New Notes Blog: `f(x)=4/(3x+2) , c=3` Find a power series for the function, centered at c and determine the interval of convergence.

Tuesday, 13 January 2015

$\displaystyle{f{{\left({x}\right)}}}=\frac{{4}}{{{3}{x}+{2}}},{c}={3}$ 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 $\displaystyle{f{{\left({x}\right)}}}=\frac{{4}}{{{3}{x}+{2}}}$ centered at $\displaystyle{c}={2}$ , we may apply Law of Exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ and Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{x}}^{{n}}={n}\cdot{{x}}^{{{n}-{1}}}$ .

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

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

Let $\displaystyle{u}={3}{x}+{2}$ then $\displaystyle\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}={3}$ .

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{c}\cdot{{\left({3}{x}+{2}\right)}}^{{n}}={c}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{{\left({3}{x}+{2}\right)}}^{{n}}$

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

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

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

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

$\displaystyle=-{12}{{\left({3}{x}+{2}\right)}}^{{-{2}}}{\quad\text{or}\quad}\frac{{2}}{{{\left({3}{x}+{2}\right)}}^{{2}}}$

$\displaystyle{{f}}^{{2}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}-{12}{{\left({3}{x}+{2}\right)}}^{{-{2}}}$

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

$\displaystyle={72}{{\left({3}{x}+{2}\right)}}^{{-{3}}}{\quad\text{or}\quad}\frac{{72}}{{{\left({3}{x}+{2}\right)}}^{{3}}}$

$\displaystyle{{f}}^{{3}}{\left({x}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{72}{{\left({3}{x}+{2}\right)}}^{{-{3}}}$

$\displaystyle={3}\cdot{\left({72}\right)}{\left(-{3}\right)}{{\left({3}{x}+{2}\right)}}^{{-{3}-{1}}}$

$\displaystyle=-{648}{{\left({3}{x}+{2}\right)}}^{{-{4}}}{\quad\text{or}\quad}-\frac{{648}}{{{\left({3}{x}+{2}\right)}}^{{4}}}$

Plug-in $\displaystyle{x}={3}$ for each $\displaystyle{{f}}^{{n}}{\left({x}\right)}$ , we get:

$\displaystyle{f{{\left({3}\right)}}}=\frac{{4}}{{{3}{\left({3}\right)}+{2}}}$

$\displaystyle=\frac{{4}}{{11}}$

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

$\displaystyle=-\frac{{12}}{{{11}}^{{2}}}$

$\displaystyle=-\frac{{12}}{{121}}$

$\displaystyle{{f}}^{{2}}{\left({3}\right)}=\frac{{72}}{{{\left({3}{\left({3}\right)}+{2}\right)}}^{{3}}}$

$\displaystyle=\frac{{72}}{{{11}}^{{3}}}$

$\displaystyle=\frac{{72}}{{1331}}$

$\displaystyle{{f}}^{{3}}{\left({3}\right)}=-\frac{{648}}{{{\left({3}{\left({3}\right)}+{2}\right)}}^{{4}}}$

$\displaystyle=-\frac{{648}}{{{11}}^{{4}}}$

$\displaystyle=-\frac{{648}}{{14641}}$

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

$\displaystyle\frac{{4}}{{{3}{x}+{2}}}={\sum_{{{n}={0}}}^{\infty}}\frac{{{{f}}^{{n}}{\left({3}\right)}}}{{{n}!}}{{\left({x}-{3}\right)}}^{{n}}$

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

$\displaystyle=\frac{{4}}{{11}}+{\left(-\frac{{12}}{{121}}\right)}{\left({x}-{3}\right)}+\frac{{\frac{{72}}{{1331}}}}{{{2}!}}{{\left({x}-{3}\right)}}^{{2}}+\frac{{-\frac{{648}}{{14641}}}}{{{3}!}}{{\left({x}-{3}\right)}}^{{3}}+\ldots$

$\displaystyle=\frac{{4}}{{11}}-{\left(\frac{{12}}{{121}}\right)}{\left({x}-{3}\right)}+\frac{{\frac{{72}}{{1331}}}}{{2}}{{\left({x}-{3}\right)}}^{{2}}-\frac{{\frac{{648}}{{14641}}}}{{6}}{{\left({x}-{3}\right)}}^{{3}}+\ldots$

$\displaystyle=\frac{{4}}{{11}}-\frac{{12}}{{121}}{\left({x}-{3}\right)}+\frac{{36}}{{1331}}{{\left({x}-{3}\right)}}^{{2}}-\frac{{108}}{{14641}}{{\left({x}-{3}\right)}}^{{3}}+\ldots$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}{4}\frac{{{\left(-{3}{\left({x}-{3}\right)}\right)}}^{{{n}-{1}}}}{{{11}}^{{n}}}$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}{4}{{\left(-{3}{\left({x}-{3}\right)}\right)}}^{{-{1}}}\frac{{{\left(-{3}{\left({x}-{3}\right)}\right)}}^{{n}}}{{{11}}^{{n}}}$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}\frac{{4}}{{-{3}{\left({x}-{3}\right)}}}{{\left(\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}\right)}}^{{n}}$

$\displaystyle={\sum_{{{n}={1}}}^{\infty}}\frac{{4}}{{-{3}{x}+{9}}}{{\left(\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}\right)}}^{{n}}$

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 $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{4}}{{-{3}{x}+{9}}}{{\left(\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}\right)}}^{{n}}$ with $\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}\cdot{{r}}^{{n}}$ , we determine: $\displaystyle{r}=\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}$ .

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

$\displaystyle{\left|\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}\right|}\lt{1}$

$\displaystyle{\left|-{1}\right|}\cdot{\left|\frac{{{3}{\left({x}-{3}\right)}}}{{11}}\right|}\lt{1}$

$\displaystyle{1}\cdot{\left|\frac{{{3}{\left({x}-{3}\right)}}}{{11}}\right|}\lt{1}$

$\displaystyle{\left|\frac{{{3}{\left({x}-{3}\right)}}}{{11}}\right|}\lt{1}$

$\displaystyle{\left|\frac{{{3}{x}-{9}}}{{11}}\right|}\lt{1}$

$\displaystyle-{1}\lt\frac{{{3}{x}-{9}}}{{11}}\lt{1}$

Multiply each sides by $\displaystyle{11}$ :

$\displaystyle-{1}\cdot{11}\lt\frac{{{3}{x}-{9}}}{{11}}\cdot{11}\lt{1}\cdot{11}$

$\displaystyle-{11}\lt{3}{x}-{9}\lt{11}$

Add 9 on each sides:

$\displaystyle-{11}+{9}\lt{3}{x}-{9}+{9}\lt{11}+{9}$

$\displaystyle-{2}\lt{3}{x}\lt{20}$

Divide each side by $\displaystyle{3}$ :

$\displaystyle-\frac{{2}}{{3}}\lt\frac{{{3}{x}}}{{3}}\lt\frac{{20}}{{3}}$

$\displaystyle-\frac{{2}}{{3}}\lt{x}\lt\frac{{20}}{{3}}$

Thus, the power series of the function $\displaystyle{f{{\left({x}\right)}}}=\frac{{4}}{{{3}{x}+{2}}}$ centered at $\displaystyle{c}={3}$ is $\displaystyle{\sum_{{{n}={1}}}^{\infty}}\frac{{4}}{{-{3}{x}+{9}}}{{\left(\frac{{-{3}{\left({x}-{3}\right)}}}{{11}}\right)}}^{{n}}$

with an interval of convergence: $\displaystyle-\frac{{2}}{{3}}\lt{x}\lt\frac{{20}}{{3}}$ .

New Notes Blog

Tuesday, 13 January 2015

$\displaystyle{f{{\left({x}\right)}}}=\frac{{4}}{{{3}{x}+{2}}},{c}={3}$ 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, 13 January 2015

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?

$\displaystyle{f{{\left({x}\right)}}}=\frac{{4}}{{{3}{x}+{2}}},{c}={3}$ 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?