d**********o
发帖数: 1321 | 1 hw3b c-.y file
上面这一楼贴了载止hw3b deadline时我match的结果（也就是老师可以以这些不match
的ERROR为借口不给后来我补上的成绩），但是因为当时我还是没有写完，后来感恩节
期间就接着又写了一些，而且hw5是based on hw3 & hw3b的基础上（当我hw5是based
on更好的hw3的结果时，我应该可以得更多的分吧）。
hw4因为写得比较顺利，就不曾保留任何交上去作业的output，没有什么一目了然的结
果是我可以贴在这里的。原本我是想要把自己最的一次作业hw5贴出来的，但那已经是
一个完整的compiler，而且以后我还需要用自己的course project来找工作，所以一定
就不贴最终结果了。那就贴一个hw3b的c-.y文件吧，它集中的hw1、hw2、hw3、 hw3b的
结果，是我自己hw3b *.y文件的最完整版本。这些作业里面也有很多机关一一人为增加
的难度，比如那六七个IO相关的function，不仅traverse tree、build syntax tree的
时候会成为一个考点（把它们作为一个node连在syntax... 阅读全帖 |
|
d**********o
发帖数: 1321 | 2 hw3b c-.y file
上面这一楼贴了载止hw3b deadline时我match的结果（也就是老师可以以这些不match
的ERROR为借口不给后来我补上的成绩），但是因为当时我还是没有写完，后来感恩节
期间就接着又写了一些，而且hw5是based on hw3 & hw3b的基础上（当我hw5是based
on更好的hw3的结果时，我应该可以得更多的分吧）。
hw4因为写得比较顺利，就不曾保留任何交上去作业的output，没有什么一目了然的结
果是我可以贴在这里的。原本我是想要把自己最的一次作业hw5贴出来的，但那已经是
一个完整的compiler，而且以后我还需要用自己的course project来找工作，所以一定
就不贴最终结果了。那就贴一个hw3b的c-.y文件吧，它集中的hw1、hw2、hw3、 hw3b的
结果，是我自己hw3b *.y文件的最完整版本。这些作业里面也有很多机关一一人为增加
的难度，比如那六七个IO相关的function，不仅traverse tree、build syntax tree的
时候会成为一个考点（把它们作为一个node连在syntax... 阅读全帖 |
|
r**a
发帖数: 536 | 3 Is this strange?
The key is as follows:
\begin{align}
P(\tau_m \leq t) &= P(\tau_m \leq t, W(t)\leq m)+P(\tau_m \leq t, W(t)\geq m
)\
&=P(W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=P(\tau_m\leq t, W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=2P(W(t)\geq m)
\end{align}
The naive understanding is that you have two possible paths letting $\tau_m<
t$ if $W(t)\geq m$ according to the reflection principle. Those two paths
correspond to the same $\tau_m$. So if the possibility getting one path is p
, t... 阅读全帖 |
|
d******e
发帖数: 7844 | 4 我感觉MB，BMW \geq Audi \geq Lexus, Caddy \geq Infiniti \geq Acura \approx
Volvo \
approx Buick \approx Lincoln |
|
m******2
发帖数: 564 | 5 The key is as follows:
\begin{align}
P(\tau_m \leq t) &= P(\tau_m \leq t, W(t)\leq m)+P(\tau_m \leq t, W(t)\geq m
)\
&=P(W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=P(\tau_m\leq t, W(t)\geq m)+P(\tau_m \leq t, W(t)\leq m)\
&=2P(W(t)\geq m)
\end{align}
我知道你这个宏观的证明，可是这个证明和概率密度之间的关系有什么联系？ |
|
q********e
发帖数: 1255 | 6 no.
it's known that A\geq 0, B\geq 0 does not imply AB+BA\geq 0.
take C=I.
matrices |
|
H****r
发帖数: 2801 | 7 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. |
|
t******u
发帖数: 449 | 8 2，部分基本词汇
上海话形容词和副词后头的相当普通话“的”跟“地”的，都是“个”。“个”
是该字的本字，但是读音发生了演变。“个”本生不是入声字，比如“个性”一词当中
，读“顾”。以往一般写作“格”，苏州话是这样（箇能）读个。不能写作
“额（“疑”母字）”或“饿”。
“此、彼、何”，在南北方言里差异很大。普通话的“这、那、哪”，原本无字，是借
来注音的字和后造字，与古无据。也就是原本是不能写的。相对来说吴语的情况要好些。
上海话里向“此”称为“搿（浊音）”，这个字的本字也是“个”。也就是“这
里”可以写作“个个”，迪能可能会引起误解，所以钱乃荣老师提出用异体字“箇”代
替，大家习惯上用“搿（浊音）”代替，不能写作“格（全清音字）”。另
外还有一个读音“迪”，又说本字是“是”，尚未有定论。苏州话里“此”或称
为“该”。
“彼”称为“哎[E]（传统上写作“哀”，中国人嘛，讲点口彩）”；不能写作“艾
（“疑”母字）”。
“何”字在上海话中音变为“嚡”。
因此上海话个“ |
|
G******i
发帖数: 163 | 9 Let B_k be a square root of A_k: (B_k)^2 =A_k, (B_k)^(-2)=(A_k)^(-1).
For any matrix S, we have
Sum_k [ (B_k +B_k^{-1}S)^* (B_k +B_k^{-1}S), k=1..K) \geq 0;
i.e.
(Sum_k A_k) +KS +KS^* +S^*(Sum_k A_k^{-1})S \geq 0.
Setting S= -K(Sum_k A_k^{-1})^{-1},
the required inequality follows. |
|
h******9
发帖数: 84 | 10 多谢各位大侠的讨论先.
二楼的大侠, 我发信到你的信箱了. 有点等不及你的回复了. 先贴到这里. 大家一起讨
论下. 莫介意.
你的这步展开:
Sum_k [ (B_k +B_k^{-1}S)^* (B_k +B_k^{-1}S), k=1..K) \geq 0;
i.e.
(Sum_k A_k) +KS +KS^* +S^*(Sum_k A_k^{-1})S \geq 0.
应该是用到了 B_k^* B_k^-1 =I and (B_k^-1)^* B_k =I. A_k 正定无法保证 B_k也正
定. 这两个identities还能成了麽? 我是学工科的, 矩阵功底很烂.
我觉得你的方法思路很好. 有没有办法改进上面这个小问题呢? 谢谢大家的帮忙. |
|
a***s
发帖数: 616 | 11 This seems impossible except some degenerated cases.
Let $Y_k := K(x^*, X_k)$. Since $\{X_k\}_{k \geq 1}$ are i.i.d., $\{Y_k\}_{k
\geq 1}$ are also i.i.d. Since the value of the kernal is between 0 and 1,
$Y_k$'s are nonnegative. Then Borel-Cantelli lemma says $\sum_{k=1}^{\infty}
Y_k = \infty$ with probability one unless $Y_k = 0$ with probability one.
Therefore for the infinite sequence $K$ to be in $l^1$ with positive probabi
lity (one, actually), we must have $Y_k = 0$ with probability one. ... 阅读全帖 |
|
B********e
发帖数: 10014 | 12 should be yes
easy to see it's true for 1D
transform the problem to 1D
let x(t)=A+t(B-A),
consider S(t)=S(x(t)) and show S(1)\geq S(0)
show first S'(t)\geq 0\forall t |
|
B********e
发帖数: 10014 | 13 f(x)=(x^{m+1}-x)/(x^{n+1}-1);
f'(x)=f1(x)/(x^{n+1}-1)^2,
where f1(x)=-(n-m)x^{n+m+1}-(m+1)x^m+nx^{n+1}+1.
want: f1(x)\leq 0 when x\geq 1.
sufficient: f1(1)=0, f1'(x)\leq 0, when x \geq 1.
f1'=x^{m-1}*f2(x), with f2=something.
then sufficient: f2(1)=0, f2'\leq 0.
which is not difficult because
f2'=-(n+m+1)(n-m)(n+1)x^n+n(n+1)(n-m+1)x^{n-m}.
homework: complete it by showing the absolute value of the coef of first
term
is larger than the coef of the second term, plus the fact n>n-m and x>1. |
|
N**G
发帖数: 392 | 14 I have an interesting question for you, it can be solved in elementary
school techniques. It comes from my own research.
begin{equation}
(2-N)a+2b+d_1+d_2+ldots+d_5=1,\
Na^2-2ab+d_1^2+d_2^2+ldots+d_5^2=1
end{equation}
Show that the number of integral solutions $(a,b,d_1,d_2,ldots,d_5)$ is
independent of $N\geq 0,N\in\mathbb{Z}$. |
|
i**n
发帖数: 1481 | 15 看看你不用google能知道多少奥运会里国家的简写 :-)
一共202个
AFG
AHO
ALB
ALG
AND
ANG
ANT
ASA
ARG
ARM
ARU
AUS
AUT
AZE
BAH
BAN
BAR
BDI
BEL
BEN
BER
BHU
BIH
BIZ
BLR
BOL
BOT
BRA
BRN
BRU
BUL
BUR
CAF
CAM
CAN
CAY
CGO
CHA
CHI
CHN
CIV
CMR
COD
COK
COL
COM
CPV
CRC
CRO
CUB
CYP
CZE
DEN
DJI
DMA
DOM
ECU
EGY
ERI
ESA
ESP
EST
ETH
FIJ
FIN
FRA
FSM
GAB
GAM
GBR
GBS
GEO
GEQ
GER
GHA
GRE
GRN
GUA
GUI
GUM
GUY
HAI
HKG
HON
HUN
INA
IND
IRI
IRL
IRQ
ISL
ISR
ISV
ITA
IVB
JAM
JOR
JPN
KAZ
KEN
KIR
KGZ
KOR
KSA
KUW
LAO
LAT
LBA
LBR
LCA
LES
LIB
LIE
LTU
LUX
MAD |
|
a*****e
发帖数: 688 | 16 \Delta x \times \Delta y \geq \hbar\div 2 |
|
a*****e
发帖数: 688 | 17 \Delta x \times \Delta y \geq \hbar\div 2 |
|
g****g
发帖数: 1828 | 18 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 |
|
l****e
发帖数: 23 | 19 Maybe BDG doesn't work. But I made it in a way similar to the proof of BDG:
Let M_{t}:=\int_{0}^{t}\exp{\beta s-t}dW_{s}.
Apply Ito Formula to |M_{t}|^{2} and integrate from 0 to T, we get
|M_{T}|^{2}=T-2\beta\int_{0}^{T}|M_{t}|^{2}dt+\int_{0}^{T}2M_{t}dW_{t}.
Then taking expectation gives
E|M_{T}|^{2}=T-2\beta E\int_{0}^{T}|M_{t}|^{2}dt
\geq T-2\beta TE\sup_{0\leq t\leq T}|M_{t}|^{2}.
That is
E\sup_{0\leq t\leq T}|M_{t}|^{2}
\leq (T-E|M_{T}|^{2})/(2\beta T)
=1/(2\beta)-(1-\exp{-2\beta T})/(4\be |
|
h**********c
发帖数: 4120 | 20 Fermat last theorem
\equiv
2^{\frac{1}{m}} is irrational; for m \in N, m \geq 2;
I am not kidding, if you are serious. |
|
q********e
发帖数: 1255 | 21 construct a sequence of unit vectors {e_i}, let x=\sum e_i/i.
show {(x,y_k)} unbounded
hint:
you expect
1. e_n perpendicular to both {e_1,e_2,...,e_{n-1}} and {y_1,y_2,...,y_{n-1}}
for all n,
2. |(e_n,y_n)|\geq n[\sum_{i=1 to n-1} sup_k{|(e_i, y_k)|}+n]
,y |
|
p******e
发帖数: 1151 | 22 when the space has infinity measure, for example, R, you need a simple
interpolation inequality to use the fact that f is L^q for some q fixed. (
actually you only need to assume that f is L^q for some q fixed, not for all
p\geq q).
\int |f|^p \leq |f|_{L^\infty}^{p-1} \int |f|, take 1/p and let p go to
infinity.
(here assume q=1, but no difference for some fixed constant q.) |
|
p***o
发帖数: 1252 | 23 max b s.t.
Ax=0,
x_i \geq b, \forall i
随便拿个LP solver跑一下就出来了。 |
|
a***s
发帖数: 616 | 24 Let $\{ X_k \}_{k \geq 1}$ be i.i.d. random variables such that they are non
negative and $\Pr( X_1 > 0 ) > 0$.
In particular, we can assume $\Pr( X_1 > c ) = \delta > 0$ with some nonrand
om positive constants $c$ and $\delta$.
Now B-C lemma tells us that with probability one, events $\{ X_k > c \}$ hap
pens infinitely many times. Thus with probability one, $\sum_{k=1}^{\infty}
X_k = \infty$. |
|
p***c
发帖数: 2403 | 25 给定一个方阵G＝(g(i,j)), 假设 对任意的一个置换p都有
sum_i g(i,i) \geq sum_i g(i, p(i))
求证不存在同阶方阵 X 满足
（1）X的第i行元素之和等于第i列元素之和
（2）所有元素非负
（3）sum_i sum_j x(i,j)g(i,i) < sum_i sum_j x(i,j)g(i,j)
2阶方阵这是显然的，对于3阶以上的，证了几个小时证不出来。我太笨了。 |
|
p***c
发帖数: 2403 | 26 我有一个非常简单的分段函数，定义在[0，\infty)上，
a>0, c>0，d>0 是三个常数
f(x)=a if x
f(x)=-d if x>=c
现在要求这样一个g函数：
g(x)=sup{ v(x) | v is convex, and v(y)\leq f(y) for all y\geq 0}
简单的说，g是所有小于f的凸函数的上确界
从图上直观我的来说 g 就应该是一条直线把(0,a)和(c,-d)连起来，在x>c的地方不变
怎么具体证明啊？ 谢谢 |
|
i****g
发帖数: 3896 | 27 http://blog.sina.com.cn/s/blog_c24597bf0101b871.html
致谢：I would like to thank Prof. Shing-Tung Yau for suggesting the title of
this article, Prof. William Dunham for information on the history of the
Twin Prime Conjecture, Prof. Liming Ge for biographic information about
Yitang Zhang, Prof. Shiu-Yuen Cheng for pointing out the paper of
Soundararajan cited in this article, Prof. Lo Yang for information about
Chengbiao Pan quoted below, and Prof. Yuan Wang for detailed information on
result... 阅读全帖 |
|
y***s
发帖数: 23 | 28 Head is counted 1; tail is counted 0.
Sum of heads for P1 is Binomial(5, 1/2) while sum of heads for P2 is
Binomial(4, 1/2).
Then use computer, which is the practice of street right?, (it might take a
while by hand) to find
Pr[ Binomial(5, 1/2) \geq Binomial(4, 1/2) ]. |
|
