When is the conjugate of a function equal to the function of the conjugate of the argument ($f(x^*) =...
$begingroup$
Is there a special property that unites all functions for which $f(x^*) = f^*(x)$ holds?
Naively I can show that it is true for any function that has a series expansion with only real coefficients, and does not hold for a function that has a series expansion with complex coefficients. But this statement is not very rigorous, at least because a series expansion might not be valid for the entire domain. Is there a more formal way to write this statement?
Just curious, no coursework involved
functions complex-numbers
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
$endgroup$
add a comment |
$begingroup$
Is there a special property that unites all functions for which $f(x^*) = f^*(x)$ holds?
Naively I can show that it is true for any function that has a series expansion with only real coefficients, and does not hold for a function that has a series expansion with complex coefficients. But this statement is not very rigorous, at least because a series expansion might not be valid for the entire domain. Is there a more formal way to write this statement?
Just curious, no coursework involved
functions complex-numbers
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
$endgroup$
$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11
add a comment |
$begingroup$
Is there a special property that unites all functions for which $f(x^*) = f^*(x)$ holds?
Naively I can show that it is true for any function that has a series expansion with only real coefficients, and does not hold for a function that has a series expansion with complex coefficients. But this statement is not very rigorous, at least because a series expansion might not be valid for the entire domain. Is there a more formal way to write this statement?
Just curious, no coursework involved
functions complex-numbers
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
$endgroup$
Is there a special property that unites all functions for which $f(x^*) = f^*(x)$ holds?
Naively I can show that it is true for any function that has a series expansion with only real coefficients, and does not hold for a function that has a series expansion with complex coefficients. But this statement is not very rigorous, at least because a series expansion might not be valid for the entire domain. Is there a more formal way to write this statement?
Just curious, no coursework involved
functions complex-numbers
functions complex-numbers
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
asked Dec 3 '18 at 22:47
Aleksejs FominsAleksejs Fomins
480211
480211
$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11
add a comment |
$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11
$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$zin Dimpliesoverline zin D.tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $mathbb R$. Then the conditions
- $(forall zin D):fleft(overline zright)=overline{f(z)}$
- ($forall zin Dcapmathbb{R}):f(z)inmathbb R$
are equivalent. In fact, if the first condition holds and if $zin Dcapmathbb{R}$, then $overline z=z$ and therefore $f(z)=fleft(overline zright)=overline{f(z)}$, which implies that $f(z)inmathbb R$. And if the second condition holds, then let $g(z)=overline{fleft(overline zright)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(forall zin Dcapmathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$zin Dimplies fleft(overline zright)=gleft(overline zright)=overline{f(z)}.$$
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$zin Dimpliesoverline zin D.tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $mathbb R$. Then the conditions
- $(forall zin D):fleft(overline zright)=overline{f(z)}$
- ($forall zin Dcapmathbb{R}):f(z)inmathbb R$
are equivalent. In fact, if the first condition holds and if $zin Dcapmathbb{R}$, then $overline z=z$ and therefore $f(z)=fleft(overline zright)=overline{f(z)}$, which implies that $f(z)inmathbb R$. And if the second condition holds, then let $g(z)=overline{fleft(overline zright)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(forall zin Dcapmathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$zin Dimplies fleft(overline zright)=gleft(overline zright)=overline{f(z)}.$$
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
add a comment |
$begingroup$
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$zin Dimpliesoverline zin D.tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $mathbb R$. Then the conditions
- $(forall zin D):fleft(overline zright)=overline{f(z)}$
- ($forall zin Dcapmathbb{R}):f(z)inmathbb R$
are equivalent. In fact, if the first condition holds and if $zin Dcapmathbb{R}$, then $overline z=z$ and therefore $f(z)=fleft(overline zright)=overline{f(z)}$, which implies that $f(z)inmathbb R$. And if the second condition holds, then let $g(z)=overline{fleft(overline zright)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(forall zin Dcapmathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$zin Dimplies fleft(overline zright)=gleft(overline zright)=overline{f(z)}.$$
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
add a comment |
$begingroup$
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$zin Dimpliesoverline zin D.tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $mathbb R$. Then the conditions
- $(forall zin D):fleft(overline zright)=overline{f(z)}$
- ($forall zin Dcapmathbb{R}):f(z)inmathbb R$
are equivalent. In fact, if the first condition holds and if $zin Dcapmathbb{R}$, then $overline z=z$ and therefore $f(z)=fleft(overline zright)=overline{f(z)}$, which implies that $f(z)inmathbb R$. And if the second condition holds, then let $g(z)=overline{fleft(overline zright)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(forall zin Dcapmathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$zin Dimplies fleft(overline zright)=gleft(overline zright)=overline{f(z)}.$$
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$zin Dimpliesoverline zin D.tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $mathbb R$. Then the conditions
- $(forall zin D):fleft(overline zright)=overline{f(z)}$
- ($forall zin Dcapmathbb{R}):f(z)inmathbb R$
are equivalent. In fact, if the first condition holds and if $zin Dcapmathbb{R}$, then $overline z=z$ and therefore $f(z)=fleft(overline zright)=overline{f(z)}$, which implies that $f(z)inmathbb R$. And if the second condition holds, then let $g(z)=overline{fleft(overline zright)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(forall zin Dcapmathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$zin Dimplies fleft(overline zright)=gleft(overline zright)=overline{f(z)}.$$
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 3 '18 at 23:02
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
add a comment |
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
So the equality holds for all analytic functions that are defined over a set of complex numbers. What about non-analytic functions? Is the converse true that for every non-analytic function defined over the set of complex numbers, there exists an argument for which the equality does not hold? Do not bother with a proof, intuition is sufficient in this case
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:41
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
$begingroup$
Right now, I don't see how no approach the problem for non-analytic functions. Since you mentioned Taylor series, I thought that you were only interested in analytic functions.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:44
add a comment |
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$begingroup$
What kind of functions are you considering. You talk about series expansions in you attempt but there is no assumption that $f$ is even continuous.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:07
$begingroup$
I talk about series expansion because it helps me to get intuition, since I do not know how to proceed otherwise. If there is a difference in how conjugation works for continuous vs non-continuous functions, please mention it in your answer
$endgroup$
– Aleksejs Fomins
Dec 4 '18 at 8:11