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Continuation of a real function into the total complex plane

  1. Mar 19, 2010 #1
    suppose i have a real function f=f(x)

    this function is smooth everywhere on the real line

    for example, f=e^x.

    The problem is, is the continuation of the function into the complex plane unique?

    if so, does it hold that f(z)=f(z*)*?
     
  2. jcsd
  3. Mar 19, 2010 #2
    i just note that for function f=1/(1+x^2)

    there must be a pole at z=i and -i
     
  4. Mar 19, 2010 #3
    dear wdlang, your function needs to be real analytic, being smooth is not enough. There is an excelent discussion of the difference by Dave Renfro at http://mathforum.org/kb/thread.jspa?forumID=13&threadID=81514&messageID=387148#387148


    if the radius of convergence is infinite, than your function is entire, and the continuation is unique by the principle of permanence. As you note in your second post, things get complicated when the radius of convergence is finite.

    let [tex]f_n = z^n[/tex]. Is [tex] f_n(z)=f_n(z*)*[/tex]?
    let [tex] p [/tex] be an real polynomial. Is [tex] p(z)=p(z*)*[/tex]
    let [tex] g=\sum_{n=0}^{\infty }a_n z^n , a_n \in R [/tex] be a real power series with radius of convergence [tex] r > 0 [/tex]
    let [tex] z \in C , |z| < r[/tex]. Is [tex] g(z)=g(z*)* [/tex]
    Hint: Use that * is an continous automorphism of C.
     
  5. Mar 19, 2010 #4
    thanks a lot for your reply

    yes, the function by cauchy is just amazing

    exp(-1 / x^2) [with f(0) = 0]

    the taylor series at x=0 does not converge to the function itself

    but how about another point?
     
  6. Mar 19, 2010 #5
    For uniqueness: yes... any two extensions agree on the real line, which is a set with a limit point, so they agree on their entire (connected) common domain.

    f(z) = f(z*)* is the "reflection principle".
     
  7. Mar 19, 2010 #6
    for any other point x the radius of convergence is |x|. exp(-1/x^2) is holomorphic on C\{0} and the radius of convergence is the distance to the next singularity (which the function has only one at the point z=0)
     
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