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(Implicit Function Theorem)
 
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See also [[06-240/Linear Algebra - Why We Care]].
 
See also [[06-240/Linear Algebra - Why We Care]].
  
==Class Notes==
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{{0708-1300/Class Notes}}
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===Differentiability===
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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:'''
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We say that <math>f</math> is differentiable (''diffable'') at <math>x</math> if there is a linear map <math>L</math> so that
 +
 
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<math>\lim_{h\rightarrow0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.</math>
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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>.
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'''Theorem'''
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If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following assertions hold:
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# <math>df_{x}</math> is unique.
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# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
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# If <math>f</math> is linear then <math>df_{x}=f</math>
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# <math>d(f\circ g)_{x}=df_{g(x)}\cdot dg_{x}</math>
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# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>
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===Implicit Function Theorem===
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'''Example'''
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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.
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'''Notation'''
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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>
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'''Definition'''
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<math>C^{p}(V)</math> will be the class of all functions defined on <math>V</math> with continuous partial derivatives up to order <math>p.</math>
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'''Theorem'''(''Implicit function theorem'')
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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:
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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>.

Latest revision as of 17:23, 1 November 2007

Announcements go here

Contents

In Small Scales, Everything's Linear

06-240-QuiltBeforeMap.png \longrightarrow 06-240-QuiltAfterMap.png
z \mapsto z^2

Code in Mathematica:

QuiltPlot[{f_,g_}, {x_, xmin_, xmax_, nx_}, {y_, ymin_, ymax_, ny_}] :=
Module[
  {dx, dy, grid, ix, iy},
  SeedRandom[1];
  dx=(xmax-xmin)/nx;
  dy=(ymax-ymin)/ny;
  grid = Table[
    {x -> xmin+ix*dx, y -> ymin+iy*dy},
    {ix, 0, nx}, {iy, 0, ny}
  ];
  grid = Map[({f, g} /. #)&, grid, {2}];
  Show[
    Graphics[Table[
      {
        RGBColor[Random[], Random[], Random[]],
        Polygon[{
          grid[[ix, iy]],
          grid[[ix+1, iy]],
          grid[[ix+1, iy+1]],
          grid[[ix, iy+1]]
        }]
      },
      {ix, nx}, {iy, ny}
    ]],
    Frame -> True
  ]
]

QuiltPlot[{x, y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
QuiltPlot[{x^2-y^2, 2*x*y}, {x, -10, 10, 8}, {y, 5, 10, 8}]

See also 06-240/Linear Algebra - Why We Care.

Class Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

Differentiability

Let U, V and W be two normed finite dimensional vector spaces and let f:V\rightarrow W be a function defined on a neighborhood of the point x.

Definition:

We say that f is differentiable (diffable) at x if there is a linear map L so that

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

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

Theorem

If f:V\rightarrow W and g:U\rightarrow V are diffable maps then the following assertions hold:

  1. df_{x} is unique.
  2. d(f+g)_{x}=df_{x}+dg_{x}
  3. If f is linear then df_{x}=f
  4. d(f\circ g)_{x}=df_{g(x)}\cdot dg_{x}
  5. For every scalar number \alpha it holds d(\alpha f)_{x}=\alpha df_{x}

Implicit Function Theorem

Example Although x^2+y^2=1 does not define y as a function of x, in a neighborhood of (0;-1) we can define g(x)=-\sqrt{1-x^2} so that x^2+g(x)^2=1. Furthermore, g is differentiable with differential dg_{x}=\frac{x}{\sqrt{1-x^2}}. This is a motivation for the following theorem.

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).

Definition

C^{p}(V) will be the class of all functions defined on V with continuous partial derivatives up to order p.

Theorem(Implicit function theorem)

Let f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m be a C^{1}(\mathbb{R}^n \times \mathbb{R}^m) function defined on a neighborhood U of the point (x_0;y_0) and such that f(x_0;y_0)=0 and suppose that d(f_{[x]})_{y} is non-singular then, the following results holds:

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