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Math Help - Power Series Expansions of Complex Valued Functions

  1. #1
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    Power Series Expansions of Complex Valued Functions

    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?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by robeuler View Post
    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}
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  3. #3
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    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.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by robeuler View Post
    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}
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    Quote Originally Posted by Mathstud28 View Post
    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
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by robeuler View Post
    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.
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    Quote Originally Posted by Mathstud28 View Post
    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)
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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    Quote Originally Posted by Mathstud28 View Post
    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:
    1. a^ba^c = a^{b+c}
    2. (a^b)^c = a^{bc}


    If a,b,c are (non-zero) complex numbers then we may ask if:
    1. a^ba^c = a^{b+c}
    2. (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.
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    Quote Originally Posted by Mathstud28 View Post
    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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