Can you expand and find the radius of convergence for this Maclaurin series?

[aruwin]

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

but I don't know how to use it for this problem.

 

[Prove It]

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

but I don't know how to use it for this problem.

$\displaystyle \begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}$

The second term is easy to write as a geometric series. For the first term, use the fact that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}z } \left( \frac{1}{ 1 - z } \right) = \frac{1}{ \left( 1 - z \right) ^2 } \end{align*}$, so when you write $\displaystyle \begin{align*} \frac{1}{1 - z } \end{align*}$ as a geometric series, when you differentiate it you'll have a series for $\displaystyle \begin{align*} \frac{1}{ \left( 1 - z \right) ^2 } \end{align*}$...

 

[chisigma]

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

but I don't know how to use it for this problem.

An idea can be to use the identity $\displaystyle \frac{2 - z}{(1 - z)^{2}} = \frac{1}{1-z} + \frac{1}{(1-z)^{2}}$ and the fact that is $\displaystyle \frac{d}{d z} \frac{1}{1-z} = \frac{1}{(1-z)^{2}}$...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

$\displaystyle \begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}$

The second term is easy to write as a geometric series. For the first term, use the fact that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x } \left( \frac{1}{ 1 - z } \right) = \frac{1}{ \left( 1 - z \right) ^2 } \end{align*}$, so when you write $\displaystyle \begin{align*} \frac{1}{1 - z } \end{align*}$ as a geometric series, when you differentiate it you'll have a series for $\displaystyle \begin{align*} \frac{1}{ \left( 1 - z \right) ^2 } \end{align*}$...

Then what about the radius of convergence? Since the series are all going to be $\frac{1}{(1-z)^n}$, I assume the answer is R=1?

 

[aruwin]

Sorry, I misunderstood before. Let me try and write out the maclaurin series first and then the radius.

 

[Prove It]

Then what about the radius of convergence? Since the series are all going to be $\frac{1}{(1-z)^n}$, I assume the answer is R=1?

Yes, any series that is obtained through differentiation or integration has the same radius of convergence as the series it came from.

The only place it might be different is at the endpoints. For example, $\displaystyle \begin{align*} \frac{1}{1 - z} = \sum_{n = 0}^{\infty}{ z^n } \end{align*}$ is convergent for $\displaystyle \begin{align*} |z| < 1 \end{align*}$ and divergent for $\displaystyle \begin{align*} |z| \geq 1 \end{align*}$, while if we integrate it we have

$\displaystyle \begin{align*} \frac{1}{1 - z} &= \sum_{n = 0}^{\infty}{z^n} \\ \int{ \frac{1}{1 - z}\,\mathrm{d}z } &= \int{ \sum_{n = 0}^{\infty}{z^n}\,\mathrm{d}z } \\ -\ln{ \left( 1 - z \right) } &= \sum_{n = 0}^{\infty} {\frac{z^{n+1}}{n+1}} \\ \ln{ \left( 1 - z \right) } &= -\sum_{n = 0}^{\infty}{ \frac{z^{n+1}}{n+1} } \end{align*}$

which is convergent where $\displaystyle \begin{align*} -1 \leq z < 1 \end{align*}$. Don't believe me? When $\displaystyle \begin{align*} z = -1 \end{align*}$ we have $\displaystyle \begin{align*} \ln{ \left( 2 \right) } = \ln{ \left[ 1 - \left( -1 \right) \right] } &= -\sum_{n = 0}^{\infty}{\frac{\left( -1 \right)^{n + 1}}{n + 1}} \end{align*}$, which is an alternating series, and since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{1}{x + 1} \right) = -\frac{1}{\left( x + 1 \right) ^2} \end{align*}$, that means the non-alternating part $\displaystyle \begin{align*} \frac{1}{n + 1} \end{align*}$ is decreasing, and thus is convergent by the Alternating Series Test.So to answer your previous question - in general any series obtained from differentiating or integrating has the same RADIUS of convergence, but may have different convergence behaviour at the ENDPOINTS of that radius.

 

[aruwin]

Yes, any series that is obtained through differentiation or integration has the same radius of convergence as the series it came from.

The only place it might be different is at the endpoints. For example, $\displaystyle \begin{align*} \frac{1}{1 - z} = \sum_{n = 0}^{\infty}{ z^n } \end{align*}$ is convergent for $\displaystyle \begin{align*} |z| < 1 \end{align*}$ and divergent for $\displaystyle \begin{align*} |z| \geq 1 \end{align*}$, while if we integrate it we have

$\displaystyle \begin{align*} \frac{1}{1 - z} &= \sum_{n = 0}^{\infty}{z^n} \\ \int{ \frac{1}{1 - z}\,\mathrm{d}z } &= \int{ \sum_{n = 0}^{\infty}{z^n}\,\mathrm{d}z } \\ -\ln{ \left( 1 - z \right) } &= \sum_{n = 0}^{\infty} {\frac{z^{n+1}}{n+1}} \\ \ln{ \left( 1 - z \right) } &= -\sum_{n = 0}^{\infty}{ \frac{z^{n+1}}{n+1} } \end{align*}$

which is convergent where $\displaystyle \begin{align*} -1 \leq z < 1 \end{align*}$. Don't believe me? When $\displaystyle \begin{align*} z = -1 \end{align*}$ we have $\displaystyle \begin{align*} \ln{ \left( 2 \right) } = \ln{ \left[ 1 - \left( -1 \right) \right] } &= -\sum_{n = 0}^{\infty}{\frac{\left( -1 \right)^{n + 1}}{n + 1}} \end{align*}$, which is an alternating series, and since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{1}{x + 1} \right) = -\frac{1}{\left( x + 1 \right) ^2} \end{align*}$, that means the non-alternating part $\displaystyle \begin{align*} \frac{1}{n + 1} \end{align*}$ is decreasing, and thus is convergent by the Alternating Series Test.So to answer your previous question - in general any series obtained from differentiating or integrating has the same RADIUS of convergence, but may have different convergence behaviour at the ENDPOINTS of that radius.

Thanks for the explanation.

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 :

$$\sum_{n=0}^{\infty}(n+2)z^n$$

or is it enough to just leave the answer as $$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$ ?

 

[Prove It]

Thanks for the explanation.

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 :

$$\sum_{n=0}^{\infty}(n+2)z^n$$

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{n\,z^{n-1}} + \sum_{n = 0}^{\infty}{z^n} &= 0 + 1 + 2z + 3z^2 + 4z^3 + \dots + 1 + z + z^2 + z^3 + \dots \\ &= 2 + 3z + 4z^2 + 5z^3 + \dots \\ &= \sum_{n = 0}^{\infty}{ \left[ \left( n + 2 \right) \, z^n \right] } \end{align*}$

 

[aruwin]

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{n\,z^{n-1}} + \sum_{n = 0}^{\infty}{z^n} &= 0 + 1 + 2z + 3z^2 + 4z^3 + \dots + 1 + z + z^2 + z^3 + \dots \\ &= 2 + 3z + 4z^2 + 5z^3 + \dots \\ &= \sum_{n = 0}^{\infty}{ \left[ \left( n + 2 \right) \, z^n \right] } \end{align*}$

Thanks!