Bsrgvty

Why is the air inside airliners so dry (low humidity)?

What does "2 fingers to Scotland" mean in Peter Grant's statement about Johnson not listening to the SNP's Westminster leader speeches?

Does the Presidential Records Act apply to the Trump/Zelensky telephone call 25 July 2019?

English equivalent of the Malayalam saying "don't stab/poke the dead body"?

What clothing should I bring to Mt. Titlis in Switzerland during December?

Can Alter Self be used to enter an enemy's body and destroy it from the inside?

Does the on'yomi of 輪 (リン) have any relation to the English "ring", or is it a coincidence?

Do dead weight 'components' exist?

Dealing with strict rules for paper submission using LaTeX

Adding coordinate constant to file in QGIS

How can I learn about unpublished VFR reporting points?

Does the original Game Boy game "Tetris" have a battery memory inside the cartridge?

Locked out of my own server

Teleportation is invented but it has nasty side-effects % of the time, how will that affect usage of the technology?

During a log backup is the data backed up to the start or end of the operation?

Employer reneged on negotiated clauses that weren't part of agreed contract, citing budget cuts - what can I do?

I was mistakenly identified as a criminal, and this has caused rumors. How can I convince my friends that it is all a mistake?

Tapping 3 times on one leg and 4 times on another leg in 4 beats. What is it called and how to do it?

If someone orders a pizza in the US and doesn't pay for it, could they be arrested?

Log user out after change of IP address?

How time is defined in astronomical science

Trying to find a short story about the representative of a city discovering the "primitive" folk are actually more advanced

There are polygons with only right angles which have an odd number of corners

Wood versus marble rolling pin 'performance'

Why should the equality of mixed partials be “intuitively obvious”?

derivatives of multi variable fuction“How many” vector fields are conservative?Can cross partial derivatives exist everywhere but be equal nowhere?The symmetry of mixed partials, for derivatives of order > 2Twice differentiable implies equality of mixed partials?Proof of Equality with Mixed PartialsShow that gravity is described by this 1-formDescribe $15/(4,3,2,1)$ as a vectorWhat is the geometric reason of why is the divergence of the curl of a vector field equal to zero?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;

6

1

$begingroup$

I am reading Ted Shifrin's excellent book Multivariable Mathematics. It claims that the equality of mixed partials is "an intuitively obvious result, but the proof is quite subtle". However, I guess I must be thinking in the wrong way, because I do not see the intuition behind this result. This is how I think about it:

Let $f:mathbbR^2 to mathbbR$. I think of $f_x$ as a "field of slopes" in the $x$-direction. If we analyze the movement in the $y$ direction in this field of slopes, we get $f_xy$. Now $f_y$ is a "field of slopes" in the $y$-direction. If we analyze movement in the $x$ direction here, we get $f_yx$.

It's unclear to me why movement in the $x$-direction in the "field of $y$-slopes" should be the same as movement in the $y$-direction in the "field of $x$-slopes".

real-analysis analysis multivariable-calculus

share|cite|improve this question

edited Jun 14 at 18:20

Ovi

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

$endgroup$

add a comment
 | 

6

1

$begingroup$

I am reading Ted Shifrin's excellent book Multivariable Mathematics. It claims that the equality of mixed partials is "an intuitively obvious result, but the proof is quite subtle". However, I guess I must be thinking in the wrong way, because I do not see the intuition behind this result. This is how I think about it:

Let $f:mathbbR^2 to mathbbR$. I think of $f_x$ as a "field of slopes" in the $x$-direction. If we analyze the movement in the $y$ direction in this field of slopes, we get $f_xy$. Now $f_y$ is a "field of slopes" in the $y$-direction. If we analyze movement in the $x$ direction here, we get $f_yx$.

It's unclear to me why movement in the $x$-direction in the "field of $y$-slopes" should be the same as movement in the $y$-direction in the "field of $x$-slopes".

real-analysis analysis multivariable-calculus

share|cite|improve this question

edited Jun 14 at 18:20

Ovi

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I am reading Ted Shifrin's excellent book Multivariable Mathematics. It claims that the equality of mixed partials is "an intuitively obvious result, but the proof is quite subtle". However, I guess I must be thinking in the wrong way, because I do not see the intuition behind this result. This is how I think about it:

Let $f:mathbbR^2 to mathbbR$. I think of $f_x$ as a "field of slopes" in the $x$-direction. If we analyze the movement in the $y$ direction in this field of slopes, we get $f_xy$. Now $f_y$ is a "field of slopes" in the $y$-direction. If we analyze movement in the $x$ direction here, we get $f_yx$.

It's unclear to me why movement in the $x$-direction in the "field of $y$-slopes" should be the same as movement in the $y$-direction in the "field of $x$-slopes".

real-analysis analysis multivariable-calculus

share|cite|improve this question

edited Jun 14 at 18:20

Ovi

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 14 at 18:20

Ovi

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

share|cite|improve this question

edited Jun 14 at 18:20

Ovi

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

asked Jun 14 at 17:49

OviOvi

13.6k1010 gold badges4848 silver badges128128 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

If you write the difference quotient for a small change $Delta x$ in $x$ and then the difference quotient for that when you change $y$ by $Delta y$ the result is the symmetric expression
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)

Delta x Delta y .
$$

share|cite|improve this answer

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
  $endgroup$
  – Ted Shifrin
  Jun 14 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin You need some smoothness assumption to justify taking the limit in either order.
  $endgroup$
  – Ethan Bolker
  Jun 14 at 20:26
  

  
  
  
  

add a comment
 | 

5

$begingroup$

I guess most people develop intuition based on examples, and most examples we pick to examine are $C^2$ functions, where the equality holds. Or, alternatively, you could say that the intuition comes from experience with Taylor's Theorem (which appears in Section 3 of Chapter 5 of my book). The intuition I guess I'm fondest of appears in Chapter 7 (exercise 19 of Section 2), just using a double integral and interchanging the order of integration. (After all, it's natural to think about $displaystyleintleft(int fracpartial^2fpartial xpartial ydyright)dx$ and its companion.) I agree that it's not obvious a priori that the $y$ rate of change of $f_x$ should agree with the $x$ rate of change of $f_y$; the $C^2$ condition is subtle, as I said.

share|cite|improve this answer

edited Jun 14 at 18:41

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the reply!
  $endgroup$
  – Ovi
  Jun 14 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was hoping that you would respond to this question!
  $endgroup$
  – Andres Mejia
  Jun 14 at 18:35
  
  

  
  
  
  

add a comment
 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3262481%2fwhy-should-the-equality-of-mixed-partials-be-intuitively-obvious%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

If you write the difference quotient for a small change $Delta x$ in $x$ and then the difference quotient for that when you change $y$ by $Delta y$ the result is the symmetric expression
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)

Delta x Delta y .
$$

share|cite|improve this answer

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
  $endgroup$
  – Ted Shifrin
  Jun 14 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin You need some smoothness assumption to justify taking the limit in either order.
  $endgroup$
  – Ethan Bolker
  Jun 14 at 20:26
  

  
  
  
  

add a comment
 | 

5

$begingroup$

If you write the difference quotient for a small change $Delta x$ in $x$ and then the difference quotient for that when you change $y$ by $Delta y$ the result is the symmetric expression
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)

Delta x Delta y .
$$

share|cite|improve this answer

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
  $endgroup$
  – Ted Shifrin
  Jun 14 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TedShifrin You need some smoothness assumption to justify taking the limit in either order.
  $endgroup$
  – Ethan Bolker
  Jun 14 at 20:26
  

  
  
  
  

add a comment
 | 

$begingroup$

If you write the difference quotient for a small change $Delta x$ in $x$ and then the difference quotient for that when you change $y$ by $Delta y$ the result is the symmetric expression
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)

Delta x Delta y .
$$

share|cite|improve this answer

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

answered Jun 14 at 18:52

Ethan BolkerEthan Bolker

58.3k55 gold badges6464 silver badges140140 bronze badges

5

$begingroup$

I guess most people develop intuition based on examples, and most examples we pick to examine are $C^2$ functions, where the equality holds. Or, alternatively, you could say that the intuition comes from experience with Taylor's Theorem (which appears in Section 3 of Chapter 5 of my book). The intuition I guess I'm fondest of appears in Chapter 7 (exercise 19 of Section 2), just using a double integral and interchanging the order of integration. (After all, it's natural to think about $displaystyleintleft(int fracpartial^2fpartial xpartial ydyright)dx$ and its companion.) I agree that it's not obvious a priori that the $y$ rate of change of $f_x$ should agree with the $x$ rate of change of $f_y$; the $C^2$ condition is subtle, as I said.

share|cite|improve this answer

edited Jun 14 at 18:41

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the reply!
  $endgroup$
  – Ovi
  Jun 14 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was hoping that you would respond to this question!
  $endgroup$
  – Andres Mejia
  Jun 14 at 18:35
  
  

  
  
  
  

add a comment
 | 

5

$begingroup$

I guess most people develop intuition based on examples, and most examples we pick to examine are $C^2$ functions, where the equality holds. Or, alternatively, you could say that the intuition comes from experience with Taylor's Theorem (which appears in Section 3 of Chapter 5 of my book). The intuition I guess I'm fondest of appears in Chapter 7 (exercise 19 of Section 2), just using a double integral and interchanging the order of integration. (After all, it's natural to think about $displaystyleintleft(int fracpartial^2fpartial xpartial ydyright)dx$ and its companion.) I agree that it's not obvious a priori that the $y$ rate of change of $f_x$ should agree with the $x$ rate of change of $f_y$; the $C^2$ condition is subtle, as I said.

share|cite|improve this answer

edited Jun 14 at 18:41

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the reply!
  $endgroup$
  – Ovi
  Jun 14 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was hoping that you would respond to this question!
  $endgroup$
  – Andres Mejia
  Jun 14 at 18:35
  
  

  
  
  
  

add a comment
 | 

$begingroup$

I guess most people develop intuition based on examples, and most examples we pick to examine are $C^2$ functions, where the equality holds. Or, alternatively, you could say that the intuition comes from experience with Taylor's Theorem (which appears in Section 3 of Chapter 5 of my book). The intuition I guess I'm fondest of appears in Chapter 7 (exercise 19 of Section 2), just using a double integral and interchanging the order of integration. (After all, it's natural to think about $displaystyleintleft(int fracpartial^2fpartial xpartial ydyright)dx$ and its companion.) I agree that it's not obvious a priori that the $y$ rate of change of $f_x$ should agree with the $x$ rate of change of $f_y$; the $C^2$ condition is subtle, as I said.

share|cite|improve this answer

edited Jun 14 at 18:41

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 14 at 18:41

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges

answered Jun 14 at 18:29

Ted ShifrinTed Shifrin

70.8k44 gold badges5050 silver badges9696 bronze badges