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

Tuesday, 31 January 2017

$\displaystyle{f{{\left({x}\right)}}}={\ln{{x}}},{c}={1}$ 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 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 apply the definition of Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({x}\right)}}}$ , we list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

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

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

...

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

To apply the definition of Taylor series for the given function $\displaystyle{f{{\left({x}\right)}}}={\ln{{\left({x}\right)}}}$ , we list $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ as:

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

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

Apply Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$

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

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

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

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

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

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

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

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

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{{x}}^{{-{3}}}$

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle={0}+{1}\cdot{\left({x}-{1}\right)}+\frac{{-{1}}}{{{2}!}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{2}}{{{3}!}}{{\left({x}-{1}\right)}}^{{3}}+\frac{{-{6}}}{{{4}!}}{{\left({x}-{1}\right)}}^{{4}}+\ldots$

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

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

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

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

Notes

Tuesday, 31 January 2017

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

Tuesday, 31 January 2017

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?

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