# Math Help - Find the power series representation of this function

1. ## 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??

2. 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.

3. $-3 \sum^ \infty_ {n=0}2(x-2)^n$ ???

4. 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

5. (-3)(x-2)^n ?

6. 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

7. Could this work:

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

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

8. 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

9. I'm sorry for asking such a mundane question but, how would I take out the (1/3)?

10. 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

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

now we have the required form.

11. 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?

12. 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, $\frac 1{1 - f(x - c)}$