- #1
Tsunoyukami
- 215
- 11
"[F]ind the power-series expansion about the given point for each of the functions; find the largest disc in which the series is valid.
10. ##e^{z}## about ##z_{o} = \pi i##" (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 133)$$f(z) = e^{z} = e^{z-a} \cdot e^{a} = e^{a} \cdot \sum \frac{(z-a)^{n}}{n!}$$
We set the center of the power series to be ##a = \pi i##; then:
$$f(z) = e^{a} \cdot \sum \frac{(z-a)^{n}}{n!} = e^{\pi i} \cdot \sum \frac{(z-\pi i)^{n}}{n!} = (-1) \sum \frac{(z-\pi i)^{n}}{n!}$$
I'm not sure if this is the "final" answer or if I'm supposed to write out the individual terms of the series. Is this correct?My second question pertains to the second half of the question, "find the largest disc in which the series is valid." The function ##f(z) = e^{z}## is entire; as a result, I would expect the disc to have infinite radius of convergence. Similarly, I would expect the series to be valid on a disk of infinite radius centered at ##z_{o} = \pi i##. Is this the correct interpretation?
Any guidance would be appreciated. Thanks!
10. ##e^{z}## about ##z_{o} = \pi i##" (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 133)$$f(z) = e^{z} = e^{z-a} \cdot e^{a} = e^{a} \cdot \sum \frac{(z-a)^{n}}{n!}$$
We set the center of the power series to be ##a = \pi i##; then:
$$f(z) = e^{a} \cdot \sum \frac{(z-a)^{n}}{n!} = e^{\pi i} \cdot \sum \frac{(z-\pi i)^{n}}{n!} = (-1) \sum \frac{(z-\pi i)^{n}}{n!}$$
I'm not sure if this is the "final" answer or if I'm supposed to write out the individual terms of the series. Is this correct?My second question pertains to the second half of the question, "find the largest disc in which the series is valid." The function ##f(z) = e^{z}## is entire; as a result, I would expect the disc to have infinite radius of convergence. Similarly, I would expect the series to be valid on a disk of infinite radius centered at ##z_{o} = \pi i##. Is this the correct interpretation?
Any guidance would be appreciated. Thanks!