- #1
MissP.25_5
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Hello.
I am stuck on this question. I'd appreciate if anyone could help me on how to do this.
The question:
Expand the following into maclaurin series and find its radius of convergence.
$$\frac{2-z}{(1-z)^2}$$
I know that we can use geometric series as geometric series is generally
$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$
So,
\begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}
$$\frac{d}{dz}\frac{1}{(1-z)}=\frac{1}{(1-z)^2}$$
Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
$$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$
How do I finish this? The answer I was given is :
$$\sum_{n=0}^{\infty}(n+2)z^n$$
I am stuck on this question. I'd appreciate if anyone could help me on how to do this.
The question:
Expand the following into maclaurin series and find its radius of convergence.
$$\frac{2-z}{(1-z)^2}$$
I know that we can use geometric series as geometric series is generally
$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$
So,
\begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}
$$\frac{d}{dz}\frac{1}{(1-z)}=\frac{1}{(1-z)^2}$$
Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
$$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$
How do I finish this? The answer I was given is :
$$\sum_{n=0}^{\infty}(n+2)z^n$$