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Can a differentiable function be real valued at infinite points and complex valued at some other intervals

Complex vs. Real DifferentiableComplex differentiable function.If $f$ is a real valued function, complex differentiable at $z_0$, then $f'(z_0)=0$Derivative of real-valued function must be real-valuedFatou Coordinate with two complex conjugate fixed points and extending Tetration to real valuescomplex valued function that is real for real arguments but also has complex roots?Derivative of real valued function w.r.t real valued variableReal-valued functions on complex numbersSubderivative of a function of a complex-valued variableExtending a complex-differentiable function

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2

$begingroup$

Can a differentiable function (in real part) be real valued at an interval (infinite points) and complex valued at some other intervals?
For example Exp[it] is real valued at countable points and complex valued at other uncountable points. I am trying to understand if a a differentiable function can have both real valued and complex valued intervals (on real axis).

complex-analysis functions

share|cite|improve this question

edited Jun 8 at 17:38

vyaman

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "Continous function", yes it can exist. "Analytic function", no it cannot. This is why you are struggling to find examples; most elementary functions are analytic.
  $endgroup$
  – Giuseppe Negro
  Jun 8 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $sqrt x$ match your requirements?
  $endgroup$
  – N74
  Jun 8 at 20:41
  

  
  
  
  

add a comment | 

2

$begingroup$

Can a differentiable function (in real part) be real valued at an interval (infinite points) and complex valued at some other intervals?
For example Exp[it] is real valued at countable points and complex valued at other uncountable points. I am trying to understand if a a differentiable function can have both real valued and complex valued intervals (on real axis).

complex-analysis functions

share|cite|improve this question

edited Jun 8 at 17:38

vyaman

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "Continous function", yes it can exist. "Analytic function", no it cannot. This is why you are struggling to find examples; most elementary functions are analytic.
  $endgroup$
  – Giuseppe Negro
  Jun 8 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does $sqrt x$ match your requirements?
  $endgroup$
  – N74
  Jun 8 at 20:41
  

  
  
  
  

add a comment | 

$begingroup$

Can a differentiable function (in real part) be real valued at an interval (infinite points) and complex valued at some other intervals?
For example Exp[it] is real valued at countable points and complex valued at other uncountable points. I am trying to understand if a a differentiable function can have both real valued and complex valued intervals (on real axis).

complex-analysis functions

share|cite|improve this question

edited Jun 8 at 17:38

vyaman

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 8 at 17:38

vyaman

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

share|cite|improve this question

edited Jun 8 at 17:38

vyaman

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

asked Jun 8 at 17:30

vyamanvyaman

5088 bronze badges

2 Answers
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4

$begingroup$

Yes, it can. We are looking at $f: mathbbR rightarrow mathbbC$ here; which can be visualised as a line in $3$-D space, where the $x$-axis is the input axis, $y$ is the real part of the output, and $z$ is the imaginary part. We simply draw a straight lines along the $z=0$ plane at some intervals, and join this to some other lines along the $y=0$ plane via more straight lines. This is still continuous as required. Other constructions can also affirm the existence of such functions even when differentiability or smoothness is required - for differentiable, simple round the joined corners in a way that preserves differentiability (for example, an arc should work). Smooth functions can also be done in this way although it is a little harder to find an explicit construction.

share|cite|improve this answer

edited Jun 8 at 17:59

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I meant to say differentiable instead of continuous. I have now amended the question. Could you let me know your comment for the differentiable functions?
  $endgroup$
  – vyaman
  Jun 8 at 17:39
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f(x)= exp(-1/x^2)$ if $x<0$, $f(x)=sqrt-1exp(-1/x^2)$ if $x>0$, and $f(0)=0$. It is infinitely differentiable.

share|cite|improve this answer

answered Jun 8 at 20:45

SomosSomos

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add a comment | 

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4

$begingroup$

Yes, it can. We are looking at $f: mathbbR rightarrow mathbbC$ here; which can be visualised as a line in $3$-D space, where the $x$-axis is the input axis, $y$ is the real part of the output, and $z$ is the imaginary part. We simply draw a straight lines along the $z=0$ plane at some intervals, and join this to some other lines along the $y=0$ plane via more straight lines. This is still continuous as required. Other constructions can also affirm the existence of such functions even when differentiability or smoothness is required - for differentiable, simple round the joined corners in a way that preserves differentiability (for example, an arc should work). Smooth functions can also be done in this way although it is a little harder to find an explicit construction.

share|cite|improve this answer

edited Jun 8 at 17:59

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I meant to say differentiable instead of continuous. I have now amended the question. Could you let me know your comment for the differentiable functions?
  $endgroup$
  – vyaman
  Jun 8 at 17:39
  

  
  
  
  

add a comment | 

4

$begingroup$

Yes, it can. We are looking at $f: mathbbR rightarrow mathbbC$ here; which can be visualised as a line in $3$-D space, where the $x$-axis is the input axis, $y$ is the real part of the output, and $z$ is the imaginary part. We simply draw a straight lines along the $z=0$ plane at some intervals, and join this to some other lines along the $y=0$ plane via more straight lines. This is still continuous as required. Other constructions can also affirm the existence of such functions even when differentiability or smoothness is required - for differentiable, simple round the joined corners in a way that preserves differentiability (for example, an arc should work). Smooth functions can also be done in this way although it is a little harder to find an explicit construction.

share|cite|improve this answer

edited Jun 8 at 17:59

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I meant to say differentiable instead of continuous. I have now amended the question. Could you let me know your comment for the differentiable functions?
  $endgroup$
  – vyaman
  Jun 8 at 17:39
  

  
  
  
  

add a comment | 

$begingroup$

Yes, it can. We are looking at $f: mathbbR rightarrow mathbbC$ here; which can be visualised as a line in $3$-D space, where the $x$-axis is the input axis, $y$ is the real part of the output, and $z$ is the imaginary part. We simply draw a straight lines along the $z=0$ plane at some intervals, and join this to some other lines along the $y=0$ plane via more straight lines. This is still continuous as required. Other constructions can also affirm the existence of such functions even when differentiability or smoothness is required - for differentiable, simple round the joined corners in a way that preserves differentiability (for example, an arc should work). Smooth functions can also be done in this way although it is a little harder to find an explicit construction.

share|cite|improve this answer

edited Jun 8 at 17:59

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 8 at 17:59

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

answered Jun 8 at 17:33

auscryptauscrypt

7,40466 silver badges1414 bronze badges

1

$begingroup$

Let $f(x)= exp(-1/x^2)$ if $x<0$, $f(x)=sqrt-1exp(-1/x^2)$ if $x>0$, and $f(0)=0$. It is infinitely differentiable.

share|cite|improve this answer

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Let $f(x)= exp(-1/x^2)$ if $x<0$, $f(x)=sqrt-1exp(-1/x^2)$ if $x>0$, and $f(0)=0$. It is infinitely differentiable.

share|cite|improve this answer

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $f(x)= exp(-1/x^2)$ if $x<0$, $f(x)=sqrt-1exp(-1/x^2)$ if $x>0$, and $f(0)=0$. It is infinitely differentiable.

share|cite|improve this answer

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges

answered Jun 8 at 20:45

SomosSomos

16.4k11 gold badge1414 silver badges3838 bronze badges