Continuation of a real function into the total complex plane

  • Thread starter wdlang
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  • #1
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Main Question or Discussion Point

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*)*?
 

Answers and Replies

  • #2
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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*)*?
i just note that for function f=1/(1+x^2)

there must be a pole at z=i and -i
 
  • #3
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suppose i have a real function f=f(x)

this function is smooth everywhere on the real line
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


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

if so, does it hold that f(z)=f(z*)*?
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.
 
  • #4
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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.
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?
 
  • #5
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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".
 
  • #6
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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?
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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