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

NDSolve with Piecewise gives the incorrect answer randomlyWhy is NHoldFirst not propagated to symbolic derivatives?Numerical errors/inaccuracies in ProductLogWhy does this integral have a complex component?Why is (-1.)^2. a complex numberWhy does FindMinimum return 'The function value Null is not a real number'?Funny behavior when integratingWhy am I getting RootSearch::numb:Magnitude is returning a complex number if I precede it with Complex Expand?How to take derivative of the argument of an interpolating function

9

1

$begingroup$

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 yesterday

xzczd

asked yesterday

xzczdxzczd

28k577258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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♦
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny Answer, Just ignore the method it uses to calculate and use your own method of calculation based on the result.. The reason for using this is to extended divided by two. Arg'[1./2]
  $endgroup$
  – Zillinium
  yesterday
  

  
  
  
  

 | 
show 2 more comments

9

1

$begingroup$

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 yesterday

xzczd

asked yesterday

xzczdxzczd

28k577258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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♦
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny Answer, Just ignore the method it uses to calculate and use your own method of calculation based on the result.. The reason for using this is to extended divided by two. Arg'[1./2]
  $endgroup$
  – Zillinium
  yesterday
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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 yesterday

xzczd

asked yesterday

xzczdxzczd

28k577258

$endgroup$

Arg'[1. + I]

(* -0.5 *)

share|improve this question

edited yesterday

xzczd

asked yesterday

xzczdxzczd

28k577258

share|improve this question

edited yesterday

xzczd

asked yesterday

xzczdxzczd

28k577258

asked yesterday

xzczdxzczd

28k577258

asked yesterday

xzczdxzczd

28k577258

2 Answers
2

active

oldest

votes

8

$begingroup$

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 evaluate Derivative.

share|improve this answer

edited yesterday

xzczd

28k577258

answered yesterday

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd I guess they protected Abs'[] from being evaluated in this way. Re'[] and Im'[] do not evaluate, also. Maybe they overlooked Arg'[]?
  $endgroup$
  – Michael E2
  yesterday
  

  
  
  
  

add a comment | 

4

$begingroup$

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 partial derivative with respect to the real part of $z$:

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

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 yesterday

answered yesterday

RomanRoman

6,29611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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8

$begingroup$

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 evaluate Derivative.

share|improve this answer

edited yesterday

xzczd

28k577258

answered yesterday

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd I guess they protected Abs'[] from being evaluated in this way. Re'[] and Im'[] do not evaluate, also. Maybe they overlooked Arg'[]?
  $endgroup$
  – Michael E2
  yesterday
  

  
  
  
  

add a comment | 

8

$begingroup$

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 evaluate Derivative.

share|improve this answer

edited yesterday

xzczd

28k577258

answered yesterday

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @xzczd I guess they protected Abs'[] from being evaluated in this way. Re'[] and Im'[] do not evaluate, also. Maybe they overlooked Arg'[]?
  $endgroup$
  – Michael E2
  yesterday
  

  
  
  
  

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 evaluate Derivative.

share|improve this answer

edited yesterday

xzczd

28k577258

answered yesterday

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

edited yesterday

xzczd

28k577258

answered yesterday

Michael E2Michael E2

151k12203483

edited yesterday

xzczd

28k577258

edited yesterday

xzczd

28k577258

answered yesterday

Michael E2Michael E2

151k12203483

answered yesterday

Michael E2Michael E2

151k12203483

4

$begingroup$

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 partial derivative with respect to the real part of $z$:

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

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 yesterday

answered yesterday

RomanRoman

6,29611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

$begingroup$

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 partial derivative with respect to the real part of $z$:

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

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 yesterday

answered yesterday

RomanRoman

6,29611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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 partial derivative with respect to the real part of $z$:

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

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 yesterday

answered yesterday

RomanRoman

6,29611132

$endgroup$

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 yesterday

answered yesterday

RomanRoman

6,29611132

answered yesterday

RomanRoman

6,29611132

answered yesterday

RomanRoman

6,29611132