Finding the Radius of Convergence for a Complex Function.

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To find the radius of convergence for the Taylor expansion of the function f(z)=e^z/((z-1)(z+1)(z-3)(z-2)), one must consider the poles introduced by the denominator, which are located at -1, 1, 2, and 3. The Taylor series converges for all z, but the radius of convergence is determined by the distance from the expansion point to the nearest singularity. If expanding around z=0, the radius is the distance to the closest pole, which is 1. For an expansion around z=i, the same principle applies, with the radius being the distance from i to the nearest pole. Understanding the relationship between the expansion point and singularities is crucial for determining the radius of convergence.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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