Jacobian matrix - Revision history

Vipul at 02:32, 13 May 2012

2012-05-13T02:32:03Z

← Older revision		Revision as of 02:32, 13 May 2012
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	Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.		Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

			<center>{{#widget:YouTube\|id=VCM4RVM09_I}}</center>

	==Definition as a function==		==Definition as a function==
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	If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.		If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

			<center>{{#widget:YouTube\|id=jTmwUMnuUec}}</center>
	==Particular cases==		==Particular cases==

Vipul: /* Definition at a point in terms of gradient vectors as row vectors */

2012-05-12T22:17:09Z

Definition at a point in terms of gradient vectors as row vectors

← Older revision		Revision as of 22:17, 12 May 2012
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	<math>\begin{pmatrix} \nabla(f_1)(a_1,a_2,\dots,a_n) \\ \nabla(f_2)(a_1,a_2,\dots,a_n) \\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m)(a_1,a_2,\dots,a_n) \\\end{pmatrix}</math>		<math>\begin{pmatrix} \nabla(f_1)(a_1,a_2,\dots,a_n) \\ \nabla(f_2)(a_1,a_2,\dots,a_n) \\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m)(a_1,a_2,\dots,a_n) \\\end{pmatrix}</math>

			<center>{{#widget:YouTube\|id=O8isoxng_9g}}</center>

	===Definition at a point in terms of partial derivatives===		===Definition at a point in terms of partial derivatives===

Vipul at 15:49, 12 May 2012

2012-05-12T15:49:52Z

← Older revision		Revision as of 15:49, 12 May 2012
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	===Definition in terms of gradient vectors as row vectors===		===Definition in terms of gradient vectors as row vectors===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>.		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>. Explicitly, it looks like this:

			<math>\begin{pmatrix} \nabla(f_1) \\ \nabla(f_2)\\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m) \\\end{pmatrix}</math>


	Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.		Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.
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	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>

	wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.		wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. Here's how the matrix looks:

			<math>\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2}& \dots & \frac{\partial f_1}{\partial x_n}\\
			\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n}\\
			\cdot & \cdot & \cdot & \cdot \\
			\frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \dots & \frac{\partial f_m}{\partial x_n}\\\end{pmatrix}</math>

			If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

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

Vipul at 15:47, 12 May 2012

2012-05-12T15:47:08Z

← Older revision		Revision as of 15:47, 12 May 2012
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	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.

			Explicitly, in terms of rows, it looks like:

			<math>\begin{pmatrix} \nabla(f_1)(a_1,a_2,\dots,a_n) \\ \nabla(f_2)(a_1,a_2,\dots,a_n) \\ \cdot \\ \cdot \\ \cdot \\ \nabla(f_m)(a_1,a_2,\dots,a_n) \\\end{pmatrix}</math>

	===Definition at a point in terms of partial derivatives===		===Definition at a point in terms of partial derivatives===
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	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>

			Here's how the matrix looks:

			<math>\begin{pmatrix} \frac{\partial f_1}{\partial x_1}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_1}{\partial x_2}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_1}{\partial x_n}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\
			\frac{\partial f_2}{\partial x_1}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_2}{\partial x_2}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_2}{\partial x_n}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\
			\cdot & \cdot & \cdot & \cdot \\
			\frac{\partial f_m}{\partial x_1}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \frac{\partial f_m}{\partial x_2}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} & \dots & \frac{\partial f_m}{\partial x_n}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)} \\\end{pmatrix}</math>

	Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.		Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

Vipul at 23:30, 11 May 2012

2012-05-11T23:30:53Z

← Older revision		Revision as of 23:30, 11 May 2012
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	===Definition at a point in terms of partial derivatives===		===Definition at a point in terms of partial derivatives===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:

	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>
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	===Definition in terms of gradient vectors as row vectors===		===Definition in terms of gradient vectors as row vectors===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,~~f_n~~(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>.		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>.

	Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.		Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.
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	===Definition in terms of partial derivatives===		===Definition in terms of partial derivatives===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,~~f_n~~(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math> is a <math>m \times n</math> matrix of ''functions'' whose <math>(ij)^{th}</math> entry is given by:		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix''' of <math>f</math> is a <math>m \times n</math> matrix of ''functions'' whose <math>(ij)^{th}</math> entry is given by:

	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>

Vipul: /* Definition at a point */

2012-05-11T23:29:43Z

Definition at a point

← Older revision		Revision as of 23:29, 11 May 2012
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	===Definition at a point in terms of gradient vectors as row vectors===		===Definition at a point in terms of gradient vectors as row vectors===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,~~f_n~~(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix''' of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.

	===Definition at a point in terms of partial derivatives===		===Definition at a point in terms of partial derivatives===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,~~f_n~~(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_m(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:

	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)\|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>

Vipul: /* Definition at a point in terms of partial derivatives */

2012-05-11T03:49:04Z

Definition at a point in terms of partial derivatives

← Older revision		Revision as of 03:49, 11 May 2012
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	Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.		Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.

	===Definition ~~at a point~~ in terms of partial derivatives===		===Definition in terms of partial derivatives===

	Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math> is a <math>m \times n</math> matrix of ''~~numbers~~'' whose <math>(ij)^{th}</math> entry is given by:		Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math> is a <math>m \times n</math> matrix of ''functions'' whose <math>(ij)^{th}</math> entry is given by:

	<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>		<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>

Vipul at 21:39, 8 May 2012

2012-05-08T21:39:35Z

← Older revision		Revision as of 21:39, 8 May 2012
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	wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.		wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

			==Particular cases==

			{\| class="sortable" border="1"
			! Case !! What happens in that case?
			\|-
			\| <matH>m = n = 1</math> \|\| <math>f</math> is a real-valued function of one variable. The Jacobian matrix is a <math>1 \times 1</math> matrix whose entry is the ordinary [[derivative]].
			\|-
			\| <math>n = 1</math>, <math>m > 1</math> \|\| <math>f</math> is a vector-valued function of one variable. We can think of it as a parametric curve in <math>\R^m</math>. The Jacobian matrix is a <math>m \times 1</math> matrix which, read as a column vector, is the parametric derivative of the vector-valued function.
			\|-
			\| <math>m = 1</math>, <math>n > 1</math> \|\| <math>f</math> is a real-valued function of multiple variables. The Jacobian matrix is a <math>1 \times n</math> matrix which, read as a row vector, is the [[gradient vector]] function.
			\|-
			\| <math>f</math> is a linear or affine map. \|\| The Jacobian matrix is the same as the matrix describing <math>f</math> (or, if <math>f</math> is affine, the matrix describing the linear part of <math>f</math>).
			\|-
			\| <math>m = n</math>, and we are identifying the spaces of inputs and outputs of <math>f</math>. \|\| The Jacobian matrix can then be thought of as a linear self-map from the <math>n</math>-dimensional space to itself. In this context, we can consider the [[Jacobian determinant]].
			\|}

Vipul: Created page with "{{multivariable analogue of|derivative}} ==Importance== The Jacobian matrix is the appropriate notion of '''derivative''' for a function that has multiple inputs (or equival..."

2012-05-08T21:34:11Z

Created page with "{{multivariable analogue of|derivative}} ==Importance== The Jacobian matrix is the appropriate notion of '''derivative''' for a function that has multiple inputs (or equival..."

New page

{{multivariable analogue of|derivative}}

==Importance==

The Jacobian matrix is the appropriate notion of '''derivative''' for a function that has multiple inputs (or equivalently, vector-valued inputs) and multiple outputs (or equivalently, vector-valued outputs).

==Definition at a point==

===Direct epsilon-delta definition===

===Definition at a point in terms of gradient vectors as row vectors===

Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math> at <math>(a_1,a_2,\dots,a_n)</math>.

===Definition at a point in terms of partial derivatives===

Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Suppose <math>(a_1,a_2,\dots,a_n)</math> is a point in the domain of <math>f</math> such that <math>f_i</math> is differentiable at <math>(a_1,a_2,\dots,a_n)</math> for <math>i \in \{ 1,2,\dots,m\}</math>. Then, the '''Jacobian matrix'''of <math>f</math> at <math>(a_1,a_2,\dots,a_n)</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:

<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)|_{(x_1,x_2,\dots,x_n) = (a_1,a_2,\dots,a_n)}</math>

Note that for this definition to be correct, it is still necessary that the gradient vectors exist. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.

==Definition as a function==

===Definition in terms of gradient vectors as row vectors===

Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math>is a <math>m \times n</math> matrix of ''functions'' whose <math>i^{th}</math> row is given by the [[gradient vector]] of <matH>f_i</math>.

Note that the domain of this function is the set of points at which all the <math>f_i</math>s individually are differentiable.

===Definition at a point in terms of partial derivatives===

Suppose <math>f</matH> is a vector-valued function with <math>n</math>-dimensional inputs and <math>m</math>-dimensional outputs. Explicitly, suppose <math>f</math> is a function with inputs <math>x_1,x_2,\dots,x_n</math> and outputs <math>f_1(x_1,x_2,\dots,x_n), f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n)</math>. Then, the '''Jacobian matrix'''of <math>f</math> is a <math>m \times n</math> matrix of ''numbers'' whose <math>(ij)^{th}</math> entry is given by:

<math>\frac{\partial f_i}{\partial x_j}(x_1,x_2,\dots,x_n)</math>

wherever all the <math>f_i</math>s individually are differentiable in the sense of the gradient vectors existing. If the gradient vectors do not exist but the partial derivatives do, a matrix can still be constructed using this recipe but it may not satisfy the nice behavior that the Jacobian matrix does.