Why should the equality of mixed partials be “intuitively obvious”?“How many” vector fields are conservative?Existence of mixed partials in Clairaut's theorem.Can cross partial derivatives exist everywhere but be equal nowhere?Conditions needed for interchanging limits and mixed partials?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-formEquality of mixed partials proofDescribe $15/(4,3,2,1)$ as a vectorWhat is the geometric reason of why is the divergence of the curl of a vector field equal to zero?
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Why should the equality of mixed partials be “intuitively obvious”?
“How many” vector fields are conservative?Existence of mixed partials in Clairaut's theorem.Can cross partial derivatives exist everywhere but be equal nowhere?Conditions needed for interchanging limits and mixed partials?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-formEquality of mixed partials proofDescribe $15/(4,3,2,1)$ as a vectorWhat is the geometric reason of why is the divergence of the curl of a vector field equal to zero?
$begingroup$
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
asked 9 hours ago
OviOvi
13k1040119
$endgroup$
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
asked 9 hours ago
OviOvi
13k1040119
$endgroup$
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
asked 9 hours ago
OviOvi
13k1040119
$endgroup$
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
real-analysis analysis multivariable-calculus
asked 9 hours ago
OviOvi
13k1040119
asked 9 hours ago
OviOvi
13k1040119
edited 8 hours ago
Ovi
asked 9 hours ago
OviOvi
13k1040119
asked 9 hours ago
OviOvi
13k1040119
asked 9 hours ago
OviOvi
13k1040119
13k1040119
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
$endgroup$
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
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
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)
Delta x Delta y .
$$
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
$endgroup$
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
$endgroup$
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
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.
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
$endgroup$
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
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.
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
$endgroup$
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.
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
edited 8 hours ago
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
answered 8 hours ago
Ted ShifrinTed Shifrin
66.5k44893
66.5k44893
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
add a comment |
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
Thank you for the reply!
$endgroup$
– Ovi
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
$begingroup$
I was hoping that you would respond to this question!
$endgroup$
– Andres Mejia
8 hours ago
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
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)
Delta x Delta y .
$$
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
$endgroup$
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
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
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)
Delta x Delta y .
$$
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
$endgroup$
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
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
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)
Delta x Delta y .
$$
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
$endgroup$
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
$$frac
f(x + Delta x, y + Delta y)
-f(x + Delta x, y )
-f( x, y + Delta y)
+f(x,y)
Delta x Delta y .
$$
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
answered 8 hours ago
Ethan BolkerEthan Bolker
50.4k558128
50.4k558128
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
This is, of course, how the proof proceeds. So why do we need $C^2$ (or something slightly weaker) at all? :)
$endgroup$
– Ted Shifrin
7 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
$begingroup$
@TedShifrin You need some smoothness assumption to justify taking the limit in either order.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
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