Power Series Expansions of Complex Valued Functions

**robeuler** · Nov 17th 2008, 03:07 PM

Here is my problem:
Let sum(a_k*z^k, k=0->infinity)
be the power series expansion of
2/((3-2z)^2).

I need to figure out a closed form expression for a_k.

My first problem is figuring out how to manipulate the function into something I know how to express (geometric series?).

Secondly (although this will probably be easier once I figure out issue number 1) is that I don't see how it will be obvious to express a_k unless you just explicitly get the power series.

any hints?

**Mathstud28** · Nov 17th 2008, 03:31 PM

Originally Posted by robeuler

Here is my problem:
Let sum(a_k*z^k, k=0->infinity)
be the power series expansion of
2/((3-2z)^2).

I need to figure out a closed form expression for a_k.

My first problem is figuring out how to manipulate the function into something I know how to express (geometric series?).

Secondly (although this will probably be easier once I figure out issue number 1) is that I don't see how it will be obvious to express a_k unless you just explicitly get the power series.

any hints?

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

So we know that $\forall{z}\backepsilon|z|<1\quad\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$

So we can ascertain that

$\forall{z}\backepsilon\left|\frac{2}{3}z\right|<1\ quad\frac{1}{3}\cdot\frac{1}{1-\frac{2}{3}z}=\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{2}{3}z\right)^n$

So on that same interval (due to unifrom convergence, ignore if you dont care or it isnt part of what your learnging) we may differentiate to obtain

$\begin{aligned}\left(\frac{1}{3}\frac{1}{1-\frac{2}{3}z}\right)'&=\frac{2}{(3-2z)^2}\\<br /> &=\frac{d}{dz}\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{2}{3}z\right)^n\\<br /> &=\frac{1}{3}\sum_{n=1}^{\infty}n\left(\frac{2}{3} \right)^nz^{n-1}<br /> \end{aligned}$

**robeuler** · Nov 17th 2008, 03:44 PM

Very nice approach. Uniform continuity is great!

Here is one more I am stuck on, if you have the time:
f(z)=sinz/(pi-z)
g(z)=(z(pi-z))/sinz

I need to find the power series representation of f(z) at pi, and its radius of convergence. Then I need to find the radius of convergence of the power series of g(z) at pi/2.

I can write out sinz as a power series no problem. And again I feel like I should write out the denominator as a geometric series, and then multiply the two series together. I can't figure out how to express it as a geometric series.

**Mathstud28** · Nov 17th 2008, 03:52 PM

Originally Posted by robeuler

Very nice approach. Uniform continuity is great!

Here is one more I am stuck on, if you have the time:
f(z)=sinz/(pi-z)
g(z)=(z(pi-z))/sinz

I need to find the power series representation of f(z) at pi, and its radius of convergence. Then I need to find the radius of convergence of the power series of g(z) at pi/2.

I can write out sinz as a power series no problem. And again I feel like I should write out the denominator as a geometric series, and then multiply the two series together. I can't figure out how to express it as a geometric series.

Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$

**robeuler** · Nov 17th 2008, 03:57 PM

Originally Posted by Mathstud28

Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$

I know the expansion for sin(z), or are you suggesting I write it out as you do and then using that to manage the 1/pi-z

**Mathstud28** · Nov 17th 2008, 03:58 PM

Originally Posted by robeuler

I know the expansion for sin(z), or are you suggesting I write it out as you do and then using that to manage the 1/pi-z

Well, let me put a disclaimer in here, I am not the best at complex analysis. I am more knowledgable of real valued series, but what I am saying is consider that $e^{iz}$ is geometric.

**ThePerfectHacker** · Nov 17th 2008, 04:44 PM

Originally Posted by Mathstud28

Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$

That is not true.

Just consider this fallacy,
$1 = 1^{\pi} = (e^{2\pi i})^{\pi} = e^{2\pi^2 i} = \cos (2\pi ^2) + i\sin (2\pi^2)$

**Mathstud28** · Nov 17th 2008, 05:12 PM

Originally Posted by ThePerfectHacker

That is not true.

Just consider this fallacy,
$1 = 1^{\pi} = (e^{2\pi i})^{\pi} = e^{2\pi^2 i} = \cos (2\pi ^2) + i\sin (2\pi^2)$

Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.

**ThePerfectHacker** · Nov 17th 2008, 05:20 PM

Originally Posted by Mathstud28

Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.

If $a,b,c$ are positive real numbers then we have the exponents rules:

$a^ba^c = a^{b+c}$
$(a^b)^c = a^{bc}$

If $a,b,c$ are (non-zero) complex numbers then we may ask if:

$a^ba^c = a^{b+c}$
$(a^b)^c = a^{bc}$

It turns out that #1 is always true. But #2 needs to be used with a lot of care. One way when #2 always works is when $c \in \mathbb{Z}$ . You certainly seen that $(e^{i\theta})^n = e^{ni\theta}$ . Well, that is another special case when $c$ is an integer.

**robeuler** · Nov 17th 2008, 05:29 PM

Originally Posted by Mathstud28

Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.

No worries, your answer to the first question was really good.

Perhaps TPH could shed some light on what I have left; I'll work on it until I go to bed in an hour or so. If I get something I'll come back.

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