User:Rii'jeg'fkep'c/mathem

Because Wikipedia blocked by GFW(the Great firewall of China), I can't creat the page User:Rii'jeg'fkep'c/mathem in my simple English Wikipedia. So I write mathematical formula in here.

Basic calculation

$-(a)=-a\ ;\quad a-0=0$

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${\begin{aligned}&a+b=(a+b)\\&(-a)+(-b)=-(a+b)\\&a+(-b)=a-b=-(b-a)\\&a+0=a-0=a\\&a+b=b+a\\&(a+b)+c=a+b+c=a+(b+c)=(a+c)+b\\&a-b=a+(-b)\end{aligned}}$${\begin{aligned}&a\cdot b=b\cdot a=ab=ba\\&a\cdot 0=0\\&-a=(-1)\cdot a\\&(-1)\cdot (-1)=(-1)^{2}=1\\&-a\cdot b=-ab\\&(-a)\cdot (-b)=(-1)\cdot (-1)\cdot ab=ab\\&(a+b)\cdot c=(a+b)c=ac+ab\\&a\cdot (b\cdot c)=abc=(ab)c=(ac)b=(bc)a\\&a\div b={\frac {a}{b}}\end{aligned}}$${\begin{aligned}&b^{n}=\underbrace {b\cdot b\cdot \mathrm {etc.} \quad b\cdot b} _{\mathrm {n} }\\&b^{2}=b\cdot b\ ,\ b^{3}=b\cdot b\cdot b\\&b^{\tfrac {1}{2}}={\sqrt {b}}\\&b^{-n}={\frac {1}{b^{n}}}\\&b^{-1}={\frac {1}{b}}=1\div b\\&b^{0}=1\end{aligned}}$${\begin{aligned}&{\frac {A}{B}}={\frac {A\cdot M}{B\cdot M}}\qquad {\frac {A}{B}}={\frac {A\div N}{B\div N}}\ ,N\neq 0\\&{\frac {b_{1}}{a_{1}}}+{\frac {b_{2}}{a_{1}}}={\frac {b_{1}+b_{2}}{a_{1}}}\\&{\frac {b_{1}}{a_{1}}}+{\frac {b_{2}}{a_{2}}}={\frac {b_{1}a_{2}}{a_{1}a_{2}}}+{\frac {a_{1}b_{2}}{a_{1}a_{2}}}={\frac {a_{1}b_{2}+a_{2}b_{1}}{a_{1}a_{2}}}\\&{\frac {b_{1}}{a_{1}}}\cdot {\frac {b_{2}}{a_{2}}}={\frac {b_{1}b_{2}}{a_{1}a_{2}}}\\&{\frac {b_{1}}{a_{1}}}\div {\frac {b_{2}}{a_{2}}}={\frac {a_{2}b_{1}}{a_{1}b_{2}}}\end{aligned}}$

${\begin{aligned}&***+**=*****\\&******-****=**\\&***\times **=***+***=******\\&*******\div ***=**{\dots \dots }*\\&******\div **=***\\&(***)!=*\times **\times ***=******\\&1^{r_{+}}=1,r\in \mathrm {R} \\&\underbrace {******\cdots \cdots **} _{n}=n\cdot *=1^{n-1}+1^{n-2}+{\cdots \cdots }+1^{1}+1^{0}=\underbrace {1+1+\cdots \cdots +1} _{n}=n\end{aligned}}$

${\begin{aligned}&{\begin{aligned}(a+b)(c+d)&=a(c+d)+b(c+d)\\&=ac+ad+bc+bd\end{aligned}}\\&\\&(a+b)(a-b)=a^{2}-b^{2}\\&(a+b)^{2}=a^{2}+2ab+b^{2}\\&(a-b)^{2}=a^{2}-2ab+b^{2}\\&(a+b)^{2}-(a-b)^{2}=4ab\\&\\&{\frac {1}{n(n+1)}}={\frac {1}{n}}+{\frac {1}{n+1}}\\&{\sqrt {a}}\cdot {\sqrt {b}}={\sqrt {ab}}\\&{\sqrt {a(b+c)}}={\sqrt {a}}{\sqrt {b+c}}\end{aligned}}$

Sequence

${\begin{aligned}&a_{n}=a_{1}+(n-1)l\ ,\ \Sigma _{n}={\frac {n(a_{1}+a_{n})}{2}}=na_{1}+{\frac {(n-1)nl}{2}}\\&a_{n}=a_{1}q^{n-1}\ ,\ \Sigma _{n}={\frac {a_{1}(1-q^{n})}{1-q}}={\frac {a_{1}(q^{n}-1)}{q-1}}\\&a_{n}=n\ ,\ \Sigma _{n}={\frac {n(n+1)}{2}}\\&a_{n}=2n-1\ ,\ \Sigma _{n}=n^{2}\\&a_{n}=n^{2}\ ,\ \Sigma _{n}={\frac {n(n+1)(2n+1)}{6}}\end{aligned}}$

Equations

${\begin{aligned}&a+(b-c)=a+b-c\\&a-(b+c)=a-b-c\\&a-(b-c)=a-b+c\\&a(b\div c)=a\cdot b\div c={\frac {ab}{c}}\\&a\div (b\cdot c)=a\div b\div c={\frac {a}{bc}}\\&a\div (b\div c)=a\div b\cdot c={\frac {ac}{b}}\end{aligned}}$${\begin{aligned}&a=b\\&\therefore \\&a+c=b+c\\&a-c=b-c\\&a\cdot c=b\cdot c\\&a\div c=b\div c\\&a^{n}=b^{n}\\&\ln {a}=\ln {b}\\&n^{a}=n^{b}\\&\mathrm {etcetera} \end{aligned}}$${\begin{aligned}&{\begin{cases}a_{1}x+b_{1}y+c_{1}=0&\\a_{2}x+b_{2}y+c_{2}=0\end{cases}}\\&y=-{\frac {a_{1}}{b_{1}}}x-{\frac {c_{1}}{b_{1}}}\\&a_{2}x+b_{2}\left[-{\frac {a_{1}}{b_{1}}}x-{\frac {c_{1}}{b_{1}}}\right]+c_{2}=0\\&a_{2}b_{1}x-a_{1}b_{2}x-b_{2}c_{1}+b_{1}c_{2}=0\\&(a_{1}b_{2}-a_{2}b_{1})x=(b_{1}c_{2}-b_{2}c_{1})\\&x_{0}={\frac {b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}\\&\\&{\frac {b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}\cdot a_{1}+b_{1}y+c_{1}=0\\&{\begin{aligned}b_{1}y&=-{\frac {a_{1}b_{1}c_{2}-a_{1}b_{2}c_{1}+a_{1}b_{2}c_{1}-a_{2}b_{1}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}\\&=-{\frac {a_{1}b_{1}c_{2}-a_{2}b_{1}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}\end{aligned}}\\&y_{0}=-{\frac {a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}\\&\\&{\begin{cases}x_{0}={\frac {b_{1}c_{2}-b_{2}c_{1}}{M}}&\\y_{0}=-{\frac {a_{1}c_{2}-a_{2}c_{1}}{M}}\end{cases}}\\&M=a_{1}b_{2}-a_{2}b_{1}\end{aligned}}$

Geology for calculation

pythagorean theorem

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${\begin{aligned}&{\begin{aligned}c^{2}&=4\cdot {\tfrac {1}{2}}ab+(b-a)^{2}\\&=2ab+b^{2}-2ab+a^{2}\\&=a^{2}+b^{2}\end{aligned}}\\&{\begin{aligned}c^{2}&=(a+b)^{2}-4\cdot {\tfrac {1}{2}}ab\\&=a^{2}+2ab+b^{2}-2ab\\&=a^{2}+b^{2}\end{aligned}}\end{aligned}}$

Descartes' geology

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${\begin{aligned}&\\&y-y_{0}=k(x-x_{0})\\&y=kx+b\\&{\frac {y-y_{1}}{y_{2}-y_{1}}}={\frac {x-x_{1}}{x_{2}-x_{1}}}\\&{\frac {x}{a}}+{\frac {y}{b}}=1\\&Ax+By+C=0\end{aligned}}$

number e

${\begin{aligned}&\mathrm {e} =\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}\\&=\lim _{n\to \infty }\sum _{k=0}^{n}\mathrm {C} _{n}^{k}1^{n-k}\left({\frac {1}{n}}\right)^{k}\\&=\lim _{n\to \infty }\left[\mathrm {C} _{n}^{0}1^{n}\left({\frac {1}{n}}\right)^{0}+\mathrm {C} _{n}^{1}1^{n-1}\left({\frac {1}{n}}\right)^{1}+\mathrm {C} _{n}^{2}1^{n-2}\left({\frac {1}{n}}\right)^{2}+\mathrm {C} _{n}^{3}1^{n-3}\left({\frac {1}{n}}\right)^{3}+...+\mathrm {C} _{n}^{n}1^{0}\left({\frac {1}{n}}\right)^{n}\right]\\&=\lim _{n\to \infty }\left[1\times 1+n\times {\frac {1}{n}}+{\frac {n!}{\left(n-2\right)!2!}}\times {\frac {1}{n^{2}}}+{\frac {n!}{\left(n-3\right)!3!}}\times {\frac {1}{n^{3}}}+...+1\times {\frac {1}{n^{n}}}\right]\\&=\lim _{n\to \infty }\left[1+1+{\frac {n\times \left(n-1\right)}{2!\times n^{2}}}+{\frac {n\times \left(n-1\right)\left(n-2\right)}{3!\times n^{3}}}+...+{\frac {1}{n^{n}}}\right]\\&={\frac {1}{\Pi (0)}}+{\frac {1}{\Pi (1)}}+{\frac {1}{\Pi (2)}}+{\frac {1}{\Pi (3)}}+...\\&=\sum _{n=0}^{\infty }{\frac {1}{\Pi (n)}}\end{aligned}}$

${\begin{aligned}&\Pi (0)=1\\&\Pi (\mathrm {n} )=1\cdot 2\cdot 3\cdot ...\mathrm {n} =\textstyle \prod _{k=1}^{n}\displaystyle k=\mathrm {n(n-1)} !\\&\mathrm {C_{n}^{m}} ={\frac {\mathrm {A_{n}^{m}} }{\mathrm {m!} }}={\frac {\mathrm {n!} }{\mathrm {(n-m)!m!} }}\end{aligned}}$

Logarithm

${\begin{aligned}&\ln(a)+\ln(b)=\ln(ab)\\&\ln(a)-\ln(b)=\ln({\frac {a}{b}})\\&\ln 1=0\\&\ln e=1\\&\ln a^{n}=n\ln a\\&\log _{a}b={\frac {\ln b}{\ln a}}\\&\lg n={\frac {\ln(n)}{\ln 10}}\\&a^{n}=e^{n\ln a}\\&\ln n=\ln {\frac {n}{n-1}}+\ln {(n-1)}=2\sum _{k=1}^{\infty }[{\frac {1}{2k-1}}({\frac {1}{2n-1}})^{2k-1}]+\ln(n-1)\\&\log _{a}{N}=x|_{a^{x}=N}\\&\log _{a}1=0\\&\log _{a}{a}=1\\&a^{\log _{a}{N}}=N\end{aligned}}$${\begin{aligned}&\log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}\\&m=\log _{a}{M},n=\log _{a}{N}\\&a^{m}\cdot a^{n}=a^{m+n}\quad \because a^{\log _{a}{N}}=N\\&\therefore MN=a^{m+n}\\&m+n=\log _{a}{(MN)}\\&\log _{a}{M}+\log _{a}{N}=\log _{a}{(MN)}\\&\log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}\\&m=\log _{a}{M},n=\log _{a}{N}\\&a^{m}\div a^{n}=a^{m-n}\quad \because a^{\log _{a}{N}}=N\\&\therefore {\frac {M}{N}}=a^{m-n}\\&m-n=\log _{a}{\frac {M}{N}}\\&\log _{a}{M}-\log _{a}{N}=\log _{a}{\frac {M}{N}}\\&\log _{a}{N^{m}}=m\log _{a}{N}\end{aligned}}$${\begin{aligned}&\log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}\\&\log _{a}{b}=m\\&a^{m}=b\\&\log _{c}{b}=\log _{c}{a^{m}}=m\log _{c}{a}\\&m=\log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}={\frac {\log _{c}{a^{m}}}{\log _{c}{a}}}={\frac {m\log _{c}{a}}{\log _{c}{a}}}=m\\&\log _{a}{b}={\frac {\log _{c}{b}}{\log _{c}{a}}}\\&\log _{m}{n}={\frac {1}{\log _{n}{m}}}\\&\log _{a^{m}}{b^{n}}={\frac {n}{m}}\log _{a}{b}\\&e^{n}=10^{\frac {n}{\ln 10}}\end{aligned}}$

Derivative series

Tends to infinitesimal

${\begin{aligned}&a_{n}={\frac {1}{n}}\\&\exists \varepsilon \\&{\frac {1}{n}}<\varepsilon \\&1<n\varepsilon \\&n\varepsilon >1\\&n>{\frac {1}{\varepsilon }}\\&\lim _{n\to \infty }{\frac {1}{n}}=0\end{aligned}}$