Difference between revisions of "0708-1300/Class notes for Tuesday, September 11"

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(Class Notes)
(Implicit Function Theorem)
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'''Theorem'''(''Implicit function theorem'')
 
'''Theorem'''(''Implicit function theorem'')
  
Let <math>f:\mathbb{R}^n x \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a ''continuous diffable'' 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>
+
Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a ''continuous diffable'' 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>

Revision as of 16:57, 11 September 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

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) if there is a linear map L so that

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

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

Theorem

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

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)}\circ 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 defines 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.

Theorem(Implicit function theorem)

Let f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m be a continuous diffable function defined on a neighborhood U of the point (x_0;y_0) and such that f(x_0;y_0)=0