本论坛支持 LaTEX 数学文本输入和显示.

elim

输入 \((ax^2+bx+c=0\,(a\ne 0))\implies(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a})\
网页显示:
\((ax^2+bx+c=0,\;(a\ne 0))\implies(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a})\)

LaTEX 要诀:

elim

新手可以跟贴试试直接键入数学式, 告别数学公式帖图.....

elim

王守恩何不率先试试 \(\Omega - \bigcup_{k=1}^n E_k = \bigcap_{k=1}^n (\Omega-E_k)\)

elim

Taylor 定理: 设开区间\(\,I\ni a,\;f\in\mathscr{C}^{k+1}(I),\underset{\,}{\;}\)则
\(\small f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\cdots+\frac{f^{(k)}(a)}{k!}(x-a)^k+{\displaystyle\int_a^x}\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt\)
证：\(\small\;f(x)=f(a)-{\small\displaystyle\int_a^x} f'(t)d(x-t)\overset{(\dagger)}{=}f(a)+\frac{f'(a)}{1!}(x-a)+\underset{\,}{\small\displaystyle\int_a^x\frac{f''(t)}{1!}}(x-t)dt\)
\(\qquad\cdots \overset{(\ddagger)}{=}\small f(a)+\frac{f'(a)}{1!}(x-a)+\cdots+\frac{f^{(k)}(a)}{k!}(x-a)^k+{\small\displaystyle\int_a^x\frac{f^{(k+1)}(t)}{k!}}(x-t)^kdt.\)

注记：\(\small(\dagger),\;\;(\ddagger)\) 依次表示“分部积分”及“以此类推”.       

elim

简单说来，博主或贴文作者递交的文字，包括LaTEX 内容会基本不变地存入网站的数据库．当读者点击链接或在浏览器输入帖子或博文网址时，网站软件会把相应的贴子，博文内容从数据库取出，嵌入被称为模版的网页代码（文本）的适当位置以完成整个网页的实时汇编，并通过网站服务器输送给读者的浏览器．浏览器根据网页代码“画出”网页．模板中如果有一段关于处理LaTEX的指令，告诉浏览器碰到某种标示，例如\(及\)之间的内容属于LaTEX内容，须到哪里获得进一步的数学符号字体和排版指令，那么我们就说相应的网站是支持LaTEX的．所以问题的关键在于服务器方微博或论坛软件使用的模板是否支持LaTEX.  而这只要写一段LaTEX内容一试就知道了．

新版的论坛软件都提供启用或关闭MathJax 的管理员设置. 想来微博软件应该也如此. 当然, 所有这一切对过于陈旧的浏览器都没有意义. 例如 Internet Explorer 8 以下的就不行.

elim

\[f(a)=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-a}dz\]

请用鼠标右键点击以上数学表达式,选择 数式显示形式 / TeX 命令 以便看 LaTEX 代码.

乘简

目前我测试了电脑上的浏览器有：IE，Firefox，Chrome，发现IE显示的效果都很不错的。Chrome 与 Firefox开始是点阵字体，后来改成用AVG显示就好很多，只是分辩率高的电脑，要右键--数学设置--绽放所有数学，输入130，放大显示就好了。。

手机浏览器试了Safari，Chrome，Firefox，夸克，UC，等都无问题。。

希望大家多打点各种各样的数学公式来进行测试。。。

fungarwai

我隨便貼過來看看能顯示多少

[hide=Chapter 0 Basic properties of finite difference]

Definition

Define \(\Delta p(x)=p(x+1)-p(x)\) which is called Finite difference

Linear operator

\(\Delta\) is a linear operator such that

\(\begin{cases}\Delta (p_1(x)+p_2(x))=\Delta p_1(x)+\Delta p_2(x)\\
\Delta (kp(x))=k\Delta p(x)\end{cases}\)

n th power of finite difference

\(\Delta^n p(x)=\Delta (\Delta^{n-1} p(x)
=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k} p(x+k)\)

General results of high power finite difference

Define the degree of polynomial \(p(k)\) as \(\deg(p)\), for the polynomial with leadiing coefficient 1 :

\(\deg(p)=n\Rightarrow \begin{cases}\Delta^{n+1}p(x)=0\\
\Delta^n p(x)=n!\end{cases}\)[/hide]

[hide=Chapter 1 Newton's series]

Every polynomial can be represented by each degree of its finite difference

\(p(x)=\sum_{m=0}^{deg(p)} \binom{x-a}{m}\Delta^m p(a)\) which is known as Newton's series

[hide=Proof]Let \(p(x)=\sum_{m=0}^{deg(p)} c_m \binom{x-a}{m}=c_0+c_1\binom{x-a}{1}+c_2\binom{x-a}{2}+\dots+c_{deg(p)}\binom{x-a}{deg(p)}\)

\(p(a)=c_0\)

\(\Delta \binom{x-a}{k}=\binom{x+1-a}{k}-\binom{x-a}{k}=\binom{x-a}{k-1}\)

\(\Delta p(x)=c_1+c_2\binom{x-a}{1}+c_3\binom{x-a}{2}+\dots+c_{deg(p)}\binom{x-a}{deg(p)-1}\)

\(\Delta p(a)=c_1\)

\(\Delta^m p(k)=c_m+c_{m+1}\binom{x-a}{1}+c_{m+2}\binom{x-a}{2}+\dots+c_{deg(p)}\binom{x-a}{deg(p)-m}\)

\(\Delta^m p(a)=c_m\)[/hide]

[hide=Example]\(p(x)=x^2,\Delta p(x)=2x+1,\Delta^2 p(x)=2\)

\(p(x)=a^2+(2a+1)\binom{x-a}{1}+2\binom{x-a}{2}\)[/hide][/hide]

[size=150]Chapter 2 Summation with polynomial only

\(\sum_{k=1}^n p(k)=\sum_{m=0}^{deg(p)} \binom{n}{m+1}\Delta^m p(1)\)

[hide=Proof]\(\sum_{k=1}^n \binom{k-1}{m}=\binom{n}{m+1}\) which is called Hockey-stick identity

\(\sum_{k=1}^n p(k)=\sum_{m=0}^{deg(p)} \Delta^m p(1) \sum_{k=1}^n  \binom{x-1}{m}=\sum_{m=0}^{deg(p)} \binom{n}{m+1}\Delta^m p(1)\)[/hide]

[hide=Example]\(p(k)=k,~\Delta p(k)=1\)

\(\sum_{k=1}^n k=\binom{n}{1}+\binom{n}{2}\)

\(p(k)=k^2,~\Delta p(k)=2k+1,~\Delta^2 p(k)=2\)

\(\sum_{k=1}^n k^2=\binom{n}{1}+3\binom{n}{2}+2\binom{n}{3}\)

\(p(k)=k^3,~\Delta p(k)=3k^2+3k+1,~\Delta^2 p(k)=6k+6,~\Delta^3 p(k)=6\)

\(\sum_{k=1}^n k^3=\binom{n}{1}+7\binom{n}{2}+12\binom{n}{3}+6\binom{n}{4}\)[/hide]

[size=150]Chapter 3 Summation with geometric sequence

\(\sum_{k=1}^n p(k)q^{k-1}=f(n)q^n-f(0)\), where \(f(n)
=\frac{p(n)}{q-1}+\frac{1}{(q-1)^2}\sum_{k=1}^{deg(p)} \frac{(-1)^kq^{k-1}}{(q-1)^{k-1}}\Delta^k(p(n))
=\frac{1}{q-1}\sum_{k=0}^{deg(p)} (\frac{-q}{q-1})^k\Delta^k p(n+1)\)

[hide=Proof]\(\Delta (\sum_{k=1}^n p(k)q^{k-1})=\Delta(f(n)q^n-f(0))\)

\(p(n+1)q^n=f(n+1)q^{n+1}-f(n)q^n\)

\(p(n+1)=qf(n+1)-f(n)\)

\((I+\Delta)p(n)=q(I+\Delta)f(n)-f(n)=[(q-1)I+q\Delta]f(n)\)

\(f(n)=\frac{I+\Delta}{(q-1)I+q\Delta}p(n)=\frac{1}{(q-1)I+q\Delta}p(n+1)\)

\(\frac{I+\Delta}{(q-1)I+q\Delta}=\frac{I+\Delta}{q-1}\sum_{k=0}^{deg(p)} (\frac{-q}{q-1})^k \Delta^k\)

\(=\frac{1}{q-1}[\sum_{k=0}^{deg(p)} (\frac{-q}{q-1})^k \Delta^k+\sum_{k=0}^{deg(p)} (\frac{-q}{q-1})^k \Delta^{k+1}]
=\frac{1}{q-1}[\sum_{k=0}^{deg(p)} (\frac{-q}{q-1})^k \Delta^k+\sum_{k=1}^{deg(p)} (\frac{-q}{q-1})^{k-1} \Delta^k]\)

\(=\frac{1}{q-1}I+\frac{1}{q-1}\sum_{k=1}^{deg(p)} [(\frac{-q}{q-1})^k+(\frac{-q}{q-1})^{k-1}] \Delta^k
=\frac{1}{q-1}I-\frac{1}{(q-1)^2}\sum_{k=1}^{deg(p)} (\frac{-q}{q-1})^{k-1} \Delta^k\)

\(=\frac{1}{q-1}I+\frac{1}{(q-1)^2}\sum_{k=1}^{deg(p)} \frac{(-1)^k q^{k-1}}{(q-1)^{k-1}} \Delta^k\)[/hide]

[hide=Example]\(p(k)=k^2,\Delta p(k)=2k+1,\Delta^2 p(k)=2\)

\(f(n)=\frac{n^2}{q-1}-\frac{2n+1}{(q-1)^2}+\frac{2}{(q-1)^3}q\)

\(\sum_{k=1}^n k^2 q^{k-1}=(\frac{n^2}{q-1}-\frac{2n+1}{(q-1)^2}+\frac{2}{(q-1)^3}q)q^n+\frac{1}{(q-1)^2}-\frac{2}{(q-1)^3}q\)

When \(|q|<1\), \(\sum_{k=1}^{\infty} k^2 q^{k-1}=\frac{1}{(q-1)^2}-\frac{2}{(q-1)^3}q=\frac{1+q}{(1-q)^3}\)
[/hide]

[hide=Chapter 4 Summation with geometric sequence and factorial]\(\sum_{n=0}^{\infty} \frac{p(n)}{n!}x^n=e^x\sum_{m=0}^{deg(p)} \frac{\Delta^m p(0)}{m!}x^m\)

[hide=Proof]\(\sum_{n=0}^\infty \frac{p(n)}{n!}x^n
=\sum_{m=0}^{deg(p)} \sum_{n=0}^\infty \binom{n}{m}\frac{\Delta^m p(0)}{n!}x^n\)

\(=\sum_{m=0}^{deg(p)} \frac{\Delta^m p(0)}{m!}x^m\sum_{n=m}^\infty \frac{1}{(n-m)!}x^{n-m}
=e^x \sum_{m=0}^{deg(p)} \frac{\Delta^m p(0)}{m!}x^m\)

This is related to a significant formula from Schaum's Outline of Calculus of Finite Differences and Difference Equations

\(\sum_{k=0}^\infty u_k v_k x^k=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)\)

\(\sum_{k=0}^\infty u_k v_k x^k
=\sum_{k=0}^\infty \sum_{l=0}^\infty \binom{k}{l}\Delta^l u_0 v_k x^k
=\sum_{k=0}^\infty \sum_{l=k}^\infty \binom{l}{k}\Delta^k u_0 v_l x^l\)
\(=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\sum_{l=k}^\infty l(l-1)\dots(l-k+1) v_l x^{l-k}\)
\(=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)\)[/hide]

[hide=Example]\(p(k)=2k+1,\Delta p(k)=2\)

\(\sum_{n=0}^\infty \frac{p(n)}{n!}x^n=(1+2x)e^x\)[/hide][/hide]

[hide=Chapter 5 Summation with Harmonic number]Define \(H_n=\sum_{k=1}^n \frac{1}{k}\) as Harmonic number

\(\sum_{k=1}^n H_k p(k)=(\sum_{m=0}^{deg(p)} \binom{n+1}{m+1} \Delta^m p(0))H_n-\sum_{m=0}^{deg(p)}\frac{1}{m+1}\binom{n}{m+1}\Delta^m p(0)\)

[hide=Proof]\(\sum_{k=1}^n H_k p(k)=\sum_{k=1}^n \sum_{t=1}^k \frac{p(k)}{t}
=\sum_{t=1}^n \sum_{k=t}^n p(k)
=\sum_{t=1}^n \frac{1}{t} (\sum_{k=1}^n p(k)-\sum_{k=1}^{t-1} p(k))\)

\(=(\sum_{m=0}^{deg(p)}\binom{n+1}{m+1}\Delta^m p(0))H_n-
\sum_{m=0}^{deg(p)}\sum_{t=1}^n \frac{1}{t}\binom{t}{m+1}\Delta^m p(0)\)

\(=(\sum_{m=0}^{deg(p)}\binom{n+1}{m+1}\Delta^m p(0))H_n-
\sum_{m=0}^{deg(p)}\frac{1}{m+1}\binom{n}{m+1}\Delta^m p(0)\)[/hide]

[hide=Example]\(p(k)=2k+1,\Delta p(k)=2\)

\(\sum_{k=1}^n H_k p(k)=(n+1+(n+1)n)H_n-(n+\frac{1}{2}n(n-1))
=(n+1)^2 H_n-\frac{1}{2}n(n+1)\)[/hide][/hide]

elim

用 LaTEX 书写数学内容被大部分专业刊物视为必须. 一般花上几个小时就可以基本掌握. 值得学习!

ysr

\((ax^2+bx+c=0\,(a\ne 0))\implies(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a})\

我帖一个试试，哈哈哈，复制1楼的！