User:Foxjwill/Math stuff

Total derivative

${\frac {\mathrm {d} f}{\mathrm {d} x}}={\frac {\partial f}{\partial x}}+\sum _{j=1}^{k}{\frac {\partial y_{j}}{\partial x}}{\frac {\partial f}{\partial y_{j}}},$

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

Example 1

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

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

Example 2

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

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

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

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

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

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

Continued fractions

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

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

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

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

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

Tetration and beyond

• $\forall a,b\in \mathbb {N}$

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

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 ${\frac {d}{dt}}$.

Wedge product

$\mathbf {u} =u_{1}\mathbf {e_{1}} +u_{2}\mathbf {e_{2}} +u_{3}\mathbf {e_{3}}$

$\mathbf {v} =v_{1}\mathbf {e_{1}} +v_{2}\mathbf {e_{2}} +v_{3}\mathbf {e_{3}}$

${\begin{aligned}\mathbf {u} \wedge \mathbf {v} &=(u_{1}v_{2}-v_{1}u_{2})\wedge \mathbf {e_{1}} +(u_{2}v_{3}-v_{2}u_{3})\wedge \mathbf {e_{2}} +(u_{3}v_{1}-v_{1}u_{3})\wedge \mathbf {e_{3}} \\&=u_{1}v_{1}\end{aligned}}$

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 $(My)''$ and setting it equal to

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

$\left\{{\begin{aligned}2M'&=Ma(x)\\M''&=Mb(x)\\\end{aligned}}\right.$

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

$u=M'$

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

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

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

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

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

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

${\begin{aligned}(My)''&=Mc(x)\\(My)'&={\int Mc(x)dx}+C_{1}\\My&={\int {\int Mc(x)dx}dx}+C_{1}x+C_{2}\\\end{aligned}}$

$y={\frac {{\int {\int {\int c(x)e^{2{\int {\frac {b(x)}{a(x)}}dx}}dx}dx}dx}+C_{1}x+C_{2}}{\int e^{2{\int {\frac {b(x)}{a(x)}}dx}}dx}}$

Differential example

The key to differentials is to think of $x$ as a function from some real number $p$ to itself; and $dx$ 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 $dx$ as

$dx:\mathbb {R} \rightarrow \mathbb {R} ^{1\times 1}$

$dx:p\mapsto [1]$

(As a side note, the value of $dx$, and similarly for all differentials, at $p$ is usually written $dx_{p}$.)

Without loss of generality, let's take the function $f(x)=x^{2}$. Differentiating, we have

${\frac {df}{dx}}=2x.$

Since we defined $dx_{p}$ as $[1]$ and $x(p)$ as $p$, we can rewrite the derivative as

$df_{p}[1]^{-1}=2x(p).\,$

Multiplying both sides by $[1]$, we have

$df_{p}=2x(p)[1].\,$

And voilà! We can say that for any function $f:\mathbb {R} \rightarrow \mathbb {R}$,

$df_{p}=\left[f'(x(p))\right]=f'(x(p))dx_{p}$