z***h
发帖数: 11 | 1 下周面试，有谁了解关于 frac water 的处理工艺、设备及发展前景等方面，网上的资
料比较少，这里高人多，谢谢！ |
|
g****g
发帖数: 1828 | 2 In probability theory, the normal (or Gaussian) distribution, is a
continuous probability distribution that is often used as a first
approximation to describe real-valued random variables that tend to cluster
around a single mean value. The graph of the associated probability density
function is “bell”-shaped, and is known as the Gaussian function or bell
curve:[nb 1]
f(x) = \tfrac{1}{\sqrt{2\pi\sigma^2}}\; e^{ -\frac{(x-\mu)^2}{2\sigma^2}
},
where parameter μ is the mean (location of the pe... 阅读全帖 |
|
X*7
发帖数: 40 | 3 谢了，但是我不太明白您给的提示，比如矩阵是这样的
begin{pmatrix}
frac{frac{1}{2tau^4}}{frac{1}{2tau^4} frac{tr[WVWV]}{2tau^4}-left (frac{tr[
WV]}{2tau^4}right )^2}&frac{-frac{1}{2tau^4}tr[WV]}{frac{1}{2tau^4} frac{tr[
WVWV]}{2tau^4}-left (frac{tr[WV]}{2tau^4}right )^2}&text{Huge{0}} \
frac{-frac{1}{2tau^4}tr[WV]}{frac{1}{2tau^4} frac{tr[WVWV]}{2tau^4}-left (
frac{tr[WV]}{2tau^4}right )^2}&frac{frac{1}{2tau^4}tr[WVWV]}{frac{1}{2tau^4}
frac{tr[WVWV]}{2tau^4}-left (frac{tr[WV]}{2tau^4}right )^2}&text{Huge{0}}\
text{Huge{0}... 阅读全帖 |
|
A**u
发帖数: 2458 | 4 应该可以解，
但我前面算错了
前面写的
dS(t,w) = (S_t + 0.5 S_xx) dt + S_x dw
ds = a dt + b S dw
所以
1 S_t + 0.5 S_xx = a
2 S_x = bS
由 2 S = exp(bx) + c(t)
1. c'(t) + 0.5 b * b * exp(bx) = a
这好象没解
这个题好想，我也看到过，有人贴出来，大概就这么做的
________________________________________________
方程还是不变.
2--> S = c(t)exp(bx)
1--> (c'(t) + 0.5 b*b*c(t) exp(bx) = a
所以 c'(t) + 0.5b*b*c(t) = a * exp(-bx).
这时 y' + p(t)y = g(t) 形式, p(t) = 0.5b*b, g(t) = a * exp(-bx)
乘以 z(t) 两边
z(t) y' + z(t)p(t) y = z(t) g(t)
有 z'(t) = z(t)p(t).
所以z(t) = exp(... 阅读全帖 |
|
g****g
发帖数: 1828 | 5 标准差（Standard Deviation），在概率统计中最常使用作为统计分布程度（
statistical dispersion）上的测量。标准差定义为方差的算术平方根，反映组内个体
间的离散程度。测量到分布程度的结果，原则上具有两种性质：
1. 为非负数值，
2. 与测量资料具有相同单位。
一个总量的标准差或一个随机变量的标准差，及一个子集合样品数的标准差之间，有所
差别。其公式如下所列。
标准差的观念是由卡尔·皮尔逊 (Karl Pearson)引入到统计中。
目录
[隐藏]
* 1 阐述及应用
* 2 标准差的定义及简易计算公式
o 2.1 标准计算公式
o 2.2 简化计算公式
o 2.3 随机变量的标准差计算公式
o 2.4 样本标准差
o 2.5 连续随机变量的标准差计算公式
o 2.6 标准差的性质
* 3 范例
* 4 正态分布的规则
* 5 标准差与平均值之间的关系
* 6 几何学解释
... 阅读全帖 |
|
p*****n
发帖数: 143 | 6 Let $X \sim N(0,1)$, $Y \sim \chi^2(n)$. Let $U=\frac{X}{\sqrt{\frac{Y}{n}}
}$, find the distribution of $U$.\vspace{1em}
$Y \sim \chi^2(n)$: \[f_{Y}(y)=\frac{1}{{\Gamma \left( {\frac{n}{2}} \right
)}}{y^{\frac{n}{2} - 1}}{\left( {\frac{1}{2}} \right)^{\frac{n}{2}}}{e^{ - \
frac{y}{2}}}\]
Let $V= Y$
\[
\begin{cases} X = U \sqrt{\frac{V}{n}}
\
Y = V
\end{cases}
\]
\[{\mathbf{J}} = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}} & {\frac{{\partial x}}{{\partial v}}}
\
|
|
o**a
发帖数: 76 | 7 这里是tex文件，可以直接latex编译:
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{align*}
&\int_0^{\infty} \frac{2\eta}{\eta^2+z^2} \cdot \frac{d\eta}{e^{2\pi\eta}-1} \
\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi}{e^{2
\pi\eta}-1} d\eta \\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi e^{-2
\pi\eta}}{1-e^{-2\pi\eta}} d\eta\\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot d(\log(1-e^{-2\
pi\eta} |
|
d******a
发帖数: 32122 | 8 ☆─────────────────────────────────────☆
SetTimer (RumorKiller) 于 (Wed Jun 20 08:59:26 2012, 美东) 提到:
一句话解释漂浮：地球引力全部用于产生速度改变(就mg=F=ma；这个速度改变，主要
指方向沿椭圆的改变）
一句话解释什么是文科生心中的“重力”：地球引力使人体对地面产生压力，其
反作用力使文科生感觉到重力；在空间站，这个重力（i.e., 地球引力）还是存在，只
是不用于小将老将废柴压地面，而是用于速度方向改变（将轨道简化为正圆）
一句话解释失重和超重：Nmg超重。N在这里是压力或拉力
一句话解释，为什么空间站不掉下来：因为mg刚好等于ma，所以没有多余的力干别的了，
这也是为什么地球不掉到太阳上，月亮不掉到地球上
一个词解释，为什么不提供能量的话最终还是要掉下来：大气。这个所需能量不比
你家空调大
☆─────────────────────────────────────☆
cellneuron (cell2) 于 (Wed Jun 20 09:59:3... 阅读全帖 |
|
e**n
发帖数: 478 | 9 Is this you want?
\begin{eqnarray*}
a & = & b+c+d+e+\sqrt{f^{2}+\frac{1}{\sqrt{g^{\frac{h^{2}+i^{2}}{2}}}}}\\
& & +\left(\sqrt{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}+\sqrt
{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}\right)^{3}\\
& & +\sqrt{ff^{2}+\frac{1}{\sqrt{gg^{\frac{h^{2}+i^{2}}{2}}}}}+\cdots\end{
eqnarray*} |
|
n***p
发帖数: 7668 | 10 让我尝试着解释一下。 It is just about symmetry.
Let r =|y-x|. Then
\frac{\partial r}{\partial y_i} = (y_i-x_i)/r
and
\frac{\partial r}{\partial x_i} = (x_i- y_i)/r.
That is
\frac{\partial r}{\partial y_i}
= - \frac{\partial r}{\partial x_i}.
For any function f(r), if you take partial derivative in
y, it is just chain rule. Say
\frac{\partial}{\partial y_i} f(r)
= f'(r) \frac{\partial r}{\partial y_i}
= - f'(r)\frac{\partial r}{\partial x_i}
= - \frac{\partial}{\partial x_i} f(r).
When you take... 阅读全帖 |
|
a*****y
发帖数: 33185 | 11 http://en.wikipedia.org/wiki/Pi
Second millennium AD
Until the second millennium AD, estimations of π were accurate to fewer
than 10 decimal digits. The next major advances in the study of π came with
the development of infinite series and subsequently with the discovery of
calculus, which permit the estimation of π to any desired accuracy by
considering sufficiently many terms of a relevant series. Around 1400,
Madhava of Sangamagrama found the first known such series:
{\pi} = 4\sum^\infty_... 阅读全帖 |
|
D**o
发帖数: 2653 | 12 注意作者 \author{YHBKJ}
Atiyah-Bott Localization 1
2012-09-05 09:24:19
\documentclass[a4paper,12pt]{article}
\usepackage{amsfonts}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{CJK,graphicx}
\usepackage{amscd}
\usepackage{amssymb}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{lemma}{Lemma}[section]
\begin{document}
\title{\textbf{\Huge{Atiyah-Bott Localization 1}}}\author{YHBKJ}\date{}\
maketitle
\begin{ab... 阅读全帖 |
|
n******d
发帖数: 244 | 13 We need to solve $x^12=2^x$, first,
$x\neq 0$, when $x>0$, we rewrite it
as $f(y)=y^y=2^{-\frac{1}{12}}$, where
$y=\frac{1}{x}$, $y^y$ has only critical point
at $y=e^{-1}$, and we have $f(0+)=1$,
$f(+\infty)=\infty$.
Notice $0<2^{-\frac{1}{12}}<1$. So since
$f(e^{-1})<2^{-\frac{1}{12}}$,
we have 2 solution in this case.
Next suppose $x<0$, let $y=-\frac{1}{x}$, we
have $f(y)=y^y=2^{\frac{1}{12}}$, notice
$2^{\frac{1}{12}}>1$, we have
1 solution in this case.
Draw the graph of $f(y)=y^y$$, you |
|
A****h
发帖数: 385 | 14 “如果你打算從二十歲到四十歲，每年都可考慮結婚。假如你每年只考慮一位對 象，
那麼最好的方法大約是M8，也就是說，自你二十八歲起，才開始要 「認真」起來。(年
青的時侯多玩玩無妨)。 ”
可见真是学好数理化，走遍天下全不怕。要想求得好姻缘，不靠摇签靠概率。
==================================================================
什么是秘书问题
在机率及博弈论上，秘书问题（类似名称有相亲问题、止步问题、见好就收问题、
苏丹的嫁妆问题、挑剔的求婚者问题等)内容是这样的：要聘请一名秘书，有n人来面试
。每次面试一人，面试过后便要即时决定聘不聘他，如果当时决定不聘他，他便不会回
来。面试时总能清楚了解求职者的适合程度，并能和之前的每个人作比较。问凭什么策
略，才使选得到最适合担任秘书的人的机率最大？
秘书问题的策略
基本解決策略如下：对于某些整数r，其中1 le r < n。先面试首r人，都不聘请他
们，在之后的n − r人中，如果任何一人比之前面试的人都更佳，便聘请他。
r的最佳值應該是rapprox frac{n}{... 阅读全帖 |
|
f**********e
发帖数: 1994 | 15 In TeXForm
\{ \{ {{y(x)}\rightarrow
{\frac{\Mfunction{C}(1)\,\Mfunction{HypergeometricU}(\frac{b\,
\left( 1 + \frac{a}{{\sqrt{a^2 - 4\,c}}} \right) }{2},b,
{\sqrt{a^2 - 4\,c}}\,x) +
\Mfunction{C}(2)\,\Mfunction{LaguerreL}(\frac{b\,
\left( -1 - \frac{a}{{\sqrt{a^2 - 4\,c}}} \right) }{2},-1 + b,
{\sqrt{a^2 - 4\,c}}\,x)}{e^
{\frac{\left( a + {\sqrt{a^2 - 4\,c}} \right) \,x}{2}}}}}\} \} |
|
b****e
发帖数: 460 | 16 我不知道算没算错，但是结果很惊人
首先
[/dP=\rho (r)g(r)dr/]
($\rho (r)$是密度，P是压强，g(r)是重力加速度，r是距地心的距离)这是Pascal定理
的变形，原因是重力加速度变了，同时空气密度也变了。
重力加速度很好算
[/g(r)=\frac{GMr}{R^2}=g_0\frac{r}{R}/]
其中$g_0$是地球表面的重力加速度，$R$是地球半径。
现在假设空气是理想气体，4/5是氮气14，1/5是氧气16，所以估算出空气的摩尔密度是
14.4g/mol。
根据理想气体状态方程$PV=NkT$(V是气体体积，k是Boltzmann常数，$k=1.38\times 10
^{-23}J/K$，T是温度)得知
[/\rho=\frac{m_0P}{N_AkT}/]
(其中$m_0$是摩尔质量，$N_A$是啊佛家的罗常数)
所以
[/dP=\frac{m_0g_0rP}{N_AkTR}dr/]
于是
[/\frac{dP}{P}=\frac{m_0g_0}{N_AkTR}rdr/]
做积分，左边从$P |
|
R*********r
发帖数: 1855 | 17 先设 0
\sum_{p=1}^{\infty}\frac{\sin px}{px}
=\frac{1}{x}\Im \sum_{p=1}^{\infty}-\frac{e^{-ipx}}{p}
=\frac{1}{x}\Im\ln (1-e^{-ix})
=\frac{\pi-x}{2x}
对其它的x，级数值可以从上式推出来。 |
|
s*****V
发帖数: 21731 | 18 http://sbseminar.wordpress.com/2013/05/30/i-just-cant-resist-th
TAO的算功还是刚刚的
Actually, it looks like there is considerable room to improve the
estimation of the quantity \Sigma_2, which is currently being controlled in
terms of the very large quantity \delta_2 (which remains rather large even
after the improved estimate given previously), which may lead to a
significant improvement to k_0 even without attempting either of (a) and (b)
above. I’ll try to write up some details soon.
27. Terence ... 阅读全帖 |
|
TN
发帖数: 1870 | 19 Oil and Gas Companies Injected Cancer-Causing Chemicals During Fracking;
Backlash Grows
Drilling companies still claim that fracking, which involves shooting
chemicals and water underground to release gas, is safe but they are
increasingly fighting an uphill battle against facts.
The AP reported today that a new report by House Dems Henry Waxman of
California, Edward Markey of Massachusetts and Diana DeGette of Colorado
found that oil and gas companies injected millions of gallons of potentially... 阅读全帖 |
|
g****g
发帖数: 1828 | 20 f^{\pm}(p_1, p_2) = f(p_1, p_2) \pm \lambda \sqrt{ (\frac{\partial f}{\
partial p_1})^2 M_{11} + (\frac{\partial f}{\partial p_2})^2 M_{22} +\frac{\
partial f}{\partial p_1}\frac{\partial f}{\partial p_2} M_{11}M_{22} } |
|
d**********o
发帖数: 1321 | 21 第一个项目report
这时偶刚到CSAC工作不久，与小A同学还不熟，我用的还是latex。随着贴的作业越来越
多，应该是用有共同爱好的小伙伴更亲密些。这次贴latex，下次才再org-mode。
\documentclass[b5paper,11pt, abstraction, titlepage]{scrartcl}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{CJKutf8}
\usepackage{multirow}
\usepackage{multicol}
\usepackage{listings}
\usepackage{geometry}
\geometry{b5paper}
\usepackage{graphicx,floatrow}
\usepackage{graphicx,subfigure}
\newsavebox{\abstractbox}
\renewenvironment{abstract}
{\begin{lrbox}{0}\begin{minipage}{\t... 阅读全帖 |
|
d**********o
发帖数: 1321 | 22 第一个项目report
这时偶刚到CSAC工作不久，与小A同学还不熟，我用的还是latex。随着贴的作业越来越
多，应该是用有共同爱好的小伙伴更亲密些。这次贴latex，下次才再org-mode。
\documentclass[b5paper,11pt, abstraction, titlepage]{scrartcl}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{CJKutf8}
\usepackage{multirow}
\usepackage{multicol}
\usepackage{listings}
\usepackage{geometry}
\geometry{b5paper}
\usepackage{graphicx,floatrow}
\usepackage{graphicx,subfigure}
\newsavebox{\abstractbox}
\renewenvironment{abstract}
{\begin{lrbox}{0}\begin{minipage}{\t... 阅读全帖 |
|
L******r
发帖数: 199 | 23 \begin{equation}\label{SOLocationWeight}
\[
Weight = \left\{\begin{array}{ll}
0.2, &\mbox{if $l \leq \frac{1}{3}$} \\
2.4\cdot l-0.6, & \mbox{if $\frac{1}{3}\leq l \leq \frac{2}
{3}$} \\
-1.8 \cdot l+2.2, & \mbox{if $\frac{2}{3}\leq l \leq 1
$}}
\end{array} \right.
\]
\end{equation}
前面用同样的都没问题，这个公式应该是2.3，但是没有显示出来
系统也把这个公式给默认标号了，因为后面
的公式是2.4都出来了。
谢谢指教 |
|
T*******n
发帖数: 493 | 24 Here is the proper way of doing this:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\label{SOLocationWeight}
\text{Weight} =
\begin{cases}
0.2, & \text{if $l \leq \frac{1}{3}$}; \\
2.4\cdot l-0.6, & \text{if $\frac{1}{3}\leq l \leq \frac{2}{3}$}; \\
-1.8\cdot l+2.2, & \text{if $\frac{2}{3}\leq l \leq 1 $}.
\end{cases}
\end{equation}
\end{document}
2}
1 |
|
B*B
发帖数: 87 | 25 我写下一个超长的矩阵，latex以后dvi只能显示前半部分，
后面就不显示了，也不换行。
我想让latex把矩阵横过来，这样可能可以显示，不知道应该怎么做。
或者可以缩小点字体，让公式变短点？也不知道该怎么做。
或者高人知道还有别的方法怎么解决这个超长公式的显示问题。
谢谢！
我的代码如下
\begin{equation}
\begin{pmatrix}
P^2_{g}+P^2_{b}+(1-w)(1-\frac{N+1}{N}w)P^2_{ig}+2P_{b}P_{g}+(2-\frac{2N+1}{N
}w)P_{g}P_{ig}+(2-\frac{2N+1}{N}w)P_{b}P_{ig} & (2N+1)wP^2_{b}-(1-w)\frac{N+
1}{N}wP^2_{ig}+(2N+1)wP_{b}P_{g} & [(N^2+N)w^2-Nw]P^2_{b}-NwP_{b}P_{g}-NwP_{
b}P_{ig} & [(N+1)w^2-w]P^2_{b}-wP_{b}P_{g}-wP_{b}P_{ig} \\ %(s1 s2 s3 s4)
(1-w)wP^2_{ig}+ |
|
d****y
发帖数: 53 | 26 AA^{-1}=I
hence (schematically)
\frac{\partial A}{\partial D} A^{-1}+A\frac{\partial A^{-1}}{\partial D}=0
this gives
\frac{\partial A^{-1}}{\partial D}=-A^{-1}\frac{\partial A}{\partial D}A^{-1} |
|
w*********i
发帖数: 77 | 27 4\times (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})
collect |
|
x********9
发帖数: 31 | 28 The answer is [-1,1]. The only requirement is that the matrix \Sigma
\begin{bmatrix}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{bmatrix} is positive definite.
For any X, X^{\top} \Sigma X is equal to \frac{a}{2}(x_1+x_2+x_3)^2 + (1-\
frac{a}{2})(x_1^2+x_2^2+x_3^2) or -\frac{a}{2}(x_1+x_2+x_3)^2 + (1+\frac{a}{
2})(x_1^2+x_2^2+x_3^2)
Both expressions are positive when a \in [-1,+1]
The reasoning using 3 dimension space is obviously wrong. A priori choosing
three random variables is to choose three ... 阅读全帖 |
|
r**a
发帖数: 536 | 29 I got another solution.
Let Y=X^{-1/2}. Then dY=3/8*Y^{-1}dt-1/2*dB
Now let Z=e^Y. Then
$$
dZ=[Z(\frac{1}{8}+\frac{3}{8\log(Z)})]dt-\frac{Z}{2}dB
$$
Notice that the above eq. regarding Z is in the form of
$$
dZ=f(Z)dt+\frac{Z}{2}dB
$$
Then we may use the result of exersice 5.16 in Oksendal's book "stochastic
differential equations" to solve it. |
|
d*b
发帖数: 4453 | 30 你个五毛傻逼能不能少羞些你先人？
1. Empire State Manufacturing Survey vs The Philadelphia Fed's manufacturing
index 是两个不同的东西。 Empire State Manufacturing Survey, The monthly
survey of manufacturers in New York State conducted by the Federal Reserve
Bank of New York.计算的是工业生产出货与订单的比较关系。同时，把出货量与订单
量×各该价格后转换为复合指数，来表示工业生产的现状和预测短期经济走向。
从10月份公布的这两个指标看，美国的工业生产和销售状况都在变好，the New York
Fed's Empire State Manufacturing Survey showed an improvement to -11.36 from
-14.67 in September. The Philadelphia Fed's manufacturing index... 阅读全帖 |
|
发帖数: 1 | 31 Naïve Bayes
$$Pleft ( frac{A}{B} right ) = frac{Pleft ( frac{B}{A} right )Pleft ( A
right )}{Pleft ( B right )}$$ |
|
r******e
发帖数: 344 | 32 其实现在已经打好了的那些井，还没frac的，足够frac两三年的。
frac本身投资并不大，而且周期相当短。
一两千万，两个月，井就开始泵油了。这点成本和风险，那些风投不会犹豫的。
60 |
|
k***y
发帖数: 17 | 33 http://www.mitbbs.com/clubarticle_t/Petroleum/31200353.html
Field Engineer/Fracturing Engineer
Employer: Amerril Energy LLC
Desired Expertise: Petroleum engineering, Fracturing engineering
Experience: 1-2 years
Education: Bachelors/3-5 yr Degree
Location: Houston, TX
Amerril focuses on oil and gas exploration and development while also
providing oilfield technical services especially in fracturing service.
Amerril is willing to recruit entry-level field engineer who are interested... 阅读全帖 |
|
t*h
发帖数: 148 | 34 \begin{equation*}
\frac{\partial N\left(x,y,t\right)}{\partial t}
= \underbrace{\biggl[- MN\biggr]}_{\text{Mortality}} +
\underbrace{D\biggl[
\frac{\partial^2 N\left(x,y,t\right)}{\partial x^2} +
\frac{\partial^2 N\left(x,y,t\right)}{\partial y^2}
\biggr]}_{\text{Diffusion}}
\end{equation*}
就是下括号高高低低的 |
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T*******n
发帖数: 493 | 35 \documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
\frac{\partial N\left(x,y,t\right)}{\partial t}
% added \vphantom{\bigg|} in next line
% and removed \biggl and \biggr from []
= \underbrace{\vphantom{\bigg|}[- MN]}_{\text{Mortality}} +
\underbrace{D\biggl[
\frac{\partial^2 N\left(x,y,t\right)}{\partial x^2} +
\frac{\partial^2 N\left(x,y,t\right)}{\partial y^2}
\biggr]}_{\text{Diffusion}}
\end{equation*}
\end{document} |
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w****g
发帖数: 78 | 36 这么多高手，没有一个人愿意说句公道话。或者底气不足，不敢说？
原题如下：
Solve the following equation:
\[\frac{\partial T}{\partial t}-\frac{\partial^2 T}{\partial x^2} +kT=0; T(x
,0)=1, T(0,t)=e^{-kt}, T(1,t)=0.\]
Where $k$ is a constant. Let $u(x,t)\triangleq e^{kt}T(x,t)$, change the
original PDE and conditions in terms of $T$ into a new one in terms of $u$.
Solve the problem in terms of $u$.\\
\textbf{Solution:}\\
Let $u(x,t)=e^{kt}T(x,t)$, then $T(x,t)=e^{-kt}u(x,t)$ and
\[\frac{\partial T}{\partial t} = -ke^{-kt}u(x,t)+e^{- |
|
p******n
发帖数: 66 | 37 the obj fun = A + \frac{C\beta+(D-A)}{1+\beta^2}
Let f(\beta)=\frac{C\beta+(D-A)}{1+\beta^2}.
f'(\beta)=\frac{-C\beta^2+2(A-D)\beta+C}{(1+\beta^2)^2}.
The denominator of f'(\beta) is always positive, so you just need to analyze
the sign of the numerator which is a quadratic function of \beta. |
|
F*******h
发帖数: 136 | 38 \int sec(x) dx = \int \frac{1}{cos(x)} dx = \int \frac{1}{cos(x)^2} d sin(x)
= \int \frac{1}{1-x^2} dx
One doesn't need to remember the formula. |
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t*********n
发帖数: 1292 | 39 At time t, the value of these holdings will be
\Pi = V - S\frac{\partial V}{\partial S}.
Then why d\Pi=dV-dS\frac{\partial V}{\partial S}? It seems we do not have to
differentiate \frac{\partial V}{\partial S}. Any good explanation? |
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w*******e
发帖数: 15912 | 40 F=h_bar Frac {{m_1}{m_2}} {R^2}
% where h_bar=Frac {h}{2\pi}
谁愿意拿去就拿去吧，随便发个Science,Nature，ＰＲＬ伍的，ｃｏ－author就不必了
，文章后边acknowledgment部分感谢一下哥就行了。 |
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H****r
发帖数: 2801 | 41 This should be O(1)?
http://en.wikipedia.org/wiki/Fibonacci_number
Computation by rounding
Since \begin{matrix}|1-\varphi|^n/\sqrt 5 < 1/2\end{matrix} for all n\geq 0,
the number F(n) is the closest integer to \varphi^n/\sqrt 5\, . Therefore
it can be found by rounding, or in terms of the floor function:
F(n)=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor,\ n
\geq 0. |
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P****i
发帖数: 12972 | 42 F_n = \frac{psi^n-{-psi}^{-n}}{\sqrt{5}}
psi=\frac{1+\sqrt{5}}{2} |
|
s**h
发帖数: 477 | 43 我们在说页岩油，近5年美国产油飙升，主要因为页岩油。页岩油开采靠hydro frac,
但每次frac, 都只能是小范围的，所以页岩油井产量会迅速下降 |
|
l*******b
发帖数: 2586 | 44 \frac{\beta}{\lambda} + \frac{\lambda c -\beta}{\lambda^2(P-cL)} |
|
H******7
发帖数: 34403 | 45 The device is 2x2x4 meters in size. It is cylindrical shaped. It has a
vacuum inside with high magnetic fields, made using electromagnets.
Uncharged deuterium gas is injected. It is heated using radio waves, in much
the same way a microwave heats food. When the gas temperature reaches over
16 electron-volts, the gas ionizes into ions and electrons. This plasma
exerts a pressure on the surrounding magnetic fields. This plasma pressure
is counterbalanced by the magnetic field pressure in a beta ra... 阅读全帖 |
|
发帖数: 1 | 46 http://rosettacode.org/wiki/24_game/Solve#C
改一下就行了。这是输入4个的，可以改成输入3个的。
6 17 3 7: No solution
……
#include
#include
#include
#define n_cards 4
#define solve_goal 29
#define max_digit 9
typedef struct { int num, denom; } frac_t, *frac;
typedef enum { C_NUM = 0, C_ADD, C_SUB, C_MUL, C_DIV } op_type;
typedef struct expr_t *expr;
typedef struct expr_t {
op_type op;
expr left, right;
int value;
} expr_t;
void show_expr(expr e, op_type prec, int is_r... 阅读全帖 |
|
j**u
发帖数: 6059 | 47 ft
第一个那个科大牛mm的才是解决问题的，其他都是依葫芦画瓢。
text1=text(x,y,'$$frac{a}{b}$$');
set(text1,'interpret','latex')
or
title1=title('$$frac{a}{b}$$')
set(title1,'interpret','latex')
还有ylabel, legend, ...，没区别。 |
|
r****c
发帖数: 2585 | 48
There are lots of different difinition about what kind of binary trees are
different,
the most famous the Cn, which is the catalan number.
Cn=\frac{1}{n+1} \frac{2n!}{n!^2} |
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c*******d
发帖数: 255 | 49 只要用xfig -specialtext -latexfont -startlatexFont default 来启动xfig.
画的图中文字写成$\frac{\alpha}{1+\beta}$等形式,然后存成test.fig
然后在TeXmacs中插入这个test.fig, 刚才的$\frac{\alpha}{1+\beta}$就被显示成了数
学公式.
当然,我一直都是用metapost来画带数学公式的示意图的. |
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