Finding a Power Series for a function

[aerospacev]

Homework Statement

Find a power series representation for the function $f(x) = \frac{(x-1)}{(3-x)^2}^2$, valid for every $x$ with $|x|<3$

Homework Equations

The equation that I think would be useful is $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$

The Attempt at a Solution

I began by just looking at the term $\frac{1}{(3-x)^2}$

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$

differentiating both sides:

$\frac{1}{(1-x)^2}= \sum_{n=1}^\infty nx^{(n-1)}$

now I'm stuck here, my train of thought is that if I can turn $\frac{1}{(3-x)^2}$ into someform of $\frac{1}{(1-x)^2}$ to obtain its summation formula, I can then multiply the polynomial ${(x-1)}^2$ to that summation and go from there.

Usually the assignment questions involve functions in the likes of $\frac{x}{(1-x^2)}$ or $\frac{x^2}{(1+x)^3}$, which I can solve, but the $3$ in front of the x in $\frac{1}{(3-x)^2}$ is really troubling me, any help? Thanks!

 

[Saitama]

Hi aerospacev!

You can factor out the 3 from the denominator and write it as following:
$$f(x)=\frac{(x-1)^2}{9}\frac{1}{(1-x/3)^2}$$
Can you proceed now?

 

[aerospacev]

I wondered about that, maybe I need to review my basic arithmetic, am I allowed to simply divide each number individually in ${(3-x)^2}$ by 3 and multiply by $3^2$? Even though the entire term is squared?

 

[Saitama]

I wondered about that, maybe I need to review my basic arithmetic, am I allowed to simply divide each number individually in ${(3-x)^2}$ by 3 and multiply by $3^2$? Even though the entire term is squared?

I am not sure what you mean there. What I did is the following:
$$(3-x)^2=\left(3\left(1-\frac{x}{3}\right)\right)^2=3^2\left(1-\frac{x}{3}\right)^2=9\left(1-\frac{x}{3}\right)^2$$
I hope that helps.

 

[Chestermiller]

I am not sure what you mean there. What I did is the following:
$$(3-x)^2=\left(3\left(1-\frac{x}{3}\right)\right)^2=3^2\left(1-\frac{x}{3}\right)=9\left(1-\frac{x}{3}\right)$$
I hope that helps.

Hi Panav-Arora.

I think you want to correct those last two versions of the right hand sides, to have an exponent of 2.

Chet

 

[Saitama]

Hi Panav-Arora.

I think you want to correct those last two versions of the right hand sides, to have an exponent of 2.

Chet

Woops, thanks Chet! :D

 

[aerospacev]

I am not sure what you mean there. What I did is the following:
$$(3-x)^2=\left(3\left(1-\frac{x}{3}\right)\right)^2=3^2\left(1-\frac{x}{3}\right)^2=9\left(1-\frac{x}{3}\right)^2$$
I hope that helps.

Oh ok thanks, I think there was just some confusion on my part.

I got the correct answer, thanks again!

 

[vanceEE]

How about
$f(x) = (\frac{x-1}{3-x})^2 = (\frac{x}{3-x}-\frac{1}{3-x})^2$
$= (\frac{-1}{1-(\frac{3}{x})}-\frac{(\frac{1}{3})}{1-(\frac{x}{3})})^2$
$so, f(x) = [-\sum\limits_{n=0}^∞ (\frac{3}{x})^n - \frac{1}{3}\sum\limits_{n=0}^∞ (\frac{x}{3})^n]^2$ where $-3 < x < 3$

 

[aerospacev]

How about
$f(x) = (\frac{x-1}{3-x})^2 = (\frac{x}{3-x}-\frac{1}{3-x})^2$
$= (\frac{-1}{1-(\frac{3}{x})}-\frac{(\frac{1}{3})}{1-(\frac{x}{3})})^2$
$so, f(x) = [-\sum\limits_{n=0}^∞ (\frac{3}{x})^n - \frac{1}{3}\sum\limits_{n=0}^∞ (\frac{x}{3})^n]^2$ where $-3 < x < 3$

I don't think so, wouldn't you then have to divide by zero?

 

[jbunniii]

Also, even if the calculation is correct,
$$[-\sum\limits_{n=0}^∞ (\frac{3}{x})^n - \frac{1}{3}\sum\limits_{n=0}^∞ (\frac{x}{3})^n]^2$$
is not a power series as written. It is a expression involving power series, but that's not the same thing.

 

[vanceEE]

I don't think so, wouldn't you then have to divide by zero?

Divide what by zero? The index of summation n = 0 means that we would first have a zero as an exponent. This series isn't convergent, an error on my part, but there is nothing being divided by zero in this series; x ≠ 0.

 

[jbunniii]

there is nothing being divided by zero in this series; x ≠ 0.

But the original problem statement says "valid for every $x$ with $|x|< 3$", so that includes $x=0$.