}$

$\displaystyle=\frac{{0}}{{1}}$

$\displaystyle={0}$

$\displaystyle{{f}}^{{2}}{\left({0}\right)}=\frac{{{2}-{2}\cdot{{0}}^{{2}}}}{{{\left({{0}}^{{2}}+{1}\right)}}^{{2}}}$

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

$\displaystyle={2}$

$\displaystyle{{f}}^{{3}}{\left({0}\right)}=\frac{{{4}\cdot{{0}}^{{3}}-{12}\cdot{0}}}{{{\left({{0}}^{{2}}+{1}\right)}}^{{3}}}$

$\displaystyle=\frac{{0}}{{1}}$

$\displaystyle={0}$

$\displaystyle{{f}}^{{4}}{\left({0}\right)}=\frac{{-{12}\cdot{{0}}^{{4}}+{72}\cdot{{0}}^{{2}}-{12}}}{{{\left({{0}}^{{2}}+{1}\right)}}^{{4}}}$

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

$\displaystyle=-{12}$

$\displaystyle{{f}}^{{5}}{\left({0}\right)}=\frac{{-{480}\cdot{{0}}^{{3}}+{28}\cdot{{0}}^{{5}}+{240}\cdot{0}}}{{{\left({0}+{1}\right)}}^{{5}}}$

$\displaystyle=\frac{{0}}{{1}}$

$\displaystyle={0}$

$\displaystyle{{f}}^{{6}}{\left({0}\right)}=\frac{{-{240}\cdot{{0}}^{{6}}+{3600}\cdot{{0}}^{{4}}-{3600}\cdot{{0}}^{{2}}+{240}}}{{{\left({0}+{1}\right)}}^{{6}}}$

$\displaystyle=\frac{{240}}{{1}}$

$\displaystyle={240}$

Applying the formula for Taylor series, we get:

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

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

$\displaystyle={f{{\left({0}\right)}}}+{f{'}}{\left({0}\right)}{x}+\frac{{{f{'}}{\left({0}\right)}}}{{{2}!}}{{x}}^{{2}}+\frac{{{f{'}}{\left({0}\right)}}}{{{3}!}}{{x}}^{{3}}+\frac{{{f{'}}{\left({0}\right)}}}{{{4}!}}{{x}}^{{4}}+\ldots$

$\displaystyle={0}+{0}\cdot{x}+\frac{{2}}{{{2}!}}{{x}}^{{2}}+\frac{{0}}{{{3}!}}{{x}}^{{3}}+\frac{{-{12}}}{{{4}!}}{{x}}^{{4}}+\frac{{0}}{{{5}!}}{{x}}^{{5}}+\frac{{{240}}}{{{6}!}}{{x}}^{{6}}+\ldots$

$\displaystyle={0}+{0}\cdot{x}+\frac{{2}}{{2}}{{x}}^{{2}}+\frac{{0}}{{6}}{{x}}^{{3}}-\frac{{12}}{{24}}{{x}}^{{4}}+\frac{{0}}{{120}}{{x}}^{{5}}+\frac{{240}}{{720}}{{x}}^{{6}}+\ldots$

$\displaystyle={0}+{0}+{{x}}^{{2}}+{0}-\frac{{1}}{{2}}{{x}}^{{4}}+{0}+\frac{{1}}{{3}}{{x}}^{{6}}$

$\displaystyle={{x}}^{{2}}-\frac{{1}}{{2}}{{x}}^{{4}}+\frac{{1}}{{3}}{{x}}^{{6}}+\ldots$

The Taylor series of the function $\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({{x}}^{{2}}+{1}\right)}}}$ centered at $\displaystyle{c}={0}$ is:

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

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

Notes

Friday, 2 June 2017

$\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({{x}}^{{2}}+{1}\right)}}},{c}={0}$ 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?

Friday, 2 June 2017

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?

$\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({{x}}^{{2}}+{1}\right)}}},{c}={0}$ 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?