New Notes Blog: `f(x) = 1/(3-x) , c=1` Find a power series for the function, centered at c and determine the interval of convergence.

Tuesday, 17 June 2014

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{3}-{x}}},{c}={1}$ 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 $\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{3}-{x}}}$ centered at $\displaystyle{c}={1}$ , we may apply Law of Exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ and Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ .

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

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

Let $\displaystyle{u}={3}-{x}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}=-{1}$

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

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

$\displaystyle=-{c}{n}{{\left({3}-{x}\right)}}^{{{n}-{1}}}$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle={24}{{\left({3}-{x}\right)}}^{{-{5}}}{\quad\text{or}\quad}\frac{{24}}{{{\left({3}-{x}\right)}}^{{5}}}$

Plug-in $\displaystyle{x}={1}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

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

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

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

$\displaystyle{{f}}^{{3}}{\left({1}\right)}=\frac{{6}}{{{\left({3}-{1}\right)}}^{{4}}}=\frac{{3}}{{8}}$

$\displaystyle{{f}}^{{4}}{\left({1}\right)}=\frac{{24}}{{{\left({3}-{1}\right)}}^{{5}}}=\frac{{3}}{{4}}$

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

$\displaystyle\frac{{1}}{{{3}-{x}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({1}\right)}}}{{{n}!}}{{\left({x}-{1}\right)}}^{{n}}$

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

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

$\displaystyle=\frac{{1}}{{2}}+\frac{{1}}{{4}}{\left({x}-{1}\right)}+\frac{{\frac{{1}}{{4}}}}{{2}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{\frac{{3}}{{8}}}}{{6}}{{\left({x}-{1}\right)}}^{{3}}+\frac{{\frac{{3}}{{4}}}}{{24}}{{\left({x}-{1}\right)}}^{{4}}+\ldots$

$\displaystyle=\frac{{1}}{{2}}+\frac{{1}}{{4}}{\left({x}-{1}\right)}+\frac{{1}}{{8}}{{\left({x}-{1}\right)}}^{{2}}+\frac{{1}}{{16}}{{\left({x}-{1}\right)}}^{{3}}+\frac{{1}}{{32}}{{\left({x}-{1}\right)}}^{{4}}+\ldots$

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

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

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

$\displaystyle{\left|\frac{{{x}-{1}}}{{2}}\right|}\lt{1}$

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

Multiply each sides by 2:

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

$\displaystyle-{2}\lt{x}-{1}\lt{2}$

Add 1 on each sides:

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

$\displaystyle-{1}\lt{x}\lt{3}$

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

New Notes Blog

Tuesday, 17 June 2014

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{3}-{x}}},{c}={1}$ 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?

Tuesday, 17 June 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?

$\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{3}-{x}}},{c}={1}$ 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?