# need help with a/(1-r) form

• Apr 12th 2009, 09:28 AM
saiyanmx89
need help with a/(1-r) form
$\displaystyle f(x)=\frac{4}{4+x^2}$, c=0

$\displaystyle f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}$

$\displaystyle \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n$

If I did this right, where do I go from here?? Thank you.
• Apr 12th 2009, 09:31 AM
TheEmptySet
Quote:

Originally Posted by saiyanmx89
$\displaystyle f(x)=\frac{4}{4+x^2}$, c=0

$\displaystyle f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}$

$\displaystyle \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n$

If I did this right, where do I go from here?? Thank you.

Almost

$\displaystyle \frac{4}{4+x^2}=\frac{1}{1+\left( \frac{x}{2}\right)^2}=...$
• Apr 12th 2009, 09:34 AM
Plato
Quote:

Originally Posted by saiyanmx89
$\displaystyle f(x)=\frac{4}{4+x^2}$, c=0

$\displaystyle f(x)=\frac{4}{4(1+(\frac{x^2}{4}))} = \frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-(\frac{-x^2}{4}))}$

$\displaystyle \sum^ \infty_ {n=0} (\frac{-x^2}{4})^n = \frac{-1}{4}\sum^ \infty_ {n=0} (x^2)^n$

You need to stop with $\displaystyle \color{blue}\sum^ \infty_ {n=0} (\frac{-x^2}{4})^n$.
• Apr 12th 2009, 09:39 AM
saiyanmx89
you mean I didn't need to go any further with it?
• Apr 12th 2009, 09:42 AM
Plato
Quote:

Originally Posted by saiyanmx89
you mean I didn't need to go any further with it?

No, I meant that you cannot factor the $\displaystyle \frac{-1}{4}$ out of the sum!
Do you see why?
• Apr 12th 2009, 09:57 AM
saiyanmx89
oh yea. the x^2, which is why I must use= (x/2)^2

But, where do I go from here:
$\displaystyle \sum^ \infty_ {n=0} (\frac{-x}{2})^{2n}$ = ??
• Apr 12th 2009, 10:01 AM
Jhevon
Quote:

Originally Posted by saiyanmx89
oh yea. the x^2, which is why I must use= (x/2)^2

well, you could distribute the power. what do you think it would look like then?
• Apr 12th 2009, 10:05 AM
saiyanmx89
could I use the Ratio Test on $\displaystyle \frac{-x^{2n}}{4^n}$
• Apr 12th 2009, 10:28 AM
Jhevon
Quote:

Originally Posted by saiyanmx89
could I use the Ratio Test on $\displaystyle \frac{-x^{2n}}{4^n}$

review how to distribute powers.

Spoiler:
$\displaystyle \sum_{n = 0}^\infty (-1)^n \frac {x^{2n}}{4^n}$

this is to highlight the fact it is an alternating series
• Apr 12th 2009, 01:36 PM
saiyanmx89
so, by using the Alternating Series, I can take the absolute value of the function leaving me with (x^2/4) or (x/4) ??? I'm not sure which...
• Apr 12th 2009, 01:53 PM
Plato
Quote:

Originally Posted by saiyanmx89
so, by using the Alternating Series, I can take the absolute value of the function leaving me with (x^2/4) or (x/4) ??? I'm not sure which...

Are you not making too much of this? Over think it?
The common ratio is $\displaystyle \frac{-x^2}{4}$.
So what is the absolute value?
• Apr 12th 2009, 01:54 PM
saiyanmx89
x^2/4