User:Foxjwill/Math stuff

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

where ${\displaystyle f}$ is a function of ${\displaystyle k}$ variables.

Example 1

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

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

Example 2

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

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

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

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

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

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

Because ${\displaystyle L}$ can't be negative, ${\displaystyle L=1}$.

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

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

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

The derivative of a polynomial,

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

can be defined as

${\displaystyle {\frac {d}{dt}}(a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})=a_{1}+a_{2}x+a_{3}x^{2}+\cdots +a_{n}x^{n-1}}$.

If we use the standard ordered basis

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

then

${\displaystyle (a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n})_{\mathbf {e} }}$

can be written as

${\displaystyle {\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}}$,

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

${\displaystyle {\frac {d}{dt}}:{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\mapsto {\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix}}}$.

Since

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

satisfies

${\displaystyle A{\begin{pmatrix}a_{0}\\a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\\0\end{pmatrix}}}$

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

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

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

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

A second degree linear ordinary differential equation is given by

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

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

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

Expanding ${\displaystyle (My)''}$ and setting it equal to

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

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

${\displaystyle M''a(x)=2M'b(x)}$

${\displaystyle u=M'}$

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

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

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

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

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

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

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

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

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

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

${\displaystyle dx:p\mapsto [1]}$

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

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

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

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

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

Multiplying both sides by ${\displaystyle [1]}$, we have

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

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

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