MHB 206.11.3.11 Find the first four nozero terms of the Maciaurin series

[karush]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
a&=0 \\
f(x)&=(-5+x^2)^{-1} \\
\\
f^0(x)&=(-5+x^2)^{-1}\therefore f^0(a) = 1 \\
f^1(x)&=\frac{-2x}{(x^2-5)^2} \therefore f^1(a) = 0 \\
f^2(x)&=\frac{2(3x^3+5)}{(x^2-5)^3} \therefore f^2(a) = \frac{-2}{25} \\
f^3(x)&=\frac{-24x(x^2+5))}{(x^2-5)^4} \therefore f^3(a) = 0 \\
f^4(x)&=\frac{120x(x^4+10x^2+5))}{(x^2-5)^5} \therefore f^4(a) = \frac{-24}{125} \\
f^5(x)&=\frac{240x(x^4+10x^2+5))}{(x^2-5)^6} \therefore f^5(a) = 0 \\
f^6(x)&=\frac{720x(7x^6+175x^4+525x^2+125))}{(x^2-5)^7} \therefore f^6(a) = \frac{-144}{125}\\
\end{align}
$\textsf{so}$
\begin{align}
f(x)&=1+\frac{-x^2}{25}-\frac{-x^4}{125}-\frac{x^6}{625}
\end{align}
$\textsf{b. write the power series using $\sigma$ notation.} \\$
\begin{align}
\displaystyle
f(x)&=\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \textsf{ for } |x|<1 \\
&=\sum_{k=0}^{\infty}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[a,b\right]
\end{align}

$\textit{not even sure if I did a. correctly.. tons of calculation which I did most on TI}\\$
$\textit{b. and c. will try if a. looks ok did a few before on other problems.}\\$
$\textit{On a. I just skipped the the ones that =0 thanks for know it is a headache!}$

 

[topsquark]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
a&=0 \\
f(x)&=(-5+x^2)^{-1} \\
\\
f^0(x)&=(-5+x^2)^{-1}\therefore f^0(a) = 1 \\
f^1(x)&=\frac{-2x}{(x^2-5)^2} \therefore f^1(a) = 0 \\
f^2(x)&=\frac{2(3x^3+5)}{(x^2-5)^3} \therefore f^2(a) = \frac{-2}{25} \\
f^3(x)&=\frac{-24x(x^2+5))}{(x^2-5)^4} \therefore f^3(a) = 0 \\
f^4(x)&=\frac{120x(x^4+10x^2+5))}{(x^2-5)^5} \therefore f^4(a) = \frac{-24}{125} \\
f^5(x)&=\frac{240x(x^4+10x^2+5))}{(x^2-5)^6} \therefore f^5(a) = 0 \\
f^6(x)&=\frac{720x(7x^6+175x^4+525x^2+125))}{(x^2-5)^7} \therefore f^6(a) = \frac{-144}{125}\\
\end{align}
$\textsf{so}$
\begin{align}
f(x)&=1+\frac{-x^2}{25}-\frac{-x^4}{125}-\frac{x^6}{625}
\end{align}
$\textsf{b. write the power series using $\sigma$ notation.} \\$
\begin{align}
\displaystyle
f(x)&=\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \textsf{ for } |x|<1 \\
&=\sum_{k=0}^{\infty}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[a,b\right]
\end{align}

$\textit{not even sure if I did a. correctly.. tons of calculation which I did most on TI}\\$
$\textit{b. and c. will try if a. looks ok did a few before on other problems.}\\$
$\textit{On a. I just skipped the the ones that =0 thanks for know it is a headache!}$

I didn't actually check all of the derivatives but you need to look at f(0) and f^(4)(0) again. It does look like you've got the idea down, though. Give b) and c) a try.

-Dan

 

[topsquark]

f^(2)(x) has a typo. It's [math]3x^2 + 5[/math] not [math]x^3[/math].

f^(5)(x) has a big typo...the polynomial in the parenthesis is just copied over from the line above. And you missed an overall minus sign.

In f^(4)(x) and f^(6)(x) the x's on the outside of the parenthesis should not be there. But these are just typos.

-Dan

 

[Prove It]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
a&=0 \\
f(x)&=(-5+x^2)^{-1} \\
\\
f^0(x)&=(-5+x^2)^{-1}\therefore f^0(a) = 1 \\
f^1(x)&=\frac{-2x}{(x^2-5)^2} \therefore f^1(a) = 0 \\
f^2(x)&=\frac{2(3x^3+5)}{(x^2-5)^3} \therefore f^2(a) = \frac{-2}{25} \\
f^3(x)&=\frac{-24x(x^2+5))}{(x^2-5)^4} \therefore f^3(a) = 0 \\
f^4(x)&=\frac{120x(x^4+10x^2+5))}{(x^2-5)^5} \therefore f^4(a) = \frac{-24}{125} \\
f^5(x)&=\frac{240x(x^4+10x^2+5))}{(x^2-5)^6} \therefore f^5(a) = 0 \\
f^6(x)&=\frac{720x(7x^6+175x^4+525x^2+125))}{(x^2-5)^7} \therefore f^6(a) = \frac{-144}{125}\\
\end{align}
$\textsf{so}$
\begin{align}
f(x)&=1+\frac{-x^2}{25}-\frac{-x^4}{125}-\frac{x^6}{625}
\end{align}
$\textsf{b. write the power series using $\sigma$ notation.} \\$
\begin{align}
\displaystyle
f(x)&=\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \textsf{ for } |x|<1 \\
&=\sum_{k=0}^{\infty}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[a,b\right]
\end{align}

$\textit{not even sure if I did a. correctly.. tons of calculation which I did most on TI}\\$
$\textit{b. and c. will try if a. looks ok did a few before on other problems.}\\$
$\textit{On a. I just skipped the the ones that =0 thanks for know it is a headache!}$

I would realize that this can be written as a geometric series, it's so much less work...

$\displaystyle \begin{align*} f(x) &= \frac{1}{-5 + x^2 } \\ &= -\frac{1}{5} \left( \frac{1}{1 - \frac{x^2}{5}} \right) \\ &= -\frac{1}{5} \sum_{n = 0}^{\infty} \left( \frac{x^2}{5} \right) ^n \end{align*}$

which is convergent where

$\displaystyle \begin{align*} \left| \frac{x^2}{5} \right| &< 1 \\ \frac{ x^2 }{ 5 } &< 1 \\ x^2 &< 5 \\ \left| x \right| &< \sqrt{5} \\ -\sqrt{5} < x &< \sqrt{5} \end{align*}$So the first four nonzero terms are

$\displaystyle \begin{align*} -\frac{1}{5}\left( 1 + \frac{x^2}{5} + \frac{x^4}{25} + \frac{x^6}{125} \right) = -\frac{1}{5} - \frac{x^2}{25} - \frac{x^4}{125} - \frac{x^6}{625} \end{align*}$

 

[karush]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
a&=0 \\
f(x)&=(-5+x^2)^{-1} \\
\\
f^0(x)&=(-5+x^2)^{-1}\therefore f^0(a) = 1 \\
f^1(x)&=\frac{-2x}{(x^2-5)^2}
\therefore f^1(a) = 0 \\
f^2(x)&=\frac{2(3x^2+5)}{(x^2-5)^3}
\therefore f^2(a) = \frac{-2}{25} \\
f^3(x)&=\frac{-24x(x^2+5))}{(x^2-5)^4}
\therefore f^3(a) = 0 \\
f^4(x)&=\frac{120(x^4+10x^2+5))}{(x^2-5)^5}
\therefore f^4(a) = \frac{-24}{125} \\
f^5(x)&=-\frac{240.(3x^4+50x^2+75))}{(x^2-75)^6}
\therefore f^5(a) = 0 \\
f^6(x)&=\frac{720x(7x^6+175x^4+525x^2+125))}{(x^2-5)^7} \therefore f^6(a) = \frac{-144}{125}\\
\end{align}
$\textsf{so}$
\begin{align}
f(x)&=1+\frac{-x^2}{25}-\frac{-x^4}{125}-\frac{x^6}{625}
\end{align}
$\textsf{b. write the power series using $\sigma$ notation.} \\$
\begin{align}
\displaystyle
f(x)&;=\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \textsf{ for } |x|<1 \\
&=\sum_{k=0}^{\infty}
\end{align}

\begin{align*}
\displaystyle
f(x) &= \frac{1}{-5 + x^2 } \\
&= -\frac{1}{5} \left( \frac{1}{1 - \frac{x^2}{5}} \right) \\
&= -\frac{1}{5} \sum_{n = 0}^{\infty} \left( \frac{x^2}{5} \right) ^n \end{align*}

$\textsf{c.Determine the interval of convergence of the series.} \\$

\begin{align*}
\displaystyle
\left| \frac{x^2}{5} \right| &< 1 \\
\frac{ x^2 }{ 5 } &< 1 \\ x^2 &< 5 \\
\left| x \right| &< \sqrt{5} \\
-\sqrt{5} < x &< \sqrt{5}
\end{align*}

 

[karush]

after correcting and plagiarising...

 

[Prove It]

after correcting and plagiarising...

Shame you didn't synthesise the information - it's pretty obvious you've made some mistakes, as the answers are supposed to match!

 

[karush]

well being confined to a small tablet doesn't help the typo problem to much. also doing so much cut and paste will intoduce mistakes too.

the answer is ussauly given anyway so the steps are of most interest to me. plus the best way to formally present the problem so it is it is understood.

I'm personally self learning this. I was auditing the class but had to sit in the back of a very large class and could not read the board or hear the teacher plus a very long jaring bus ride round trip.

by far most of the learning I receilve is gratfully thru MHB. I try other sites but the LateX has no live review. plus the replies here are much more helpful. Mahalo