- #1

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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*)*?

- Thread starter wdlang
- Start date

- #1

- 307

- 0

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

- 307

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

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*)*?

there must be a pole at z=i and -i

- #3

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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#387148suppose i have a real function f=f(x)

this function is smooth everywhere on the real line

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

let [tex]f_n = z^n[/tex]. Is [tex] f_n(z)=f_n(z*)*[/tex]?if so, does it hold that f(z)=f(z*)*?

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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thanks a lot for your replydear 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.

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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f(z) = f(z*)* is the "reflection principle".

- #6

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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)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?

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