0708-1300/Class notes for Tuesday, September 11

From Drorbn
Jump to navigationJump to search
Announcements go here

In Small Scales, Everything's Linear

06-240-QuiltBeforeMap.png [math]\displaystyle{ \longrightarrow }[/math] 06-240-QuiltAfterMap.png
[math]\displaystyle{ z }[/math] [math]\displaystyle{ \mapsto }[/math] [math]\displaystyle{ z^2 }[/math]

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 [math]\displaystyle{ U }[/math], [math]\displaystyle{ V }[/math] and [math]\displaystyle{ W }[/math] be two normed finite dimensional vector spaces and let [math]\displaystyle{ f:V\rightarrow W }[/math] be a function defined on a neighborhood of the point [math]\displaystyle{ x }[/math].

Definition:

We say that [math]\displaystyle{ f }[/math] is differentiable (diffable) at [math]\displaystyle{ x }[/math] if there is a linear map [math]\displaystyle{ L }[/math] so that

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

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

Theorem

If [math]\displaystyle{ f:V\rightarrow W }[/math] and [math]\displaystyle{ g:U\rightarrow V }[/math] are diffable maps then the following asertions holds:

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

Implicit Function Theorem

Example Although [math]\displaystyle{ x^2+y^2=1 }[/math] does not defines [math]\displaystyle{ y }[/math] as a function of [math]\displaystyle{ x }[/math], in a neighborhood of [math]\displaystyle{ (0;-1) }[/math] we can define [math]\displaystyle{ g(x)=-\sqrt{1-x^2} }[/math] so that [math]\displaystyle{ x^2+g(x)^2=1 }[/math]. Furthermore, [math]\displaystyle{ g }[/math] is differentiable with differential [math]\displaystyle{ dg_{x}=\frac{x}{\sqrt{1-x^2}} }[/math]. 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

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

Theorem(Implicit function theorem)

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

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