Clairaut's theorem on equality of mixed partials - Revision history

Vipul: /* General statement */

2013-01-25T22:49:58Z

General statement

← Older revision		Revision as of 22:49, 25 January 2013
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	Note: The video below will probably be replaced by a better video that covers the subtleties better, but it's provisionally okay.		Note: The video below will probably be replaced by a better video that covers the subtleties better, but it's provisionally okay.

	<center>{{#widget:YouTube\|id=~~ppYpzL8Nnf0~~}}</center>		<center>{{#widget:YouTube\|id=IAtCXqzBH3Q}}</center>

	==Particular cases==		==Particular cases==

Vipul at 02:17, 25 January 2013

2013-01-25T02:17:45Z

← Older revision		Revision as of 02:17, 25 January 2013
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	We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.		We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.

	<center>{{#widget:YouTube\|id=~~dN8r-FPihRU~~}}</center>		<center>{{#widget:YouTube\|id=xTBxv4RKq1w}}</center>

Vipul: /* Proof */

2012-04-22T04:07:09Z

Proof

← Older revision		Revision as of 04:07, 22 April 2012
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	We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.		We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.

			<center>{{#widget:YouTube\|id=dN8r-FPihRU}}</center>

Vipul: /* Proof */

2012-04-21T00:39:22Z

Proof

← Older revision		Revision as of 00:39, 21 April 2012
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	A similar calculation for the other mixed partial yields that:		A similar calculation for the other mixed partial yields that:

	<math>f_{yx}(x_0,y_0) = \lim_{x \to x_0} \lim_{y\to y_0} \left[\frac{f(x,y) - f(x,y_0) - f(x_0,y) + f(x_0,y_0)}{(x -~~x_0~~)(y - ~~y_0~~)}\right]</math>		<math>f_{yx}(x_0,y_0) = \lim_{x \to x_0} \lim_{y\to y_0} \left[\frac{f(x,y) - f(x,y_0) - f(x_0,y) + f(x_0,y_0)}{(y -y_0)(x - x_0)}\right]</math>

	We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.		We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.

Vipul at 00:31, 21 April 2012

2012-04-21T00:31:38Z

← Older revision		Revision as of 00:31, 21 April 2012
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	* [[Failure of Clairaut's theorem where only one of the mixed partials is defined]]		* [[Failure of Clairaut's theorem where only one of the mixed partials is defined]]
	* [[Failure of Clairaut's theorem where both mixed partials are defined but not equal]] (this happens because one or both of them is not continuous)		* [[Failure of Clairaut's theorem where both mixed partials are defined but not equal]] (this happens because one or both of them is not continuous)

			==Proof==

			===Key computation as a double limit at a point===

			Based on the [[second-order mixed partial derivative#Definition as a double limit at a point\|definition as a double limit at a point]], we have:

			<math>f_{xy}(x_0,y_0) = \lim_{y \to y_0} \frac{\lim_{x\to x_0} \frac{f(x,y) - f(x_0,y)}{x- x_0} - \lim_{x \to x_0} \frac{f(x,y_0) - f(x_0,y_0)}{x- x_0}}{y - y_0}</math>

			Now, conditional to the existence of this expression, we can use that [[limit is linear]] to combine the two limits in the inner sum on the right side and get:

			<math>f_{xy}(x_0,y_0) = \lim_{y \to y_0} \frac{\lim_{x\to x_0} \left(\frac{f(x,y) - f(x_0,y)}{x- x_0} - \frac{f(x,y_0) - f(x_0,y_0)}{x- x_0}\right)}{y - y_0}</math>

			With some simple algebraic tidying up, this becomes:

			<math>f_{xy}(x_0,y_0) = \lim_{y \to y_0} \lim_{x\to x_0} \left[\frac{f(x,y) - f(x_0,y) - f(x,y_0) + f(x_0,y_0)}{(x -x_0)(y - y_0)}\right]</math>

			A similar calculation for the other mixed partial yields that:

			<math>f_{yx}(x_0,y_0) = \lim_{x \to x_0} \lim_{y\to y_0} \left[\frac{f(x,y) - f(x,y_0) - f(x_0,y) + f(x_0,y_0)}{(x -x_0)(y - y_0)}\right]</math>

			We notice that, although the terms are written in a slightly different order, ''the expressions whose limit is being computed is the same in both cases''. The only difference between the two cases is the order in which we are taking limits. In the <math>f_{xy}</math> case, we take the limit with respect to <math>x \to x_0</math> on the inside, and then take the limit with respect to <math>y \to y_0</math> on the outside. In the <math>f_{yx}</math> case, we take the limit with respect to <math>y \to y_0</math> on the inside, and then take the limit with respect to <math>x \to x_0</math> on the outside.

Vipul: /* Statement */

2012-04-08T19:26:00Z

Statement

← Older revision		Revision as of 19:26, 8 April 2012
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	{{quotation\|'''CAUTION''': This general version of Clairaut's theorem does not merely state that if two higher-order mixed partials that involve the same number of differentiations in each variable are continuous, they must be equal. ''Rather'', they require ''all'' the higher-order mixed partials of that type to be defined and continuous. The reason is that in the proof of this using Clairaut's theorem for second-order mixed partials, we may need to go ''via'' the other higher-order mixed partials and hence need those to exist and be continuous as well. Thus, the generalization does ''not'' say, for instance, that if <math>f_{xxy}</math> and <math>f_{yxx}</math> exist and are continuous, they must be equal.<br>We can formulate a slightly stronger version of this general statement, that only requires the continuity of certain specific other partial derivatives, but this would get too complicated.}}		{{quotation\|'''CAUTION''': This general version of Clairaut's theorem does not merely state that if two higher-order mixed partials that involve the same number of differentiations in each variable are continuous, they must be equal. ''Rather'', they require ''all'' the higher-order mixed partials of that type to be defined and continuous. The reason is that in the proof of this using Clairaut's theorem for second-order mixed partials, we may need to go ''via'' the other higher-order mixed partials and hence need those to exist and be continuous as well. Thus, the generalization does ''not'' say, for instance, that if <math>f_{xxy}</math> and <math>f_{yxx}</math> exist and are continuous, they must be equal.<br>We can formulate a slightly stronger version of this general statement, that only requires the continuity of certain specific other partial derivatives, but this would get too complicated.}}

			Note: The video below will probably be replaced by a better video that covers the subtleties better, but it's provisionally okay.

			<center>{{#widget:YouTube\|id=ppYpzL8Nnf0}}</center>

	==Particular cases==		==Particular cases==

Vipul: /* General statement */

2012-04-08T19:06:52Z

General statement

← Older revision		Revision as of 19:06, 8 April 2012
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	* Suppose <math>f</math> is a function of two variables <math>x</math> and <math>y</math>, and the three mixed partials <math>f_{xxy}, f_{xyx}, f_{yxx}</math> exist and are continuous on an open subset <math>U</math> of <math>\R^3</math>. Then, all three of them are equal on <math>U</math>. (Note that these mixed partials all involve differentiating twice with respect to <math>x</math> and once with respect to <math>y</math>).		* Suppose <math>f</math> is a function of two variables <math>x</math> and <math>y</math>, and the three mixed partials <math>f_{xxy}, f_{xyx}, f_{yxx}</math> exist and are continuous on an open subset <math>U</math> of <math>\R^3</math>. Then, all three of them are equal on <math>U</math>. (Note that these mixed partials all involve differentiating twice with respect to <math>x</math> and once with respect to <math>y</math>).
	* Suppose <math>f</math> is a function of three variables <math>x,y,z</math>, and the six mixed partials <math>f_{xyz}, f_{xzy}, f_{yxz}, f_{yzx}, f_{zxy}, f_{zyx}</math> exist and are continuous on an open subset <math>U</math> of <math>\R^3</math>. Then, all six of them are equal on <math>U</math>.		* Suppose <math>f</math> is a function of three variables <math>x,y,z</math>, and the six mixed partials <math>f_{xyz}, f_{xzy}, f_{yxz}, f_{yzx}, f_{zxy}, f_{zyx}</math> exist and are continuous on an open subset <math>U</math> of <math>\R^3</math>. Then, all six of them are equal on <math>U</math>.

			{{quotation\|'''CAUTION''': This general version of Clairaut's theorem does not merely state that if two higher-order mixed partials that involve the same number of differentiations in each variable are continuous, they must be equal. ''Rather'', they require ''all'' the higher-order mixed partials of that type to be defined and continuous. The reason is that in the proof of this using Clairaut's theorem for second-order mixed partials, we may need to go ''via'' the other higher-order mixed partials and hence need those to exist and be continuous as well. Thus, the generalization does ''not'' say, for instance, that if <math>f_{xxy}</math> and <math>f_{yxx}</math> exist and are continuous, they must be equal.<br>We can formulate a slightly stronger version of this general statement, that only requires the continuity of certain specific other partial derivatives, but this would get too complicated.}}

	==Particular cases==		==Particular cases==

Vipul: /* Related facts */

2012-04-08T19:03:39Z

Related facts

← Older revision		Revision as of 19:03, 8 April 2012
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	==Related facts==		==Related facts==

	* [[Failure of Clairaut's theorem ~~when second-order~~ mixed partials are not ~~continuous~~]]		* [[Failure of Clairaut's theorem where only one of the mixed partials is defined]]
			* [[Failure of Clairaut's theorem where both mixed partials are defined but not equal]] (this happens because one or both of them is not continuous)

Vipul at 19:01, 8 April 2012

2012-04-08T19:01:55Z

@@ Line 2: / Line 2: @@
 ==Statement==
-Suppose <math>f</math> is a real-valued function of two variables <math>x,y</math> and <math>f(x,y)</math> is defined on an open subset <math>U</math> of <math>\R^2</math>. Suppose further that both the [[second-order mixed partial derivative]]s <math>f_{xy}(x,y)</math> and <math>f_{yx}</math> exist and are continuous on <math>U</math>. Then, we have:
+===Statement for second-order mixed partial of function of two variables===
 <math>\! f_{xy} = f_{yx}</math>
 on all of <math>U</math>.

Vipul at 00:28, 13 February 2012

2012-02-13T00:28:09Z

← Older revision		Revision as of 00:28, 13 February 2012
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			{{commuting operators}}
	==Statement==		==Statement==