Sfdwhf

Should the Product Owner dictate what info the UI needs to display?

As an international instructor, should I openly talk about my accent?

Combinatorics problem, right solution?

Apply a different color ramp to subset of categorized symbols in QGIS?

My bank got bought out, am I now going to have to start filing tax returns in a different state?

Can I criticise the more senior developers around me for not writing clean code?

Retract an already submitted recommendation letter (written for an undergrad student)

How long after the last departure shall the airport stay open for an emergency return?

Multiple fireplaces in an apartment building?

How to have a sharp product image?

What does MLD stand for?

A Paper Record is What I Hamper

Negative Resistance

Unknown code in script

Which big number is bigger?

Why did C use the -> operator instead of reusing the . operator?

How do I reattach a shelf to the wall when it ripped out of the wall?

Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?

How can I practically buy stocks?

Crossed out red box fitting tightly around image

Magical attacks and overcoming damage resistance

All ASCII characters with a given bit count

Nails holding drywall

Why is the underscore command _ useful?

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

8

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 17 hours ago

xzczd

asked 18 hours ago

xzczdxzczd

28k577258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can shed some light: Trace[ Arg'[1. + I], TraceInternal -> True ]
  $endgroup$
  – Kuba♦
  17 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
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  17 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♦
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

8

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 17 hours ago

xzczd

asked 18 hours ago

xzczdxzczd

28k577258

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This can shed some light: Trace[ Arg'[1. + I], TraceInternal -> True ]
  $endgroup$
  – Kuba♦
  17 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
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kuba Oh blinding light…
  $endgroup$
  – xzczd
  17 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♦
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  13 hours ago
  

  
  
  
  

 | 
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 17 hours ago

xzczd

asked 18 hours ago

xzczdxzczd

28k577258

$endgroup$

Arg'[1. + I]

(* -0.5 *)

share|improve this question

edited 17 hours ago

xzczd

asked 18 hours ago

xzczdxzczd

28k577258

share|improve this question

edited 17 hours ago

xzczd

asked 18 hours ago

xzczdxzczd

28k577258

asked 18 hours ago

xzczdxzczd

28k577258

asked 18 hours ago

xzczdxzczd

28k577258

2 Answers
2

active

oldest

votes

7

$begingroup$

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 14 hours ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  11 hours ago
  

  
  
  
  

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 13 hours ago

answered 17 hours ago

RomanRoman

6,08611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  17 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
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f196998%2fwhy-does-arg1-i-return-0-5%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

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 14 hours ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  11 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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 14 hours ago

Michael E2Michael E2

151k12203483

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  11 hours 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 14 hours 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 14 hours ago

Michael E2Michael E2

151k12203483

answered 14 hours ago

Michael E2Michael E2

151k12203483

answered 14 hours ago

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 13 hours ago

answered 17 hours ago

RomanRoman

6,08611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  17 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
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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 13 hours ago

answered 17 hours ago

RomanRoman

6,08611132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Er… how is derivative of $ln(z^*)$ defined here?
  $endgroup$
  – xzczd
  17 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
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser they do give the same result when you apply N: Arg'[1 + I] // N also gives -0.5.
  $endgroup$
  – Roman
  16 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 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 13 hours ago

answered 17 hours ago

RomanRoman

6,08611132

$endgroup$

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 13 hours ago

answered 17 hours ago

RomanRoman

6,08611132

answered 17 hours ago

RomanRoman

6,08611132

answered 17 hours ago

RomanRoman

6,08611132