Finding the Radius of Convergence for a Complex Function.

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Homework Help Overview

The discussion revolves around finding the radius of convergence for the Taylor expansion of the complex function \( f(z)=\frac{e^z}{(z-1)(z+1)(z-3)(z-2)} \). Participants are exploring the implications of the function's poles on convergence.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the method of differentiation and the ratio test for determining the Taylor expansion, questioning the complexity of this approach. There is also mention of expanding around a specific point, such as \( z = 0 \) or \( z = i \), and the need to consider the distance to the nearest singularity.

Discussion Status

The conversation is ongoing, with some participants providing insights about the nature of the poles and their influence on the radius of convergence. There is an acknowledgment of the complexity involved in the expansion process and the potential for different interpretations regarding the point of expansion.

Contextual Notes

Participants are considering the implications of expanding around different points and the associated singularities, which are critical for determining the radius of convergence. The original poster's assumption about the expansion point is questioned, indicating a need for clarity in the problem setup.

shedrick94
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< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color]

How would you find the radius of convergence for the taylor expansion of:

\begin{equation} f(z)=\frac{e^z}{(z-1)(z+1)(z-3)(z-2)} \end{equation}

I was thinking that you would just differentiate to find the taylor expansion and then use the ratio test but this seems far too tedious to be the right way to do it! Any help?
 
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The Taylor expansion of ez converges for all z. The denominator introduces poles at -1, 1, 2 and 3, so you need to be sufficiently far away from those values. Now determine "sufficiently far away"...
 
shedrick94 said:
< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

How would you find the radius of convergence for the taylor expansion of:

\begin{equation} f(z)=\frac{e^z}{(z-1)(z+1)(z-3)(z-2)} \end{equation}

I was thinking that you would just differentiate to find the taylor expansion and then use the ratio test but this seems far too tedious to be the right way to do it! Any help?

Should we assume you want to expand around ##z = 0##?
 
shedrick94 said:
< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

How would you find the radius of convergence for the taylor expansion of:

\begin{equation} f(z)=\frac{e^z}{(z-1)(z+1)(z-3)(z-2)} \end{equation}

I was thinking that you would just differentiate to find the taylor expansion and then use the ratio test but this seems far too tedious to be the right way to do it! Any help?

If (as I asked in #3 but did not receive an answer!) you are expanding about ##z = 0##, you can just expand each factor separately, then multiply the series together. The resulting series will be messy and not easy to write explicitly, but at least you can say something about the radius of convergence, since a lower bound on the radius of convergence of a product of series is known in terms of the individual radii of convergence. Google 'product of power series' or something similar.
 
Sorry I understand this now. The expansion was about z=i, but I understand you would just the distance between the place you are expanding around and the closest singularity.
 

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