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Homework Help: Power Series for log z

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Does there exist a power series expansion of log z around z=0? If so, what is it? If not, demonstrate that it is impossible.

    2. Relevant equations

    [PLAIN]http://img839.imageshack.us/img839/4839/eq1.gif [Broken]

    Here is the expansion of log z around z=1.

    3. The attempt at a solution

    I'm actually trying to find the value of the singularity at z=0 (the coefficient of 1/(z-a) around z=a), if it's possible. I have no idea what to do.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 5, 2011 #2

    I like Serena

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    Welcome to PF, Harrisonized! :smile:

    It is impossible to find the coefficient of 1/(z-a) if a=0, with z=0.
    This means it is not possible to make an expansion at 0.
  4. Aug 5, 2011 #3
    Thank you for the welcome.

    Why is it impossible? I tried putting the function into Wolfram|Alpha and shrinking the results to 0. Here's what happened:

    [PLAIN]http://img6.imageshack.us/img6/7266/msp180819gf50a73e94c05h.gif [Broken]

    [PLAIN]http://img814.imageshack.us/img814/6152/msp355719gf4d225i9ahfee.gif [Broken]

    [PLAIN]http://img37.imageshack.us/img37/200/msp218319gf52gfh2fadihb.gif [Broken]

    Here's from the imaginary axis:

    [PLAIN]http://img833.imageshack.us/img833/9853/msp347219gf4f1dgd6a2e17.gif [Broken]

    [PLAIN]http://img607.imageshack.us/img607/7232/msp210519gf4ifefdd61bgd.gif [Broken]

    [PLAIN]http://img232.imageshack.us/img232/3484/msp145319gf5304af95gbb8.gif [Broken]

    [PLAIN]http://img811.imageshack.us/img811/3433/msp103419gf53i0b5cd6h31.gif [Broken]

    This isn't a homework problem by the way. I just felt like the question wasn't worthy of real discussion.
    Last edited by a moderator: May 5, 2017
  5. Aug 5, 2011 #4

    I like Serena

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    The expansion around 0 is:
    ln(z) = ln(0) + z * 1/0 + ...

    Neither ln(0) nor 1/0 are defined.
    As you can see in your results, the coefficients approach infinity.
  6. Aug 5, 2011 #5
    But e^(1/z) isn't defined at 0, yet a power series can still be found for e^(1/z) about z=0. In fact, e^(-∞)=0, so that's the equivalent of e^(1/z) approached to 0 from the negative real axis. Maybe there's some way of inverting e^(1/z) = 1+z^(-1)+z^(-2)/2!+z^(-3)/3!+... ?
    Last edited: Aug 5, 2011
  7. Aug 5, 2011 #6

    I like Serena

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    I am not aware of an expansion of e^(1/z) at 0.
    I believe the one you mention is at z=infinity.
    The expansion of e^(1/z) at 0 has a coefficient for the first derivative that is undefined (approaches infinity).
  8. Aug 5, 2011 #7

    Ray Vickson

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    The function f(z) = ln(z) has a *branch point* at z = 0. For any r > 0 (no matter how small) we get different results for ln(-r), depending on whether we approach x=-r from positive imaginary values or negative imaginary values. This could not happen if f(z) had a convergent power series around zero.

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