d***s
发帖数: 55 | 1 不太明白，尤其是 x goes to infty, h(x) goes to 0 是怎么得到的？
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( |
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d*z
发帖数: 150 | 2 \lim_{n->+\infty}(\sum_{i=1}^{n}\frac{(-i)^{n-i}e^i}{(n-i)!})-2n=\frac{2}{3}
如附件的图片 |
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R*********r
发帖数: 1855 | 3 公式是对的，你仔细看看。
事实上准确的E(s)的表达式也很简单。
对于“不太好”的分布函数f(x)，上面那个近似表达式也是靠谱的。
比如f(x)=delta(x-x0)，很显然准确值应该是[s/x0]（向上取整），上面的公式给出s/x0+1/2。
完整的公式给出E(s)=s/x0+1/2+2\sum_{n=1}^{\infty}sin(2nπs/x0)/(2nπ)=[s/x0]（向上取整） |
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h***1
发帖数: 40 | 4 For x/1+|x|, when x goes to +infty, it goes to 1. Thus a cauchy sequence is
easy to get. |
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m*******s
发帖数: 3142 | 5 For subsets A and B of R, define A+B={a+b|a in A, b in B}
suppose that B is a Borel set, Prove that A+B is a Borel set if A is open
我的思路 大概是 這樣的. 利用 Lindelof定理 ,
A+B= \bigcup_{b\in B}(A+b)=\bigcup_{i=1}^{\infty}(A+{b}_{i}) where {b}_{i}\
in B
再利用Borel set countable union的封閉性,似乎已經完成了證明,里頭根本沒有用到B
is a Borel set. 請問我的 證明是否正確? |
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m*******s
发帖数: 3142 | 6 Let I be a nonempty set and {x_l}l∈I an indexed collection of nonegative
real numbers, that is, x_l≥0 for each l∈I ,Define
\sum_{l\in I}x_l=sup \left \{\sum_{l\in F}x_l: F finite, F\subset I \right
\}
where each sum in the set on the right is the ordinary sum of a finite
collection of real numbers
Show that if \sum_{l\in I}x_l<\infty, then {l∈I:x_l>0} is countable.
这题看起来好像很显然，但是我说不清理由，请大家说说证明关键的地方。 |
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R*********r
发帖数: 1855 | 7 记E_n={l∈I:x_l>1/n}，每个E_n都必须是有限的，否则\sum_{i\in E_n}x_i>|E_n|/n=
\infty，于是{l∈I:x_l>0}=UE_n是可数的。 |
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n******t
发帖数: 4406 | 10 This question is effectively equal to show for a poisson process N(t) with
parameter 1,
lim N(t)/t = 1 as t-> infty, something called elementary renewal theorem.
a |
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v**i
发帖数: 50 | 11 你这个问题，我先试着简化一下。我假设$z_s$ 恒等于-2.22, 这样你的问题就是个确
定性的优化问题。重新写目标函数（把已知的y_s 从目标函数分离出来），得到
E_t[\int_{t}^{\infty} [e^{-2.22}y_s-x_s]ds],
你的约束条件还是以前那个 dy_s=(x_s-0.1y_s)ds. 你现在面临的是线性约束，目标函
数也是线性的，这种问题一般都存在技术性的问题，具体说，如果参数稍微不对，优化
以后目标函数就是正无穷, 所以这类问题本身就没设计好。你从哪儿拿到的这个题？是
从文献，教科书里面看到的？ 还是自己设计的？ |
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a*******h
发帖数: 123 | 12 可以写成这么一个公式, 不知道是不是你要的,
\sum_{i=1}^\infty \lfloor \frac{k}{p^i}\rfloor
用k |
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a***n
发帖数: 3633 | 13 请问一个定义在正实轴上的函数f。|f|_\infty=1.
f的傅立叶级数的系数们要满足什么条件啊？
什么书是说这个问题的啊？谢谢啦！ |
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a***n
发帖数: 3633 | 14 傅立叶展开可以看作用三角函数为基对一个函数在L2意义下的
最佳近似。请问如果求一个函数对于三角函数基在无穷模下的
近似？换言之，如果使得最大近似误差达到最小？
谢谢。 |
|
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D**u
发帖数: 204 | 16 A guy starts walking from position x_0 on earth (we treat earth as a sphere
with radius R). He randomly chooses a direction and walk 1 meter (this is
the spherical distance) and reaches position x_1. He repeats the walk n
times and arrives at position x_n.
Question:
(1) what is E((x_n - x_0)^2)? (here ((x_n - x_0)^2 is the square of the
Euclidean distance, not the spherical distance).
(2) does E((x_n - x_0)^2) converge when n --> \infty? |
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q********e
发帖数: 1255 | 17 analytically, let f(x)=\sum_1^\infty (x/2)^n, then your sum=f'(1)
certainly you need to analyze the convergence |
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Q***5
发帖数: 994 | 18 This can be proved:
For any epsilon >0, there exists K, such that \sum_K^\infty i*P(i)<\epsilon
Let N big enough, such that K/N<\epsilon.
Now, for any n>N,
\sum_1^n i*P(i)/n <= (\sum_1^(K-1) K*P(i)+ n*\sum_K^n P(i))/n
epsilon< 2*\epsilon |
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l****8
发帖数: 77 | 19 这个复积分的解能用closed form表示出来吗？
$\int_0^\infty e^{(-1/6+i)y}dy$
谢谢 |
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l******r
发帖数: 18699 | 20 我觉得应该可以证明这个弧上的积分收敛到零当R->infty |
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l******r
发帖数: 18699 | 21 for any e>0,A:={x:|f(x)|>a-e} is not zero measured
so ||f||_p>(a-e) (measure(A))^{1/p}
letting p go to infty |
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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.) |
|
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a***g
发帖数: 2761 | 24 Actually,Let X be a non empty set and Y be a real Banach space; then the
real normed space of bounded functions from X into Y is a real Banach space.
I also consider of L_\infty(R). But I'm not sure it's a Banach space. I don'
t know how to proof the completeness of it, since the whole real line cann't
fit a finite measure in common case. |
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j***i
发帖数: 1278 | 25 \frac{d\left(\int_{-\infty}^{x+a}f(x+a-t)p(t)d t\right)}{d x}
其中 a 是常数， |
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z****e
发帖数: 702 | 26 用borel-cantelli引理，似乎证明的是事件概率的和吧，
此处是随机变量的和。
{k
,
infty}
probabi
mean |
|
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E*****T
发帖数: 1193 | 28 你看看对不对。在全空间是L^2 Hilbert space的前提下。
Let y_2=x_1-x_2, y_3=x_2-x_3,...,y_n=x_{n-1}-x_n, D=span{y_2,y_3,...y_n}
D' be D's orthogonal complement w.r.t L^2 inner product. y_1= projection of
x_1 in D'.
Then your question is equivalent to find sup_{y \in D'} =sup_{y \in
(D'+D)} = ||y_1||_{L^\infty}. |
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l******r
发帖数: 18699 | 29 The normaization is only under L^2 norm, i.e., all the eigenfunctions have L
^2 norm equal to one. This will make their H^r norms inflating with the
corresponding eigenvalues since the eigenvalues must approach +infty.
I agree with you about mimicking SL's theory, probably that is the right
track. So little about this in literature:-)
>n
2
configuration |
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a***n
发帖数: 3633 | 30 如果对于平稳随机过程x(t), Ex(t)=0. R(tau)是它的autocovariance
我知道 R(0)可以用 r=\int_0^\infty{x(t)x(t)dt}来估计，我想知道
如果x(t)是ergodic的，那么r和R(0)之间有没有类似Chebyshev之类的
不等式，就是类似
P(|r-R(0)|
谢谢。 |
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l****8
发帖数: 77 | 31 Let $\alpha\in (0,1)$. A function $f:[0,\infty)\to \R$ is called $\alpha$-
concave if $f(kx)\ge k^\alpha f(x)$ for any $k\in (0, 1)$ and $x\ge 0$.
最简单的例子当然是 $f(x)=cx^\alpha$, c is a constant;以及这种函数的线性组合. 我的问
题是有没有其他函数满足这个条件？
另外，这个函数必须是非减的。 |
|
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p***c
发帖数: 2403 | 33 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 |
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f**********d
发帖数: 4960 | 34 p(x) is a probability measure, q(z=k|x) is a likelihood, where k=1,2,...,K,
with \sum_k q(z=k|x)=1. The integral \int q(z=k|x) dx =infinity, for
k=1,2,...K. Questions:
1. Since q(z=k|x), k=1,2,...,K are measures in the l^\infty space, if they
are mutually singular, can we say that they are 'orthoganal to each other'?
I have seen that the sign to denote the mutually
singular is the same as the orthoganal sign. But I don't know if we can just
say the 'mutually singular' as 'orthogonal'?
2. Althoug... 阅读全帖 |
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f**********d
发帖数: 4960 | 35 或者这样描述：
x是l^2空间的，但很可惜y不是l^2的，但y是l^\infty的。
但是\int xy 总是好的（即小于无穷，这是l^2空间的内积定义。)
现在我的问题是：我真的很想把\int xy说成x和y的内积（就好像y也在l^2里一样），
而且
想把这个积分记做（这是通常的内积表达式），我想知道这样是否可以？
,
?
just |
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B********e
发帖数: 10014 | 36 given x in l^2, the fact y in l^\infty does not imply ' \sum x_ny_n 是好的'
x_n=1/(1+|n|), y_n=1 |
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n***p
发帖数: 7668 | 38 其实是微积分里面的例题。
We need only consider S_n = sqrt( 2 + sqrt(2 + sqrt(2+ ...) ) )
with n level of sqrt and consider its limit as nto infty.
S_n is increasing as n increases, and Sn is bounded above (by 2). Then
S_n has a limit, say S.
Recursively S_n = sqrt(2 + S_{n-1}). Taking limit on both sides, we have
S = sqrt(2+S). Solving it we obtain S=2.
What LZ wants is just sqrt(1+S) = sqrt(3). |
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w***w
发帖数: 84 | 39 任何有限集上的metric 可以等距嵌入l_\infty. Bourgain证明任何n个点可以嵌入欧式
空间
with O(log n) multiplicative distortion. 这些有不少算法上的应用。 |
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w***w
发帖数: 84 | 40 任何有限集上的metric 可以等距嵌入l_\infty. Bourgain证明任何n个点可以嵌入欧式
空间
with O(log n) multiplicative distortion. 这些有不少算法上的应用。 |
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l*3
发帖数: 2279 | 41 是有多个这样的lambda, 你要找的, 是最小的那个lambda (这里以 表达式为 (H -
lambda I) x = b 为准, 其中H对称)
我学的trust region 是教可以用牛顿法/拟牛顿法去找.
也可以对问题先进行分析:
你可以先对H做特征矩阵分解 (特征向量正交), 然后把 |x|^2的精确表达式写出来, 是
一个关于lambda的分段凸函数, 其值在每个特征值处去向infty (如果b在对应的特征向
量上分量不为0的话).
其实我的意思是, 这个问题我好像不知道有啥精确求解方法. 如果你是出于优化或是什
么的实际需要, 那有问题可以继续留言问. 如果是理论表达式的话, 那就无力了. |
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B********e
发帖数: 10014 | 42 no.
let c=1, F1=t^2, F2=t. You can show b->1/6, as a->\infty.
, |
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B********e
发帖数: 10014 | 43 \int_x^\infty (e^(i*a*t) - 1)*t^(-1)*e^(-b*t)dt=
Ei(-b*x)-Ei(-b*x+ia*x)+i\pi
combine with simon's post, see if that's what you want |
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l******r
发帖数: 18699 | 44 尼玛，你告诉不用复分析怎么算exp(x)/(1+x^2)在0 to infty上的积分？ |
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h******y
发帖数: 26 | 45 Some friend asked me the following question:
For a real scalar field \phi, assume that H = H_free - \int d^3\ x J \phi. J
(x, t) is just some real number, source, or background field, without second
quantization. Now, what is the amplitude \psi(x, t) for finding a particle
at time t(before, during, or after source is on/off) at position x? The J(x,
t) is nonzero only for finite period of time. And the initial state is
vacuum, when t --> -\infty
This question looks simple. However, I cannot find |
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i*******n
发帖数: 188 | 46 I don't know the name of this theorem, when we learn the course of complex
analysis there was no name for that.
But Jordan' lemma is a different and stronger theorem, saying that if
f(z) -> 0 as z-> \infty, then
\int f(z) e^{i p z} dz =0 for the infinite contour.
See this:
http://en.wikipedia.org/wiki/Jordan%27s_lemma |
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c****e
发帖数: 2097 | 47 not familiar with any numerical methods, but generally speaking, you have to
know the asymptotic behavior of your integrand to decide what to do with
your -\infty.
if it dies away over there, for example, i guess you can take some large
negative value.
however, if the case is not so, then you should manipulate your integrand
and integration domain a bit doing some mapping so as not to lose the
important stuff.
实函数，而是比较复杂的含时算符（矩阵）的函数， 不能直接积分得到解析表达式，
而且积分表达式随着里头算符的具体表达式不同而不同。
数来代替负无穷？ |
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v**g
发帖数: 20 | 48 =\int_0^T\int_0^T dtdt'
=v^2 \int_0^T\int_0^T dtdt'
\in [0,1],assume =exp{-|t-t'|/t_c},
where t_c is the correlation time of the velocity. Physically, t_c means we
have to wait such a time before we find different velocities. Then
=2*v^2*t_c*[T + t_c*exp(-T/t_c) - t_c].
As T->\infty, scales as 2*v^2*t_c*T, by definition,
the diffusion coefficient D = v^2*t_c/2. |
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N***m
发帖数: 4460 | 49 cohen tannoudji
atom photon interaction
我记得里面好像有讲这个的。
大意是观测时间T->infty，基本上就是个等于号。不过本来也就是估算，
没必要太精细。 |
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x***1
发帖数: 197 | 50
you can define a hilbert space by the direct sum
H = \sum_{n=0}^\infty H_n
to avoid such "difference". |
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