f******k
发帖数: 297 | 1 given a square matrix A, what is the minimum of
abs((x'Ax)/(|Ax||x|)) over all nonzero x, where |.| is
Euclidean norm and x' is the transpose of x? Or
equivalently what is the minimum angle between Ax and x? |
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F******n
发帖数: 160 | 2 Kind of interesting. I'll give a quick shot here - just some quick clues. I'll
also along give my reasonings, which should lead to some rigorous proofs or
dis-proofs. No much rigorous analysis has been conducted, so I may miss things
here. Besides, not sure about your motivation for this problem, so I cannot
determine if the answer gives what you want.
First, some basic corollaries can be established:
1. L1 and L2 must have the same rank (the same number of nonzero diagonal
elements)
2. without |
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b*****e
发帖数: 474 | 3 现在的 PUTNAM 好象简单了. 比如这几个:
Problem A1
Show that every positive integer is a sum of one or more numbers
of the form (2^r)(3^s), where r and s are nonnegative integers and
no summand divides another.
Problem B1
Find a nonzero polynomial P(x,y) such that P( floor(a), floor(2a) )=0
for all real numbers a. (Note: floor(v) is the greatest integer less
than or equal to v.)
这道题初中生应该能做, 如果时间够:
Problem B2
Find all positive integers n, k_1, k_2, ..., k_n such that
Sum_{1<=i<=n} k_i = 5n-4, and
Sum_{1<=i<=n} |
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c*******h
发帖数: 1096 | 4 if alpha and beta are linearly independent, the two nonzero eigenvalues
are
/----------------------
(ax+by) + / (ax-by)^2 + 4(az)(bz)
-/ |
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B********e
发帖数: 10014 | 5 一个貌似应该很easy的题怎末也搞不定
a(t) bounded continuous, y(t) is a nonzero solution of y''+a(t)y=0 such that
lim_{t->\infty} y=0. show there is a solution on [0,\infty) which is not b
ounded.
证明或者证伪
3x |
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G******i
发帖数: 163 | 6 Let z(t) be another solution that is linearly independent from y(t).
y(t)->0 => y''(t)=-a(t)y(t)->0 => y'(t) ->0 as t->infinity.
Suppose z(t) is bounded on [0,infinity).
We have z''(t)=-a(t)z(t) bounded and hence z'(t) bounded.
but (yz' -y'z)'=0 => yz'-y' z=C (nonzero).
Contradiction! |
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G******i
发帖数: 163 | 7 Yes. That's enough for us to see your original claim was wrong.
Consider this one:
U(x,y)=[1-g(x)]*[y-y(1-y)h(y)], with
g(x)=exp(-1/x) *sin(1/x),
h(y)=exp(-1/y^2).
In this example, whatever nonzero k you take,
the desired inequality does not hold for a near 0. |
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o**a
发帖数: 86 | 8 U is a unitary matrix associated with an euler rotation.
A is a 3x3 matrix.
U.A.U(-1) converts A to another matrix. U(-1) is the inverse of U.
Here I need to treat A as a vector of 9-dimensional space, e.g., the basis
vectors can be matrices with only one nonzero element. An operator Q can be
defined that converts A to another vector just as U.A.U(-1) does. Apparently
Q is a 9x9 matrix.
Is there any simple way of showing Q is a unitary matrix? It seems so by my
brutel calculation on reduced dime |
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c******m
发帖数: 41 | 9 Let v be a non-zero column vector in R^(n)( i.e n维列向量空间） and v ^(T) (
i.e v 的转置）be its transpose.
(a) Show that the n*n matrix A defined by v*v^(T) ( i.e v乘以v的转置） is of
rank one and is a diagonalizable matrix.
(b) If u is a unit vector in R^(n), show that the matrix H=I-2u*u^(T) is an
invertible matrix. Here I is the identity matrix. ( Hint: Note that u^(T)*v
is a scalar for any column vector v. You may show that the null space of H
is trivial)
(c) If v and w are any two nonzero orthogonal vecto |
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Q***5
发帖数: 994 | 10 64 Let V be the real vector space of real-valued functions defined on the
real numbers and having derivatives of all orders. If D is the mapping from
V into V that maps every function in V to its derivative, what are all the
eigenvector of D?
answer: All nonzero functions of the form (ke)^(lamda*x), where k and lamda
are real numbers
don't know how to get this?
This seems to simple: you are looking for all functions f\in V, such that D
(f) = \lambda f, where \lambda is some constant. From the d |
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l*******t
发帖数: 3 | 11 来自主题: Mathematics版 - 关于不等式 如何用Farkas Lemma推Gordan's theorem
Gordan's theorem: Either Ax < 0 has a solution x, or ATy = 0 has a nonzero
solution y with y ≥ 0.
严格不等号怎么处理呢？ |
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a***n
发帖数: 3633 | 12 Thanks you two. I don't need x(t) to be nonzero throughout the interval.
BTW: in what course/books such knowledge was taught?
]?
t
0 |
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c**********w
发帖数: 1746 | 13 请原谅简单问题，我刚开始看dynamic systems，正在看saddle fixed point。这个地
方讲到，discrete-time system的unstable invariant manifolds和stable invariant
manifolds可以intersect at nonzero angle，但是continuous-time就不行。我完全
没搞懂，这个intersect是指相互无限接近类似limit point，还是指有非空交集? 为什
么discrete-time可以有transversal intersection, 而连续系统不行？区别在哪里？
刚刚学，问题很弱，谢谢指教！ |
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z***c
发帖数: 102 | 14 OK I was not being precise. But I think the OP was asking about the same
fixed point.
The concept of transversality in differential topology is the following: Two
submanifolds M, N of the manifold W are transversal at x if
TxM + TxN = TxW
the sum is in the sense of vector subspace. This is somewhat different from
the naive sense of transversality. If two submanifolds intersects at at a
point, then the transversality condition is the same as having "nonzero
angle".
As I explained, the intersec... 阅读全帖 |
|
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s*****V
发帖数: 21731 | 16 https://www.quantamagazine.org/20160628-peter-scholze-arithmetic-geometry-
profile/
2010, a startling rumor filtered through the number theory community and
reached Jared Weinstein. Apparently, some graduate student at the University
of Bonn in Germany had written a paper that redid “Harris-Taylor” — a
288-page book dedicated to a single impenetrable proof in number theory —
in only 37 pages. The 22-year-old student, Peter Scholze, had found a way to
sidestep one of the most complicated parts of... 阅读全帖 |
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a******n
发帖数: 98 | 17 Thank you for the time and efforts in answering my questions!
I tried the case of elastic plate you used. The result is better in elastic
plates. As for my problem, I have the coupled electromechanical field. My
result is not so good. The nonzero transverse normal stress Szz on the free
surface is more than 10% of the maximum value in the whole plate. I guess it
may be caused be the large electric field at that point Ez, which is
related to the calculation of stresses.
In addition, I found out t |
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h******y
发帖数: 26 | 18 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*o
发帖数: 202 | 19 刚才好像出问题了，再发一次。
SIM是用一个 one-dimensional grid pattern作为illumination的trick来achieve
optical sectioning的。因为zero-order (in- and out-of-focus light) won't
attenuate with defocus, but nonzero-orders do, 所以去掉DC项和AC项中的
modulation (communication 中的square-law detection)就得到optically sectioned
images.
我不明白的是为什么 zero-order doesn't attenuate with defocus 呢？又或者为什
么其他项就会随着defocus而越来越弱？
谢谢！ |
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j**p
发帖数: 53 | 20 if you'd like a bit more rigor, here it is :
(let's drop a and b; that wouldn't change the essense of the discussion)
z=c-c is actually c_1 - c_2, where c_1 and c_2 are independent random
variables with identical distribution. So, even though z has zero mean, its
variance is nonzero (Var[z]=Var[c_1]+Var[c_2]=2Var[c]; i.e. the std of z is
sqrt(2)*(std of c)) |
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s*****i
发帖数: 467 | 21 I partly agree with you. Anapole moment is from the multipole decomposition,
but it is nonzero only due to weak interaction.
dovich |
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w****1
发帖数: 4931 | 22 Actually, at genus 3 it is still okay - the moduli space of Jacobians cover
the entire Siegel upper half space
except for some singular locus of nonzero codimension. So one should be able
to express genus 3 string
amplitudes in terms of ratios of entire Siegel modular forms, I suppose.
Starting at genus 4 it really becomes
hopeless because little is known about the Teichmuller modular forms of
genus 4 and higher.
Z)
functions |
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A****t
发帖数: 69 | 23
It's nonobvious how does your second sentence follow from the first.
There are many many cases where one could stand to benefit from gain
of your "opponent" (a classical case being iterated prisoners' dilemma).
"Nonzero sum games" have certainly been studied extensively in Game Theory.
I'm not certain what's meant here. Multiple outcomes can often (if
not always?) be reduced to single outcome. It's only a matter of
formulation, not a fundamental limitation of Game Theory.
Two points:
1. As po |
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p*****k
发帖数: 318 | 24 not sure what exactly DuGu had in mind, but im guessing that
he wants the radii of these circles nonzero and bounded.
otherwise, e.g., the infinite str8 line R could be considered
as a circle with infinite radius:
(x-R)^2+y^2=R^2 with R->infty |
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p*****k
发帖数: 318 | 25
native, i like your problem 3, and nice sol.
for your new problem, what do you exactly mean by "tie"?
the minimum # among three guesses >0, so nonzero chance
that everyone loses? or the case that at least two ppl
guess the same #? |
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p*****k
发帖数: 318 | 26 ilovekl, some hints:
(this ought to be easy to find the answer by yourself)
what is the definition of "martingale"?
a martingale does not have to have expectation of zero, you can
add in any nonzero constant. so in the definition,
which quantity is zero? |
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L*****e
发帖数: 169 | 27 For k=1
E(W(t+delata)^3-W(t)^3)=E((W(t+delta)-W(t))^3)
=0 Due to symmetry.
How did you get it nonzero? |
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p*****k
发帖数: 318 | 28 DuGu, thanks for the hint. one thing i could think of is to use
contour integral (or in general, line integral), though i have
no idea what could vanish for an arbitrary rectangular contour,
while nonzero for some other particular contour.
(or in terms of the line integral, a field not doing work for
rectangular paths, but not for some other paths) |
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p*****k
发帖数: 318 | 29 needlaw, assume standard BM with no drift, i interpret your
question as the prob. of first-time hitting "a" while the path
has hit "b" earlier.
(instead of the conditional prob of hitting "a" given the path
already hit "b", which is just 1)
assume a and b are not equal and both nonzero. couple of cases:
if a>b>0 or a
if b>a>0 or b
if a<0
event, then as Taikonaut said, one just needs the prob of never
hitting "b" |
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p*****k
发帖数: 318 | 30 MsPiggy, im probably saying something stupid again:
if the call is priced at 10, the arbitrage i had in mind is
to short 1 stock, long 1 call and deposit the rest of 100
into the bank account.
seems the nonzero prob of the stock being 0 tmr results
the arbitrage, no? |
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p*****k
发帖数: 318 | 31 solarquant, if the prob of ever reaching 10 is strictly <1,
then there is nonzero prob of the first passage time being
infinity, hence the infinite average.
the nontrivial case is, as you said, when p=1/2. all points
on the line would be hit infinitely many times with prob of 1,
while the average first passage time is infinite.
one argument (which i personally feel misleading) is to consider
the well-known two-barrier version then take one of the barrier
to the infinity on the opposite side |
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x******a
发帖数: 6336 | 32 11. a. law of the iterated logarithm.
b. weak law of large numbers.
c. P(|B_t/t|> x)=P(|B_1|>\sqrt{t}x)\to 0 for any x>0.
13 E(S_t)=E(S_0). lim(S_t)=0 (assuming \sigma is a nonzero constant) because
sigma*B_t-\sigam^2t/2 ->-\infty. |
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t**********a
发帖数: 166 | 33 Don't think so,
for the new process V(t) = W(t-r), V(t+d)-V(t)=W(t-r+d)-W(t-r) for d
r+d)-W(t-r) is known to {F_t}, thus expectation is nonzero. |
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l**********n
发帖数: 72 | 34 In Mermin and Wagner's paper (PRL, vol. 17, No.22, 1966) they rigorously
proved that "At nonzero temperature, there is neither ferromagnetic nor
antiferromagnetic phase transition in 1D or 2D isotropic Heisenberg model."
The key words are "isotropic Heisenberg model". For 2D Ising model, the
assumption is that the spin either points to the + or - z direction,
therefore, the Ising model contains only discrete symmetry. While for
Heisenberg model, the Hamiltonian looks like H=J[S(r)*S(r')], where |
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b****d
发帖数: 1311 | 35
(1) rk(A) = rk(QR) =< min(rk(Q), rk(R)) . You can use linearly dependence of
row or column vector to prove this.
Also, any invertible matrix can be presented as products of fundamental
transformation matrixes. so if diagonal entries of R are nonzero, it's just
doing some basic transformation on column vectors of Q. so it's rk n.
(2) rk(A) = rk(Q) >=Q . first we can know rk(A) =< min(rk(Q),rk(R))=rk(Q) .
second, Let Q' be the transpose of Q. rk(A) >= rk(Q'A) = rk(Q'QR) = rk(DR)=
rk(Q), D is a nx |
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y**t
发帖数: 50 | 36 I dunno which version of stone-w theorem you know
the one I know is that if A contains a nonzero
constant function,then A is dense in C(X).And if you
know the proof of the s-w theorem,it should not be
hard for you to prove this exercise.
Use contradiction to prove that there is a point p
in X such that any f in A f(p)=0 if A is not dense
in C(X).if any point x in X there is a function such that
f(x)!=0then f!=0 in a neiborhood of x,using compactness
you can find f1 through fl such that f=f1^2+.. |
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k*******y
发帖数: 56 | 37 please forget about what I said, this is the standard proof:
consider the following matrix product
sorry for the messy notation
1
matrix(I,0,-B,xI, nrow=2)*matrix(xI,A,B,xI)
and 2
matrix(xI, -A,0,I) *(xI,A,B,xI)
their det differ by only a factor of a power of x
on the other hand the products are nothing but
matrix(xI,A,0,x^2I-BA)
and
matrix(x^2I-AB,0,B,xI)
now let x=sqrt(\lambda)
can get that the characteristic polymonials of AB and BA
differ only by a vector of \lambda^{m-n}
ie, the nonzero eig |
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w****o
发帖数: 20 | 38 Only when the laser energy is below the bandgap of these two materials.
Because refractive index will have imaginary part when above the bandgap Eg,
ie, N(E)=n(E)+i*k(E), k>0 when E>Eg; and k_material1 can be diffrent from
k_material2 if they are nonzero.
Reflection is
R=((1-n)^2+k^2)/((1+n)^2+k^2). |
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f**d
发帖数: 768 | 39 通过这种方法怎么计算entropy?
感谢！
或如何理解下面导出的Nc and S?
Suppose there n particles randomly distributed in a region of N positions.
Even if n is much smaller than N, there is a nonzero chance
for coincidence, i.e., some positions will get more than one particles.
Since the probability for a particle to fall in any given position is 1/N,
the number of coincidences is easily estimated by:
Nc=n(n-1)/2 *1/N
(why?)
Let us define R to be the coincidence rate which is the probability of
finding a coincidence per tri |
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s*********e
发帖数: 1051 | 40 in a standard poisson, zero outcome has nonzero probability.
the zero-truncated poisson is a standard poisson with zero probability of
zero outcome. |
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x********u
发帖数: 64 | 41 thanks
那结论就是x1和x2这2个variable都对y没有影响吗？
Summary Statistics for col by mis
Controlling for row
Cochran-Mantel-Haenszel Statistics (Based on Table Scores)
Statistic Alternative Hypothesis DF Value Prob
1 Nonzero Correlation 1 0.9728 0.3240
2 Row Mean Scores Differ 2 3.7930 0.1501
3 General Association 2 3.7930 0.1501 |
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m*********7
发帖数: 343 | 42 你的问题就看两个东西之间的关系，应该可以不用regression model吧，对于每个问题
，把数据写成
Q \GPA A B C D
student1 1
2
3
4
5
student2 1
2
3
4
5
.....
然后用 proc freq; table student*Q*GPA / cmh; run; 用nonzero correlation那行
的p-value就行了，这样就知道对于每个问题，选什么level是不是和GPA高低有关了。
或者在proc logitistic， proc catmod和proc genmod里面用proportional odds
model, 但是每个问题的1，2，3，4，5要写成dummy |
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S******y
发帖数: 1123 | 43 should be --
...
else if level=2 or level= 3 then expertise='Medium';
...
3 is always true sinceany nonzero, nonmissing constant is always true |
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l****u
发帖数: 529 | 44 No, there will not be such misleading conclusions. After fitting the
model, you go to
check the slope. If there are nonzero slopes, you make a contrast to
demonstrate unequal slope. If the interaction exists, the p-value of the
contrast must be significant.
and A |
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c***z
发帖数: 6348 | 45 We use a two stage approach, i.e. add a dummy for zero/nonzero cases,
together with the original variable. |
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D**u
发帖数: 288 | 46 matrix 是square的？nonzero 多不多？是不是sparse的？
anyway, 可以考虑Partial SVD + Lanczos diagonalization。
简单说：
Partial SVD 是用k dimension 去 approximate n dimesion （k
后类似 cross validation 看看是不是converge。
具体对每个k dimension的用Lanczos就是一个block接一个block的diagonalization，
然后算eigen value。
相同道理，类似的方法也有不少，不过不知道是不是你想要的答案
这些都是可以stream line的， 程序写起来很麻烦，不过如果用R早就有人package好了
。强烈推荐
R 的 irlba package参见
http://cran.r-project.org/web/packages/irlba/vignettes/irlba.pd
可以跟bigmemory package 一起用，100GB data no problem。 |
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