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Complex Analysis - Taylor series of 1/(1+exp(z))

  1. Dec 5, 2013 #1

    LoA

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    1. The problem statement, all variables and given/known data

    Compute the first four terms of the Taylor series of [itex] \frac{1}{1+e^{z}} [/itex] at [itex] z_{0} = 0[/itex] and give it's radius of convergence.

    2. Relevant equations

    [itex] e^{z} = \sum\frac{z^{n}}{n!} = 1 + z +\frac{z^{2}}{2!} + \frac{z^{3}}{3!} + o(z^{3}) [/itex]

    [itex] \frac{1}{1+w} = \sum(-1)^{n}w^{n} = 1 - w + w^{2} - w^{3} + o(w^{3}) [/itex]

    3. The attempt at a solution

    First let's look at the radius of convergence. Since [itex] e^{z} [/itex] converges for all z, the only thing that matters is [itex] \left|w\right| < 1 [/itex] where [itex] w = e^{z} [/itex].

    So I have to solve [itex] \left|e^{z}\right| < 1 [/itex]. Easy enough. (or just observe that exp(z) = -1 when z = iπ)

    [itex] -1 < e^{z} < 1 \Rightarrow Log(-1) < z < Log(1) \Rightarrow -i\pi < z < 0 \Rightarrow \left|z\right| < \left|i\pi\right| = \pi [/itex]

    So the series will converge for [itex] \left|z\right| < \pi [/itex]

    So much for that. Now down to brass tacks, computing the terms. Working out the derivatives at [itex] z_{0} = 0 [/itex] (and double checking with wolfram) gives me:

    [itex] \frac{1}{2} - \frac{z}{4} + \frac{z^{3}}{48} - \frac{z^{5}}{480} +... [/itex]

    But when I try to arrive at the same form by subbing the series for exp(z) into the geometric series, I get something A) different, and B) messy. I'll write it out upon request, but not tonight, I've been at this way too long. My suspicion is that in order to get the first four terms that way would require way more algebra than it's worth, but I'd like to at least figure out if I'm on the right track...

    Questions: Do I need to expand the terms to a higher order? When substituting the series expansion for one function into the series expansion of another, is it necessary to expand each series out to the exact same order? We worked a few problems like this in class the other day, computing things like [itex] e^{sin(z^{2})} [/itex] and I thought I understood at the time, but now I'm not quite so sure...
     
  2. jcsd
  3. Dec 6, 2013 #2

    ShayanJ

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    I think its better to expand [itex] \frac{1}{1+e^z} [/itex] directly!
     
  4. Dec 6, 2013 #3

    vela

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    That doesn't make sense. Complex numbers can't be ordered like that.

    Think about this. You're saying, for example, that if ##z=3 < \pi##, then ##e^z = e^3 < 1##.

    Try using ##e^z = e^{x+iy} = e^x e^{iy}##.

    I suspect you're writing ##1+e^z+(e^z)^2+\cdots## and substituting in the series for ##e^z##. That won't work because every term will contribute to all orders. You want to expand ##e^z## as a series in ##\frac{1}{1+e^z}##, and then expand the result as a geometric series.

     
  5. Dec 6, 2013 #4

    HallsofIvy

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    As for the radius of convergence, a power series for a function of a complex variable will converge as long as there "is no problem". Here, [itex]e^{i\pi}= -1[/itex] so that the denominator is 0. This function is not analytic at [itex]z= i\pi[/itex] but is for all z closer to 0 so its radius of convergence is [itex]\pi[/itex].
     
  6. Dec 6, 2013 #5
    You can't just muscle-through long division huh?
     
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