1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radius of convergence

  1. Nov 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the power serie ##\sum_{n\ge 0} a_n z^n##, with radius of convergence ##R##, if there exists a complex number ##z_0## such that the the serie is semi-convergent at ##z_0##, show that ##R = |z_0|##.

    2. Relevant equations


    3. The attempt at a solution

    Firstly, since ##\sum_{n\ge 0} a_n z_0^n## is not absolutely convergent, then ## R \le |z_0|##.
    Secondly, ##\sum_{n\ge 0} a_n z_0^n## is convergent so ##(a_n z_0^n)_n## is bounded as it tends to zero, so ## |z_0| \le R##.
    So ##R = |z_0|##.

    Is it correct ?
     
    Last edited: Nov 20, 2015
  2. jcsd
  3. Nov 20, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The proof may be missing something.
    What is the justification for this statement? Failure to absolutely converge does not entail failure to converge, and radius of convergence delimits the disc on which ordinary convergence happens, not absolute convergence.
    I don't recognise the symbol string ##(a_n z_0^n)_n##. What do you intend it to mean?
     
  4. Nov 20, 2015 #3
    The radius of convergence delimits the disc where absolute convergence happens, and absolute convergence implies ordinary convergence.
    Therefore, if the serie fails to converge absolutely in ##z_0##, which is the case because it is semi-convergent in ##z_0##, then ##R\le |z_0|##

    I meant the sequence with general term ##u_n = a_nz_0^n##
     
  5. Nov 21, 2015 #4

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Agreed.
    I don't understand why the fact that ##(a_n z_0^n)_n## is bounded implies ## |z_0| \le R##.
    But the above quote is correct if you remove part of it.
     
  6. Nov 21, 2015 #5
    My definition of radius of convergence is the supremum, if it exists, of the set ## S= \{ r \ge 0 \text{ s.t the sequence } (a_n r^n)_n \text{ is bounded }\} ##. Also, for any ##r\in S##, ##[0,r[## is in this set.
    With that definition, the serie converges absolutely inside the disc of convergence, and outside the disc of convergence, the sequence ##(a_n z^n)_n## is not bounded .
    So if ##(a_n z_0^n)_n## is bounded, then by contraposition, ##|z_0| \le R##.
     
  7. Nov 21, 2015 #6

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you are right. I never saw that definition for radius of convergence, but it is equivalent to the one I knew.
     
  8. Nov 21, 2015 #7
    Thank you for looking at it
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Radius of convergence
  1. Radius of Convergence (Replies: 4)

  2. Radius of Convergence (Replies: 4)

  3. Radius of convergence (Replies: 2)

  4. Radius of convergence (Replies: 7)

  5. Radius of convergence (Replies: 6)

Loading...