0708-1300/Class notes for Tuesday, September 11 - Revision history

Aroesler: /* Implicit Function Theorem */

2007-11-01T21:23:40Z

Implicit Function Theorem

← Older revision		Revision as of 17:23, 1 November 2007
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	'''Example'''		'''Example'''
	Although <math>x^2+y^2=1</math> does not ~~defines~~ <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.		Although <math>x^2+y^2=1</math> does not define <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.

	'''Notation'''		'''Notation'''

Aroesler: /* Differentiability */

2007-11-01T21:20:55Z

Differentiability

← Older revision		Revision as of 17:20, 1 November 2007
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	'''Theorem'''		'''Theorem'''

	If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following ~~asertions~~ ~~holds~~:		If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following assertions hold:

	# <math>df_{x}</math> is unique.		# <math>df_{x}</math> is unique.
	# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>		# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
	# If <math>f</math> is linear then <math>df_{x}=f</math>		# If <math>f</math> is linear then <math>df_{x}=f</math>
	# <math>d(f\circ g)_{x}=df_{g(x)}\~~circ~~ dg_{x}</math>		# <math>d(f\circ g)_{x}=df_{g(x)}\cdot dg_{x}</math>
	# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>		# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

Aptikuis: /* Implicit Function Theorem */

2007-09-20T01:51:55Z

Implicit Function Theorem

← Older revision		Revision as of 21:51, 19 September 2007
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	Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:		Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

	There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a ''diffable'' function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.		There is an open neighborhood of <math>x_0</math>, <math>V\subset U</math>, and a ''diffable'' function <math>g:V\rightarrow\mathbb{R}^m</math> such that <math>g(x_0)=y_0</math> and for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.

Aptikuis at 04:59, 18 September 2007

2007-09-18T04:59:35Z

← Older revision		Revision as of 00:59, 18 September 2007
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	'''Notation'''		'''Notation'''

	If f:X\times Y\rightarrow Z then given x\in X we will define f_{[x]}:Y\rightarrow Z by f_{[x]}(y)=f(x;y)		If <math>f:X\times Y\rightarrow Z</math> then given <math>x\in X</math> we will define <math>f_{[x]}:Y\rightarrow Z</math> by <math>f_{[x]}(y)=f(x;y).</math>

	'''Definition'''		'''Definition'''

Drorbn at 12:12, 12 September 2007

2007-09-12T12:12:25Z

← Older revision		Revision as of 08:12, 12 September 2007
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	See also [[06-240/Linear Algebra - Why We Care]].		See also [[06-240/Linear Algebra - Why We Care]].

	==Class Notes==		{{0708-1300/Class Notes}}
	===Differentiability===		===Differentiability===
	Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.		Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.

Drorbn: /* Implicit Function Theorem */

2007-09-12T02:51:16Z

Implicit Function Theorem

← Older revision		Revision as of 22:51, 11 September 2007
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	Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:		Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

	There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a ~~<math>~~diffable~~</math>~~ function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.		There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a ''diffable'' function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.

Megan: /* Differentiability */

2007-09-12T02:10:42Z

Differentiability

← Older revision		Revision as of 22:10, 11 September 2007
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	==Class Notes==		==Class Notes==
	===Differentiability===		===Differentiability===
	Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>		Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.

	'''Definition:'''		'''Definition:'''

	We say that <math>f</math> is differentiable (''diffable'') if there is a linear map <math>L</math> so that		We say that <math>f</math> is differentiable (''diffable'') at <math>x</math> if there is a linear map <math>L</math> so that

	<math>\lim_{h\rightarrow0}\frac{\|f(x+h)-f(x)-L(h)\|}{\|h\|}=0.</math>		<math>\lim_{h\rightarrow0}\frac{\|f(x+h)-f(x)-L(h)\|}{\|h\|}=0.</math>

	In this case we will say that <math>L</math> is a differential of <math>f</math> and will denote it by <math>df_{x}</math>.		In this case we will say that <math>L</math> is a differential of <math>f</math> at <math>x</math> and will denote it by <math>df_{x}</math>.

	'''Theorem'''		'''Theorem'''

Megan: /* Differentiability */

2007-09-12T02:05:22Z

Differentiability

← Older revision		Revision as of 22:05, 11 September 2007
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	We say that <math>f</math> is differentiable (''diffable'') if there is a linear map <math>L</math> so that		We say that <math>f</math> is differentiable (''diffable'') if there is a linear map <math>L</math> so that

	<math>\lim_{h\rightarrow0}\frac{\|f(x+h)-f(x)-L(h)\|}{\|h\|}.</math>		<math>\lim_{h\rightarrow0}\frac{\|f(x+h)-f(x)-L(h)\|}{\|h\|}=0.</math>

	In this case we will say that <math>L</math> is a differential of <math>f</math> and will denote it by <math>df_{x}</math>.		In this case we will say that <math>L</math> is a differential of <math>f</math> and will denote it by <math>df_{x}</math>.

Drorbn: /* Differentiability */

2007-09-11T22:36:53Z

Differentiability

← Older revision		Revision as of 18:36, 11 September 2007
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	# <math>df_{x}</math> is unique.		# <math>df_{x}</math> is unique.

	# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>		# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>

	# If <math>f</math> is linear then <math>df_{x}=f</math>		# If <math>f</math> is linear then <math>df_{x}=f</math>

	# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>		# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>

	# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>		# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

Drorbn: /* Differentiability */

2007-09-11T22:36:42Z

Differentiability

← Older revision		Revision as of 18:36, 11 September 2007
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	If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:		If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:

	1)<math>df_{x}</math> is unique.		# <math>df_{x}</math> is unique.

	2)<math>d(f+g)_{x}=df_{x}+dg_{x}</math>		# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>

	3)If <math>f</math> is linear then <math>df_{x}=f</math>		# If <math>f</math> is linear then <math>df_{x}=f</math>

	4)<math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>		# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>

	5)For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>		# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

	===Implicit Function Theorem===		===Implicit Function Theorem===