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Why does Arg'[1. + I] return -0.5?

4

$begingroup$

From the document we know that

Arg[z] gives the gives the argument of the complex number z.

Then how about Arg'[z]? This seems to be meaningless, but Mathematica returns something if z is a non-exact number, for example

Arg'[1. + I]

(* -0.5 *)

So my question is:

1. How is the numeric value of Arg'[z] defined?




2. Why does Arg' behave like this? What's the potential usage of this behavior?

calculus-and-analysis numerics complex implementation-details

share|improve this question

edited 3 hours ago

xzczd

asked 3 hours ago

xzczdxzczd

27.9k576258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can shed some light: Trace[ Arg'[1. + I], TraceInternal -> True ]
  $endgroup$
  – Kuba♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is impossible to understand what the output from Trace[ Arg'[1. + I], TraceInternal -> True ] mean. May be numerics gone mad or something :) so just change 1.0 to 1 in the example given and then Arg will no longer do what you show.
  $endgroup$
  – Nasser
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser and xzczd it looks like it calculates derivative value from set of points f or Arg, but I wasn't paying too much attention.
  $endgroup$
  – Kuba♦
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

From the document we know that

Arg[z] gives the gives the argument of the complex number z.

Then how about Arg'[z]? This seems to be meaningless, but Mathematica returns something if z is a non-exact number, for example

Arg'[1. + I]

(* -0.5 *)

So my question is:

1. How is the numeric value of Arg'[z] defined?




2. Why does Arg' behave like this? What's the potential usage of this behavior?

calculus-and-analysis numerics complex implementation-details

share|improve this question

edited 3 hours ago

xzczd

asked 3 hours ago

xzczdxzczd

27.9k576258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can shed some light: Trace[ Arg'[1. + I], TraceInternal -> True ]
  $endgroup$
  – Kuba♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is impossible to understand what the output from Trace[ Arg'[1. + I], TraceInternal -> True ] mean. May be numerics gone mad or something :) so just change 1.0 to 1 in the example given and then Arg will no longer do what you show.
  $endgroup$
  – Nasser
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser and xzczd it looks like it calculates derivative value from set of points f or Arg, but I wasn't paying too much attention.
  $endgroup$
  – Kuba♦
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

From the document we know that

Arg[z] gives the gives the argument of the complex number z.

Then how about Arg'[z]? This seems to be meaningless, but Mathematica returns something if z is a non-exact number, for example

Arg'[1. + I]

(* -0.5 *)

So my question is:

1. How is the numeric value of Arg'[z] defined?




2. Why does Arg' behave like this? What's the potential usage of this behavior?

calculus-and-analysis numerics complex implementation-details

share|improve this question

edited 3 hours ago

xzczd

asked 3 hours ago

xzczdxzczd

27.9k576258

$endgroup$

Arg'[1. + I]

(* -0.5 *)

share|improve this question

edited 3 hours ago

xzczd

asked 3 hours ago

xzczdxzczd

27.9k576258

share|improve this question

edited 3 hours ago

xzczd

asked 3 hours ago

xzczdxzczd

27.9k576258

asked 3 hours ago

xzczdxzczd

27.9k576258

asked 3 hours ago

xzczdxzczd

27.9k576258

2 Answers
2

active

oldest

votes

5

$begingroup$

The internal Trace[] Kuba advises shows calculations consistent with the numeric approximation of the partial derivative with respect to the real part:

D[ComplexExpand[Arg[x + I y], TargetFunctions -> {Re, Im}], x] /. 

  x -> 1 /. y -> 1

(*  -(1/2)  *)

This is what Mathematica does with the derivative of a numeric function with approximate input.

Other examples:

ClearAll[f, g];

f[x_?NumericQ] := Re[x]^2;

g[x_?NumericQ] := Im[x]^2;



f'[1. + I]

g'[1. + I]

(*

  1.999999999999995`  

  -2.7506672371246275`*^-15 

*)

It seems like the wrong way to evalutate Derivative.

share|improve this answer

answered 27 mins ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny, Abs'[2. + I] returns the input.
  $endgroup$
  – xzczd
  10 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The definition of the argument is $arg(z)=text{Im}(ln(z))$. Its partial derivative with respect to $z$ would then be

$$
frac{partial arg(z)}{partial z}=
frac{partial}{partial z}frac{ln(z)-ln(z^*)}{2i}
= -frac{i}{2z}.
$$

What you see looks like twice the real part of this expression:

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)]}]



(* {-0.172414, -0.172414} *)

I don't know in which sense this is the "correct" answer. It could be that what is actually calculated is not the partial derivative with respect to $z$, but rather the sum of partial derivatives

$$
frac{partial arg(z)}{partial z}
+frac{partial arg(z)}{partial z^*}
= -frac{i}{2z}+frac{i}{2z^*}
= -frac{text{Im}(z)}{|z|^2}
$$

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2}]



(* {-0.172414, -0.172414, -0.172414} *)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

RomanRoman

6,06511132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any idea why then With[{z = 1.0 + I}, Arg'[z]] not same as With[{z = 1 + I}, Arg'[z]]? Should not these give same result?
  $endgroup$
  – Nasser
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd $z$ and $z^*$ are independent variables in complex analysis, so $partial z^*/partial z=0$ etc.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser they do give the same result when you apply N: Arg'[1 + I] // N also gives -0.5.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

The internal Trace[] Kuba advises shows calculations consistent with the numeric approximation of the partial derivative with respect to the real part:

D[ComplexExpand[Arg[x + I y], TargetFunctions -> {Re, Im}], x] /. 

  x -> 1 /. y -> 1

(*  -(1/2)  *)

This is what Mathematica does with the derivative of a numeric function with approximate input.

Other examples:

ClearAll[f, g];

f[x_?NumericQ] := Re[x]^2;

g[x_?NumericQ] := Im[x]^2;



f'[1. + I]

g'[1. + I]

(*

  1.999999999999995`  

  -2.7506672371246275`*^-15 

*)

It seems like the wrong way to evalutate Derivative.

share|improve this answer

answered 27 mins ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny, Abs'[2. + I] returns the input.
  $endgroup$
  – xzczd
  10 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

The internal Trace[] Kuba advises shows calculations consistent with the numeric approximation of the partial derivative with respect to the real part:

D[ComplexExpand[Arg[x + I y], TargetFunctions -> {Re, Im}], x] /. 

  x -> 1 /. y -> 1

(*  -(1/2)  *)

This is what Mathematica does with the derivative of a numeric function with approximate input.

Other examples:

ClearAll[f, g];

f[x_?NumericQ] := Re[x]^2;

g[x_?NumericQ] := Im[x]^2;



f'[1. + I]

g'[1. + I]

(*

  1.999999999999995`  

  -2.7506672371246275`*^-15 

*)

It seems like the wrong way to evalutate Derivative.

share|improve this answer

answered 27 mins ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny, Abs'[2. + I] returns the input.
  $endgroup$
  – xzczd
  10 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

The internal Trace[] Kuba advises shows calculations consistent with the numeric approximation of the partial derivative with respect to the real part:

D[ComplexExpand[Arg[x + I y], TargetFunctions -> {Re, Im}], x] /. 

  x -> 1 /. y -> 1

(*  -(1/2)  *)

This is what Mathematica does with the derivative of a numeric function with approximate input.

Other examples:

ClearAll[f, g];

f[x_?NumericQ] := Re[x]^2;

g[x_?NumericQ] := Im[x]^2;



f'[1. + I]

g'[1. + I]

(*

  1.999999999999995`  

  -2.7506672371246275`*^-15 

*)

It seems like the wrong way to evalutate Derivative.

share|improve this answer

answered 27 mins ago

Michael E2Michael E2

151k12203483

$endgroup$

D[ComplexExpand[Arg[x + I y], TargetFunctions -> {Re, Im}], x] /. 

  x -> 1 /. y -> 1

(*  -(1/2)  *)

ClearAll[f, g];

f[x_?NumericQ] := Re[x]^2;

g[x_?NumericQ] := Im[x]^2;



f'[1. + I]

g'[1. + I]

(*

  1.999999999999995`  

  -2.7506672371246275`*^-15 

*)

share|improve this answer

answered 27 mins ago

Michael E2Michael E2

151k12203483

answered 27 mins ago

Michael E2Michael E2

151k12203483

answered 27 mins ago

Michael E2Michael E2

151k12203483

3

$begingroup$

The definition of the argument is $arg(z)=text{Im}(ln(z))$. Its partial derivative with respect to $z$ would then be

$$
frac{partial arg(z)}{partial z}=
frac{partial}{partial z}frac{ln(z)-ln(z^*)}{2i}
= -frac{i}{2z}.
$$

What you see looks like twice the real part of this expression:

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)]}]



(* {-0.172414, -0.172414} *)

I don't know in which sense this is the "correct" answer. It could be that what is actually calculated is not the partial derivative with respect to $z$, but rather the sum of partial derivatives

$$
frac{partial arg(z)}{partial z}
+frac{partial arg(z)}{partial z^*}
= -frac{i}{2z}+frac{i}{2z^*}
= -frac{text{Im}(z)}{|z|^2}
$$

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2}]



(* {-0.172414, -0.172414, -0.172414} *)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

RomanRoman

6,06511132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any idea why then With[{z = 1.0 + I}, Arg'[z]] not same as With[{z = 1 + I}, Arg'[z]]? Should not these give same result?
  $endgroup$
  – Nasser
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd $z$ and $z^*$ are independent variables in complex analysis, so $partial z^*/partial z=0$ etc.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser they do give the same result when you apply N: Arg'[1 + I] // N also gives -0.5.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The definition of the argument is $arg(z)=text{Im}(ln(z))$. Its partial derivative with respect to $z$ would then be

$$
frac{partial arg(z)}{partial z}=
frac{partial}{partial z}frac{ln(z)-ln(z^*)}{2i}
= -frac{i}{2z}.
$$

What you see looks like twice the real part of this expression:

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)]}]



(* {-0.172414, -0.172414} *)

I don't know in which sense this is the "correct" answer. It could be that what is actually calculated is not the partial derivative with respect to $z$, but rather the sum of partial derivatives

$$
frac{partial arg(z)}{partial z}
+frac{partial arg(z)}{partial z^*}
= -frac{i}{2z}+frac{i}{2z^*}
= -frac{text{Im}(z)}{|z|^2}
$$

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2}]



(* {-0.172414, -0.172414, -0.172414} *)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

RomanRoman

6,06511132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any idea why then With[{z = 1.0 + I}, Arg'[z]] not same as With[{z = 1 + I}, Arg'[z]]? Should not these give same result?
  $endgroup$
  – Nasser
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd $z$ and $z^*$ are independent variables in complex analysis, so $partial z^*/partial z=0$ etc.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser they do give the same result when you apply N: Arg'[1 + I] // N also gives -0.5.
  $endgroup$
  – Roman
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The definition of the argument is $arg(z)=text{Im}(ln(z))$. Its partial derivative with respect to $z$ would then be

$$
frac{partial arg(z)}{partial z}=
frac{partial}{partial z}frac{ln(z)-ln(z^*)}{2i}
= -frac{i}{2z}.
$$

What you see looks like twice the real part of this expression:

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)]}]



(* {-0.172414, -0.172414} *)

I don't know in which sense this is the "correct" answer. It could be that what is actually calculated is not the partial derivative with respect to $z$, but rather the sum of partial derivatives

$$
frac{partial arg(z)}{partial z}
+frac{partial arg(z)}{partial z^*}
= -frac{i}{2z}+frac{i}{2z^*}
= -frac{text{Im}(z)}{|z|^2}
$$

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2}]



(* {-0.172414, -0.172414, -0.172414} *)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

RomanRoman

6,06511132

$endgroup$

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)]}]



(* {-0.172414, -0.172414} *)

With[{z = 2. + 5 I},

  {Arg'[z], 2Re[-I/(2z)], -Im[z]/Abs[z]^2}]



(* {-0.172414, -0.172414, -0.172414} *)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

RomanRoman

6,06511132

answered 3 hours ago

RomanRoman

6,06511132

answered 3 hours ago

RomanRoman

6,06511132