Is This Maclaurin Series Expansion Correct?

[aruwin]

Hello.
Could someone check if my solution is correct, please?

Question:
Expand the function into maclaurine series and find its radius of convergence.

$$f(z)=\frac{2z}{(1-z)^2}$$

My solution:
$$f(z)=\frac{2z}{(1-z)^2}=\frac{1}{(1-z)}\frac{2}{(1-z)}\frac{2}{(1-z)^2}$$

Since $$\frac{d}{dz}\frac{1}{1-z}=\frac{1}{(1-z)^2},$$ then

$$f(z)=\sum_{n=0}^{\infty}z^n\times\sum_{n=0}^{\infty}2z^n\times\sum_{n=1}^{\infty}2nz^{n-1}$$

$$f(z)=\sum_{n=0}^{\infty}z^n\times\Bigg(\sum_{n=0}^{\infty}2z^n\times\Bigg(\sum_{n=0}^{\infty}2(n+1)z^n\Bigg)\Bigg)$$

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

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

As for Radius,
$$R=\lim_{{n}\to{\infty}}\Bigg|\frac{n+2}{n+3}\Bigg|=1$$

 

[chisigma]

Hello.
Could someone check if my solution is correct, please?

Question:
Expand the function into maclaurine series and find its radius of convergence.

$$f(z)=\frac{2z}{(1-z)^2}$$

My solution:
$$f(z)=\frac{2z}{(1-z)^2}=\frac{1}{(1-z)}\frac{2}{(1-z)}\frac{2}{(1-z)^2}$$

Since $$\frac{d}{dz}\frac{1}{1-z}=\frac{1}{(1-z)^2},$$ then

$$f(z)=\sum_{n=0}^{\infty}z^n\times\sum_{n=0}^{\infty}2z^n\times\sum_{n=1}^{\infty}2nz^{n-1}$$

$$f(z)=\sum_{n=0}^{\infty}z^n\times\Bigg(\sum_{n=0}^{\infty}2z^n\times\Bigg(\sum_{n=0}^{\infty}2(n+1)z^n\Bigg)\Bigg)$$

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

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

As for Radius,
$$R=\lim_{{n}\to{\infty}}\Bigg|\frac{n+2}{n+3}\Bigg|=1$$

As You say is...

$\displaystyle \frac{d}{d z} \frac{1}{1-z} = \frac{1}{(1-z)^{2}}\ (1)$

... and because is also...

$\displaystyle \frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}\ (2)$

... we conclude that...

$\displaystyle \frac{2\ z}{(1-z)^{2}} = 2\ \sum_{n=0}^{\infty} n\ z^{n}\ (3)$

The (2) has radius R=1 and the same is for (3)...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

As You say is...

$\displaystyle \frac{d}{d z} \frac{1}{1-z} = \frac{1}{(1-z)^{2}}\ (1)$

... and because is also...

$\displaystyle \frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}\ (2)$

... we conclude that...

$\displaystyle \frac{2\ z}{(1-z)^{2}} = 2\ \sum_{n=0}^{\infty} n\ z^{n}\ (3)$

The (2) has radius R=1 and the same is for (3)...

Kind regards

$\chi$ $\sigma$

I am sorry, I made a mistake in the question. The function is actually
to the power of 3, not 2. Would my answer be right, then?

 

[aruwin]

Let me rewrite it.

$$f(z)=\frac{2z}{(1-z)^3}$$

My solution:

$$f(z)=\frac{2z}{(1-z)^2}=\frac{1}{(1-z)}\bigg(\frac{2}{(1-z)}+\frac{2}{(1-z)^2}\bigg)$$

$$f(z)=\sum_{n=0}^{\infty}z^n\times\bigg(\sum_{n=0}^{\infty}2z^n+\sum_{n=1}^{\infty}2nz^{n-1}\bigg)$$

$$f(z)=\sum_{n=0}^{\infty}z^n\times\Bigg(\sum_{n=0}^{\infty}2z^n+\Bigg(\sum_{n=0}^{\infty}2(n+1)z^n\Bigg)\Bigg)$$

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

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

 

[chisigma]

I am sorry, I made a mistake in the question. The function is actually
to the power of 3, not 2. Would my answer be right, then?

In this case is...

$\displaystyle f(z) = z\ \frac{d}{d z} \frac{1}{(1 - z)^{2}} = z\ \frac{d^{2}}{d z^{2}} \frac{1}{1 - z} = \sum_{n=0}^{\infty} n\ (n - 1)\ z^{n-1}\ (1)$

Kind regards

$\chi$ $\sigma$