New Notes Blog: `f(x)=2/(6-x) ,c=-2` Find a power series for the function, centered at c and determine the interval of convergence.

Friday, 31 October 2014

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}},{c}=-{2}$ Find a power series for the function, centered at c and determine the interval of convergence.

To determine the power series centered at c, we may apply the formula for Taylor series:

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

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

To list the $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ for the given function $\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}}$ centered at $\displaystyle{c}=-{2}$ , we may apply Law of Exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ and Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ .

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}}$

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

Let $\displaystyle{u}={6}-{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=-{1}$

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

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

$\displaystyle=-{c}{n}{{\left({6}-{x}\right)}}^{{{n}-{1}}}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{{\left({6}-{x}\right)}}^{{-{1}}}$

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

$\displaystyle={2}{{\left({6}-{x}\right)}}^{{-{2}}}{\quad\text{or}\quad}\frac{{2}}{{{\left({6}-{x}\right)}}^{{2}}}$

$\displaystyle{{f}}^{{2}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{2}{{\left({6}-{x}\right)}}^{{-{2}}}$

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

$\displaystyle={4}{{\left({6}-{x}\right)}}^{{-{3}}}{\quad\text{or}\quad}\frac{{4}}{{{\left({6}-{x}\right)}}^{{3}}}$

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

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

$\displaystyle={12}{{\left({6}-{x}\right)}}^{{-{4}}}{\quad\text{or}\quad}\frac{{12}}{{{\left({6}-{x}\right)}}^{{4}}}$

$\displaystyle{{f}}^{{4}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{12}{{\left({6}-{x}\right)}}^{{-{4}}}$

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

$\displaystyle={48}{{\left({6}-{x}\right)}}^{{-{5}}}{\quad\text{or}\quad}\frac{{48}}{{{\left({6}-{x}\right)}}^{{5}}}$

Plug-in $\displaystyle{x}=-{2}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

$\displaystyle{f{{\left(-{2}\right)}}}=\frac{{2}}{{{6}-{\left(-{2}\right)}}}$

$\displaystyle=\frac{{2}}{{8}}$

$\displaystyle=\frac{{1}}{{4}}$

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

$\displaystyle=\frac{{2}}{{{8}}^{{2}}}$

$\displaystyle=\frac{{1}}{{32}}$

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

$\displaystyle=\frac{{4}}{{{8}}^{{3}}}$

$\displaystyle=\frac{{1}}{{128}}$

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

$\displaystyle=\frac{{12}}{{{8}}^{{4}}}$

$\displaystyle=\frac{{3}}{{1024}}$

$\displaystyle{{f}}^{{4}}{\left(-{2}\right)}=\frac{{48}}{{{\left({6}-{\left(-{2}\right)}\right)}}^{{5}}}$

$\displaystyle=\frac{{48}}{{{8}}^{{5}}}$

$\displaystyle=\frac{{3}}{{2048}}$

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

$\displaystyle\frac{{2}}{{{6}-{x}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left(-{2}\right)}}}{{{n}!}}{{\left({x}-{\left(-{2}\right)}\right)}}^{{n}}$

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

$\displaystyle=\frac{{1}}{{4}}+\frac{{1}}{{32}}{\left({x}+{2}\right)}+\frac{{\frac{{1}}{{128}}}}{{{2}!}}{{\left({x}+{2}\right)}}^{{2}}+\frac{{\frac{{3}}{{1024}}}}{{{3}!}}{{\left({x}+{2}\right)}}^{{3}}+\frac{{\frac{{3}}{{2048}}}}{{{4}!}}{{\left({x}+{2}\right)}}^{{4}}+\ldots$

$\displaystyle=\frac{{1}}{{4}}+\frac{{1}}{{32}}{\left({x}+{2}\right)}+\frac{{\frac{{1}}{{128}}}}{{2}}{{\left({x}+{2}\right)}}^{{2}}+\frac{{\frac{{3}}{{1024}}}}{{6}}{{\left({x}+{2}\right)}}^{{3}}+\frac{{\frac{{3}}{{2048}}}}{{24}}{{\left({x}+{2}\right)}}^{{4}}+\ldots$

$\displaystyle=\frac{{1}}{{4}}+\frac{{1}}{{32}}{\left({x}+{2}\right)}+\frac{{1}}{{256}}{{\left({x}+{2}\right)}}^{{2}}+\frac{{1}}{{2048}}{{\left({x}+{2}\right)}}^{{3}}+\frac{{1}}{{16384}}{{\left({x}+{2}\right)}}^{{4}}+\ldots$

$\displaystyle=\frac{{1}}{{{2}}^{{2}}}+\frac{{1}}{{{2}}^{{5}}}{\left({x}+{2}\right)}+\frac{{1}}{{{2}}^{{8}}}{{\left({x}+{2}\right)}}^{{2}}+\frac{{1}}{{{2}}^{{11}}}{{\left({x}+{2}\right)}}^{{3}}+\frac{{1}}{{{2}}^{{14}}}{{\left({x}+{2}\right)}}^{{4}}+\ldots$

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

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

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}\frac{{{2}{{\left({x}+{2}\right)}}^{{n}}}}{{{{2}}^{{{3}{n}}}{\left({x}+{2}\right)}}}$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}{\left(\frac{{2}}{{{x}+{2}}}\right)}{{\left(\frac{{{x}+{2}}}{{{2}}^{{3}}}\right)}}^{{n}}$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}{\left(\frac{{2}}{{{x}+{2}}}\right)}{{\left(\frac{{{x}+{2}}}{{8}}\right)}}^{{n}}$

Note: Exponents of 2 as 2,5,8,11,14,... follows arithmetic sequence $\displaystyle{a}_{{n}}={a}_{{0}}+{\left({n}-{1}\right)}{d}.$

$\displaystyle{a}_{{n}}={2}+{\left({n}-{1}\right)}{3}$

$\displaystyle={2}+{3}{n}-{3}$

$\displaystyle={3}{n}-{1}$

To determine the interval of convergence, we may apply geometric series test wherein the series $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}$ is convergent if $\displaystyle{\left|{r}\right|}\lt{1}{\quad\text{or}\quad}-{1}\lt{r}\lt{1}$ . If $\displaystyle{\left|{r}\right|}\geq{1}$ then the geometric series diverges.

By comparing $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\left(\frac{{2}}{{{x}+{2}}}\right)}{{\left(\frac{{{x}+{2}}}{{8}}\right)}}^{{n}}$ with $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}$ , we determine: $\displaystyle{r}=\frac{{{x}+{2}}}{{8}}.$

Apply the condition for convergence of geometric series: $\displaystyle{\left|{r}\right|}\lt{1}$ .

$\displaystyle{\left|{\left(\frac{{{x}+{2}}}{{8}}\right)}\right|}\lt{1}$

$\displaystyle-{1}\lt\frac{{{x}+{2}}}{{8}}\lt{1}$

Multiply each sides by 8:

$\displaystyle-{1}\cdot{8}\lt\frac{{{x}+{2}}}{{8}}\cdot{8}\lt{1}\cdot{8}$

$\displaystyle-{8}\lt{x}+{2}\lt{8}$

Subtract $\displaystyle{2}$ from each sides:

$\displaystyle-{8}-{2}\lt{x}+{2}-{2}\lt{8}-{2}$

$\displaystyle-{10}\lt{x}\lt{6}$

Thus, the power series of the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}}$ centered at $\displaystyle{c}=-{2}$ is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{{\left({x}+{2}\right)}}^{{{n}-{1}}}}{{{2}}^{{{3}{n}-{1}}}}$ with an interval of convergence: $\displaystyle-{10}\lt{x}\lt{6}$ .

New Notes Blog

Friday, 31 October 2014

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}},{c}=-{2}$ Find a power series for the function, centered at c and determine the interval of convergence.

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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?

Friday, 31 October 2014

Find a power series for the function, centered at c and determine the interval of convergence.

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?

$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{{6}-{x}}},{c}=-{2}$ Find a power series for the function, centered at c and determine the interval of convergence.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?