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Homework Help: Power Series Expansion

  1. Jul 10, 2013 #1
    "[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!
  2. jcsd
  3. Jul 10, 2013 #2
    It seems correct to me.
  4. Jul 10, 2013 #3
    That's right. But can you also prove it by refering only to the series, and not to the function ##e^z##? That is, can you explicitely show the series converges for all ##z##?
  5. Jul 10, 2013 #4
    Thanks for the prompt reply. The fact that ##e^{\pi i} = -1## does not have any effect on the radius of convergence of the power series, correct (ie. it does not effect where the power series is valid)?

    Would it be sufficient to attempt to show that the radius of convergence of the power series approached infinity? I would try to show this by using a test for convergence of a (power) series and show that the series converges to 0.

    This would follow from the assertion that

    ##\frac {1}{R} = lim |\frac{a_{n+1}}{a_{n}}|##, as n approaches infinity (and likewise for the root test)
  6. Jul 10, 2013 #5
    It certainly doesn't. The validity of the power series can be determined by the ratio test.
  7. Jul 10, 2013 #6
    Excellent! Thank you very much, both of you!

    By applying the ratio test I would be able to show that the radius of convergence is infinity, and thus, the power series is valid for all z in the complex plane. Thanks!
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