How do I expand $f(z)$ into a power series?

[Dustinsfl]

$\displaystyle f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}$

How do I find the power series?

I know that

$\displaystyle\frac{1}{z+1} = \frac{1}{1-(-z)} = \sum_{n}^{\infty}(-z)^n$

and

$\displaystyle\frac{1}{(z+2)^2} = \frac{d}{dz} \frac{-1}{z+2} = \frac{d}{dz} \frac{-1}{1 - (-z-1)} = \sum_{n}^{\infty} -n(-z-1)^{n-1}$

But how do I do the above expression?

 

[ThePerfectHacker]

You should specify at what point do you want the power series to be centered at.

 

[Dustinsfl]

z = 0

 

[ThePerfectHacker]

Okay, I suggest to start with partial fractions. (Wink)

 

[Sherlock1]

Note that $4+3z = 2(z+1)+(z+2)$ and $1 = (z+2)-(z+1)$.

 

[Dustinsfl]

Okay, I suggest to start with partial fractions. (Wink)

Set like Real partial fractions?

$\displaystyle\frac{A}{z+1}+\frac{B}{z+2}+\frac{C+Dz}{(z+2)^2}$

 

[ThePerfectHacker]

Set like Real partial fractions?

$\displaystyle\frac{A}{z+1}+\frac{B}{z+2}+\frac{C+Dz}{(z+2)^2}$

No, be careful,
$$ \frac{A}{z+1} + \frac{B}{z+2} + \frac{C}{(z+2)^2}$$

Or you can be more creative like Sherlock.

 

[Fernando Revilla]

$\displaystyle f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}$ How do I find the power series?

Perhaps you have to find all Laurent series expansions centered at $0$ . In such a case we have three regions: $C_1\equiv\;0<|z|<1$ , $C_2\equiv\;1<|z|<2$ and $C_3\equiv\;2<|z|<+\infty$ . For example for the first addend we have $\dfrac{1}{1+z}=\displaystyle\sum_{n=0}^{+\infty}(-1)^nz^n$ if $|z|<1$ and $\dfrac{1}{1+z}=\dfrac{1}{z}\dfrac{1}{1+1/z}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{z^{n+1}}$ if $|z|>1$ etc.

 

[Dustinsfl]

From partial fractions,

$\displaystyle\frac{1}{1-(-z)}-\frac{1}{1-(-z-1)}+2\frac{d}{dz}\frac{-1}{1-(-z-1)}$

By theorem,

The derivative of the sum converges to same L of the sum. So can I disregard taking the derivative of the sum and have this:

$\displaystyle\sum_n^{\infty}\left[(-z)^n-3(-z-1)^n\right]$

 

[Dustinsfl]

Is this ok to do or do I need to take the derivative and then combine?

$\displaystyle\sum_n^{\infty}\left[(-z)^n-(-z-1)^n-2n(-z-1)^{n-1}\right]$

 

[Sherlock1]

Is this ok to do or do I need to take the derivative and then combine?

$\displaystyle\sum_n^{\infty}\left[(-z)^n-(-z-1)^n-2n(-z-1)^{n-1}\right]$

Yes, this is correct. I don't know what you meant about a theorem saying the derivative of the sum is the limit of the sum in the other post, and I don't think it's true.

 

[Dustinsfl]

How can I find the radius of convergence for this series?

 

[Dustinsfl]

How can I find the radius of convergence for this series?

Can I find each radius of convergence separately? Or is there a way to combine this? Or is there a way to directly do this sum?