- #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*)*?

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- Thread starter wdlang
- Start date

In summary, the conversation discusses the uniqueness of the continuation of a real function into the complex plane. The function must be real analytic, smoothness is not enough. The radius of convergence plays a crucial role, with an infinite radius leading to a unique continuation by the principle of permanence. The reflection principle states that f(z)=f(z*)* for any extension of the function. The function exp(-1/x^2) is an example of a function that is holomorphic on C\{0} and has a singularity at z=0.

- #1

- 307

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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*)*?

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- #2

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wdlang said:

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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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#387148wdlang said:suppose 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 continuous automorphism of C.

- #4

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dalle said: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 continuous 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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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)wdlang said: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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