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LoA
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Homework Statement
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.
Homework 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]
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...