User:Foxjwill/Math stuff

From Wikipedia, the free encyclopedia

1 Total derivative
- 1.1 Example 1
- 1.2 Example 2
2 Continued fractions
3 Tetration and beyond
4 Polynomials and their derivatives
5 Wedge product
6 General second degree linear ordinary differential equation
7 Differential example

[edit] Total derivative

$\frac{\mathrm df}{\mathrm dx}= \frac{\partial f}{\partial x}+\sum_{j=1}^k \frac{\partial y_j}{\partial x}\frac{\partial f}{\partial y_j},$

where $f$ is a function of $k$ variables.

[edit] Example 1

$\frac{\mathrm df}{\mathrm dx}=\frac{\partial f}{\partial x} + \frac{\partial y}{\partial x} \frac{\partial f}{\partial y},$

where $f: \Bbb{R}, \Bbb{R} \to \Bbb{R}$ .

[edit] Example 2

Given $f(x,y) = x^3 - y^2\,,$ find $\textstyle \frac{\mathrm df}{\mathrm dx}.$

Begin with the definition of the total derivative: ${\textstyle \frac{\mathrm df}{\mathrm dx}=\frac{\partial f}{\partial x} + \frac{\partial y}{\partial x} \frac{\partial f}{\partial y} }$ . Notice that in order to continue, we need to calculate $\textstyle \frac{\partial f}{\partial x},\, \textstyle \frac{\partial y}{\partial x},\,$ and $\textstyle \frac{\partial f}{\partial y}.$

$\begin{align} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}\left( x^3 \right)\\ &= 3x^2 \end{align}$

$\begin{align} \frac{\partial f}{\partial y} &= \frac{\partial}{\partial x}\left( y^2 \right)\\ &= -2y \end{align}$

$\begin{align} y^2 &= x^3\\ 2y\frac{\partial y}{\partial x} &= 3x^2\\ \frac{\partial y}{\partial x} &= \frac{3x^2}{2y} \end{align}$

Plugging the results into the definition, ${\textstyle \frac{\mathrm df}{\mathrm dx} = 3x^2 + \frac{3x^2}{-2y} \left( 2y \right)}$ , we find that ${\textstyle \frac{\mathrm df}{\mathrm dx} = 0}.$

[edit] Continued fractions

$L = 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{\ddots \,} } } }$

$\frac{1}{L} = \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{\ddots \,} } } } } = L$

$\begin{align} L^2 &= 1\\ L &= \pm \sqrt{1} \end{align}$

Because $L$ can't be negative, $L = 1$ .

$\therefore\ 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{ 0 + \cfrac{1}{\ddots \,} } } } = 1$

[edit] Tetration and beyond

$\forall a,b \in \mathbb{N}$

$\begin{array}{lcll} a\cdot b &=& a + a\cdot (b-1) &,a\cdot 0 = 0\\ a^b &=& a\cdot a^{b-1} &,a^0 = 1\\ a\uparrow\uparrow b &=& a^{a\uparrow\uparrow (b-1)} &,a\uparrow\uparrow 0 = 1 \end{array}$

[edit] Polynomials and their derivatives

The derivative of a polynomial,

$\frac{d}{dt}: \mathbf{P}^n \to \mathbf{P}^n$ ,

can be defined as

$\frac{d}{dt}(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = a_1 + a_2 x + a_3 x^2 + \cdots + a_{n} x^{n-1}$ .

If we use the standard ordered basis

$\mathbf{e} = \{1,x,x^2,\ldots,x^n\}$ ,

then

$(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n)_\mathbf{e}$

can be written as

$\begin{pmatrix} a_0\\ a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix}$ ,

and $\frac{d}{dt}$ as

$\frac{d}{dt}: \begin{pmatrix} a_0\\ a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix} \mapsto \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n\\ 0 \end{pmatrix}$ .

Since

$A=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1\\ 0 & 0 & 0 & \cdots & 0\\ \end{pmatrix}$

satisfies

$A \begin{pmatrix} a_0\\ a_1\\ a_2\\ \vdots\\ a_n \end{pmatrix} = \begin{pmatrix} a_1\\ a_2\\ \vdots\\ a_n\\ 0 \end{pmatrix}$

, $A$ represents $\frac{d}{dt}$ .

[edit] Wedge product

$\mathbf{u} = u_1 \mathbf{e_1} + u_2 \mathbf{e_2} + u_3 \mathbf{e_3}$

$\mathbf{v} = v_1 \mathbf{e_1} + v_2 \mathbf{e_2} + v_3 \mathbf{e_3}$

$\begin{align} \mathbf{u}\wedge\mathbf{v} &= (u_1 v_2 - v_1 u_2)\wedge\mathbf{e_1} + (u_2 v_3 - v_2 u_3)\wedge\mathbf{e_2} + (u_3 v_1 - v_1 u_3)\wedge\mathbf{e_3}\\ &= u_1 v_1 \end{align}$

[edit] General second degree linear ordinary differential equation

A second degree linear ordinary differential equation is given by

$y'' + a(x) y' + b(x) y = c(x).\,$

One way to solve this is to look for some integrating factor, $M$ , such that

$My'' + Ma(x)y' + Mb(x)y = (My)'' = Mc(x).\,$

Expanding $(M y)''$ and setting it equal to

$My'' + Ma(x)y' + Mb(x)y = My'' + 2My' + M''y\,$

$\left \{ \begin{align} 2M' &= M a(x)\\ M'' &= M b(x)\\ \end{align} \right.$

$M'' a (x) = 2 M' b (x)$

$u = M'$

$\frac{du}{dx} a(x) = 2u b(x)$

$\frac{1}{2}{\int \frac{du}{u} } = {\int \frac{b(x)}{a(x)} dx}$

$\frac{1}{2} \ln u = {\int \frac{b(x)}{a(x)} dx}$

$u = e^{2{\int \frac{b(x)}{a(x)} dx}}$

$M' = e^{2{\int \frac{b(x)}{a(x)} dx}}$

$M = {\int e^{2{\int \frac{b(x)}{a(x)} dx}} dx}$

$\begin{align} (My)'' &= Mc(x)\\ (My)' &= {\int Mc(x) dx} + C_1\\ My &={\int {\int Mc(x) dx}dx} + C_1 x + C_2\\ \end{align}$

$y = \frac{{\int {\int {\int c(x) e^{2{\int \frac{b(x)}{a(x)} dx}} dx} dx}dx} + C_1 x + C_2}{{\int e^{2{\int \frac{b(x)}{a(x)} dx}} dx}}$

[edit] Differential example

The key to differentials is to think of $x$ as a function from some real number $p$ to itself; and $d x$ as a function of some that same real number $p$ to a linear map $\mathbf{P}: \mathbb{R} \mapsto \mathbb{R}.$ Since all linear maps from $\mathbb{R}$ to $\mathbb{R}$ can be written as a $1\times 1$ matrix, we can define $\mathbf{P}$ as $[p]$ and $d x$ as