g***i
发帖数: 50 | 1 X: X轮把所有靶子击中
对每个靶子,i轮击中的概率 1-(1-p)^i
P(X=i)=(1-(1-p)^i)^n
so E(X)=\sum_{i=1}^{\infty} i*((1-(1-p)^i)^n)
不知道化简出来时不是yixiu的答案。 |
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b******y
发帖数: 2729 | 2 【 以下文字转载自 EE 讨论区 】
【 原文由 bjyyj 所发表 】
the pdf of the time interval between two buses is exponential
f(t)=exp(-t/15)/15
the avg number of passengers who arrives between time interval t is 2t
the avg number per bus:
\int_0^{\infty}2t\cdot f(t) dt
answer of b should be the same as |
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f*********g
发帖数: 632 | 3 Chomsky–Schützenberger theorem. If L is a context-free language admitting
an unambiguous context-free grammar, and a_k := | L \ \cap \Sigma^k | is the
number of words of length k in L, then \sum_{k = 0}^\infty a_k x^k is a
power series over \mathbb{N} that is algebraic over \mathbb{Q}(x). |
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p*********g
发帖数: 226 | 4 pac 显然不能和 SVM 比。
但Valient在别的领域也颇有建树。
归根结底，还是 norm 的选取问题。你想说的是比 L_{\infty} norm，可问题人家用 L
_1 或 L_2，你不能因此就说人家不要脸。 |
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n******t
发帖数: 4406 | 5 Then just assign deletion and insertion an -infty penalty.
deletion |
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T*******n
发帖数: 493 | 6 The following allows you to create an operator
name that typesets the same way as \lim_{x\to\infty}
(the subscribed p shifts positions depending on math
mode).
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator*{\Min}{Minimize}
\begin{document}
in-line math $\Min_p D(p)$ in-line math\par
display math display math display math display math
\begin{equation}
\Min_p D(p)
\end{equation}
\end{document} |
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l*****e
发帖数: 238 | 7 e.g.
\Sigma\limits_{n=1}^{+\infty} |
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r****t
发帖数: 10904 | 12 CLT assures your that 没有什么实际意义的差别 asymtotically -> 0 as N->\infty. So 发现差别
显著了就是有意义的显著（就 distribtion mean 而言） |
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r****t
发帖数: 10904 | 13
你这种看法犯的错误正好是我要大家警醒的，我说的正是不要动不动就用 u-test, 很
多时候 t test 是非常适用的。t-test 对产生 sample 的 distribution 没有任何假
设，不需要是 normal 的，我想 DaShagen 对这个说法没有意见。t test normality
假设是对 test statistics 而言的，sample mean -> true mean (asymtotically) as
N->\infty 的根据是 CLT, 前面有别的同学提过。
我看来这话没有道理，均数为啥没意义？（这个话有啥出处没，能不能 detail 给大家
听听？）
你用正态分布 model 身高，我没有意见。我没有说“不能用normality的假设”，在这
种情况下身高没有必要是正态分布的，你也能用 t test. 因为均值的 normalty 是
CLT 确保了的。 |
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h**********c
发帖数: 4120 | 14 correction:
stiff problem:
x'=\lambda x, when Re(\lambda)-> -\infty
x could be extended to n-vector, \lambda thus could be a matrix D, thus one
of the eigenvalues of D has its real part approaches negative infinity.
This is a totally different thing from singular Jaconbian of ODEs.
Stiff problem can be attacked by stability analysis. Normally a multi-step
implicit mesh scheme should help.
Just for discussion. |
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m*********a
发帖数: 2000 | 15 \infty. The only thing is they created many very original things, you can
not really compare. |
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g****t
发帖数: 31659 | 16 你懂得很多。确实牛。
这帖子这么多做信息论的，我看没几个知道啥是H-infty的。(包括我) |
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g****t
发帖数: 31659 | 17 你懂得很多。确实牛。
这帖子这么多做信息论的，我看没几个知道啥是H-infty的。(包括我) |
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i********n
发帖数: 53 | 18 第二个问题没有第一个严重，用interference alignment,理论上每个用户的degrees
of freedom 应该是n/(2n+1), 其中n是diversity 的数量，如果信道本身没有
diversity，degrees of freedom 就是1/3，因为n=1。而理论上最大值是1/2，当n=\
infty. 不过jafar在distributed interference alignment 里面对这个问题有解释，
我没仔细看。 |
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i********n
发帖数: 53 | 19 是啊，每个用户都用symbol extension的话。但是super symbol 对应的是每个symbol
必须经历independent fading，暗含的假设就是需要code across n diversity
branch。
不知道是否有其他解释，因为印象中Jafar 在distributed interference alignment里
谈到了接收端只需知道interfernce covariance matrix 即可用algorithm找到
precoding matrix, 并且不需要n =\infty即可实现d.o.f.=1/2 per user, 不知道这个
理解是否正确。 |
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s*****t
发帖数: 987 | 20
很多方向都这样吧，其实啊
呵呵
搞无线的也这样吧
控制的问题是，没有一个产业来支持，学术界自娱自乐 无线比这个乐观多了
近几年我看有人在高network control system ，这个玩意是整啥的，咋还n多h2 h\
infty啊 |
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B***y
发帖数: 83 | 21 M is finite 的时候,你需要设定边界条件。不过对这个题目，一般的象Dirichlet 边界
条件,
y(0) = y(M) = 0,
或者 Neumann 边界条件, y'(0) = y'(M) = 0，或者混合型，都没关系，
结论是只要q(x) 有界， | q(x) | <= C, for some finite C > 0， q(x) 连续 (或者
更弱些，Lebesgue 可测)，then
the spectrum is discrete.
\lambda_1 < \lambda_2 < ... < \lambda_n ---> +\infty.
is |
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T*******x
发帖数: 8565 | 22
漂亮！这个构造我深感佩服。还有上次那个f \notin L^p的构造我也深感佩服。
我一直没想到离散的measure。
下面还有。
不过E=[a,b)和E=(a,b]是在不同的measure上取得的，
E=[a,b)是X=(2,\infty)，Lebesgue measure，f=1/(x logx logx)
这两个measure space和函数怎么能组合到一起我想不出来。
你想到例子了吗？ |
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c******m
发帖数: 599 | 23 上下标的问题吧
有些人就喜欢吧那个\infty写在下面 |
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G**D
发帖数: 1 | 24 Does anybody have experience on computing transition probability of
semi-Markov process? I have several questions. Hope to receive some comments
or suggestions.
1. How to define the diagonal elements of semi-Markov kernel when setting up
the model? I set them to zeros. Is it correct? Any condition on
\sum_j{Q(i,j,t) other than \sum_j{Q(i,j,\infty)} = 1 and Q(i,j,0)=0 ?
2. When computing Markov renewal function, I use R(t) = inv(I-Q) for all
discretized t. I have tried numerical integration and F |
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x******n
发帖数: 24 | 25 f:[0,\infty) --> R
f(0)=1, f(t)=0, for t>0 |
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m**a
发帖数: 10 | 26 来自主题: Mathematics版 - 做道题吧。 定义泛函 f(y,h(.))=\int_0^\infty x^y h(x)dx
假设,任给y>=1, 有f(y,h(.))=g(y)*f(y-1,h(.))
f(0,h(.))=1
其中g(y)=y*(a_1-y)*(a_2+y)/[2*(y-b_1)*(y-b_2)*(y+b_3)*(y+b_4)]
a_1,a_2,b_1,b_2,b_3,b_4皆为正常数，并满足
b_1
问 满足上面条件的f(y,h(.))作为y得函数是否唯一确定？ |
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o**a
发帖数: 76 | 27 Ahlfors said that the convergence of \sum a_n is neither sufficient nor
necessary condition for the convergence of \prod (1+a_n).
(a_n are complex numbers)
I've worked out the example that \prod (1+a_n) converges but \sum a_n doesn't.
For example, a_n=-1/(n+1), \sum a_n -> -\infty, but \prod (1+a_n) -> 0
Actually, as long as for all a_n, -1 < a_n < 0, then \prod (1+a_n) is positive
and decreasing, so converges, but obviously \sum a_n may diverge.
My question is, can you think of a series {a_n} s |
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o**a
发帖数: 76 | 28 我想我找到反例了
(1) \sum a_n 收敛，而 \prod (1+a_n) 不收敛
取 a_n = (-1)^(n+1) i/sqrt(n)
于是 a_n 是一个alternating series乘上i，容易看出\sum a_n收敛到某个有限的虚数
而\prod (1+a_n)不收敛，否则假设它收敛，则它的模也收敛
但是|1+a_n| = (1+1/n) 从而\prod_{n <= N} |1+a_n| = N+1 -> \infty 矛盾
(2) \prod (1+a_n) 收敛，而 \sum a_n不收敛
取 a_n = -1/n+i*(-1)^(n+1)*sqrt(2/n)
显然 \sum a_n的实部趋于负无穷，所以它不收敛
然而
(1+a_n)*(1+a_{n+1})
=1-2/sqrt(n+1)*[1/sqrt(n)-1/sqrt(n+1)]
+i*\sqrt(2)*(1+1/[sqrt(n)*sqrt(n+1)])*(1/sqrt(n)-1/sqrt(n+1))
容易看出上式右端去掉1后模被 1/n^{1.5}所控制，所以
\prod (1+a_n)收敛
t. |
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o**a
发帖数: 76 | 29 嗯，应该是要求onto的
看了Gamelin的复变函数上的定义conformal mapping必须是one-to-one and onto的
如果这样，这道题可以这样证：
首先，无穷不可能是本性奇点
否则任给常数C,我们可以找到z_n->\infty，w_n=f(z_n)->C
考虑f的逆f^{-1}，则它在w=C处连续，所以w_n->C => z_n -> f^{-1}(C)，矛盾
其次，有个结论说如果f在全平面解析，并且无穷是f的一个非本性奇点的话，
则f是多项式
由于f是单射，所以f只能是一次多项式 a*z+b，证毕。 |
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m*********s
发帖数: 368 | 30 ☆─────────────────────────────────────☆
okla (IP~IC~IQ卡，统统告诉我密码) 于 (Sun Aug 14 17:07:39 2005) 提到:
Ahlfors said that the convergence of \sum a_n is neither sufficient nor
necessary condition for the convergence of \prod (1+a_n).
(a_n are complex numbers)
I've worked out the example that \prod (1+a_n) converges but \sum a_n doesn't.
For example, a_n=-1/(n+1), \sum a_n -> -\infty, but \prod (1+a_n) -> 0
Actually, as long as for all a_n, -1 < a_n < 0, then \prod (1+a_n) is positive
and decreasi |
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s**********r
发帖数: 5 | 31 【 以下文字转载自 Physics 讨论区 】
发信人: Scintillator (闪烁的人), 信区: Physics
标 题: 請教一個積分
发信站: BBS 未名空间站 (Mon Dec 12 21:33:49 2005)
\int_0^\infty d k (k^2+m1^2)^a (k^2+m2^2)^b(k^2+m3^2)^c
where m1, m2, m3 are complex numbers, a, b, c real.
能不能指点一下应该怎么入手？ |
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l*****e
发帖数: 238 | 32 \lim_{n\to\infty}F_n/F_{n+1}=w
w=golden ratio |
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l*****e
发帖数: 238 | 33 可以考虑\sum_{n=1}^{\infty} cos(nx)/n^2
or 无穷乘积
or some Fourier series
or ... |
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r***w
发帖数: 35 | 34 无意中看到这个，好多年前的，我有一个idea
since f\in C^1, let's assume x_0=0 by translation, then
f(x)=\int_{x_0}^x f'(t)dt-f(x_0)
so \lim f(x)/x=\lim average of f', by Besicovitch-Lebesque theorem, you get
the results directly. (actually you may use L'Hospital now since the accumulation has
linear growth as $f'\in L^\infty$)
RMK: you are not able to apply L'Hospital since f\in C^1 and it does not enjoy linear
growth |
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H****h
发帖数: 1037 | 35 怀疑没有无穷级数展开。考虑各阶导数。一次导数在各点都存在，等于
sum 4sin(mx)^3cos(mx)/m^2。求二次导数就不能简单地逐项求导再求和了。
只求零点的二次导数，就是计算 lim_{x->0} sum 4sin(mx)^3cos(mx)/(m^2x)。
令h(x)=4sin(x)^3cos(x)/x^2。则上式等于 lim_{x->0} sum h(mx)x.
不严格地看，结果应该等于h在[0,\infty)上的积分。 |
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B********e
发帖数: 10014 | 36 有形式证明（如果用laplace变换的话）
把1/x写成 int_0^\infty{exp(-sx)ds}交换积分号变成先对sin(x)的
laplace变换得到1/(1+s^2)然后在R上积分得到pi
只是形式证明因为需要证明积分可交换先 |
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A****e
发帖数: 44 | 37 Let f and f_n be functions from R to R. Assume f_n(x_n) --> f(x) as n-->
infty and for any sequence x_n --> x. Show that f is continuous.
书后的一道练习题，想了好久还没什么招，请指教。
多谢了! |
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B****n
发帖数: 11290 | 38 Assume X is a brownian motion in R^d
Does a radnom variable M exist such that for almost every w
|X(w,t)-X(w,s)|<=M(w)||s-t||^{1/2} for every s,t in R^d and
EM^2<\infty?
Here, ||.|| is the Euclidean distance in R^d
簡單的說就是Brownion Motion是不是滿足Holder condition with constant 1/2 我印
象中好像是對的 請大俠指點一下 謝謝 |
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A*******r
发帖数: 768 | 39 Definitely Not!
It is a two-stage problem.
Actually,
D(y)={min_{x>=0} c'x+y'(Ax-b)}
= y'b, if c'-y'A \ge 0,
-\infty, otherwise |
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a*****k
发帖数: 704 | 40 【 以下文字转载自 Quant 讨论区 】
发信人: artwork (嘿嘿), 信区: Quant
标 题: how to show this
发信站: BBS 未名空间站 (Sun Jul 1 16:00:46 2007), 转信
lim_{\beta -> \infty} \sup_{0 \leq t\leq T}|\exp(-\beta t) \int^t_0 \exp(\
beta s)dW_s| = 0, a.s.
anybody has any idea? |
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l****e
发帖数: 23 | 41 又想了一下，好像结论不对啊，但是没有把握，请批评指正。：）
Problem: \lim_{\beta\to\infty}\sup_{t\in[0,T]}|\int_{0}^{t}e^{\beta(s-t)}dW_
{s}|=0, a.s.?
As before we denote M_{t}:=\int_{0}^{t}e^{\beta(s-t)}dW_{s}.
Integration by parts gives
M_{t}=W_{t}-\int_{0}^{t}\beta e^{\beta(s-t)}W_{s}ds.
Clearly there exists a subset \Omega_{1} of \Omega with probability 1 such
that:
1, for all \omega\in\Omega_{1}, W is continuous on [0,T] and
2, for all \omega\in\Omega_{1}, there exists a t*\in[0,T] such that W_{t*}\
neq0, namely, |W_{t* |
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l****e
发帖数: 23 | 42 呵呵，我翻看原题了，还是没想明白。如果你弄清楚了恳请告诉我，谢谢！：）
另外，前面第一章“停时”一节开头有个小问题，我们当初读的时候有些迷惑，一并请
教：
Let X be a stochastic process and T a stopping time of (the filtration
generated by X) {F^{X}_{t}}.
Suppose that for any \omega, \omega' in \Omega, we have X_{t}(\omega)=X_{t}(
\omega') for all t in the
intersection of [0,T(\omega)] and [0,\infty). Show that T(\omega)=T(\omega'). |
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B********e
发帖数: 10014 | 43 the second way is, take it as principal integral,so
2\pi*the integral=lim_{N->\infty}\int_N^N{cos(kx)dk}
which can be explained(or proved) in the third definition there
descri |
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j******w
发帖数: 690 | 44 no
Here is just a hint:
If \Sum_i \mu(E_i)<\infty, then the index set of positive measure sets must
be countable.
Here I just simply assume ``the sum" is the least upper bound of all
countable sums.
Actually one can show that there exists no uncountably many mutually
disjoint positive measure sets. |
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H****h
发帖数: 1037 | 45 来自主题: Mathematics版 - 序列和级数 假设（X_n)是一个实数序列，(S_n)是部分和序列。即，S_n=X_1+X_2+...+X_n。
求证：如果级数 \sum_{n=1}^\infty (X_n/n) 收敛，那么 S_n/n -> 0。 |
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G******i
发帖数: 163 | 46 Dentoe by N(epsilon) the epsilon neighborhood of S.
For each small \epsilon>0, choose a smooth
f_\epsilon : R^3 -N(1/2*epsilon) -> (1/2*epsilon,\infty)
and uniformly approximates d_\epsilon(x) on R^3 -N(1/2*epsilon)
Define d_\epsilon(x) on R^3 as follows:
d_\epsilon(x) =d(x)^2 in N(epsilon);
d_\epsilon(x) =f_\epsilon in R^3 -N(2*epsilon);
in between, make a smooth and positive transition.
R^
,
continuous
(i
=0 |
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d***s
发帖数: 55 | 47 x^T * A * x =0, A 对称矩阵， x 列相量
怎么得出 as ||x|| goes to \infty, ||A|| goes to 0? |
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c*******h
发帖数: 1096 | 48 不能得出，如
+- -+ +- -+
x = | n |, A = | 1/n^2 0 |
| 1 | | 0 -1 |
+- -+ +- -+
as n goes to infty, both 2-norm and F-norm of A tends to 1. |
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d***s
发帖数: 55 | 49 其实题目最初是要证明
if f(kx)=kf(x) for any k
then as ||x|| goes to \infty, the Hessian matrix of f goes to 0
大家帮忙看看吧 |
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c*******h
发帖数: 1096 | 50 ok. the hessian need not be zero.
the proof of your original problem:
df(x)
let ------- = g(x), and let y = kx.
d(xi)
kdf(x) d(kf(x)) df(y)
then -------- = ---------- = ------- = g(y) = g(kx).
kd(xi) d(kxi) d(yi)
hence g(x) = g(kx).
dg(x)
let ------- = h(x).
d(xj)
dg(x) dg(x) dg(y)
then -------- = -------- = ------- = h(y) = h(kx).
kd(xj) d(kxj) d(yj)
hence h(x) = kh(kx).
thus when x->infty, h(x)->0.
therefore hessia |
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