MHB 206.11.3.12 write the power series

[karush]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
f^0(x)&=\ln{ (6 x + 1)} &\therefore f^0(a)&=0\\
f^1(x)&=\frac{6}{(6 x + 1)} &\therefore f^1(a)&=6\\
f^2(x)&= \frac{-36}{(6 x + 1)^2} &\therefore f^2(a)&=-36\\
f^3(x)&= \frac{432}{(6 x + 1)^3} &\therefore f^3(a)&=432\\
f^4(x)&= \frac{-7776}{(6 x + 1)^4} &\therefore f^3(a)&=-7776\\
\end{align}
\begin{align}
&=\frac{0}{0!}(x-a)^0 +\frac{6}{1!}(x-a)^1+
\frac{-36}{2!}(x-a)^2+\frac{432}{3!}(x-a)^3
+\frac{-7776}{4!}(x-a)^4\\
\ln{ (6 x + 1)}&\approx 6x-18x^{2}+72x^{3}-324x^{4}
\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}(6x)^{(k+1)}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[-\frac{1}{6},\frac{1}{6}\right]
\end{align}

$\textit{not sure about this too many steps !}$

 

[topsquark]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
f^0(x)&=\ln{ (6 x + 1)} &\therefore f^0(a)&=0\\
f^1(x)&=\frac{6}{(6 x + 1)} &\therefore f^1(a)&=6\\
f^2(x)&= \frac{-36}{(6 x + 1)^2} &\therefore f^2(a)&=-36\\
f^3(x)&= \frac{432}{(6 x + 1)^3} &\therefore f^3(a)&=432\\
f^4(x)&= \frac{-7776}{(6 x + 1)^4} &\therefore f^3(a)&=-7776\\
\end{align}
\begin{align}
&=\frac{0}{0!}(x-a)^0 +\frac{6}{1!}(x-a)^1+
\frac{-36}{2!}(x-a)^2+\frac{432}{3!}(x-a)^3
+\frac{-7776}{4!}(x-a)^4\\
\ln{ (6 x + 1)}&\approx 6x-18x^{2}+72x^{3}-324x^{4}
\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}(6x)^{(k+1)}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[-\frac{1}{6},\frac{1}{6}\right]
\end{align}

$\textit{not sure about this too many steps !}$

I'm assuming a = 0.

For b) work out a few terms of your series. They don't match the approximation.

For c) x = -1/6 gives f(0) = ln(0). Can we do that?

-Dan

 

[karush]

didn't seem to say what a= was so i used 0,

not sure how to do b. thot i was following an example but

didn't understand your Q on c.

 

[topsquark]

didn't seem to say what a= was so i used 0,

not sure how to do b. thot i was following an example but

didn't understand your Q on c.

b) Your summand doesn't give the correct series. Let's look at it this way. We have the series (which we don't know in advance if it's arithmetic, geometric, or something else. The coefficients are 6, -18, 72, -324,... What's the pattern here? Once you have that, you should be able to prove it pretty easily.

c) Your interval of convergence contains the point x = -1/6. This means your function value f(-1/6) = ln(0). But can we calculate ln(0)? So is -1/6 in the interval of convergence? (By the way I haven't checked to see if your interval is correct in general. I've been lazy.)

-Dan

 

[karush]

b) The coefficients are 6, -18, 72, -324,...

\begin{array}
\displaystyle
& (6)(1)(1)&+(6)(3)(-1)&+(6)(12)(1)&+(6)(54)(-1) \\
&6 &-18 &+72 &-324
\end{array}
$\textsf{got this far but...}$

 

[Greg]

a. is correct.

b. It's not a geometric series. It has coefficients (before you divided out the factorial(s))

$$6\cdot0!\over1!$$
$$-6^2\cdot1!\over2!$$
$$6^3\cdot2!\over3!$$
$$-6^4\cdot3!\over4!$$
$$6^5\cdot4!\over5!$$
$$\vdots$$

In sigma notation,

$$\sum_{n=1}^\infty\frac{x^n\cdot6^n\cdot(-1)^{n+1}\cdot(n-1)!}{n!}$$

(Nerd) I'm wondering about a proof by induction. (Wondering)

I'll let you refine the above (if you are so inclined) and complete c.

 

[Prove It]

$\textsf{a. Find the first four nozero terms of the Maciaurin series for the given function} \\$
\begin{align}
f^0(x)&=\ln{ (6 x + 1)} &\therefore f^0(a)&=0\\
f^1(x)&=\frac{6}{(6 x + 1)} &\therefore f^1(a)&=6\\
f^2(x)&= \frac{-36}{(6 x + 1)^2} &\therefore f^2(a)&=-36\\
f^3(x)&= \frac{432}{(6 x + 1)^3} &\therefore f^3(a)&=432\\
f^4(x)&= \frac{-7776}{(6 x + 1)^4} &\therefore f^3(a)&=-7776\\
\end{align}
\begin{align}
&=\frac{0}{0!}(x-a)^0 +\frac{6}{1!}(x-a)^1+
\frac{-36}{2!}(x-a)^2+\frac{432}{3!}(x-a)^3
+\frac{-7776}{4!}(x-a)^4\\
\ln{ (6 x + 1)}&\approx 6x-18x^{2}+72x^{3}-324x^{4}
\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}(6x)^{(k+1)}
\end{align}
$\textsf{c.Determine the interval of convergence of the series.} \\$
\begin{align}
IOC&=\left[-\frac{1}{6},\frac{1}{6}\right]
\end{align}

$\textit{not sure about this too many steps !}$

Notice that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \ln{ \left( 6\,x + 1 \right) } \right] = \frac{6}{6\,x + 1} \end{align*}$ and

$\displaystyle \begin{align*} \frac{6}{6\,x + 1} &= 6\,\left( \frac{1}{1 + 6\,x} \right) \\ &= 6\,\left[ \frac{1}{1 - \left( -6\,x \right) } \right] \\ &= 6\sum_{n = 0}^{\infty}{ \left( -6\,x \right) ^n } \textrm{ for } \left| -6\,x \right| < 1 \\ &= 6\sum_{n = 0}^{\infty}{ \left( -1 \right) ^n\, 6^n\,x^n } \textrm{ for } 6\left| x \right| < 1 \\ &= \sum_{n = 0}^{\infty}{\left( -1 \right) ^n \,6^{n+1}\,x^n} \textrm{ for } \left| x \right| < \frac{1}{6} \end{align*}$

so that means

$\displaystyle \begin{align*} \int{ \frac{6}{6\,x + 1}\,\mathrm{d}x } &= \int{ \sum_{n = 0}^{\infty}{ \left( -1 \right) ^n\,6^{n+1}\,x^n }\,\mathrm{d}x } \\ \ln{ \left( 6\,x + 1 \right) } + C &= \sum_{n = 0}^{\infty}{ \frac{\left( -1 \right) ^n\,6^{n+1}\,x^{n+1}}{n+1} } \end{align*}$

If we let $\displaystyle \begin{align*} x = 0 \end{align*}$ it should be obvious that $\displaystyle \begin{align*} C = 0 \end{align*}$, thus

$\displaystyle \begin{align*} \ln{ \left( 6\,x + 1 \right) } &= \sum_{n = 0}^{\infty}{ \frac{\left( -1 \right) ^n\,6^{n+1}\,x^{n+1}}{n+1} } \textrm{ for } \left| x \right| < \frac{1}{6} \\ &= 6\,x - 18\,x^2 + 72\,x^3 - 324\,x^4 + \dots \textrm{ for } \left| x \right| < \frac{1}{6} \end{align*}$Now a small side note, even though we know the radius of convergence of this series is the same as the geometric series we started with, we know nothing about the endpoints yet (as this is where the ratio test would give the inconclusive answer of 1) so now you need to substitute in $\displaystyle \begin{align*} x = -\frac{1}{6} \end{align*}$ and $\displaystyle \begin{align*} x = \frac{1}{6} \end{align*}$ and test the convergence of each resulting series.

 

[karush]

wow, thanks everyone that was great help

i asked some others but they couldn't help
get my best input here

we had to move on to the next section but still need to do more of these series.