# Find the power series representation of this function

• April 11th 2009, 04:01 PM
saiyanmx89
Find the power series representation of this function
$f(x)=\frac{3}{2x-1}$ , c=2

This is what I did:

$\frac{a}{1-r} = \frac{-3}{1-2x} = -3\sum^\infty_ {n=0} (2x)^n$

Now, what do I do next since its centered at c=2??
• April 11th 2009, 04:05 PM
Jhevon
Quote:

Originally Posted by saiyanmx89
$f(x)=\frac{3}{2x-1}$ , c=2

This is what I did:

$\frac{a}{1-r} = \frac{-3}{1-2x} = -3\sum^\infty_ {n=0} (2x)^n$

Now, what do I do next since its centered at c=2??

Hint: you need to write it in the form $\sum_{n = 0}^\infty C_n (x - 2)^n$

can you figure that out? you do that manipulation before you even get to the first series.
• April 11th 2009, 04:11 PM
saiyanmx89
$-3 \sum^ \infty_ {n=0}2(x-2)^n$ ???
• April 11th 2009, 04:17 PM
Jhevon
Quote:

Originally Posted by saiyanmx89
$-3 \sum^ \infty_ {n=0}2(x-2)^n$ ???

all you did was replace the x in your series with (x - 2)

i said you make the change BEFORE getting to that series.

$\frac 3{2x - 1} = \frac {-3}{1 - 2x} = \frac {-3}{4 - 3 - 2x} = \frac {-3}{-3 -2(x - 2)}$

now go on to forming your series
• April 11th 2009, 04:29 PM
saiyanmx89
(-3)(x-2)^n ?
• April 11th 2009, 04:44 PM
Jhevon
Quote:

Originally Posted by saiyanmx89
(-3)(x-2)^n ?

question: what form for the power series are you trying to use? you are trying to use the fact that $\frac 1{1 - x} = \sum_{n = 0}^\infty x^n$, for $|x| < 1$, am i right?

now look at what we have. does it look like it is in that form? is the constant in the denominator a 1? is it 1 - (some function of x) in the denominator? no it isn't. get it into that form before you make the power series
• April 11th 2009, 04:51 PM
saiyanmx89
Could this work:

$\frac{-3}{1(-3)-2(x-2)}$

But the -3 still needs to be taken out right??
• April 11th 2009, 04:56 PM
Jhevon
Quote:

Originally Posted by saiyanmx89
Could this work:

$\frac{-3}{1(-3)-2(x-2)}$

But the -3 still needs to be taken out right??

yes, you need to factor -1/3 out of the fraction
• April 11th 2009, 05:09 PM
saiyanmx89
I'm sorry for asking such a mundane question but, how would I take out the (1/3)?
• April 11th 2009, 05:14 PM
Jhevon
Quote:

Originally Posted by saiyanmx89
I'm sorry for asking such a mundane question but, how would I take out the (1/3)?

you could start by factoring out -3 from the denominator and then you can just remove it and write -1/3 in front of the fraction. you should review factoring :p

$\frac {-3}{-3 -2(x - 2)} = \frac {-3}{-3 \left[ 1 + \frac {2(x - 2)}3\right]} = \frac 1{1 + \frac {2(x - 2)}3} = \frac 1{1 - \left[ - \frac {2(x - 2)}3\right]}$

now we have the required form.
• April 12th 2009, 09:51 AM
saiyanmx89
Why is this form acceptable then for c=0??

$\frac{-3}{1-2x}$ , c=0

I understand that $\frac{-3}{4-3-2x}$ is the same thing, but is this just something you have to do when the function isn't centered at 0?
• April 12th 2009, 10:46 AM
Jhevon
Quote:

Originally Posted by saiyanmx89
Why is this form acceptable then for c=0??

$\frac{-3}{1-2x}$ , c=0

I understand that $\frac{-3}{4-3-2x}$ is the same thing, but is this just something you have to do when the function isn't centered at 0?

if c = 0 it means you want your function part to look like (x - 0) which happens to just be x. thus, having that x there (possibly times some constant) is good enough. then, you have the constant in front being a "+1", so we have the form. you would NOT change it to 4 - 3 - 2x, because that is getting away from the form you want, which is $\frac 1{1 - f(x)}$, where $f(x)$ is some function in $x$. in general, you would want in this case, $\frac 1{1 - f(x - c)}$