0708-1300/Class notes for Tuesday, September 11

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In Small Scales, Everything's Linear

 ${\displaystyle \longrightarrow }$ ${\displaystyle z}$ ${\displaystyle \mapsto }$ ${\displaystyle z^{2}}$

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}]


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

Definition:

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

${\displaystyle \lim _{h\rightarrow 0}{\frac {|f(x+h)-f(x)-L(h)|}{|h|}}=0.}$

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

Theorem

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

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

Implicit Function Theorem

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

Notation

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

Definition

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

Theorem(Implicit function theorem)

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

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