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This is an excellent and deep question.
While traditional textbooks (like mine) tend to promote Bayes factors as equivalent to posterior probabilities of the null and alternative hypotheses or of two models under comparison, which is formally correct as detailed in the following extract from my Bayesian Choice, I now tend to think that the Bayes factor
per se should not be used for decision-making but rather as a measure of relative evidence of one model versus the other. For instance, using $\mathfrak{B}^\pi_{01}(x)=1$ as the dividing line between null and alternative (or between model a and model b) does not strike me as a natural choice. Furthermore, I do not think the 0-1 loss advocated by Neyman and Pearson and later adopted by almost everyone is making much sense and brings any support to the decisional interpretation of the Bayes factor.
My current perspective on the Bayes factor is more in a prior or posterior predictive mode where the behaviour of $\mathfrak{B}^\pi_{01}(x)$ is assessed under both models, in order to calibrate the observed value $\mathfrak{B}^\pi_{01}(x)$ against both prior or posterior distributions of $\mathfrak{B}^\pi_{01}(x)$. This gets us away from the decisional perspective.
[From The Bayesian Choice, 2007, Section 5.2.2, page 227]
From a decision-theoretic point of view the
Bayes
factor is only a one-to-one transform of
the posterior probability, but this notion came out to be considered
on its own ground in Bayesian testing.
The
Bayes factor is the ratio of the posterior probabilities of the
null and the alternative hypothesis over the ratio of the prior
probabilities of the null and the alternative hypothesis, i.e., $$
\mathfrak{B}^\pi_{01}(x) = {\mathbb{P}(\theta \in \Theta_ 0\mid x)
\over \mathbb{P}(\theta \in \Theta_1\mid x)} \bigg/ {\pi(\theta \in
\Theta_ 0) \over \pi(\theta \in \Theta_ 1)}. $$
This ratio evaluates the modification of the odds of $\Theta_0$ against $\Theta_1$ due to the observation(s) and can naturally be compared to $1$, although an exact comparison scale can only be based upon a loss function.
The Bayes factor is, from a Bayesian decision-theoretic point of view,
completely equivalent to the posterior probability of the null hypothesis as $H_0$ is accepted when
$$
B^\pi_{01} (x) \ge {a_1\over a_0} \big/ {\rho_0 \over \rho_1}
= {a_1\rho_1 \over a_0\rho_0},
$$
where
$$
\begin{align*}
\rho_0 &= \pi(\theta\in\Theta_0) \quad \hbox{ and } \nonumber\\
\rho_1 &= \pi(\theta\in\Theta_1)\\
&=1-\rho_0.
\end{align*}
$$
and where $a_0$ and $a_1$ are the penalties for
wrongly selecting the alternative and null hypotheses or the models $\mathfrak{M}_0$ and $\mathfrak{M}_1$. respectively, in Neyman-Pearson formulation:$$\mathfrak{L}(\theta, \varphi) = \begin{cases} 0 &\text{if $\varphi=\mathbb{I}_{\Theta_0}(\theta)$,} \cr a_0 &\text{if $\theta\in\Theta_0$ and $\varphi=0$,} \cr a_1 &\text{if $\theta\not\in\Theta_0$ and $\varphi=1$,}\cr\end{cases}$$ |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
Minimum 31) Let x,y are 2 positive real integers such that \(\sqrt{xy}\cdot\left(x-y\right)=x+y\)
Find the minimum value of \(P\) = x + y
FacuFeri 10/05/2019 at 11:48
We have : \(\sqrt{xy\left(x-y\right)}=x+y\Leftrightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)
\(xy\left(x-y\right)^2=\dfrac{1}{4}.4xy\left[\left(x+y\right)^2-4xy\right]\le\dfrac{\left(x+y\right)^4}{16}\)
so \(\left(x+y\right)^4\ge16\left(x+y\right)^2\) \(\Leftrightarrow p^4-16p^2\ge0\Leftrightarrow p\ge4\)
Equal sign occurs \(\Leftrightarrow x=2+\sqrt{2};b=2-\sqrt{2}\)
32) Let x,y > 0 such that x + y \(\ge\) 10.
Find the minimum value of \(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}\)
Huy Toàn 8A (TL) 01/03/2019 at 04:38
\(P=\left(\dfrac{30}{x}+\dfrac{6x}{5}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\dfrac{4}{5}\left(x+y\right)\ge2.\sqrt{\dfrac{30}{x}+\dfrac{6x}{5}}+2.\sqrt{\dfrac{y}{5}+\dfrac{5}{y}}+\dfrac{4}{5}.10\)
\(P=2.6+2.1+8=22\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{30}{x}=\dfrac{6x}{5}\\\dfrac{y}{5}=\dfrac{5}{y}\\x+y=10\end{matrix}\right.=>\left\{{}\begin{matrix}x^2=25\\y^2=25\\x+y=10\end{matrix}\right.\)
\(=>x=y=5\)
\(Pmin=22< =>x=y=5\)
Huy Toàn 8A (TL) 28/02/2019 at 14:06
\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\left(x+y\right)\left(\dfrac{5}{x}+\dfrac{5}{y}\right)+\left(\dfrac{25}{x}+x\right)\)
\(\left\{{}\begin{matrix}x>0=>x=\left(\sqrt{x}\right)^2\\y>0=>y=\left(\sqrt{y}\right)^2\end{matrix}\right.\)
\(P1=x+y\ge10\)
\(P2=\dfrac{5}{x}+\dfrac{5}{y}=5.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge5.\dfrac{4}{x+y}=\dfrac{5.4}{10}=2\) khi \(x=y\)
\(P3=\dfrac{25}{x}+x\ge2.\sqrt{\dfrac{25}{x}.x}=2.5=10\) khi \(x=5\)
\(P=\sum P\ge10+2+10=22\) khi \(\left(x;y\right)=\left(5;5\right)\)
Study well
Vui Ghét Nét 07/03/2019 at 03:53
Reference Lê Anh Duy
Niushi has 13 packs of milk. She remember that: Her mom said her drinks milk every day and each day drinks 2 packs of milk in minimum, and 3 packs of milk in maximum. What is the minimum of days Niushi can drink with these pack of milk?
Alone 23/10/2018 at 13:01
\(x^2+5y^2-4xy-x+2y-6=0\)
\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)
\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)
\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)
\(\Leftrightarrow-2\le x-2y\le3\)Lê Anh Duy selected this answer.
Sam, Taylor and Pat counted the number of fish in each of their fish tanks. They noticed that Sam’s tank had exactly 25% more fish than Taylor’s tank, and Pat’s tank had exactly 24% more fish than Sam’s tank. If each tank had at least one fish, what is the minimum combined number of fish that could have been in the three tanks?
23) There are two numbers: one has 6 divisors and one has 4 divisors. What is the minimum divisors of their product?
Tôn Thất Khắc Trịnh 31/07/2018 at 14:11
Mininum divisors? That is the number 1, obviously
Or did you mean the minimum NUMBER of divisors?
If so than it's 9, how you ask? To create less permutations with the divisors, the factors should overlap the most, hence, the 6 divisor number is a 5and the 4 divisor number is a 3, so the product would be a 8, having 9 divisors.
Let a,b are numbers which satisfy \(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\)
Find minimum value of the expression : \(P=a^4+b^4+6a^2b^2\)?
Kaya Renger Coodinator 16/08/2017 at 21:35
\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=9\\a^4+2a^2b^2+b^4\ge25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2ab=9-\left(a^2+b^2\right)\\a^4+b^4+2a^2b^2\ge25\end{matrix}\right.\)
We have : a
2+ b 2\(\ge\) 5
=> \(-\left(a^2+b^2\right)\le5\)
=> \(9-\left(a^2+b^2\right)\ge9-5=4\)
=> \(2ab\ge4\)
=> \(ab\ge2\)
<=> \(a^2b^2\ge4\)
<=> \(4a^2b^2\ge16\)
Plus \(4a^2b^2\ge16\) into \(a^4+b^4+2a^2b^2\ge25\)
=> \(a^4+b^4+6a^2b^2\ge41\)
=> Min = 41
That is my opinion :vSelected by MathYouLike
¤« 03/04/2018 at 13:32
{a+b=3a2+b2≥5⇒{a2+2ab+b2=9a4+2a2b2+b4≥25⇒{2ab=9−(a2+b2)a4+b4+2a2b2≥25
We have : a2 + b2 ≥
5
=> −(a2+b2)≤5
=> 9−(a2+b2)≥9−5=4
=> 2ab≥4
=> ab≥2
<=> a2b2≥4
<=> 4a2b2≥16
Plus 4a2b2≥16
into a4+b4+2a2b2≥25
=> a4+b4+6a2b2≥41
=> Min = 41
That is my opinion :v
There is something is wrong here
Change to :
We have a2+b2≥5
=> 9−(a2+b2)≤4
=> 2ab≤4
=> ab≤2
<=> a2b2≥4
=> 4a2b2≥16
Kaya Renger Coodinator 16/08/2017 at 22:06
There is something is wrong here
Change to :
We have \(a^2+b^2\ge5\)
=> \(9-\left(a^2+b^2\right)\le4\)
=> \(2ab\le4\)
=> \(ab\le2\)
<=> \(a^2b^2\ge4\)
=> \(4a^2b^2\ge16\)
¤« 03/04/2018 at 13:33
(*) 1+N=.....=x2/(x-2)2+4
We have x2 ≥
0 , (x-2)2+4 ≥
4 > 0
So 1+N ≥
0 => N ≥
-1 ;equality : x=0
(*)1-N=....=(x-4)2/(x-2)2+4
....... -> N ≤
1 , equality : x=4
¤« 03/04/2018 at 13:32
There is something is wrong here
Change to :
We have a2+b2≥5
=> 9−(a2+b2)≤4
=> 2ab≤4
=> ab≤2
<=> a2b2≥4
=> 4a2b2≥16
Phan Văn Hiếu 28/03/2017 at 21:30
trình bày = tiếng việt đc ko tiếng anh ngại viết lắm |
Bondarenko,Andriy and Ivić,Aleksandar and Saksman,Eero and Seip,Kristian-On certain sums over ordinates of zeta-zeros II
hrj:5110 -Hardy-Ramanujan Journal,January 23, 2019
On certain sums over ordinates of zeta-zeros II
Authors: Bondarenko,Andriy and Ivić,Aleksandar and Saksman,Eero and Seip,Kristian
Let γ denote the imaginary parts of complex zeros ρ = β + iγ of ζ(s). The problem of analytic continuation of the function $G(s) :=\sum_{\gamma >0} {\gamma}^{-s}$ to the left of the line $\Re{s} = −1 $ is investigated, and its Laurent expansion at the pole s = 1 is obtained. Estimates for the second moment on the critical line $\int_{1}^{T} {| G (\frac{1}{2} + it) |}^2 dt $ are revisited. This paper is a continuation of work begun by the second author in [Iv01]. |
In Season 1 Episode 2 of The Big Bang Theory, “The Big Bran Hypothesis”, Penny (Kaley Cuoco) asks Leonard (Johnny Galecki) to sign for a furniture delivery if she isn’t home. Unfortunately for Leonard and Sheldon, they are left with the task of getting a huge (and heavy) box up to Penny’s apartment.
To solve this problem, Leonard suggest using the stairs as an inclined plane, one of the six classical simple machines defined by Renaissance scientists. Both Leonard and Sheldon have the right idea here. Not only are inclined planes used to raise heavy loads but they require less effort to do so. Though this may make moving a heavy load easier the tradeoff is that the load must now be moved over a greater distance. So while, as Leonard correctly calculates, the effort required to move Penny’s furniture is reduced by half, the distance he and Sheldon must move Penny’s furniture twice the distance to raise it directly.
Mathematics of the Inclined Plane Effort to lift block on Inclined Plane
Now we got an inclined plane. Force required to lift is reduced by the sine of the angle of the stairs… call it 30 degrees, so about half.
To analyze the forces acting on a body, physicists and engineers use rough sketches or free body diagrams. This diagram can help physicists model a problem on paper and to determine how forces act on an object. We can resolve the forces to see the effort needed to move the block up the stairs.
If the weight of Penny’s furniture is \(W\) and the angle of the stairs is \(\theta\) then
\[\angle_{\mathrm{stairs}}\equiv\theta \approx 30^\circ\] and \[\Rightarrow\sin 30^\circ = \frac{1}{2}\] So the effort needed to keep the box in place is about half the weight of the furniture box or \(\frac{1}{2}W\), just as Leonard says. Distance moved along Inclined Plane
While the inclined plane allows Leonard and Sheldon to push the box with less effort, the tradeoff is that the distance they move along the incline is twice the height to raise the box vertically. Geometry shows us that
\[\sin \theta = \frac{h}{d}\] We again assume that the angle of the stairs is approximately \(30^\circ\) and \(\sin 30^{\circ} = 1/2\) then we have \(d=2h\). Uses of the Inclined Plane
We see inclined planes daily without realizing it. They are used as loading ramps to load and unload goods. Wheelchair ramps also allow wheelchair users, as well as users of strollers and carts, to access buildings easily. Roads sometimes have inclined planes to form a gradual slope to allow vehicles to move over hills without losing traction. Inclined planes have also played an important part in history and were used to build the Egyptian pyramids and possibly used to move the heavy stones to build Stonehenge.
Lombard Street (San Francisco)
Lombard Street in San Francisco is famous for its eight tight hairpin turns (or switchbacks) that have earned it the distinction of being the crookedest street in the world (though this title is contested). These eight switchbacks are crucial to the street’s design as the reduce the hills natural 27° grade which is too steep for most vehicles. It is also a hazard to pedestrians, who are more accustomed to a more reasonable 4.86° incline due to wheel chair navigability concerns.
Technically speaking, the “zigzag” path doesn’t make climbing or coming down the hill any easier. As we have seen, all it does is change how various forces are applied. It just requires less effort to move up or down but the tradeoff is that you travel a longer distance. This has several advantages. Car engines have to be less powerful to climb the hill and in the case of descent, less force needs to be applied on the brakes. There are also safety considerations. A car will not accelerate down the switch back path as fast than if it was driven straight down, making speeds safer and more manageable for motorists.
This idea of using zigzagging paths to climb steep hills and mountains is also used by hikers and rock climbers for very much the same reason Lombard Street zigszags. The tradeoff is that the distance traveled along the path is greater than if a climber goes straight up.
The Descendants of Archimedes
We don’t need strength, we’re physicists. We are the intellectual descendants of Archimedes. Give me a fulcrum and a lever and I can move the Earth. It’s just a matter of… I don’t have this, I don’t have this!
We see that Leonard had the right idea. If we were to assume are to assume — based on the size of the box — that the furniture is approximately 150 lbs (65kg) and the effort is reduced by half, then they need to push with at least 75 lbs of force. This is equivalent to moving a 34kg mass. If they both push equally, they are each left pushing a very manageable 37.5 lbs, the equivalent of pushing a 17kg mass.
Penny’s apartment is on the fourth floor and we if we assume a standard US building design of ten feet per floor, this means a 30 foot vertical rise. The boys are left with the choice of lifting 150 lbs vertically 30 feet or moving 75lbs a distance of 60 feet. The latter is more manageable but then again, neither of our heroes have any upper body strength. |
Let's assume your model is sufficient for your application, and we can really describe the behavior of the device simply in additive terms, one for the per-measurement stochastic behavior and one for the overall bias. Three observations before we get started:
I assume you are running the sensor on a 3V supply. If not, you can use the values in the datasheet to adjust the calculations. These terms are expressed in the same units as \$\bar{a}\$, which in this case is volts, not \$g\$. You will actually need three such equations, one for each axis. So 6 terms in total.
For the bias \$b_a\$ we can turn to the charts in figures 5,6, and 7 on page 6 of the datasheet, titled "{X,Y,Z}-axis zero g bias". In a perfect world the zero g output would be 1.5V, but as we can see from the charts the actual value varies between parts. To select your \$b_a\$ for a particular simulated device for a particular axis, you can draw a random sample from that distribution, and use the offset from the expected value of 1.5 as your value for \$b_a\$ for that axis.
Let's look for example at the X-axis term for a particular device. Eyeballing the distribution's parameters I would model it as a Gaussian with \$\mu = 1.53V\$ and \$\sigma=0.01V\$. This means that the distribution for your bias \$b_a\$ for that axis (0g output - expected 0g output of 1.5V) is also a Gaussian, but with \$\mu = 0.03V\$ and \$\sigma=0.01V\$.
In order to assess the random noise, we need to stipulate some sort of output filtering. As is mentioned in the data sheet, by reducing the bandwidth you also significantly reduce noise on the output. I am going to assume a bandwidth of 100Hz just to make the math easier, but feel free to substitute your own values. There is a fairly extensive treatment of this topic in the datasheet under the heading "Design trade-offs for selecting filter characteristics".
With a bandwidth of 100HZ we can expect, according to the datasheet, a noise around 280*10 \$\mu g\$= 2.8 \$mg\$ RMS for the x-axis. We need to convert this to volts in order be able to add it to the formula. The expected sensitivity is about 300 mV/g, so we're execpting a noise of about 0.8 mV RMS. Note that RMS is exactly equal to the standard deviation of the distribution, so you can draw your per-measurement noise samples \$\mu_a\$ directly from a gaussian with \$\mu=0\$ and \$\sigma=0.0008 V\$.
So, for an output filtering of 100HZ: \$\mu_a\ \sim \mathcal{N}(0,0.0008)\$ and \$b_a \sim \mathcal{N}(0.03,0.01)\$, with the stipulation that \$\mu_a\$ is sampled at every measurement, and \$b_a\$ is sampled once for every device.
A factor that we neglected to consider is the variation in sensitivity between devices. This can be accounted for in a manner similar to our treatment of \$b_a\$, but since it's a multiplicative factor, it is not easily captured in your additive model. |
What are Penrose Tiles?I first came across Penrose tiles in Martin Gardner's "Mathematical Games" column in Scientific American in 1977. They were invented shortly before that by Professor Roger Penrose who was a brilliant British mathematician and cosmologist. Among many other things he "revolutionised the mathematical tools that we use to analyse the properties of spacetime".
There are two types of tiles: a kite and a dart:
A kite and dart on the left An ace or fool's kite on the right.
There are four other ways to tile kites and darts round a vertex without coming unstuck.
You can also see from the cut rhombus that the lengths of the sides as either Short or Long. It turns our that the ratio of the long side to the short side is the golden ration known as ##\phi = 1.61803...## (phi) to the ancient Greeks. They would have written it$$\phi=\frac{\left(1+\sqrt5\right)}{2}$$They probably got it from considering a rectangle that contained another rectangle inside it with the same aspect ratio. The smaller rectangle is drawn with a dashed line and the equation for ##\phi## follows directly:
My adventures with Penrose tiles.
Back in 1977, computers were a rarity and the best way to play with Penrose tiles was to cut out bits of paper and arrange them on a table. But the corners curled up, the slightest breath of wind would misalign them and any pattern would soon go wonky. I had some paper tiles in a tin for a long time and imagined having a computer program where I could really keep them neat and tidy. I never really studied deflation. I have now. PowerPoint came along in 1987 and in 1990 I met Microsoft Windows for the first time. In those days it used to crash about once every ten minutes. I was writing graphics programs for the beast. PowerPoint 7.0 in 1995 had VBA programming. I caught on to that in 2019 and realised a month ago that it would be handy for drawing Penrose tiles. It is very. All the above pictures and the movie are done with VBA in PowerPoint (screen capture by Debut). The biggest tiling so far is that one at the end of the movie: a whopping 13,854 kites, 8,669 darts - 22,523 tiles in total. That's very near the limit. These big tilings are produced by deflation which was also discovered by the great professor Penrose back in ~1974 about 45 years ago.
More to come - watch this space! |
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Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Elsevier, 2016-02)
The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ...
Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2013-11)
We present the first wide-range measurement of the charged-particle pseudorapidity density distribution, for different centralities (the 0-5%, 5-10%, 10-20%, and 20-30% most central events) in Pb-Pb collisions at $\sqrt{s_{NN}}$ ...
Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays
(Elsevier, 2014-11)
The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity |y|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ... |
Natarajan, Srinivasan (2005)
Hydrothermal synthesis and structures of two zero-dimensional zinc phosphate polymorphs. In: Solid State Sciences, 7 (12). pp. 1542-1548.
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Abstract
Two polymorphs of zero-dimensional zinc phosphate with the formula, $0_{\infinity}[Zn(2,2'-bipy)(H_{2}PO_{4})_{2}]$, have been synthesized employing hydrothermal technique and their structure determined by single crystal X-ray diffraction. Both the structures consists of $Zn0_{3}N_{2}$ distorted trigonal-bipyramidal and $PO_{2}(OH)_{2}$ tetrahedral units linked through their vertices giving rise to a zero-dimensional molecular zinc phosphate. The structures are stabilized by extensive hydrogen bond interactions between zero-dimensional monomers. The structures display subtle differences in their packing created by hydrogen bond interactions. Crystal data: polymorph 1, triclinic, space group P1 over bar (No. 2), a = 7.5446(15) angstrom, b = 10.450(2) angstrom, c = 10.750(2) angstrom, \alpha = 67.32(3)$^{o}$, \beta = 81.67(3)$^{o}$, gamma = 69.29(3), V = 731.4(3) $A^{3}$, Z = 2; polymorph II, triclinic, space group P (1) over bar (No. 2), a = 8.664(1) angstrom, b = 8.849(2) angstrom, c = 10.113(2) angstrom, \alpha = 97.37$(2)^{o}$, \beta = 100.54(2)$^{o}$, \gamma = 100.98$(2)^{o}$, V = 737.5(3) $A^{3}$, Z = 2.
Item Type: Journal Article Additional Information: Copyright for this article belongs to Elsevier. Keywords: Zinc; Phosphorus; N ligands; Structure elucidation Department/Centre: Division of Chemical Sciences > Solid State & Structural Chemistry Unit Depositing User: B.S Priyanka Date Deposited: 23 Jan 2006 Last Modified: 19 Sep 2010 04:23 URI: http://eprints.iisc.ac.in/id/eprint/5202 Actions (login required)
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Maciej carries a bookmark with this table on it to help him remember his secret four digit number. If his secret number is 8526, all he has to do is to remember the words HELP. To retrieve his number, he looks up the letters of the word HELP and finds the corresponding digits in the top row of the table. Another example: The word LOVE can be used to help Maciej remember the secret number 2525. Maciej has to remember a new secret number. Only three of the following words produce this new number. Which one does not?
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I. Rewrite the sentences below so that it has a similar meaning to the first, beginning with the words given
1. I find his handwriting very hard to read
-> I have....................
2. He got down to writing a letter as soon as he returned from his work
-> No sooner.................................
3. "If I were you, I wouldn't accept his marriage proposal", said Nam to Lan
-> Nam................
4. No matter how hard I tried, I could not open the window
-> Try.......................
5. Please don't ask me that question
-> I'd rather.............
II. Finish the second sentence so that it has the same meaning as the first one, using the given words. Do not change the given word
1. The fridge is complete empty (LEFT)
->
2. It is pointless to have that old typewriter repaired (WORTH)
->
3. Frank never pays any attention to my advice. (NOTICE)
->
4. John only understood very little of what the teacher said (HARDLY_
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5. Her ability to run a company really impresses me (IMPRESSED)
->
For a ; b ; c is a non-negative real number: a + b + c = 3. Prove that : \(\left(3abc+1\right)\left(a^2b+b^2c+c^2a\right)\ge12abc\)
NGƯỜI TA GIẢM CHIỀU RỘNG MẢNH ĐẤY HÌNH CHỮ NHẬT 8M ĐƯỢC MỘT MẢNH ĐẤT HÌNH CHỮ NHẬT MỚI SAU ĐÓ TĂNG CHÌỀN DÀI THÊM 8M . TÍNH DIỆN TÍCH MẢNH ĐẤT SAU KHI THAY ĐỔI CHIỀU DÀÌ VÀ CHIỀU RỘNG BIẾT RẰNG PHẦN ĐẤT HÌNH CHỮ NHẬT BỊ CẮT ĐI GẤP ĐÔI DIỆN TÍCH HÌNH CHỮ NHẬT ĐƯỢC THÊM VÀO.
Given two function \(f\left(x\right)=\sqrt{25x^2-30x+9}\), \(g\left(y\right)=y\). How many values of a are there such that \(f\left(a\right)=g\left(a\right)+7\)?
Nguyễn Linh Chi 08/08/2019 at 09:44
\(f\left(a\right)=\sqrt{25a^2-30a+9}\)
\(g\left(a\right)=a\)
We have the following:
\(f\left(a\right)=g\left(a\right)+7\)
\(\Leftrightarrow\sqrt{25a^2-30a+9}=a+7\)
\(\Leftrightarrow\left\{{}\begin{matrix}25a^2-30a+9=\left(a+7\right)^2\\a+7\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\24a^2-44a-40=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\)
Finally, there are two values of a.Uchiha Sasuke selected this answer.
Find all positive integer values of n such that \(\dfrac{n-17}{n+23}\) is a square of a quotient number.
Lux Arcadia 14/06/2019 at 14:20
\(\dfrac{n-17}{n+23}=\dfrac{n+23-40}{n+23}=1-\dfrac{40}{n+23}\)
\(\Rightarrow\left\{{}\begin{matrix}40⋮\left(n+23\right)\\n+23\ge23\left(n\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(n+23\right)\in\left\{1;2;4;5;8;10;20;40\right\}\\n+23\ge23\end{matrix}\right.\)
\(\Leftrightarrow n+23=40\Leftrightarrow n=17\)
Given two functions \(f\left(x\right)=5x+1\) and \(g\left(x\right)=ax+3\). Find the value of g(1) if \(a=f\left(2\right)-f\left(-1\right)\)
How many values of the whole number m such that the function \(y=\left(2016-m^2\right)x+3\) is increasing?
Tôn Thất Khắc Trịnh 13/06/2019 at 03:46
So as to have the function be increasing, 2016-m
2must be positive. Therefore, 2016 must be greater than m 2. Since 2016 is a positive number, m has to range from \(-\sqrt{2016}\) to \(\sqrt{2016}\). In other words, m has to be between -44.899 and 44.899. Since m is a whole number, the minimal value of m is -44 and the maximum is 44. To calculate the number of values, we use this formula: \(N=\frac{44-(-44)}{1}+1=89\)
To conclude, there are 89 values of m such that the function is increasing.Uchiha Sasuke selected this answer. |
Deshouillers,Jean-Marc-A remark on cube-free numbers in Segal-Piatestki-Shapiro sequences
hrj:5114 -Hardy-Ramanujan Journal,January 23, 2019
A remark on cube-free numbers in Segal-Piatestki-Shapiro sequences
Authors: Deshouillers,Jean-Marc
Using a method due to G. J. Rieger, we show that for $1 < c < 2 $ one has, as $x$ tends to infinity $$\textrm{Card}{n \leq x : \lfloor{n^c}\rfloor} \ \textrm{ is cube-free} } = \frac{x}{\zeta(3)} + O (x^{ (c+1)/3} \log x)$$ , thus improving on a recent result by Zhang Min and Li Jinjiang.
Source : oai:HAL:hal-01986712v1
Published on: January 23, 2019
Submitted on: January 23, 2019
Keywords: Segal-Piatetski-Shapiro sequences,cube-free numbers,estimation of trigonometric sums,discrepancy 2010 Mathematics Subject Classification 11B75,11N37,11N56,11L03,11L07,[MATH]Mathematics [math],[MATH.MATH-NT]Mathematics [math]/Number Theory [math.NT] |
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Now showing items 1-7 of 7
The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
π0 and η meson production in proton-proton collisions at √s=8 TeV
(Springer, 2018-03-26)
An invariant differential cross section measurement of inclusive π^0 and η meson production at mid-rapidity in pp collisions at √s=8 TeV was carried out by the ALICE experiment at the LHC. The spectra of π^0 and η mesons ...
Longitudinal asymmetry and its effect on pseudorapidity distributions in Pb–Pb collisions at √sNN = 2.76 TeV
(Elsevier, 2018-03-22)
First results on the longitudinal asymmetry and its effect on the pseudorapidity distributions in Pb–Pb collisions at √sNN = 2.76 TeV at the Large Hadron Collider are obtained with the ALICE detector. The longitudinal ...
Production of deuterons, tritons, 3He nuclei, and their antinuclei in pp collisions at √s=0.9, 2.76, and 7 TeV
(American Physical Society, 2018-02)
Invariant differential yields of deuterons and antideuterons in p p collisions at √ s = 0.9, 2.76 and 7 TeV and the yields of tritons, 3 He nuclei, and their antinuclei at √ s = 7 TeV have been measured with the ALICE ... |
Deformation theory tag info https://mathoverflow.net/tags/deformation-theory/info is very minimal and says almost nothing:
deformation theory is the study of infinitesimal conditions associated with varying a solution P of a problem to slightly different solutions $P_\epsilon$, where $\epsilon$ is a small number, or vector of small quantities.
Can some experienced user consider editing the tag info to make it better?
What I understood (from nlab) is the following.
A typical problem in deformation theory is the following :
Given a map $f:X\rightarrow Y$ (in a category), injective maps (monomorphisms) of special kind $i_X:X\rightarrow \tilde{X}$ and $i_Y:Y\rightarrow \tilde{Y}$ can we find a map $g:\tilde{X}\rightarrow \tilde{Y}$ such that $g\circ i_X=i_Y\circ f$. This map $g$ is called an infinitesimal deformation of $f$.
Does adding this content for tag info make it any better? |
Answer
The length of the train is approximately 200 m
Work Step by Step
We can express the angle in degrees: $\theta = 3^{\circ}20' = (3 + \frac{20}{60})^{\circ} = 3.33^{\circ}$ We can convert the angle to radians: $\theta = (3.33^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 0.05812~rad$ We can approximate the length $L$ of the train: $L \approx \theta ~r$ $L \approx (0.05812~rad)(3.5~km)$ $L \approx 0.20~km$ $L \approx 200~m$ The length of the train is approximately 200 m. |
I am now studying Quantum Mechanics, and I am facing some issue about the convergence of Fourier Transform. I want to know when I can have the following equality: $$ \lim_{n \to \infty}(\mathcal{F}f_n)(\omega)= \mathcal{F}(\lim_{n \to \infty}f_n)(\omega) $$ Where $\mathcal{F}f(\omega)=\int_\mathbb{R}f(x)e^{-2\pi ix\omega}$. Both the limit on the LHS and RHS means convergent pointwise.
Of course, I can apply Lebesgue Dominated convergence theorem here when $f(x)e^{-2\pi ix\omega}$ is integrable. However, even if it is not integrable, the fourier transformation could exist. I try to find a theorem about whether we can swap lim and $\mathcal{F}$, but I have not get anything.
Let's take an example. Take $f_n(x)=\cos{x}$ when $x\in [-\frac{(2n-1)\pi}{2},\frac{(2n-1)\pi}{2}]$, and $f_n(x)=0$ otherwise. Then $f_n \to \cos$ pointwise. The fourier transform of $f_n$ is something like $(-1)^n\frac{2\cos(\frac{(2n-1)\pi}{2}x)}{x^2-1}$. But the Fourier transform of $\cos x$ is $\sqrt{\frac{\pi}{2}}\delta(\omega-1)+\sqrt{\frac{\pi}{2}}\delta(\omega+1)$. Clearly $\lim_{n \to \infty}(\mathcal{F}f_n)(\omega)= \mathcal{F}(\lim_{n \to \infty}f_n)(\omega)$ is not true in this case. Why it is not true?
Could anyone list some general theorems about this topic? |
Slow to Fast Infinitely Extended Reservoirs for the Symmetric Exclusion Process with Long Jumps
2019, v.25, Issue 2, 217-274
ABSTRACT
We consider an exclusion process with long jumps in the box $\Lambda_N=\{1, \ldots,N-1\}$, for $N \ge 2$, in contact with infinitely extended reservoirs on its left and on its right. The jump rate is described by a transition probability $p(\cdot)$ which is symmetric, with infinite support but with finite variance. The reservoirs add or remove particles with rate proportional to $\kappa N^{-\theta}$, where $\kappa>0$ and $\theta \in\bb R$. If $\theta>0$ (resp.\ $\theta<0$) the reservoirs slowly (resp.\ fastly) add and remove particles in the bulk. According to the value of $\theta$ we prove that the time evolution of the spatial density of particles is described by some reaction-diffusion equations with various boundary conditions.
%has a distinct behavior according to the range of the parameter $\theta$.The hydrodynamic limit is given by: 1) a reaction-diffusion equation with inhomogeneous Dirichlet boundary conditions when $\theta<1$, 2) a heat equation with Robin boundary conditions if $\theta = 1$ and 3) a heat equation with Neumann boundary conditions if $\theta>1$.
Keywords: hydrodynamic limit, reaction-diffusion equation, boundary conditions, exclusion with long jumps
COMMENTS |
The idea is the following: If $U$ is an open subset of $\mathbb{R}^k$, then we know that $T_0U\cong \mathbb{R}^k$ as vector spaces. If we use the map $\phi:U\to X$ as our parametrization, then we can see that $d\phi_0$ acts on $T_0U$ as a linear isomorphism. In order to get the best linear approximation to $\phi$, we should use the first order Taylor expansion $\phi(u)\approx\phi(0)+d\phi_0(u).$The best linear approximation to $X$ as a submanifold of $\mathbb{R}^n$ is given by the image of the tangent plane $T_0U\cong \mathbb{R}^k$. This is exactly the set of points $\phi(0)+d\phi_0(u)=x+d\phi_0(u)$ for all $u\in T_0U$.
So, we apply a linear transformation $d\phi_0:T_0U\to T_xX\subseteq \mathbb{R}^n.$ Then we add $\phi(0)+x$ to
shift this linear space $T_xX$ to be tangent to $X$ at $x$. As a very concrete example, take the manifold $S^1\subseteq \mathbb{R}^2$. Near $(0,1)$ it has graph coordinates $(x,\sqrt{1-x^2})$. That is, we can parametrize a neighborhood of $(0,1)\in S^1$ by $(-1,1)\to S^1$ given by $\phi:t\mapsto (t,\sqrt{1-t^2})$. If we visually inspect $S^1$ at $(0,1)$ we expect its best linear approximation line to be given by a horizontal line $y=1$ passing through $(0,1)$, call this $A$.
The recipe given in Guillemin and Pollack says that we can find this plane by calculating the Jacobian of the parametrization, $d\phi_0$, then writing $A=(0,1)+d\phi_0 T_0(-1,1).$ $d\phi_0$ is the $2\times 1$ matrix $$ d\phi_0=\begin{bmatrix}\frac{\partial x}{\partial t}\\\frac{\partial y}{\partial t}\end{bmatrix}_{t=0}.$$This is $$ d\phi_0=\begin{bmatrix}1\\0\end{bmatrix}.$$The moral is that the image of $T_0(-1,1)=\mathbb{R}$ is the $x-$axis. $$ d\phi_0T_{0}(-1,1)=\{(x,y)\in \mathbb{R}^2:y=0\}.$$Then if we add $(0,1)$ we get that $A$ is precisely the horizontal line passing through $(0,1)$. Indeed, here $T_x(X)$ is the $x-$axis, and the best linear approximation is the shifted $x-$axis.
In response to the second question, $x+T_x(X)$ is
literally a shifted subspace tangent to the manifold $X$ at the point $x$. $T_x(X)$ can be visualized as $x+T_x(X)$ but shifted in an affine manner so it passes through the origin. The advantage to calling $T_x(X)$ the tangent space is that when it passes through the origin it is a bona fide linear subspace of $\mathbb{R}^n$. |
The strength of Hamilton's principle is obvious to me and I see the advantage. Now, for conservative systems we also have Maupertuis' principle that says:
$$ \delta \int p dq =0$$
and I am not sure how to derive an equation of motion from this? Is this of any use in practical computations? So, can one apply this principle for example to the harmonic oscillator?- I have never seen anybody using it.
Further, I read in Goldstein's classical Mechanics that the variation in Maupertuis' principle is not the one in Hamilton's principle, since we have constant Hamiltonian and changing time, whereas Hamilton's principle has constant time and varying Hamiltonian (in general).
I am a little bit wondering about this, since you could easily get Maupertuis' principle from Hamilton's principle: $$ \delta \int L dt = \delta \int p \dot{q} - H dt = \delta \int p \dot{q} dt = \delta \int p dq =0,$$ if $H$ is constant. Can anybody here explain to me, why we have to use a different variation and how one can use this principle? |
Difference between revisions of "Group cohomology of dihedral group:D8"
(→Cohomology groups for trivial group action)
(→Baer invariants)
(33 intermediate revisions by the same user not shown) Line 3: Line 3:
group = dihedral group:D8|
group = dihedral group:D8|
connective = of}}
connective = of}}
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==Homology groups for trivial group action==
==Homology groups for trivial group action==
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===Over the integers===
===Over the integers===
−
The homology groups
+
The homology groups the integers are as follows:
−
<math>
+
<math>(D_8;\mathbb{Z}) = \left \lbrace \begin{array}{rl} \mathbb{Z}, & = 0 \\\mathbb{Z}/2\mathbb{Z}, & \equiv 1 \pmod 4\\ \mathbb{Z}/\mathbb{Z}, & \equiv 3 \pmod 4 \\\\, \{ even }\\ \end{array}\right.</math>
− +
first few homology groups are :
−
{| class="
+
{| class="" border="1"
− +
<math></math> 012345678
|-
|-
−
| <math>
+
| <math>\mathbb{Z}</math> || <math>\mathbb{Z}</math> || <math>\mathbb{Z}/2\mathbb{Z}</math> || || <math>\mathbb{Z}/\mathbb{Z}</math> || <math>\mathbb{Z}/2\mathbb{Z}</math> || || <math>\mathbb{Z}/\mathbb{Z}</math> ||
|}
|}
===Over an abelian group===
===Over an abelian group===
−
==
+ + +
= =
+
an abelian group
− + + + +
<math>M</math>
+ −
==Cohomology groups for trivial group action==
==Cohomology groups for trivial group action==
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===Over the integers===
===Over the integers===
−
The cohomology groups
+
The cohomology groups the integers are as follows:
−
<math>H^
+
<math>H^(D_8;\mathbb{Z}) = \left \lbrace \begin{array}{rl} \mathbb{Z}, & =0 \\\mathbb{Z}/2\mathbb{Z}, & \equiv \pmod 4\\ \mathbb{Z}/\mathbb{Z}, & \\ne 0, \equiv \pmod 4 \\
+ + + + +
0
+ +
\\{}\\\{}\</math>
+
===Over an abelian group===
===Over an abelian group===
−
The cohomology groups
+
The cohomology groups an abelian group <math>M</math> are as follows:
−
<math>H^
+
<math>H^(D_8;M) = \left \lbrace \begin{array}{rl} M, & = 0 \\\operatorname{Ann}(M), & \equiv 1 \pmod 4\\ M/2M& \equiv 2 \pmod 4 \\\operatorname{Ann}(M), & \equiv 3 \pmod 4 \\M/, & \equiv 0 \pmod 4\\ \end{array}\right.</math>
− +
<math>
+ + + + +
\operatorname{Ann}(M)</math> 2<math>M</math> <math>\operatorname{Ann}(M)</math> <math>M</math>
+
==Cohomology ring with coefficients in integers==
==Cohomology ring with coefficients in integers==
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==Second cohomology groups and extensions==
==Second cohomology groups and extensions==
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===Second cohomology groups for trivial group action===
===Second cohomology groups for trivial group action===
{| class="sortable" border="1"
{| class="sortable" border="1"
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! Group acted upon !! Order !! Second part of GAP ID !! [[Second cohomology group for trivial group action]] !! Extensions !! Cohomology information
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! Group acted upon !! Order !! Second part of GAP ID !! [[Second cohomology group for trivial group action]] !! Extensions !! Cohomology information
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| [[group ]] || 2[[]] ||
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+ + + + + + + + + + + + + + + + + + + + + + + + + + + + Latest revision as of 00:27, 29 May 2013
Contents This article gives specific information, namely, group cohomology, about a particular group, namely: dihedral group:D8. View group cohomology of particular groups | View other specific information about dihedral group:D8 Family contexts
Family name Parameter value Information on group cohomology of family dihedral group of degree , order degree , order group cohomology of dihedral groups Homology groups for trivial group action FACTS TO CHECK AGAINST(homology group for trivial group action): First homology group: first homology group for trivial group action equals tensor product with abelianization Second homology group: formula for second homology group for trivial group action in terms of Schur multiplier and abelianization|Hopf's formula for Schur multiplier General: universal coefficients theorem for group homology|homology group for trivial group action commutes with direct product in second coordinate|Kunneth formula for group homology Over the integers
The homology groups over the integers are given as follows:
The first few homology groups are given below:
Over an abelian group
The homology groups over an abelian group are given as follows:
The first few homology groups with coefficients in an abelian group are given below:
Cohomology groups for trivial group action FACTS TO CHECK AGAINST(cohomology group for trivial group action): First cohomology group: first cohomology group for trivial group action is naturally isomorphic to group of homomorphisms Second cohomology group: formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization In general: dual universal coefficients theorem for group cohomology relating cohomology with arbitrary coefficientsto homology with coefficients in the integers. |Cohomology group for trivial group action commutes with direct product in second coordinate | Kunneth formula for group cohomology Over the integers
The cohomology groups over the integers are given as follows:
The first few cohomology groups are given below:
0 Over an abelian group
The cohomology groups over an abelian group are given as follows:
The first few cohomology groups with coefficients in an abelian group are:
Cohomology ring with coefficients in integers PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] Second cohomology groups and extensions Schur multiplier
This has implications for projective representation theory of dihedral group:D8.
Schur covering groups
The three possible Schur covering groups for dihedral group:D8 are: dihedral group:D16, semidihedral group:SD16, and generalized quaternion group:Q16. For more, see second cohomology group for trivial group action of D8 on Z2, where these correspond precisely to the stem extensions.
Second cohomology groups for trivial group action
Group acted upon Order Second part of GAP ID Second cohomology group for trivial group action (as an abstract group) Order of second cohomology group Extensions Number of extensions up to pseudo-congruence, i.e., number or orbits under automorphism group actions Cohomology information cyclic group:Z2 2 1 elementary abelian group:E8 8 direct product of D8 and Z2, SmallGroup(16,3), nontrivial semidirect product of Z4 and Z4, dihedral group:D16, semidihedral group:SD16, generalized quaternion group:Q16 6 second cohomology group for trivial group action of D8 on Z2 cyclic group:Z4 4 1 elementary abelian group:E8 8 direct product of D8 and Z4, nontrivial semidirect product of Z4 and Z8, SmallGroup(32,5), central product of D16 and Z4, SmallGroup(32,15), wreath product of Z4 and Z2 6 second cohomology group for trivial group action of D8 on Z4 Klein four-group 4 2 elementary abelian group:E64 64 [SHOW MORE] 11 second cohomology group for trivial group action of D8 on V4 Baer invariants
Subvariety of the variety of groups General name of Baer invariant Value of Baer invariant for this group abelian groups Schur multiplier cyclic group:Z2 groups of nilpotency class at most two 2-nilpotent multiplier groups of nilpotency class at most three 3-nilpotent multiplier any variety of groups containing all groups of nilpotency class at most three -- GAP implementation Computation of integral homology
The homology groups for trivial group action with coefficients in can be computed in GAP using the GroupHomology function in the
HAP package, which can be loaded by the command
LoadPackage("hap"); if it is installed but not loaded. The function outputs the orders of cyclic groups for which the homology or cohomology group is the direct product of these (more technically, it outputs the elementary divisors for the homology or cohomology group that we are trying to compute).
Here are computations of the first few homology groups:
Computation of first homology group gap> GroupHomology(DihedralGroup(8),1); [ 2, 2 ]
The way this is to be interpreted is that the first homology group (the abelianization) is the direct sum of cyclic groups of the orders listed, so in this case we get that is , which is the Klein four-group.
Computation of second homology group gap> GroupHomology(DihedralGroup(8),2); [ 2 ] Computation of first few homology groups
To compute the first eight homology groups, do:
gap> List([1,2,3,4,5,6,7,8],i->[i,GroupHomology(DihedralGroup(8),i)]); [ [ 1, [ 2, 2 ] ], [ 2, [ 2 ] ], [ 3, [ 2, 2, 4 ] ], [ 4, [ 2, 2 ] ], [ 5, [ 2, 2, 2, 2 ] ], [ 6, [ 2, 2, 2 ] ], [ 7, [ 2, 2, 2, 2, 4 ] ], [ 8, [ 2, 2, 2, 2 ] ] ] |
I'd start off by saying it's been many years since I was in chemistry lessons and have forgotten much, but recently I have had to relearn a lot of basic chemistry. The following doesn't really impede me in my work, but it irritates me that I can't figure out what is basically two sides of the same coin.
Let's imagine I want to calculate the pH of a 0.1M acetic acid ($c_\ce{CH3COOH} = \pu{0.1M}$)
Acetic acid is a weak acid and dissociates in water as follows: $\ce{CH_3COOH + H_2O <<=> CH_3COO^- + H_3O^+}$. Expressing this in terms of acid dissociation constant (and ignoring the water) yields $K_\mathrm{a} = {\large{\ce{\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}}}}$
We can look up the $K_\mathrm{a}$ of acetic acid, which is $1.8 \times 10^{-5}$. If we assume that the amount of $\ce{CH_3COOH}$ change is negligible (due to it being a weak acid and only slightly dissociating), and that $\ce{CH_3COO^-}$ and $\ce{H^+}$ concentrations are equal, we come up with the following equation:
$1.8 \times 10^{-5} = \large{\frac{[\ce{H^+}]^2}{\pu{0.1M}}}$
Rearranging and solving for $\ce{[H^+]}$ gives us $\sqrt{1.8 \times 10^{-5} \times \pu{0.1M}} = \pu{1.34E-3 M}$
Since $\mathrm{pH} = -\log([\ce{H+}])$ (or $\mathrm{pH} = \log(1/[\ce{H+}])$, whichever you prefer), $\mathrm{pH} = -\log(1.34 \times 10^{-3}) \approx 2.88$
I've checked this and I know it's correct. But I don't know how to do it from the $K_\mathrm{b}$. Or rather, I get the answer, but I don't fully understand the process.
So the opposite reaction, $\ce{CH_3COO^- + H_2O <=>> CH_3COOH + OH^-}$, also has a equilibrium constant which is called $K_\mathrm{b}$, defined as follows (again ignoring the water):
${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}}}$
We can look it up or calculate it from the relationship that $K_\mathrm{a} \times K_\mathrm{b} = 1\times10^{-14}$
$K_\mathrm{b} = {\large \frac{1\times10^{-14}}{K_\mathrm{a}}}$
In either case, $K_\mathrm{b} = 5.6 \times 10^{-10}$
Now, I don't know what to do with the $\ce{[OH^-]}$. My basic assumption is that the change in the amount of $\ce{CH_3COOH}$ is negligible, so that stay as 0.1M. But unlike previously where $\ce{CH_3COO^-}$ and $\ce{[H^+]}$ were formed in equal amounts due to the dissociation of the acid, how much of $\ce{[OH^-]}$ is there? My (wrong) logic was to assume that $\ce{[CH_3COO^-] = -[OH^-]}$, but then they cancel out, leaving me with $K_\mathrm{b} = \ce{[CH3COOH]}$, which can't be right. So yeah, I'm lost.
I solved this by substituting $\ce{[OH^-]}$ with $\large{\frac{K_\mathrm{w}}{[\ce{H+}]}}$ (because $\ce{[H^+][OH^-]} = K_\mathrm{w}$)
This yields ${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH] K_w}{[CH_3COO^-][H^+]}}}$. Again assuming that $\ce{[CH_3COO-]} = \ce{[H+]}$, and knowing that $K_w = 1 \times 10^{-14}$, I come up with ${5.6 \times 10^{-10}} = {\large \ce{\frac{0.1M \times 1 \times 10^{-14}}{[H^+]^2}}}$
Solving for $\ce{[H^+]}$ gives us $\sqrt{{\large \frac{\pu{0.1M} \times 1 \times 10^{-14}}{5.6 \times 10^{-10}}}} = \pu{1.33 \times 10^{-3}M}$
This is the same concentration of $\ce{[H^+]}$ that I got using the $K_\mathrm{a}$ above, making the $\mathrm{pH} = -\log(1.33 \times 10^{-3}) \approx 2.88$
So I can get to the same answer ignoring (or rather substituting) the $\ce{[OH^-]}$, but I'd really like to know how to do it using that. This is probably a really simple answer but after having spent many hours staring my many pieces of paper, my mind is coming up with a blank. |
Suppose $ch(F)=0$ and let $\phi: \mathbb{Z} \to F$ with $\phi:n \mapsto n \cdot 1_F$. Since the characteristic is $0$, we have $\ker \phi = \{0 \}$. By the First Isomorphism Theorem, we have $\mathbb{Z} / \{ 0\}\cong \text{Im}(\phi)$. But the ring on the right is just $\mathbb{Z}$, and the ring on the left is the prime "subfield" of $F$. Could someone point out the problem please?
Your $\phi$ is the morphism of the initial object in the category of unital rings to the ring $F$. If the codomain is a field of prime characteristic, then the image of the morphism is not just a subring, but a subfield. This is due to the fact that the prime non-zero ideals of $\Bbb Z$ are also maximal. If not, then it's just a subring.
The image of $\varphi$ is
not the prime subfield of $F$, which is isomorphic to $\mathbb{Q}$. The problem is that $\phi$ is a morphism of rings, and not a morphism of fields. Therefore, you could think of $\operatorname{im}(\varphi)$ as being a sort of "prime subring" associated to $F$, but it is not a field. In general, for any field of characteristic 0, there is a unique field homomorphism $\psi:\mathbb{Q} \to F$ whose image is the prime subfield associated to $F$, and which is isomorphic to $\mathbb{Q}$. |
Zaid Alyafeai
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Last seen Mar 21 at 12:31 Communities (18) Top network posts 66 Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$ 42 Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$ 28 Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$ 26 Calculating alternating Euler sums of odd powers 23 Really advanced techniques of integration (definite or indefinite) 20 A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ 20 Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ View more network posts → Top tags (12)
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Some time ago I came up with my own idea and I'm posting it now. Everything is done in an ordered plane i.e. within axioms of connection and order (no congruence, continuity or parallel axioms).
Points will be denoted by small letters $a,b,\ldots$. Sets like lines, rays, halfplanes etc. will be denoted by capital letters $A,B,K,L,M,N,\ldots$.If $a\neq b$, then the only line passing through $a,b$ will be denoted by $\overleftrightarrow{ab}$. If $A$ is a ray, then we use the symbol $o(A)$ to denote the origin of $A$ and $L(A)$ to denote the line, in which $A$ is contained. A ray complementary to $A$ will be denoted by $A^*$. If $M$ is a half-plane,the complementary half-plane will be denoted by $M^*$.
Then we introduce the notion that rays $A$ and $B$ have the same orientation (or direction) and the notion of orientation (direction) of the line. Roughly speaking rays $A$ and $B$ have the same orientation (direction) iff they are contained in the same line and are directed similarly. The orientation (direction) of a line is a set of all similarly directed rays contained in this line (formally an equivalence class). For each line there are exactly two directions associated with this line. If $\mathcal{D}$ is an orientation of a line, we denote the opposite orientation by $\mathcal{D}^*$. For the detailed definition see the book "Foundations of geometry" by Borsuk and Szmielew (1960) (pages 37,38).
By parallel lines we mean disjoint or equal lines.
From here definitions are invented by me. For example the definition of parallel rays is not the same as usual definition of this notion (sometimes called limiting parallel). As a sidenote I'll say that these definitions are equivalent under the assumption of the parallel postulate (which would be Playfair's axiom in this setting) but I don't assume it.
We say that rays $A,B$ are parallel, what we denote $A\parallel B$, iff lines $L(A),L(B)$ are parallel and $A,B$ have the same direction if $L(A)=L(B)$ or $A$ and $B$ lie on the same side of $\overleftrightarrow{o(A)o(B)}$ if $L(A)\cap L(B)=\emptyset$
We say that directions $\mathcal{D}$, $\mathcal{D}_1$ of some lines are parallel, what we denote by $\mathcal{D}\parallel\mathcal{D}_1$, iff for all rays $D\in\mathcal{D},D_1\in\mathcal{D}_1$, rays $D,D_1$ are parallel.
Let's head to the actual orientation definition:
Consider the family $\mathcal{R}$ of all triples of the form $(L,\mathcal{D},M)$, where $L$ is a line, $\mathcal{D}$ is an orientation of $L$ and $M$ is a halfplane with boundary $L$. We index this family the following way: $\mathcal{R}=\{O_A\}_{A\in\mathcal{A}}$, where $O_A=(L_A,\mathcal{D}_A,M_A)$ for all $A\in\mathcal{A}$. We define the relation $\sim$ on the set $\mathcal{R}$ the following way: Fix $A,B\in\mathcal{A}$ and consider cases:
$L_A=L_B$
If $\mathcal{D}_A=\mathcal{D}_B$, then $O_A\sim O_B :\iff M_A=M_B$.
If $\mathcal{D}_A=\mathcal{D}_B^*$, then $O_A\sim O_B :\iff M_A=M_B^*$
$L_A\cap L_B=\emptyset$.
If $\mathcal{D}_A\parallel \mathcal{D}_B$, then $$O_A\sim O_B :\iff\left( L_A\subset M_B\wedge L_B\subset M_A^*\right)\vee\left( L_A\subset M_B^*\wedge L_B\subset M_A\right)$$
If $\mathcal{D}_A\parallel \mathcal{D}_B^*$, then $$O_A\sim O_B :\iff\left( L_A\subset M_B\wedge L_B\subset M_A\right)\vee\left( L_A\subset M_B^*\wedge L_B\subset M_A^*\right)$$
$L_A\cap L_B=\{o\}$
Then $$O_A\sim O_B :\iff\left(P_A\subset M_B \wedge P_B\subset M_A^*\right)\vee\left(P_A\subset M_B^* \wedge P_B\subset M_A\right)$$, where $P_A\in\mathcal{D}_A,o(P_A)=o, P_B\in\mathcal{D}_B,o(P_B)=o$.
I managed to prove that $\sim$ is an equivalence relation with exactly two equivalence classes. I suppose my proof is correct although it's extremely long and tedious. We may call the equivalence classes orientations.
As for the original problem with relation between triangles we can easily introduce the orientation determined by three noncollinear points in the following way:
$$\mathcal{O}(a,b,c):=[(\overleftrightarrow{ab},[\overrightarrow{ab}],M(\overleftrightarrow{ab},c))]_{\sim}$$
where $M(\overleftrightarrow{ab},c)$ is a halfplane with boundary $\overleftrightarrow{ab}$ having point c. |
By exploring connections among the binomial, normal, and chi-squared distributions, one can answer.
$Y∼Bin(n,p)$ i.e. random variable $Y$ is number of successes in $n$ trials, with probability of success in a trial being $p$.
Assume $n$ and $p$ are such that $Y$ is well approximated by a normal distribution with mean = $np$ and variance = $np(1-p)$
Let's say $Y$ to be number of successes then,
$(Y−E(Y))^2/Var(Y)$ (i.e. square of standard normal.) will be approximately $\sim\chi^2_1$, where $E(Y)=np$ and $Var(Y)=np(1−p)$.
So $(Y−np)^2/np(1−p)$ will be approximately $\sim\chi^2_1$.
Using facts that $(Y−np)^2=[(n−Y)−n(1−p)]^2$ and the algebraic relation $1/p+1/(1−p)=1/p(1−p)$.
So,
$\frac{(Y-np)^2}{np(1-p)}$=$\frac{(Y-np)^2}{np}+\frac{(Y-np)^2}{n(1-p)}$$\quad= \frac{(Y-np)^2}{np}+\frac{[(n-Y)-n(1-p)]^2}{n(1-p)} \\\quad= \frac{(O_S-E_S)^2}{E_S}+\frac{(O_F-E_F)^2}{E_F}$
So, chi-square statistic $\frac{(O_S-E_S)^2}{E_S}+\frac{(O_F-E_F)^2}{E_F}$will have the distribution of the square of an approximately standard-normal random variable ex - $(Y-np)/\sqrt{np(1-p)}$ |
Let $\alpha>0$. Given the cylinder $$B_1=\{(x,y,z)\in\mathbb R^3\;|\;x^2+y^2\leq\alpha x\}$$ and the sphere $$B_2=\{(x,y,z)\in\mathbb R^3\;|\;x^2+y^2+z^2\leq\alpha^2\},$$ how can one find the volume of the intersection $B_1\cap B_2$?
The following image shows the situation (for $\alpha=1$).
I tried the substition $x'=x-\frac\alpha2$ to center the coordinate system $(x',y,z)$ around the center of the cylinder and using cylindrical coordinates to get $$V=2\int_0^{2\pi}\mathrm d\varphi\int_0^{\alpha/2} \rho\,\mathrm d\rho\int_0^{z(\varphi,\rho)}\mathrm d z$$ where $z(\varphi,\rho)=\sqrt{\alpha^2-x^2-y^2}=\sqrt{\alpha^2-\left(x'+\frac\alpha2\right)^2-y^2}$ from the equation of the sphere. In cylindrical coordinates, this would be equal to $z(\varphi,\rho)=\sqrt{\alpha^2-\left(\rho\cos\varphi+\frac\alpha2\right)^2-\rho^2\sin^2\varphi}=\sqrt{\frac34\alpha^2-\alpha\rho\cos\varphi-\rho^2}$
This results in the integral $$V=2\int_0^{2\pi}\mathrm d\varphi\int_0^{\alpha/2} \rho\sqrt{\frac34\alpha^2-\alpha\rho\cos\varphi-\rho^2}\,\mathrm d\rho.$$ This integral is pretty difficult to solve, however.
Is there an easier way to solve this, by using symmetries I did not see etc.? |
In the opening scene of the “The Euclid Alternative”, we see Sheldon (Jim Parsons) demanding that Leonard (Johnny Galecki)needs to drive him around to run various errands. Leonard, after spending a night in the lab using the new Free Electron Laser to perform X-ray diffraction experiments. In the background, we can see equations that describe a rolling ball problem on the whiteboard in the background.
Rolling motion plays an important role in many familiar situations so this type of motion is paid considerable attention in many introductory mechanics courses in physics and engineering. One of the more challenging aspects to grasp is that rolling (without slipping) is a combination of both translation and rotation where the point of contact is instantaneously at rest.The equations on the white board describe the velocity at the point of contact on the ground, the center of the object and at the top of the object.
Pure Translational Motion
When an object undergoes pure translational motion, all of its points move with the same velocity as the center of mass– it moves in the same speed and direction or \(v_{\textrm{cm}}\).
Pure Rotational Motion
In the case of a rotating body, the speed of any point on the object depends on how far away it is from the axis of rotation; in this case, the center. We know that the body’s speed is \(v_{\textrm{cm}}\) and that the speed at the edge must be the same. We may think that all these points moving at different speeds poses a problem but we know something else — the object’s angular velocity.
The angular speed tells us how fast an object rotates. In this case, we know that all points along the object’s surface completes a revolution in the same time. In physics, we define this by the equation: \begin{equation} \omega=\frac{v}{r} \end{equation} where \(\omega\) is the angular speed. We can use this to rewrite this equation to tell us the speed of any point from the center: \begin{equation} v(r)=\omega r \end{equation} If we look at the center, where \(r=0\), we expect the speed to be zero. When we plug zero into the above equation that is exactly what we get: \begin{equation} v(0)= \omega \times 0 = 0 \label{eq:zero} \end{equation} If we know the object’s speed, \(v_{\textrm{cm}}\) and the object’s radius, \(R\), using a little algebra we can define \(\omega\) as: \[\omega=\frac{v_{\textrm{cm}}}{R}\] or the speed at the edge, \(v_{\textrm{cm}}\) to be \(v(R)\) to be: \begin{equation} v_{\textrm{cm}}=v(R) = \omega R \label{eq:R} \end{equation} Putting it all Together
To determine the absolute speed of any point of a rolling object we must add both the translational and rotational speeds together. We see that some of the rotational velocities point in the opposite direction from the translational velocity and must be subtracted. As horrifying as this looks to do, we can reduce the problem somewhat to what we see on the whiteboard. Here we see the boys reduce the problem and look at three key areas, the point of contact with the ground (\(P)\), the center of the object, (\(C\)) and the top of the object (\(Q\)).
We have done most of the legwork at this point and now the rolling ball problem is easier to solve.
At point \(Q\)
At point \(Q\), we know the translational speed to be \(v_{\textrm{cm}}\) and the rotational speed to be \(v(R)\). So the total speed at that point is
\begin{equation} v = v_{\textrm{cm}} + v(R) \label{eq:Q1} \end{equation} Looking at equation \eqref{eq:R}, we can write \(v(R)\) as \begin{equation} v(R) = \omega R \end{equation} Putting this into \eqref{eq:Q1} and we get, \begin{aligned} v & = v_{\textrm{cm}} + v(R) \\ & = v_{\textrm{cm}} + \omega R \\ & = v_{\textrm{cm}} + \frac{v_{\textrm{cm}}}{R}\cdot R \\ & = v_{\textrm{cm}} + v_{\textrm{cm}} = 2v_{\textrm{cm}} \end{aligned} which looks almost exactly like Leonard’s board, so we must be doing something right. At point \(C\)
At point \(C\) we know the rotational speed to be zero (see equation \eqref{eq:zero}).
Putting this back into equation \eqref{eq:Q1}, we get \begin{aligned} v & = v_{\textrm{cm}} + v(r) \\ & = v_{\textrm{cm}} + v(0) \\ & = v_{\textrm{cm}} + \omega \cdot 0 \\ & = v_{\textrm{cm}} + 0 \\ & = v_{\textrm{cm}} \end{aligned} Again we get the same result as the board. At point \(P\)
At the point of contact with the ground, \(P\), we don’t expect a wheel to be moving (unless it skids or slips). If we look at our diagrams, we see that the rotational speed is in the opposite direction to the translational speed and its magnitude is
\begin{aligned} v(R) & = -\omega R \\ & = -\frac{v_{\textrm{cm}}}{R}\cdot R \\ & = -v_{\textrm{cm}} \end{aligned} It is negative because the speed is in the opposite direction. Equation \eqref{eq:Q1}, becomes \begin{aligned} v & = v_{\textrm{cm}} + v(r) \\ & = v_{\textrm{cm}} – \omega R \\ & = v_{\textrm{cm}} – \frac{v_{\textrm{cm}}}{R}\cdot R \\ & = v_{\textrm{cm}} – v_{\textrm{cm}} = 0 \end{aligned} Not only do we get the same result for the rolling ball problem we see on the whiteboard but it is what we expect. When a rolling ball, wheel or object doesn’t slip or skid, the point of contact is stationary. Cycloid and the Rolling ball
If we were to trace the path drawn by a point on the ball we get something known as a cycloid. The rolling ball problem is an interesting one and the reason it is studied is because the body undergoes two types of motion at the same time — pure translation and pure rotation. This means that the point that touches the ground, the contact point, is stationary while the top of the ball moves twice as fast as the center. It seems somewhat counter-intuitive which is why we don’t often think about it but imagine if at the point of contact our car’s tires wasn’t stationary but moved. We’d slip and slide and not go anywhere fast. But that is another problem entirely. |
Paul Dirac (see TRF biography) died exactly 30 years ago, on October 20th, 1984.
He was an eminent physicist, a co-father of quantum mechanics, the author of the Dirac equation, and a man who was convinced that the Physical law should have mathematical beauty.He wrote down this important sentence in capital letters on a blackboard during his 1955 lecture in Moscow. ;-)
Later, the English physicist would move to Florida where he was sort of happy – the place reminded him of England.
The video (well, picture+audio) at the top boasts Paul Dirac who describes how special physical constants are if they are dimensionless. The numerical values of those don't depend on the units – so even other (extraterrestrial) civilizations will unavoidably agree about these numbers. Due to this objectivity, there should be a calculation that clarifies why the numbers are what they are.
It's been an extremely important motivator for my efforts to get "deeper" to the laws of Nature, too. At some moment, I understood why string theory does guarantee that there are no continuously adjustable non-dynamical parameters in Nature. It means that in principle, every such dimensionless constant we observe in Nature is calculable with an arbitrary accuracy – as long as you learn a finite amount of discrete information about the right compactification of string theory.
Most laymen don't appreciate how incredible progress this provable characteristic of string theory represents. For the first time in the history of science, string theory gives us a framework where there doesn't exist a single "lever" that could be adjusted or fudged. Even if the number of relevant string theory vacua were really \(10^{500}\), one would still need just 500 digits of information (it's not that far from things I was able to memorize!) to understand really everything, with an arbitrary precision.
Independently of string theory, particle physics also reparameterized the examples of the dimensionless numbers that Dirac used in his monologue. The proton-electron mass ratio, about \(6\pi^5\sim 1836.15\), should be explainable from the basic laws. But it is not really the "simplest" thing. The electron is a pretty simple elementary particle but we know that the proton isn't too simple. It is a somewhat messy (even though the lightest [nearly?] stable) bound state of three valence quarks and lots of extra gluons and quark-antiquark pairs etc. In lattice QCD, a computational approach to the theory of the strong force, it's damn difficult to calculate the mass of the proton.
Moreover, the result depends on many other "more fundamental" constants such as the bare mass quarks, the QCD coupling (which runs as a function of energy, so we may replace this information by the QCD confinement scale), and other things. Why the proton mass is a rather messy quantity is easy to see: the proton itself became messy in the late 1960s and early 1970s.
Dirac's other example is the fine structure constant,\[
\alpha=\frac{e^2}{4\pi \epsilon_0\hbar c} \approx \frac{1}{137.036}.
\] That's arguably the most popular number among the physics-oriented numerologists. The idea has been that this is so simple to be defined that the calculation of the number in the final theory should be rather "straightforward". However, the electroweak theory showed us one reason why it isn't. The electrostatic force itself isn't quite "elementary". The \(U(1)_{em}\) gauge group behind it is really just a combination (diagonal group) of a more fundamental "hypercharge" \(U(1)_Y\) group and the \(Spin (2)\) subgroup of the \(z\)-axis rotations in the \(Spin(3)\sim SU(2)_W\) non-Abelian group producing the W-bosons (and their neutral friend, a combination of the photon and the Z-boson).
Because of this "mixed nature" of electromagnetism, the electromagnetic coupling \(e\) is really a function of "more fundamental" couplings \(g_1,g_2\) of the \(U(1)_Y,SU(2)_W\) groups, respectively:\[
e = \frac{g_1 g_2}{\sqrt{g_1^2+g_2^2}}
\] Now, another complication is that even these "more fundamental" couplings \(g_1,g_2\) are "problematic" because they are (logarithmically) running – they have an extra slow, quantum-mechanics-induced dependence on the energy scale and their values at "very low energies" \(g_{1},g_2(E\to 0)\) are actually relevant for the fine-structure constant \(1/137.036\dots\). However, physicists have understood that the values of the coupling constants that have a reason to result from a "simpler" calculation are, on the contrary, the values at very high energies,\[
g_1,g_2(E\to \infty)
\] or at the GUT or Planck scale, and so on. That's where you might reduce \(g_1,g_2\) to the knowledge of some other constants in a grand unified theory, or calculate them from "nothing or pure rigid geometry" in string theory. Once you do that, you may extrapolate them to \(E\to 0\) via the renormalization group equations and finally you compute the fine-structure constant using the formulae written above.
My main general point is that the dream of calculating the constants "out of nothing" is still here – and we already "surrendered a bit" so that (almost?) no one thinks it's too likely that someone will actually present the complete calculation by Christmas 2014 – but the precise idea about "which quantities should be given a truly elegant calculation of the value" is developing along with physics. The electroweak theory, the renormalization group, and (with some uncertainties surviving) grand unification and string theory imply that the previously "simple, elementary" dimensionless constants are actually a bit contrived and other constants are more fundamental even though it is not so straightforward to extract them from simple observations.
The second part of Dirac's video is about his large number hypothesis. Some large dimensionless constants such as the electron mass over the Planck scale may be extreme – much smaller than one or (which is the same because one may invert numbers) much greater than one – may be so extreme because they are functions of the "elementary" large number, the age of the Universe in the Planck units.
That's a wonderful idea to solve all these problems. Fortunately or unfortunately (it depends on your perspective), it predicts that the values of the dimensionless constants should actually change with time because the age of the Universe is surely changing with time, by its definition. Because this evolution of the constants seems to contradict the observations, the whole hypothesis by Dirac seems to be ruled out.
Note that the experiments don't exclude merely one particular, special version of the Dirac's "large number hypothesis" applied to one number. Because the constants were demonstrably the same many billions years ago and some evolution would have had a much bigger impact on all the values, we may really say that the whole
philosophyor paradigmthat Dirac offered in the talk has been killed by the experimental evidence.
I am emphasizing this point because most laymen tend to think that a theory may be ruled out only if one
exactlysays what the theory is, what are the exact values of the parameters, and what are the precise sets of problems where it should apply – and if one calculates the exact particular predictions for these quantities. But as this example (and many other examples) shows, this just isn't the case. Theories and frameworks, even incomplete and somewhat vaguely defined ones, predict rather specific patterns, regularities, or at least inequalities even before they are fully well-defined. And in most cases, they just contradict the evidence, so even the very general ideas and even "philosophies" may be falsified by the evidence long before someone manages to make the theory "truly well-defined and particular".
It's extremely unusual for a sufficiently robust, natural, and ambitious theory – one that is clearly capable of reproducing the quantitative success of other successful theories – to withstand the many possibilities by which it could have been falsified. String theory's competitors proposed as unified theories of all interactions die within minutes – and that's why the 45-year-long survival of string theory is incredibly strong circumstantial evidence that string theory is an extremely important set of ideas to be considered, even in the absence of a rigorous or irreversible proof that it applies to Nature. |
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My question is quite simple: what is the fundamental definition of extrinsic curvature of an hypersurface?Let me explain why I have not just copied one definition from the abundant literature. The specific structure on the Lorentzian manifold that I'm considering does not imply that an...
This is a refinement of a previous thread (here). I hope I am following correct protocol.1. Homework StatementI am studying Spacetime and Geometry : An Introduction to General Relativity by Sean M Carroll and have a question about commutators of vector fields. A vector field on a manifold...
1. Homework StatementI am studying Spacetime and Geometry : An Introduction to General Relativity by Sean M Carroll and have a question about commutators of vector fields. A vector field on a manifold can be thought of as differential operator which transforms smooth functions to smooth...
I am reading "Road to Reality" by Rogen Penrose. In chapter 15, Fibre and Gauge Connection ,while going through Clifford Bundle, he says:.""""........Of course, this in itself does not tell us why the Clifford bundle has no continuous cross-sections. To understand this it will be helpful to...
Hello!I was thinking about the Riemann curvature tensor(and the torsion tensor) and the way they are defined and it seems to me that they just need a connection(not Levi-Civita) to be defined. They don't need a metric. So, in reality, we can talk about the Riemann curvature tensor of smooth...
Hello!!Since connections in general do not require that we have a Riemannian manifold, but only a smooth manifold, I find it kind of weird that the only examples of connections that I find in the internet are those which use the Levi-Civita connection.So, I wanted to know of any examples of...
I am applying a Green's probabilistic elastodynamic tensor with relativistic manifold extensions to solve a pull out of a smoothly shaped deformable spheroid from a stiff inhomogenous deformable quasi-brittle host. This involves a Hooke's law tensor, a relativistic manifold Ricci tensor, a...
In my ignorance, when first learning, I just assumed that one pushed a vector forward to where a form lived and then they ate each other.And I assumed one pulled a form back to where a vector lived (for the same reason).But I see now this is idiotic: for one does the pullback and pushforward...
So I understand that the integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω.And I understand that one can pull back the integral of a 1-form over a line to the line integral between the...
I am really struggling with one concept in my study of differential geometry where there seems to be a conflict among different textbooks. To set up the question, let M be a manifold and let (U, φ) be a chart. Now suppose we have a curve γ:(-ε,+ε) → M such that γ(t)=0 at a ∈ M. Suppose further...
As I understand it, the notion of a distance between points on a manifold ##M## requires that the manifold be endowed with a metric ##g##. In the case of ordinary Euclidean space this is simply the trivial identity matrix, i.e. ##g_{\mu\nu}=\delta_{\mu\nu}##. In Euclidean space we also have that...
I'm fairly new to differential geometry (learning with a view to understanding general relativity at a deeper level) and hoping I can clear up some questions I have about coordinate charts on manifolds.Is the reason why one can't construct global coordinate charts on manifolds in general...
I've recently been studying a bit of differential geometry in the hope of gaining a deeper understanding of the mathematics of general relativity (GR). I have come across the notion of a topology and whilst I understand the mathematical definition (in terms of endowing a set of points with the...
No question this time. Just a simple THANK YOUFor almost two years years now, I have been struggling to learn: differential forms, exterior algebra, calculus on manifolds, Lie Algebra, Lie Groups.My math background was very deficient: I am a 55 year old retired (a good life) professor of...
As many of you know, using the stereographic projection one can construct a homeomorphism between the the complex plane ℂ1 and the unit sphere S2∈ℝ3. But the stereographic projection can be extended tothe n-sphere/n-dimensional Euclidean space ∀n≥1. Now what I am talking about is the the...
Why are tangent spaces on a general manifold associated to single points on the manifold? I've heard that it has to do with not being able to subtract/ add one point from/to another on a manifold (ignoring the concept of a connection at the moment), but I'm not sure I fully understand this - is...
I'm currently working through chapter 7 on Riemannian geometry in Nakahara's book "Geometry, topology & physics" and I'm having a bit of trouble reproducing his calculation for the metric compatibility in a non-coordinate basis, using the Ricci rotation coefficients...
I am trying to finish the last chapter of Spivak's Calculus on Manifolds book. I am stuck in trying to understand something that seems like it's supposed to be trivial but I can't figure it out.Suppose M is a manifold and \omega is a p-form on M. If f: W \rightarrow \mathbb{R}^n is a... |
In the TRPO paper, the objective to maximize is (equation 14) $$ \mathbb{E}_{s\sim\rho_{\theta_\text{old}},a\sim q}\left[\frac{\pi_\theta(a|s)}{q(a|s)} Q_{\theta_\text{old}}(s,a) \right] $$
which involves an expectation over states sampled with some density $\rho$, itself defined as $$ \rho_\pi(s) = P(s_0 = s)+\gamma P(s_1=s) + \gamma^2 P(s_2=s) + \dots $$
This seems to suggest that later timesteps should be sampled less often than earlier timesteps, or equivalently sampling states uniformly in trajectories but adding an importance sampling term $\gamma^t$.
However, the usual implementations simply use batches made of truncated or concatenated trajectories, without any reference to the location of the timesteps in the trajectory.
This is similar to what can be seen in the PPO paper, which transforms the above objective into (equation 3) $$ \mathbb{E}_t \left[ \frac{\pi_\theta(a_t|s_t)}{\pi_{\theta_\text{old}}(a_t|s_t)} \hat A_t \right] $$
It seems that something is missing in going from $\mathbb{E}_{s\sim \rho}$ to $\mathbb{E}_t$ in the discounted setting. Are they really equivalent? |
WHY?
Some tasks such as Sudoku require serial steps of relational inference.
WHAT?
Recurrent relational network operates on a graph representation of objects. Message passing method is used to pass the relational information to neighbor nodes to solve the task. The loss is minimized at every step.
m_{ij}^t = f(h_i^{t-1}, h_j^{t-1})\\m_j^t = \sum_{i\in N(j)}m^t_{ij}\\h_j^t = g(h_j^{t-1}, x_j, m_j^t)\\o_i^t = r(h_i^t)\\l^t = -\sum_{i=1}^I \log o_i^t[y_i]
So?
This module showed stablly better performance in bAbI task. RRN showed good performance in the newly defined Pretty-CLEVER task. This paper also showed this network solving Sudoku well.
Critic
It is clever to use message passing algorithm to solve relational tasks, but I think traing data is the issue(requires sequential answers for the task). |
EDIT, January 2015: Conway's little book is available at http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
I also put four related excerpts, all with the prefix indefinite_binary, at OTHER. Dmitry says the computer zakuski is being decommissioned, hope it continues to work through late January. I especially like Stillwell's presentation. Put it all together, for a Pell form, indeed any form $a x^2 + b xy + c y^2$ with $a > 0, \; b \geq 0, \; c < 0,$ but $b^2 - 4ac$ not a perfect square, we get a diagram that shows all of Conway's information, along with the $(x,y)$ pairs as column vectors, with an explicit illustration of the (proper) automorphism group generator, that being the mapping $(x,y) \mapsto (9x+20y,4x+9 y). $
Did not notice this one ten days ago. There is an explicit structure for representing a number by an indefinite quadratic form. This is chapter one in Conway's
The Sensual Quadratic Form. I wrote a little program recently, and no longer make simple arithmetic mistakes in these.
It turns out that all occurrences of $\pm 4$ happen along the "river" for $x^2 - 5 y^2. $
Given any solution to $x^2 - 5 y^2 = \pm 4,$ we gat the same value by switching $(x,y)$ to $$ (9x+20y,4x+9 y). $$ The two by two matrix causing this transformation (on column vectors) is $$ A \; = \; \left( \begin{array}{rr} 9 & 20 \\ 4 & 9 \end{array} \right) ,$$which you can see towards the right of the diagram as the coordinates of the final $1$ and then the final $-5,$ placed side by side. The big theorem is that the entire diagram is periodic. I find the finite set of representatives within one cycle, apply the transformation I wrote arbitrarily many times, and i get all. As there is no $xy$ term in $x^2 - 5 y^2,$ there is a simple $\pm$ symmetry as well.
So, all solutions to $x^2 - 5 y^2 = \pm 4 $ are:
Imprimitive:
+4:$$(2,0), (18,8), (322,144), (5778,2584), (103682,46368), (1860498,832040),\ldots, $$
-4:$$(-4,2), (4,2), (76,34), (1364,610), (24476,10946), (439204,196418),\ldots, $$
Primitive:
+4:$$(3,-1), (7,3), (123,55), (2207,987), (39603,17711), (710647,317811), \ldots, $$
+4:$$(3,1), (47,21), (843,377), (15127,6765), (271443,121393), \ldots, $$
-4:$$(-1,1), (11,5), (199,89), (3571,1597), (64079,28657), (1149851,514229), \ldots, $$
-4:$$(1,1), (29,13), (521,233), (9349,4181), (167761,75025), \ldots, $$
For any position in these sequences, there is a degree two recursion given by
$$ a_{n+2} = 18 a_{n+1} - a_n. $$ For example, $18 \cdot 29 - 1 = 521,$ then $18 \cdot 521 - 29 = 9349. $
Let's see, 3:21 pm. Both Fibonacci and Lucas do the same thing (by six positions), as $$ F_{n+12} = 18 F_{n+6} - F_n, $$$$ L_{n+12} = 18 L_{n+6} - L_n. $$ So, if the six orbits above satisfy the desired Fibonacci/Lucas conditions, that is a complete proof. If so, one could, carefully, interleave the six orbits in numerical order, perhaps using only the ones with strictly positive entries. See whether that works:
$$ (1,1),(3,1),(4,2),(7,3),(11,5), (18,8),$$$$ (29,13),(47,21),(76,34),(123,55),(199,89), (322,144),$$$$(521,233),(843,377),(1364,610),(2207,987),(3571,1597),(5778,2584), $$$$(9349,4181),(15127,6765),(24476,10946),(39603,17711),(64079,28657),(103682,46368), $$$$ (167761,75025),(271443,121393),(439204,196418),(710647,317811),(1149851,514229),(1860498,832040), $$Yep. The only miss is $(2,0),$ as $2$ is not a Lucas number. CORRECTION, FEB. 2015: as is commented elsewhere, it appears fairly common for people to define Lucas number $L_0 = 2,$ http://en.wikipedia.org/wiki/Lucas_number
Ummm; as you can see, $(x,y)$ and $(x,-y)$ may be distinct as far as the orbits, the six lists i wrote.
There is plenty more that could be said; anyway, these give all solutions. Oh, the other business, the "climbing lemma," says that values only increase (in absolute value) when leaving the river. The next layers of values are $\pm 11$ at the continuation of each edge with a light blue $6,$ and $\pm 19$ at the continuation of each edge with a light blue $10.$ So we have done enough to catch all $\pm 4$ already. |
Can you use only two instances of the digit $2$, along with the mathematical operations below, to create an expression that evaluates to $2015$?
Allowed operations:
arithmetic operations: addition ($+$), subtraction ($-$), multiplication ($\times$), division ($\div$, or $\frac{x}{y}$), exponentiation (like $2^2$);
factorial ($!$), absolute value ($|...|$);
extraction of the root of any degree in a form $\sqrt[a]{b}$ or the square root in a form $\sqrt{b}$;
trigonometric functions: sine, cosine, tangent, cotangent, secant, cosecant
inverse trigonometric functions: arcsine, arccosine, arctangent, arccotangent, arcsecant, arccosecant
natural logarithm ($\ln b$), or logarithm with any base ($\log_a b$)
Parentheses $()$ are also allowed. What is not allowed:
digits other than $2$, or more than two instances of the digit $2$
named constants such as $\pi, e$, etc...
defining and using your own functions
other variables
Example with three $2$s:
A couple of years ago, I managed to solve the same kind of problem, with three $2$s:
We can express any natural number $A$ using three $2$s and the above operations, like this: $$-\log_2\log_2\underbrace{\sqrt{\sqrt{...\sqrt{2}}}}_{A\,\mathrm{square\,roots}}$$
But I couldn't crack the problem using only two $2$s... |
In Season 1 Episode 2 of The Big Bang Theory, “The Big Bran Hypothesis”, Penny (Kaley Cuoco) asks Leonard (Johnny Galecki) to sign for a furniture delivery if she isn’t home. Unfortunately for Leonard and Sheldon, they are left with the task of getting a huge (and heavy) box up to Penny’s apartment.
To solve this problem, Leonard suggest using the stairs as an inclined plane, one of the six classical simple machines defined by Renaissance scientists. Both Leonard and Sheldon have the right idea here. Not only are inclined planes used to raise heavy loads but they require less effort to do so. Though this may make moving a heavy load easier the tradeoff is that the load must now be moved over a greater distance. So while, as Leonard correctly calculates, the effort required to move Penny’s furniture is reduced by half, the distance he and Sheldon must move Penny’s furniture twice the distance to raise it directly.
Mathematics of the Inclined Plane Effort to lift block on Inclined Plane
Now we got an inclined plane. Force required to lift is reduced by the sine of the angle of the stairs… call it 30 degrees, so about half.
To analyze the forces acting on a body, physicists and engineers use rough sketches or free body diagrams. This diagram can help physicists model a problem on paper and to determine how forces act on an object. We can resolve the forces to see the effort needed to move the block up the stairs.
If the weight of Penny’s furniture is \(W\) and the angle of the stairs is \(\theta\) then
\[\angle_{\mathrm{stairs}}\equiv\theta \approx 30^\circ\] and \[\Rightarrow\sin 30^\circ = \frac{1}{2}\] So the effort needed to keep the box in place is about half the weight of the furniture box or \(\frac{1}{2}W\), just as Leonard says. Distance moved along Inclined Plane
While the inclined plane allows Leonard and Sheldon to push the box with less effort, the tradeoff is that the distance they move along the incline is twice the height to raise the box vertically. Geometry shows us that
\[\sin \theta = \frac{h}{d}\] We again assume that the angle of the stairs is approximately \(30^\circ\) and \(\sin 30^{\circ} = 1/2\) then we have \(d=2h\). Uses of the Inclined Plane
We see inclined planes daily without realizing it. They are used as loading ramps to load and unload goods. Wheelchair ramps also allow wheelchair users, as well as users of strollers and carts, to access buildings easily. Roads sometimes have inclined planes to form a gradual slope to allow vehicles to move over hills without losing traction. Inclined planes have also played an important part in history and were used to build the Egyptian pyramids and possibly used to move the heavy stones to build Stonehenge.
Lombard Street (San Francisco)
Lombard Street in San Francisco is famous for its eight tight hairpin turns (or switchbacks) that have earned it the distinction of being the crookedest street in the world (though this title is contested). These eight switchbacks are crucial to the street’s design as the reduce the hills natural 27° grade which is too steep for most vehicles. It is also a hazard to pedestrians, who are more accustomed to a more reasonable 4.86° incline due to wheel chair navigability concerns.
Technically speaking, the “zigzag” path doesn’t make climbing or coming down the hill any easier. As we have seen, all it does is change how various forces are applied. It just requires less effort to move up or down but the tradeoff is that you travel a longer distance. This has several advantages. Car engines have to be less powerful to climb the hill and in the case of descent, less force needs to be applied on the brakes. There are also safety considerations. A car will not accelerate down the switch back path as fast than if it was driven straight down, making speeds safer and more manageable for motorists.
This idea of using zigzagging paths to climb steep hills and mountains is also used by hikers and rock climbers for very much the same reason Lombard Street zigszags. The tradeoff is that the distance traveled along the path is greater than if a climber goes straight up.
The Descendants of Archimedes
We don’t need strength, we’re physicists. We are the intellectual descendants of Archimedes. Give me a fulcrum and a lever and I can move the Earth. It’s just a matter of… I don’t have this, I don’t have this!
We see that Leonard had the right idea. If we were to assume are to assume — based on the size of the box — that the furniture is approximately 150 lbs (65kg) and the effort is reduced by half, then they need to push with at least 75 lbs of force. This is equivalent to moving a 34kg mass. If they both push equally, they are each left pushing a very manageable 37.5 lbs, the equivalent of pushing a 17kg mass.
Penny’s apartment is on the fourth floor and we if we assume a standard US building design of ten feet per floor, this means a 30 foot vertical rise. The boys are left with the choice of lifting 150 lbs vertically 30 feet or moving 75lbs a distance of 60 feet. The latter is more manageable but then again, neither of our heroes have any upper body strength. |
Abstract
Let ${\rm Sym}_3\, \mathbf{C} \longrightarrow \mathbf{P}_*(k \oplus {\rm Sym}_3\,k \oplus {\rm Sym}_3\,k \oplus k) = {\bf P}^{13}, A \mapsto (1: A: A’: \det A) $ be the Veronese embedding of the space of symmetric matrices of degree 3, where $A’$ is the cofactor matrix of $A$. The closure $\operatorname{SpG}(3, 6)$ of this image is a 6-dimensional homogeneous variety of the symplectic group $\operatorname{Sp}(3)$. A canonical curve $C_{16} \subset {\bf P}^8$ of genus 9 over a perfect field $k$ is isomorphic to a complete linear section of this projective variety $\operatorname{SpG}(3, 6) \subset {\bf P}^{13}$ unless $C \otimes_k \bar k$, $\bar k$ being the algebraic closure, is a covering of degree at most 5 of the projective line. We prove this by means of linear systems of higher rank.
[ACGH] E. Arbarello, M. Cornalba, P. A. Griffiths, and J. Harris, Geometry of Algebraic curves, I, New York: Springer-Verlag, 1985, vol. 267.
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@incollection {nagata, MRKEY = {977768},
AUTHOR = {Mukai, Shigeru}, TITLE = {Curves, {$K3$} surfaces and {F}ano {$3$}-folds of genus {$\leq 10$}}, BOOKTITLE = {Algebraic Geometry and Commutative Algebra, {V}ol. {\rm I}}, PAGES = {357--377}, PUBLISHER = {Kinokuniya}, ADDRESS = {Tokyo}, YEAR = {1988}, MRCLASS = {14J10 (14J28 14J30 32G13 32J15 32M10)}, MRNUMBER = {90b:14039}, MRREVIEWER = {Peter Nielsen}, ZBLNUMBER = {0701.14044}, } [Bergen] S. Mukai, "Polarized $K3$ surfaces of genus $18$ and $20$," in Complex Projective Geometry, Cambridge: Cambridge Univ. Press, 1992, vol. 179, pp. 264-276.
@incollection {Bergen, MRKEY = {1201388},
AUTHOR = {Mukai, Shigeru}, TITLE = {Polarized {$K3$} surfaces of genus {$18$} and {$20$}}, BOOKTITLE = {Complex Projective Geometry}, VENUE={{T}rieste, 1989/{B}ergen, 1989}, SERIES = {London Math. Soc. Lecture Note Ser.}, VOLUME = {179}, PAGES = {264--276}, PUBLISHER = {Cambridge Univ. Press}, ADDRESS = {Cambridge}, YEAR = {1992}, MRCLASS = {14J28 (14J10 32J15)}, MRNUMBER = {94a:14039}, MRREVIEWER = {Shigeyuki Kondo}, ZBLNUMBER = {0774.14035}, } [PJA] S. Mukai, "Curves and symmetric spaces," Proc. Japan Acad. Ser. A Math. Sci., vol. 68, iss. 1, pp. 7-10, 1992.
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@incollection {inha, MRKEY = {1285374},
AUTHOR = {Mukai, Shigeru}, TITLE = {Curves and {G}rassmannians}, BOOKTITLE = {Algebraic Geometry and Related Topics}, VENUE={{I}nchon, 1992}, SERIES = {Conf. Proc. Lecture Notes Algebraic Geom.}, VOLUME={I}, PAGES = {19--40}, PUBLISHER = {Int. Press, Cambridge, MA}, YEAR = {1993}, MRCLASS = {14H45 (14M15)}, MRNUMBER = {95i:14032}, MRREVIEWER = {Raquel Mallavibarrena}, ZBLNUMBER = {0846.14030}, } [7] S. Mukai, "Curves and symmetric spaces. I," Amer. J. Math., vol. 117, iss. 6, pp. 1627-1644, 1995.
@article {7, MRKEY = {1363081},
AUTHOR = {Mukai, Shigeru}, TITLE = {Curves and symmetric spaces. {I}}, JOURNAL = {Amer. J. Math.}, FJOURNAL = {American Journal of Mathematics}, VOLUME = {117}, YEAR = {1995}, NUMBER = {6}, PAGES = {1627--1644}, ISSN = {0002-9327}, CODEN = {AJMAAN}, MRCLASS = {14H45 (14J45 14M17)}, MRNUMBER = {96m:14040}, MRREVIEWER = {Raquel Mallavibarrena}, DOI = {10.2307/2375032}, ZBLNUMBER = {0871.14025}, } [ron] S. Mukai, "New developments in Fano manifold theory related to the vector bundle method and moduli problems," Sūgaku, vol. 47, iss. 2, pp. 125-144, 1995.
@article {ron, MRKEY = {1364825},
AUTHOR = {Mukai, Shigeru}, TITLE = {New developments in {F}ano manifold theory related to the vector bundle method and moduli problems}, JOURNAL = {Sūgaku}, FJOURNAL = {Mathematical Society of Japan. Sūgaku (Mathematics)}, VOLUME = {47}, YEAR = {1995}, NUMBER = {2}, PAGES = {125--144}, ISSN = {0039-470X}, CODEN = {SUGKAQ}, MRCLASS = {14J45 (14J10 14J60)}, MRNUMBER = {96m:14059}, MRREVIEWER = {Takao Fujita}, ZBLNUMBER={0889.14010}, } |
Like I mentioned in the comments, all I am trying suggests that what you are suggesting actually does work.
As you mentioned, intuitively it makes sense if this would work, if $X_i$ represents the posterior draw for some probability $p_i$ for event $i$ happening, you should indeed be able to sum multiple $X_i$ to get the probability of multiple events $i$, if there is no possibility of both events happening at the same time. Since this is a multinomial setting, this is not the case, so we're good.
So let's show my simulations:
library('gtools')
K <- 10
alpha <- c(rpois(K, 50)) #randomly generated alphas, just cause
k <- 2 # the number of alphas we are summing together
sim <- rdirichlet(10000, alpha)
plot(density(rowSums(sim[, 1:k]))) # the density of the summed variable
lines(density(rdirichlet(10000, c(sum(alpha[1:k]), alpha[-(1:k)]))[,1]), col = 'blue')
# the density of the variable drawn from the Dirichlet distribution with summed alphas
Let's start with $\alpha = \{10, 10, 10\}$. Summing $\alpha_1$ and $\alpha_2$ should get us $Dir(2, \{20, 10\})$:
These marginal densities look pretty similar. According to wikipedia and this random lecture I found on the internet (through wikipedia), the marginal distribution of $X_i$ to the Dirichlet distribution is as follows:
$$X_i = Beta(\alpha_i, \sum_{k=1}^K\left[\alpha_k\right] - \alpha_i)$$
This relies on actually the same principle: summing all the $\alpha$ that are
not $\alpha_i$ together, turning the corresponding multinomial to a binomial distributions with the outcomes $i$ and $not\textrm{-}i$. And indeed, if we fit the marginal distribution we would expect from the sum over the density in the previous picture, we see that it looks the same:
So theoretically, we should be able to take a Dirichlet distribution with a high $K$, sum all but one together and end up with a Beta distribution. Heck, let's try:
(99 times $\alpha \sim Pois(50)$ and 1 $\alpha = 1000$, summing together the random $\alpha\mathrm{s}$.)
To show that it also works for the joint densities, this is an example with $K=4$ and $\alpha = \{10,10,10,10\}$:
And this is with $K=3$ and $\alpha=\{20,10,10\}$:
$$\ddot{\smile}$$
So where does the confusion about $x_1^{\alpha_1-1}x_2^{\alpha_2-1}\neq (x_1 + x_2)^{\alpha_1+\alpha_2 -1 }$ come from?
Because when we sum $x_1$ and $x_2$ together, we don't just care about the density at $x_1$ and $x_2$, but for any combination of two $x$ that sums to $x_1+x_2$. I am not that strong in integration, so I'll not try and burn myself with that, but I really suggest reading this (page 3-4) for more information.
EDIT:
As @whuber correctly remarked, here is an example with low alphas, $K=4$, summing the first two $X_1$ and $X_2$: |
Let the model $$y_t=\beta_0+\beta_1t+z_t\qquad t=1,2,\dots$$ $$z_t=\epsilon_t+\theta\epsilon_{t-1}$$ where $\epsilon_t$ is White noise with zero mean and variance $\sigma^2$ and $\beta_0,\beta_1,\theta$ constants.
a) Is $y_t$ stationary?
b) Is $(1-B)y_t$ stationary? (where $B$ is a lag operator)
This model can be written as $$y_t=\delta_t+\epsilon_t+\theta\epsilon_{t-1}$$
a) It is not a stationary process, this process have trend and clearly this have no constant mean, because $E[y_t]$ varies according to $t$.
b) $$(1-B)y_t=\beta_0+\beta_1t+\epsilon_t+\theta\epsilon_{t-1}-\beta_0-\beta_1(t-1)+\epsilon_{t-1}-\theta\epsilon_{t-2}$$ $$=\beta_1+\epsilon_t+\epsilon_{t-1}(\theta-1)+\theta\epsilon_{t-2}$$ $$=\beta_1+\epsilon_t[1+B(\theta-1)-\theta B^2]$$
$$B=\frac{(1-\theta)\pm \sqrt{(\theta-1)^2+4\theta}}{2}$$
The roots are $1$ and $-\theta$, so this process is not stationary?
I have other doubts too:
1) The process $y_t$ is a $MA(1)$ with non-zero mean?
2) When I look the roots of polynomial, both roots need to be $>1$ in modulus or just one is enough? |
WHY?
Former CNNs were not spatially invariant.
WHAT?
Spatial Transformer Module can make a former CNN model invariant to cropping, translation, scale, rotation and skew without greatly increasing the computation cost. ST module consists of 3 parts: localisation network, grid generator, and sampler. Localisation network takes input feature map (
U \in \mathbb{R}^{H X W X C} and output
\theta the parameter of the transformation
\tau_{\theta}. Next, grid generator transform the coordinate of target pixel into the corresponding coordinate of source pixel. This is like designating the location of input pixels in the space of output feature map. For example, affine transformation need six parameter. Lastly, the sampler is used to get the value of each pixel of output feature map. Sampling kernel uses the resulted coordinate of source pixel to get the output value. Backpropagation is possible since we can calculate the partial derivative of sampling kernel.
\theta = f_{loc}(U)\\ \left( \begin{array}{c} x_i^s \\ y_i^s \end{array} \right) = \tau_{\theta}(G_i) = A_{\theta} \left( \begin{array}{c} x_i^t \\ y_i^t \\ 1 \end{array} \right) = \left[ \begin{array}{ccc} \theta_{11} & \theta_{12} & \theta_{13} \\ \theta_{21} & \theta_{22} & \theta_{23} \end{array} \right] \left( \begin{array}{c} x_i^t \\ y_i^t \\ 1 \end{array} \right)\\ V_i^c = \Sigma_n^H \Sigma_m^W U^c_{nm} k(x_i^s - m; \Phi_x)k(y_i^s - m; \Phi_y) Multile spatial transformer can be used in a single CNN model to detect multiple objects.
So?
ST-CNN achieved state of the art performance on Distorted MNIST, SVHN and fine-grained fine classification.
Critic
Since I am not familiar with traditional graphics and vision, the idea of mapping seems surprising. I wonder this idea can be applied to 3d object detection. |
Background: In homological algebra, a quasi-isomorphism of chain complexes is a chain map$\phi:(C,d) \to (C',d')$ so that the induced map on homology $\phi_\ast:H_\ast(C,d) \to H_\ast(C',d')$ is an isomorphism.
This is clearly a special instance of the following much more general phenomenon: given a functor $\mathcal{F}:\mathcal{C} \to \mathcal{D}$, between categories $\mathcal{C}$ and $\mathcal{D}$, a morphism in $\mathcal{C}$ is taken to an isomorphism in $\mathcal{D}$ by $\mathcal{F}$.
What is the standard terminology for a morphism of $\mathcal{C}$ so that its image under $\mathcal{F}$ is a (mono, epi, iso) morphism?
I couldn't find a term for this in MacLane's book, but I'm sorry if this is standard stuff. |
Open Boundary Conditions for the Primitive and Boussinesq Equations
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The $\ce{N2^+}$ molecule has the molecular orbitals $$\ce{\sigma(1s)^2\; \sigma^*(1s)^2\; \sigma(2s)^2\; \sigma^*(2s)^2\; \pi(2p_x)^2 \; \pi(2p_y)^2 \; \sigma(2p_z)^1,}$$ which can be seen on the following picture:
So, the ground state has $\mathrm{A_g}$ symmetry in $D_\mathrm{2h}$ point group, i.e. $\Sigma_\mathrm{g}^+$ in $D_\mathrm{\infty h}$.
But what will the excited states look like?
I'd expect, that the first excited electrons will be the ones, which "need" the smallest amount of energy to get into a higher state.
So,
the first electron will be excited from $\ce{\pi(2p_x)}$ or $\ce{\pi(2p_y)}$ orbitals to $\ce{\sigma(2p_z)}$ like this:
The first excited state will have $\mathrm{B_{3u}}$ or $\mathrm{B_{2u}}$ symmetry in $D_\mathrm{2h}$.
The second excited state would be
I.e. it would have the $\mathrm{B_{3g}}$ or $\mathrm{B_{2g}}$ symmetry.
Is my expectation correct or will the electrons be excited in a different order? And what will the 3rd and 4th excited states look like? |
I'm trying to find out if the partial derivatives of the following function exist at $(0,0)$.
$$f(x,y)=\left\{ \begin{array}{c} |x| \quad \text{ if }y = x^2 \\ 0 \quad \text{ if }y \neq x^2 \end{array} \right.$$
My attempt:
The x-derivative at zero is: $$\frac{\partial f}{\partial x}(0,0)= \lim_{t \rightarrow 0}\frac{f(t,0)-f(0,0)}{t}$$
So it will exist if the side derivatives exist and are equal.
$$\lim_{t \rightarrow 0^{-}}\frac{f(t,0)-f(0,0)}{t}=\lim_{t \rightarrow 0^{-}}\frac{|t|}{t} = \frac{-t}{t}=-1$$
$$\lim_{t \rightarrow 0^{+}}\frac{f(t,0)-f(0,0)}{t}=\lim_{t \rightarrow 0^{+}}\frac{|t|}{t} = \frac{t}{t}=1$$
So the x-derivative doesn't exist. However my book says both $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ exist and are equal to zero. Have I done something wrong? |
This article is aimed at relatively new users. It is written particularly for my own students, with the aim of helping them to avoid making common errors. The article exists in two forms: this WordPress blog post and a PDF file generated by , both produced from the same Emacs Org file. Since WordPress does not handle very well I recommend reading the PDF version.
1. New Paragraphs
In a new paragraph is started by leaving a blank line.
Do not start a new paragraph by using
\\ (it merely terminates a line). Indeed you should almost never type
\\, except within environments such as
array,
tabular, and so on.
2. Math Mode
Always type mathematics in math mode (as
$..$ or
\(..\)), to produce “” instead of “y = f(x)”, and “the dimension ” instead of “the dimension n”. For displayed equations use
$$,
\[..\], or one of the display environments (see Section 7).
Punctuation should appear outside math mode, for inline equations, otherwise the spacing will be incorrect. Here is an example.
Correct:
The variables $x$, $y$, and $z$ satisfy $x^2 + y^2 = z^2$.
Incorrect:
The variables $x,$ $y,$ and $z$ satisfy $x^2 + y^2 = z^2.$
For displayed equations, punctuation should appear as part of the display. All equations
must be punctuated, as they are part of a sentence. 3. Mathematical Functions in Roman
Mathematical functions should be typeset in roman font. This is done automatically for the many standard mathematical functions that supports, such as
\sin,
\tan,
\exp,
\max, etc.
If the function you need is not built into , create your own. The easiest way to do this is to use the
amsmath package and type, for example,
\usepackage{amsmath} ... % In the preamble. \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\inert}{Inertia}
Alternatively, if you are not using the
amsmath package you can type
\def\diag{\mathop{\mathrm{diag}}} 4. Maths Expressions
Ellipses (dots) are never explicitly typed as “…”. Instead they are typed as
\dots for baseline dots, as in
$x_1,x_2,\dots,x_n$ (giving ) or as
\cdots for vertically centered dots, as in
$x_1 + x_2 + \cdots + x_n$ (giving ).
Type
$i$th instead of
$i'th$ or
$i^{th}$. (For some subtle aspects of the use of ellipses, see How To Typeset an Ellipsis in a Mathematical Expression.)
Avoid using
\frac to produce stacked fractions in the text. Write flops instead of flops.
For “much less than”, type
\ll, giving , not
<<, which gives . Similarly, “much greater than” is typed as
\gg, giving . If you are using angle brackets to denote an inner product use
\langle and
\rangle:
incorrect: <x,y>, typed as
$<x,y>$.
correct: , typed as
$\langle x,y \rangle$
5. Text in Displayed Equations
When a displayed equation contains text such as “subject to ”, instead of putting the text in
\mathrm put the text in an
\mbox, as in
\mbox{subject to $x \ge 0$}. Note that
\mbox switches out of math mode, and this has the advantage of ensuring the correct spacing between words. If you are using the amsmath package you can use the
\text command instead of
\mbox.
Example $$ \min\{\, \|A-X\|_F: \mbox{$X$ is a correlation matrix} \,\}. $$ 6. BibTeX
Produce your bibliographies using BibTeX, creating your own bib file. Note three important points.
“Export citation” options on journal websites rarely produce perfect bib entries. More often than not the entry has an improperly cased title, an incomplete or incorrectly accented author name, improperly typeset maths in the title, or some other error, so always check and improve the entry. If you wish to cite one of my papers download the latest version of
njhigham.bib(along with
strings.bibsupplied with it) and include it in your
\bibliographycommand.
Decide on a consistent format for your bib entry keys and stick to it. In the format used in the Numerical Linear Algebra group at Manchester a 2010 paper by Smith and Jones has key
smjo10, a 1974 book by Aho, Hopcroft, and Ullman has key
ahu74, while a 1990 book by Smith has key
smit90.
7. Spelling Errors and Errors
There is no excuse for your writing to contain spelling errors, given the wide availability of spell checkers. You’ll need a spell checker that understands syntax.
There are also tools for checking syntax. One that comes with TeX Live is
lacheck, which describes itself as “a consistency checker for LaTeX documents”. Such a tool can point out possible syntax errors, or semantic errors such as unmatched parentheses, and warn of common mistakes.
8. Quotation Marks
has a left quotation mark, denoted here
\lq, and a right quotation mark, denoted here
\rq, typed as the single left and right quotes on the keyboard, respectively. A left or right double quotation mark is produced by typing two single quotes of the appropriate type. The double quotation mark always itself produces the same as two right quotation marks. Example: is typed as
\lq\lq hello \rq\rq.
9. Captions
Captions go
above tables but below figures. So put the
caption command at the start of a
table environment but at the end of a
figure environment. The
\label statement should go after the
\caption statement (or it can be put inside it), otherwise references to that label will refer to the subsection in which the label appears rather than the figure or table.
10. Tables
makes it easy to put many rules, some of them double, in and around a table, using
\cline,
\hline, and the
| column formatting symbol. However, it is good style to minimize the number of rules. A common task for journal copy editors is to remove rules from tables in submitted manuscripts.
11. Source Code
source code should be laid out so that it is readable, in order to aid editing and debugging, to help you to understand the code when you return to it after a break, and to aid collaborative writing. Readability means that logical structure should be apparent, in the same way as when indentation is used in writing a computer program. In particular, it is is a good idea to start new sentences on new lines, which makes it easier to cut and paste them during editing, and also makes a diff of two versions of the file more readable.
Example:
Good:
$$ U(\zbar) = U(-z) = \begin{cases} -U(z), & z\in D, \\ -U(z)-1, & \mbox{otherwise}. \end{cases} $$
Bad:
$$U(\zbar) = U(-z) = \begin{cases}-U(z), & z\in D, \\ -U(z)-1, & \mbox{otherwise}. \end{cases}$$ 12. Multiline Displayed Equations
For displayed equations occupying more than one line it is best to use the environments provided by the amsmath package. Of these,
align (and
align* if equation numbers are not wanted) is the one I use almost all the time. Example:
\begin{align*} \cos(A) &= I - \frac{A^2}{2!} + \frac{A^4}{4!} + \cdots,\\ \sin(A) &= A - \frac{A^3}{3!} + \frac{A^5}{5!} - \cdots, \end{align*}
Others, such as
gather and
aligned, are occasionally needed.
Avoid using the standard environment
eqnarray, because it doesn’t produce as good results as the amsmath environments, nor is it as versatile. For more details see the article Avoid Eqnarray.
13. Synonyms
This final category concerns synonyms and is a matter of personal preference. I prefer
\ge and
\le to the equivalent
\geq
\leq\ (why type the extra characters?).
I also prefer to use
$..$ for math mode instead of
\(..\) and
$$..$$ for display math mode instead of
\[..\]. My preferences are the original syntax, while the alternatives were introduced by . The slashed forms are obviously easier to parse, but this is one case where I prefer to stick with tradition. If dollar signs are good enough for Don Knuth, they are good enough for me!
I don’t think many people use ‘s verbose
\begin{math}..\end{math}
or
\begin{displaymath}..\end{displaymath}
Also note that
\begin{equation*}..\end{equation*} (for unnumbered equations) exists in the amsmath package but not in in itself. |
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Background: Pediatric craniocerebral gunshot wounds occur in the context of both accidental and intentional trauma. Unique physiologic factors merited research into prognostic factors and treatment priorities in the pediatric population. Methods: A systematic search of MEDLINE, EMBASE, Web of Science, LILACS, Cochrane Registered Trials and Systematic Reviews, ISRCTN, and ClinicalTrials.gov was conducted. Selection criteria included all studies published in any language since 2000 which described intracranial isolated gunshot wounds in a civilian individual or population of pediatric age. Post-mortem and epidemiological studies were excluded. Screening was conducted through Covidence. Results: Initial database search revealed 349 unique studies for abstract and title screening. Fifty studies were selected for full text screening. Nine studies were included in the final review. Study quality was assessed with the Newcastle-Ottawa Scale. Case series noted bullet migration, pituitary deficiency, neurovascular and neuropsychological concerns. Three single-center retrospective studies of 71, 30, and 48 pediatric patients suggested multiple negative prognostic signs on initial presentation. Early aggressive surgical treatment was recommended by some authors. Conclusions: This systematic review analyzed the best current understanding of evidence for prognostic factors and treatment considerations of intracranial gunshot wounds in the pediatric neurotrauma context. Areas for future research with larger multi-center studies were highlighted.
We argue that How Foraging Works sketches a good foundational model, but it needs expanding to incorporate hierarchical and multiscale conceptions of uncertainty and to incorporate inference of environmental controllability. Most pressingly, its algorithmic implementation needs to be better justified in terms of its functional forms and, ultimately, to be more heavily constrained by survival optimality.
The generalized Gauss circle problem concerns the lattice point discrepancy of large spheres. We study the Dirichlet series associated to$P_{k}(n)^{2}$, where$P_{k}(n)$is the discrepancy between the volume of the$k$-dimensional sphere of radius$\sqrt{n}$and the number of integer lattice points contained in that sphere. We prove asymptotics with improved power-saving error terms for smoothed sums, including$\sum P_{k}(n)^{2}e^{-n/X}$and the Laplace transform$\int _{0}^{\infty }P_{k}(t)^{2}e^{-t/X}\,dt$, in dimensions$k\geqslant 3$. We also obtain main terms and power-saving error terms for the sharp sums$\sum _{n\leqslant X}P_{k}(n)^{2}$, along with similar results for the sharp integral$\int _{0}^{X}P_{3}(t)^{2}\,dt$. This includes producing the first power-saving error term in mean square for the dimension-3 Gauss circle problem.
Various ontology visualization tools using different visualization methods exist and new ones are being developed every year. The goal of this paper is to follow up on previous surveys with an updated classification of ontology visualization methods and a comprehensive survey of available tools. The tools are analyzed for the used visualization methods, interaction techniques and supported ontology constructs. It shows that most of the tools apply two-dimensional node-link visualizations with a focus on class hierarchies. Color and shape are used with little variation, support for constructs introduced with version 2 of the OWL Web Ontology Language is limited, and it often remains vague what tasks and use cases are supported by the visualizations. Major challenges are the limited maturity and usability of many of the tools as well as providing an overview of large ontologies while also showing details on demand. We see a high demand for a universal ontology visualization framework implementing a core set of visual and interactive features that can be extended and customized to respective use cases.
Interactions between Antarctic sea ice and synoptic activity in the circumpolar trough have been investigated using meteorological data from European Centre for Medium-Range Weather Forecasts (ECMWF) Interim Re-analysis and sea-ice data from passive-microwave measurements. Total Antarctic sea-ice extent does not show large interannual variations. However, large differences are observed on a regional/monthly scale, depending on prevailing winds and currents, and thus on the prevailing synoptic situations. the sea-ice edge is also a preferred region for cyclogenesis due to the strong meridional temperature gradient (high baroclinicity) in that area. the motivation for this study was to gain a better understanding of the interaction between sea-ice extent and the general atmospheric flow, particularly the frequency of warm-air intrusions into the interior of the Antarctic continent, since this influences precipitation seasonality and must be taken into account for a correct climatic interpretation of ice cores. Two case studies of extraordinary sea-ice concentration anomalies in relation to the prevailing atmospheric conditions are presented. However, both strong positive and negative anomalies can be related to warm biases in ice cores (indicated by stable-isotope ratios), especially in connection with the negative phase of the Southern Annular Mode.
Analyses of shallow cores obtained at the European Project for Ice Coring in Antarctica (EPICA) drilling site Kohnen station (75°00′ S, 00°04′ E; 2892 m a.s.l.) on the plateau of Dronning Maud Land reveal the presence of conserved snow dunes in the firn. In situ observations during three dune formation events in the 2005/06 austral summer at Kohnen station show that these periods were characterized by a phase of 2 or 3 days with snowdrift prior to dune formation which only occurred during high wind speeds of >10 m s-1 at 2 m height caused by the influence of a low-pressure system. The dune surface coverage after a formation event varied between 5% and 15%, with a typical dune size of (4 ± 2) m × (8 ± 3) m, a maximum height of 0.2 ± 0.1 m and a periodicity length of about 30 m. The mean density within a snow dune varied between 380 and 500 kg m-3, whereas the mean density at the surrounding surface was 330 ± 5 kgm-3. The firn cores covering a time-span of 22 ± 2 years reveal that approximately three to eight events per year occurred, during which snow dunes had been formed and were preserved in the firn.
The zeta function of a curve C over a finite field may be expressed in terms of the characteristic polynomial of a unitary matrix ΘC. We develop and present a new technique to compute the expected value of tr(ΘCn) for various moduli spaces of curves of genus g over a fixed finite field in the limit as g is large, generalising and extending the work of Rudnick [Rud10] and Chinis [Chi16]. This is achieved by using function field zeta functions, explicit formulae, and the densities of prime polynomials with prescribed ramification types at certain places as given in [BDF+16] and [Zha]. We extend [BDF+16] by describing explicit dependence on the place and give an explicit proof of the Lindelöf bound for function field Dirichlet L-functions L(1/2 + it, χ). As applications, we compute the one-level density for hyperelliptic curves, cyclic ℓ-covers, and cubic non-Galois covers.
The policies of Republican Governor Scott Walker have come to symbolize a resurgent assault on the public sector, and on public employee unions in particular, by the Republican Party. The fact that this is happening in Wisconsin, the state that in the last century was considered the “laboratory of Progressivism,” makes the politics surrounding these policies all the more compelling. In The Politics of Resentment: Rural Consciousness in Wisconsin and the Rise of Scott Walker, Katherine J. Cramer analyzes the “politics of resentment” surrounding these developments. Employing an ethnographic “method of listening,” Cramer furnishes thick description of the political language employed by rural Wisconsinites, and proceeds to develop an interpretive theory of “political resentment” that illuminates the reasons why lower-class citizens so strongly oppose public policies seeking to offset social and economic inequality. The book is important methodologically and politically. We have thus invited a range of social and political scientists to comment on the book as a work of political science and as a diagnosis of the current political moment.
Lipid biomarkers have been extensively applied for tracing organisms and evolutionary processes through Earth's history. They have become especially important for the reconstruction of early life on Earth and, potentially, for the detection of life in the extraterrestrial realm. However, it is not always clear how exactly biomarkers reflect a paleoecosystem as their preservation may be influenced by increasing temperatures (T) and pressures (P) during burial. While a number of biomarker indices reflecting thermal maturity have been established, it is often less well constrained to which extent biomarker ratios used for paleoreconstruction are compromised by T and P processes. In this study we conducted hydrous pyrolysis of Green River Shale (GRS) kerogen in gold capsules for 2–2400 h at 300°C to assess the maturation behaviour of several compounds used as life tracers and for the reconstruction of paleoenvironments (n-alkanes, pristane, phytane, gammacerane, steranes, hopanes and cheilanthanes). Lignite samples were maturated in parallel with the GRS kerogen to obtain exact vitrinite reflectance data at every sampling point. Our experiment confirms the applicability of biomarker-based indices and ratios as maturity indicators (e.g. total cheilanthanes/hopanes ratio; sterane and hopane isomerization indices). However, several biomarker ratios that are commonly used for paleoreconstructions (e.g. pristane/phytane, pristane/n-C17, phytane/n-C18 and total steranes/hopanes) were considerably affected by differences in the thermal degradation behaviour of the respective compounds. Short-term experiments (48 h) performed at 400°C also revealed that biomarkers >C15 (especially steranes and hopanes) and ‘biological’ chain length preferences for n-alkanes are vanished at a vitrinite reflectance between 1.38 and 1.83% RO. Our data highlight that ‘thermal taphonomy’ effects have to be carefully considered in the interpretation of biomarkers in ancient rocks and, potentially, extraterrestrial materials.
Depression and borderline personality disorder (BPD) are both thought to be accompanied by alterations in the subjective experience of environmental rewards. We evaluated responses in women to sweet, bitter and neutral tastes (juice, quinine and water): 29 with depression, 17 with BPD and 27 healthy controls. The BPD group gave lower pleasantness and higher disgust ratings for quinine and juice compared with the control group; the depression group did not differ significantly from the control group. Juice disgust ratings were related to self-disgust in BPD, suggesting close links between abnormal sensory processing and self-identity in BPD.
The legislative processes of legal acts and their promulgation have undoubtedly entered a new area, which is based on the electronic form of creation, storage and information sharing. In the European Union member states are obliged to share public information by Directive 2003/98/EC on the re-use of public sector information (with amendments enacted in 2013). Currently, all countries provide access to national law in the form of HTML pages or PDF files, whereas a widely adopted and accepted format for information exchange on the Internet is presently based on the XML language. The main disadvantage of the traditional approach is the lack of semantic information about the real structure of a legal act and its content which makes a more advanced search in legal texts hardly possible, let alone the application of some knowledge engineering tools. Legal documents are stored locally in national databases, recorded in accordance with the individual, national data structure, and this further complicates the reuse of such documents, especially in a multinational environment. The solution to the above problems is the application of XML language in national legislative processes and the development of a common transnational standard describing the structure of XML documents from national legal systems.
The syndrome now known as involuntary emotional expression disorder (IEED) is a condition characterized by uncontrollable episodes of laughing and/or crying. It has been known for more than a century, but confusing and conflicting terminology may have hampered the progress of physicians in recognizing this condition. IEED is associated with various neurological disorders and neurodegenerative diseases, including amyotrophic lateral sclerosis, multiple sclerosis, Parkinson's disease, Alzheimer's disease and other dementias, and neurological injuries such as stroke and traumatic brain injury. It is hoped that better defined terminology for IEED may help in the future diagnosis of this debilitating condition, the establishment of accurate prevalence rates for IEED in the varying underlying conditions, and also in removing blame and stigma from sufferers by providing reassurance about the nature of their condition.
To evaluate the efficacy of a new monochloramine generation system for control of Legionella in a hospital hot water distribution system
Setting.
A 495-bed tertiary care hospital in Pittsburgh, Pennsylvania. The hospital has 12 floors covering approximately 78,000 m2.
Methods.
The hospital hot water system was monitored for a total of 29 months, including a 5-month baseline sampling period prior to installation of the monochloramine system and 24 months of surveillance after system installation (postdisinfection period). Water samples were collected for microbiological analysis (Legionella species, Pseudomonas aeruginosa, Stenotrophomonas maltophilia, Acinetobacter species, nitrifying bacteria, heterotrophic plate count [HPC] bacteria, and nontuberculous mycobacteria). Chemical parameters monitored during the investigation included monochloramine, chlorine (free and total), nitrate, nitrite, total ammonia, copper, silver, lead, and pH.
Results.
A significant reduction in Legionella distal site positivity was observed between the pre- and postdisinfection periods, with positivity decreasing from an average of 53% (baseline) to an average of 9% after monochloramine application (P > .05). Although geometric mean HPC concentrations decreased by approximately 2 log colony-forming units per milliliter during monochloramine treatment, we did not observe significant changes in other microbial populations.
Conclusions.
This is the first evaluation in the United States of a commercially available monochloramine system installed on a hospital hot water system for Legionella disinfection, and it demonstrated a significant reduction in Legionella colonization. Significant increases in microbial populations or other negative effects previously associated with monochloramine use in large municipal cold water systems were not observed.
Infect Control Hosp Epidemiol 2014;35(11):1356–1363 |
WHY?
Information retrieval from search engine becomes difficult when the query is incomplete or too complex. This paper suggests a query reformulation system that rewrite the query to maximize the probability of relevant documents returned.
WHAT?
Since the output of query reformulator is discrete, REINFORCE algorithm is used to train the model.
The query and the candidate term vectors are converted to a fixed-sized vector using CNN with max pooling or RNN. To generate a reformulated query sequentially, LSTM is used.
\phi_a(v)\\\phi_b(e_i)\\e_i \in q_0 \cup D_0\\P(t_i|q_0) = \sigma(U^{\top}tanh(W(\phi_a(v)\|\phi_b(e_i))+b))\\P(t_i^k|q_0) \propto exp(\phi_b(e_i)^{\top}h_k\\h_k = tanh(W_a\phi_a(v)+W_b\phi_b(t^{k-1}+W_h h_{k-1})
To train the model, REINFORCE algorithm is used. Entropy regularization loss is add to encourage diversity.
C_a = (R - \bar{R})\sum_{t\in T} - log P(t|q_0)\\\bar{R} = \sigma(S^{\top}tanh(V(\phi_a(v)\|\bar{e})+b))\\\bar{e} = \frac{1}{N}\sum_{i=1}^N \phi_b(e_i)\\N = |q_0 \cup D_0|\\C_b = \alpha\|R - \bar{R}\|^2\\C_H = -\lambda \sum_{t\in q_0\cup D_0} P(t|q_0) log P(t|q_0)
So?
Query reformulation showed better performance than raw retrieval, PRF models, or SL methods in TREC-CAR, Jeopardy and MSA data measured in recall, precision and mean average precision. |
The equation of motion for something under constant acceleration $\vec{a}$ is
$$\vec{x}(t) = \vec{x}(0) + \vec{v}(0)t + \frac12 \vec{a}t^2,$$
where $\vec{x}, \vec{v}, \vec{a}, t$ are displacement, velocity, acceleration, and time, respectively.
If you launch at an angle $\theta$ with respect to horizontal, then the $x$ and $y$ components of the initial velocity are $v_x(0) = v \cos\theta$ and $v_y(0) = v \sin\theta,$ where $v = |\vec{v}(0)|.$
Let's take the directions to be $+x$ to the right and $+y$ up.
Let's also define $\vec{x}(0) = \vec{0}.$
If your constant acceleration is gravity, then $a_x = 0$ and $a_y = -g$.
Then, your two equations for $x$ and $y$ become
$$x(t) = v(\cos \theta)t, \\ y(t) = v(\sin \theta)t - \frac12 g t^2.$$
So, having these,
How do you figure out what time the object hits the ground again? (It starts on the ground because of the initial condition I assumed.) Once you have that time, how do you figure out how far it traveled horizontally?
The second answer will depend on $g$, so this will tell you how gravity affects the range. |
I am interested in the following:
Let $G$ be a finite group of order $n$. Is there an explicit function $f$ such that $|s(G)| \leq f(n)$ for all $G$ and for all natural numbers $n$, where $s(G)$ denotes the set of subgroups of $G$?
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A non-identity subgroup of a group of order $n$ can be generated by $\log_{2}(n)$ or fewer elements. There is quite a lot of duplication, but if you count the number of subsets of $G$ of cardinality at most $\log_{2}(n),$ you will have an upper bound (which could be made more precise with more care) and the case of elementary Abelian $2$-groups show that such a bound is of the right general shape if it is to cover all groups of all orders.
To be more precise, this gives an upper bound of approximately $\log_{2}(n)n^{\log_{2}(n)}$ for the total number of subgroups of a group of order $n,$ which is generically rather generous. On the other hand, the number of subgroups of an elementary Abelian group of order $n= 2^{r}$ is close to $\sum_{i=0}^{r}2^{i(r-i)},$ so generally larger that $n^{log_{2}(n)/4}.$
A theorem of Borovik, Pyber and Shalev (Corollary 1.6) shows that the number of subgroups of a group $G$ of order $n=\lvert G\rvert$ is bounded by $n^{(\frac{1}{4}+o(1)) \log_2(n)}$. This is essentially best possible, cf. Geoff Robinsons answer above.
There is a variant of this question which has received a lot of attention and which may be of interest here: namely how many
maximal subgroups a finite group may have. In this context the relevant conjecture is due to Wall:
ConjectureThe number of maximal subgroups of a finite group G is less than the order of G.
This has been the subject of much study with the landmark work (until recently) being the result of Liebeck, Pyber and Shalev which states that the number of maximal subgroups is at most $c|G|^{3/2}$ where $c$ is an absolute constant. They also show that the conjecture is true if the group G is simple, up to a finite number of exceptions.
In very recent work it has now been shown that Wall's conjecture is
not true in general. An account of the demise of the conjecture can be found here. (This is not a paper, rather it's a very engaging description of the research which resulted in counter-examples being found.)
In light of this development, the bound $c|G|^{3/2}$ mentioned above assumes greater importance. Although, as the linked document mentions, it is likely that the index $\frac32$ can be reduced a great deal.
Whether one can deduce bounds on the number of non-maximal subgroups from the results of Liebeck, Pyber and Shalev, I don't know...
Can refine Stefan Kohl's suggestion by taking subsets containing the identity element of $G$ and of cardinality dividing $n$.
So the upper bound is $\sum_{d>1, d| n}^n {n-1\choose d-1}$
In a similar vein to Geoff Robinson's answer, observe that any proper subgroup of a group of order $n$ can be generated by at most $\Omega(n) - 1$ elements, where $\Omega$ counts the number of prime factors (with multiplicity) of $n$. This is tight, with equality in the case of a product of prime cyclic groups.
This gives an upper bound of:
$$ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{\Omega(n) - 1} + 1$$
subgroups, where the final $+ 1$ includes $G$ itself. Now we can invoke Michael Lugo's strict upper bound from https://mathoverflow.net/a/17236/39521 to obtain a closed-form upper bound:
$$ |s(G)| \leq \left\lceil \binom{n}{\Omega(n) - 1} \dfrac{n - \Omega(n) + 2}{n - 2 \Omega(n) + 2} \right\rceil \leq \dfrac{2 \times n^{\Omega(n) - 1}}{(\Omega(n) - 1)!}$$
Note that this is tight infinitely often: in the case where $n = p$ is prime (so $G = C_p$), we get:
$$ \dfrac{2 \times p^0}{0!} = 2 $$
where the two subgroups are the trivial group and $G$ itself. |
The implication here is that each individual discoverer must start from nothing but a bag of crying cells, and build up knowledge in a linear order before making a discovery in a vacuum.
In reality, I find we have an entire interwoven society trying to make the discoveries, not independent individuals. There is an entire section of society dedicated to distilling the human essence into teaching. There is an entire section devoted to building infrastructure to make it easier to step beyond. There is an entire section devoted to getting discoverers together, so that they don't ALL have to learn ALL of the knowledge; they merely need to have all of the knowledge when they put their minds together.
Consider that the trade knowledge needed to run a particle accelerator is equally essential to discovery as the quantum physics models used to point the accelerator in new and exciting directions. The physicists probably don't know how to correctly shim the hundreds of segments of the accelerator to be in a perfect shape (and doesn't have the time to learn). The physics probably hasn't spent enough time with high voltage to wire up thousands of electromagnets without a short taking the entire accelerator down. This knowledge, held in the minds of the tradesmen who support the physicists, is equally essential
but the physicists never had to learn them; these skills were learned in parallel by all of humanity.
The only thing I have found which can leave us with no time to discover is society itself. If society dulls, and our lives suddenly require an entire lifetime of learning just to survive, that could be the cusp where humanity simply cannot learn any further.
However, even then there is a light at the end of the tunnel. The poets have a long list of skills like "how to love" which take a lifetime to learn, and yet we keep working on them day after day. Perhaps one day, discovery will simply take the form of loving the universe and seeing what it wishes to tell us today. Oh fine! Lets see some math
Lets try to put some mathematical equations down to make sure we're all on the same page. I'll use it to show how a rather boring society resembling the Vulcans could go about never ending discover
First off, I am going to assume there is a never ending supply of things to discover in the universe. If there is a finite number of things to discover, then it is trivial to show that the number of discoveries humankind can make is finite. Let us define the universe of potential discoveries to be $\mathbb{D}$
I am going to assume the only thing in our brain that matters in the long run are structures. These are structures you have to learn over time in order to effectively do a task, such as discovering a new direction. I believe there is more to the brain, but I think this is close enough to model your question of learning and technology. Let us define these structures to be $\mathbb{S}$, the set of all helpful structures that the human brain can possibly organize into, and let $\text{Fits}(S), S\in \mathbb{S}$ to be a predicate that returns true if the set of structures $S$ would fit into a single human brain, and false otherwise. Because entering the world with new structures makes it trivial to prove we can keep discovering, we can assume $S$ of a newborn is $\emptyset$.
Now we need a notation for learning. I will assume, for simplicity, that people learn at a constant rate through their entire lives. I leave it to the reader to show that handling the case where learning rate is variable is a trivial transform from this simpler case. Because I am arguing that we will never run out of things to learn, I can assume the worst case of "you can only learn one thing at a time" without loss of generality. Consider the universe of learning activities, $\mathbb{L}$. For any learning activity $l \in \mathbb{L}$, we can define a function $\text{cost}_{\text{learn}}(l, S)$ which defines the cost (in time) of doing learning activity $l$ given that you already have all of the structures $S$ in your head. Let $\text{results}_{\text{learn}}(l, S)$ be a function which returns a set of structures in your brain after doing a learning activity.
Finally, we need a notation for discovery. $\text{cost}_{\text{discover}}(d, S)$ is the cost of discovering a particular element of $\mathbb{D}$.
Now we can define the goals. Let us define $\text{cost}_{\text{schooling}}(L)$ and $\text{results}_{\text{schooling}}(L)$ where $L$ is an ordered set of learning activities to be the cost and results of raising an individual up from $S = \emptyset$ through a sequence of learning activities. Thus $\text{cost}_{\text{schooling}}$ will be the sum of $\text{cost}_{\text{learn}}$, and $\text{results}_{\text{schooling}}$ will be the final result at the end of iterating $\text{results}_{\text{learn}}$. Our goal is to prove that there can always be a $\text{cost}_{\text{schooling}}(L) + \text{cost}_{\text{discover}}(d, \text{results}_{\text{schooling}}(L)) < \text{lifespan}$. Let us assign this a predicate: $\text{DiscoveryCapable}(L, D_{prev} \Leftrightarrow \exists_{d\in\mathbb{D},L^\prime}[(\forall{l\in L^\prime} l\in L)\land d\notin D_{prev}]$, which is a mouthful to day "A society is DiscoveryCapable if, for their set of known learning activities, and previously discovered disoveries, there exists a discoverable thing." Let us also add $\text{Discoverable}(L, d) \Leftrightarrow \exists_{L^\prime} \text{cost}_{\text{schooling}}(L^\prime) + \text{cost}_{\text{discover}}(d, \text{results}_{L^\prime}) < \text{lifespan}$, or "A discovery is discoverable if, given the known set of learning activities, someone can discover it in a lifetime."
Now here we will note that $\forall_{l\in\mathbb{L}}l \in \mathbb{D}$, or in words, every learning activity is something which can be discovered. This leads to a "Lotus Eaters" situation, where could simply continuously develop new ways to learn without going anywhere, so lets fix that. Lets define $\text{Trivial}(l)$ to be true if $\forall_{S\in populace}\exists_{L_0} (\forall_s s\in \text{results}_{\text{learn}}(l, S) \to \text{results}_{\text{learn}}(l, S_0)) \land \text{cost}_{\text{learn}}(L) \ge \text{cost}_{\text{schooling}}(L_0) $. In other words, its trivial to develop a new learning activity which doesn't teach anything new and costs more than an existing schooling!
Now we do a proof by contradiction. We assume $\text{DiscoveryCapable}(L, D_{prev})$ is false for our society. We will prove this is contradictory, meaning there is no such society that cannot find a discovery.
If $\text{DiscoveryCapable}$ is false, then that means there are no new non-trivial learning activities which are discoverable. If we find that there must be a non-trivial learning activity to discover, we have a proof by contradiction. This means we must prove $\forall_{L, D_{prev}}\exists_l \lnot \text{Trivial}(l) \land \text{Discoverable}(L, l)$
Consider the Turing machine, which is accepted to be far simpler than even a human. If we can prove that, at this time, a Turing machine can develop a new useful learning activity for us, then we can make a discovery simply by following that program. We are, after all, at least as impressive as computers.
Let us devise a turing machine to help. Select a subset of $L$ called $L_T$ which is the learning activities which can be analyzed by a Turing machine. We want to find a program which finds a $l \notin L_T$ such that $\lnot \text{Trivial}(l)$. The first step is easy. It is trivial for computers to find an activity $\exists_{l\in 2^{L_T}} l \notin L$. Such power set behaviors occur all the time in NP problems.
Now what if the computer can't do this? The next step is to gather some data about the universe. If we can't find any new data, then we are literally out of things to discover. If we find new data, we can have the computers crunch it harder, to find things that we don't understand, but computers can find. If they cannot, then all Turing-capable learning methods are exhausted, and we have covered the universe with our computational prowess. We, in effect, used computers to extend our life, crunching a subset of our possible learning activities, in hopes of finding a new one.
And now we sit back and look at the non Turing learning activities. It is not easy to tell if there is a faster way to learn such things. In fact, the only limit seems to be creativity.
The only limit for our capacity to discover is our own creativity. |
The initial interest in the zeros is their connection with the distribution of primes, which is often done via asymptotic statements about the prime counting function. In analytic number theory, it is standard fare to have an arithmetic function defined by a summation formula, and then modify it into a form that is easier to manipulate and obtain results for, in such a way that the asymptotic results about the modified function can be translated into results about the original function very easily. This is certainly the case with $\pi(x)$, which is why I bring this up. Most of the information that is relevant here can be found on the Explicit formula article at Wikipedia, for explicit formulas for the $\pi(x)$ function using the zeros of the Riemann zeta function. Two key highlights:
$(1)$
"This formula says that the zeros of the Riemann zeta function control the oscillations of primes around their 'expected' positions."
$(2)$
"Roughly speaking, the explicit formula says the Fourier transform of the zeros of the zeta function is the set of prime powers plus some elementary factors."
With the very basics of complex numbers we see that $x^\rho$, as a function of $x$, has a magnitude given by $x^{\Re (\rho)}$ and its argument by $\Im(\rho)\cdot\log x$. The imaginary parts thus contribute oscillatory behavior to the explicit formulas, while the real parts say which imaginary parts dominate over others and by how much - this is some meaning behind the 'Fourier transform' description. Indeed, given the dominant term in an asymptote for $\pi$ we have roughly the primes' "expected positions" (we are taking some license in referring to positioning when we are actually speaking of distribution in the limit), and the outside terms will speak to how much $\pi$
deviates from the expected dominant term as we take $x$ higher and higher in value. If one of the real parts differed from the others, it would privilege some deviation over others, changing our view of the regularity in the primes' distribution.
Eventually it also became clear that more and more results in number theory - even very accessible results that belie how deep the Riemann Hypothesis really has come to be - were equivalent to or could only be proven on the assumption of RH. See for example here or here or here. I'm not sure if any truly comprehensive list of the consequences or equivalences actually exists!
Moreover it is clear now that RH is not an isolated phenomenon, and instead exists as a piece in a much bigger puzzle (at least as I see it). The $\zeta$ function is a trivial case of a Dirichlet $L$-function as well a case of a trivial case of a Dedekind $\zeta$ function, and there is respectively a Generalized Riemann Hypothesis (GRH) and Extended Riemann Hypothesis for these two more general classes of functions. There are numerous analogues to the zeta function and RH too - many of these have already gained more ground or already had the analogous RH proven!
It is now wondered what the appropriate definition of an $L$-function "should" be, that is, morally speaking - specifically it must have some analytic features and of course a functional equation involving a reflection, gamma function, weight, conductor etc. but the precise recipe we need to create a slick theory is not yet known. (Disclaimer: this paragraph comes from memory of reading something a long time ago that I cannot figure out how to find again to check. Derp.)
Finally, there is the spectral interpretation of the zeta zeros that has arisen. There is the Hilbert-Pólya conjecture. As the Wikipedia entry describes it,
In a letter to Andrew Odlyzko, dated January 3, 1982, George Pólya said that while he was in Göttingen around 1912 to 1914 he was asked by Edmund Landau for a physical reason that the Riemann hypothesis should be true, and suggested that this would be the case if the imaginary parts of the zeros of the Riemann zeta function corresponded to eigenvalues of an unbounded self adjoint operator.
This has spurred quantum-mechanical approaches to the Riemann Hypothesis. Moreover, we now have serious empirical evidence of a connection between the zeta zeros and random matrix theory, specifically that their pair-correlation matches that of Gaussian Unitary Ensembles (GUEs)...
The year: 1972. The scene: Afternoon tea in Fuld Hall at the Institute for Advanced Study. The camera pans around the Common Room, passing by several Princetonians in tweeds and corduroys, then zooms in on Hugh Montgomery, boyish Midwestern number theorist with sideburns. He has just been introduced to Freeman Dyson, dapper British physicist.
Dyson: So tell me, Montgomery, what have you been up to?
Montgomery: Well, lately I've been looking into the distribution of the zeros of the Riemann zeta function.
Dyson: Yes? And?
Montgomery: It seems the two-point correlations go as... (turning to write on a nearby blackboard): $$1-\left(\frac{\sin\pi x}{\pi x}\right)^2$$
Dyson: Extraordinary! Do you realize that's the pair-correlation function for the eigenvalues of a random Hermitian matrix?
(Source:
The Spectrum of Riemannium.)
If so inclined one can see the empirical evidence in pretty pictures e.g. here. |
I was reading the paper device independent outlook on quantum mechanics. The author defines a generic two qubit density matrix as $$ \rho=\frac{1}{4}\left( I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma}\otimes I + I \otimes \vec{s_{\rho}} \cdot \vec{\sigma} + \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \right) \, . \tag{1} $$
How is it obtained and what are the constraints over $T^{ij}_{\rho}$ ? Also, seeing that it has some symmetry can a general 3 qubit density matrix be written as
\begin{align} \rho = \frac{1}{8} &\left( I \otimes I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma} \otimes I \otimes I + I \otimes \vec{s_{\rho}}\cdot \vec{\sigma} \otimes I + I \otimes I \otimes \vec{t_{\rho}}.\vec{\sigma} \right. \\ &+ \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes I + \sum_{i,j=x,y,z}U^{ij}_{\rho} \sigma_i \otimes I \otimes \sigma_j \\ &\left. + \sum_{i,j=x,y,z}W^{ij}_{\rho} I \otimes \sigma_i \otimes \sigma_j +\sum_{i,j,k=x,y,z}X^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes \sigma_k \right) ? \end{align} Here $\vec{r_{\rho}}, \vec{s_{\rho}}, and \vec{t_{\rho}}$ are 3 dimensional vectors with real components and each having magnitude $\le 1$.
EDIT:
I get the fact that tensor of pauli matrices acts as a basis but can't get the condition on $T_{ij}$. I was able to work backwards to see that $T_{\rho}^tT_{\rho}$ has to be such that its maximum eigen value is $\le 1$ so that CHSH inequality is only violated at maximum upto $2\sqrt{2}$. So if this condition is not followed then $(1)$ should not be a valid density matrix. The form given in $(1)$ is already hermitian and has trace 1. So for $T_{\rho}^tT_{\rho}$ having maximum eigen value $\ge1$ $(1)$ might not be a positive operator but I am unable to prove that. |
Maximum
Alone 23/10/2018 at 13:01
\(x^2+5y^2-4xy-x+2y-6=0\)
\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)
\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)
\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)
\(\Leftrightarrow-2\le x-2y\le3\)Lê Anh Duy selected this answer.
Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treatrous and selfish (especially the captain). The captain always proposes a distribution of the loot. All pirates vote on the proposal and if at least half of the crew (which includes himself) approve, the loot will be divided as proposal, as no pirates would be willing to take on the captain without superior force on their side. Otherwise he will face a mutiny: all pirates will turn against him and make him walk the plank. The pirates will start over again with the next senior pirate as captain. What's the maximum number of coins the captain can keep without risking his life?
Searching4You 26/07/2017 at 11:53
Use Cauchy's inequality for positive numbes a,b,c.
\(a+b\ge2\sqrt{ab},b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab}\cdot\sqrt{bc}\cdot\sqrt{ca}=8abc\)
\(\Rightarrow A=\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)
\(MaxA=\dfrac{1}{8}\Leftrightarrow a=b=c>0\)Selected by MathYouLike
¤« 03/04/2018 at 13:32
Use Cauchy's inequality for positive numbes a,b,c.
a+b≥2ab−−√,b+c≥2bc−−√,c+a≥2ca−−√
⇒(a+b)(b+c)(c+a)≥8ab−−√⋅bc−−√⋅ca−−√=8abc
⇒A=abc(a+b)(b+c)(c+a)≤18
MaxA=18⇔a=b=c>0
Kayasari Ryuunosuke Coodinator 22/07/2017 at 12:10
We have :
\(x^2-2x+5=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\)
\(\Rightarrow H=\dfrac{1}{x^2-2x+5}\le\dfrac{1}{4}\)
\(\Rightarrow Max_H=\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)Selected by MathYouLike
¤« 03/04/2018 at 13:33
We have :
x2−2x+5=x2−2x+1+4
=(x−1)2+4≥4
⇒H=1x2−2x+5≤14
⇒MaxH=14
⇔(x−1)2=0
⇔x=1
¤« 03/04/2018 at 13:33
(*) 1+N=.....=x2/(x-2)2+4
We have x2 ≥
0 , (x-2)2+4 ≥
4 > 0
So 1+N ≥
0 => N ≥
-1 ;equality : x=0
(*)1-N=....=(x-4)2/(x-2)2+4
....... -> N ≤
1 , equality : x=4
¤« 03/04/2018 at 13:32
There is something is wrong here
Change to :
We have a2+b2≥5
=> 9−(a2+b2)≤4
=> 2ab≤4
=> ab≤2
<=> a2b2≥4
=> 4a2b2≥16
Phan Văn Hiếu 28/03/2017 at 21:30
trình bày = tiếng việt đc ko tiếng anh ngại viết lắm |
TIPS FOR SOLVING QUESTIONS RELATED TO SIMPLE INTEREST:
1.
Principal: The money borrowed or lent out for a certain period is called the Principal.
2.
Interest: Interest is the sum which is paid for the use of other's money.
3.
Simple Interest:Simple Interest is the interest on a sum borrowed for a certain period reckoned uniformly i.e. on initial principal only.
4. Formula for calculating Simple Interest:
If Principal = P, Rate = R% per annum, and Time = T years, then
\begin{aligned} \text{Simple Interest (S.I.)} = \left( \frac{P\times R\times T}{100}\right) \\ \end{aligned} \begin{aligned} \text{Principal (P) } = \left( \frac{100\times S.I.}{R\times T}\right) \\ \text{Rate (R)} = \left( \frac{100\times S.I.}{P\times T}\right) \\ \text{Time (T)} = \left( \frac{100\times S.I.}{P\times R}\right) \\ \end{aligned}
5. If the rate percent is given half yearly or quarterly, then to find the rate percent per annum multiply by 2 or 4 respectively.
6. One ordinary year (1 year) = 365 days.
7. One leap year = 366 days. |
It can be done!
Approaching it on a Surface
Before we try to do this with a 3-dimensional texture, let's attempt it in 2-dimensions. This is actually fairly straightforward. We can create a proxy mesh by duplicating our object (
Shift +
D), removing the modifiers from our duplicated object, then setting the
Texture Mesh setting for our original object to the mesh of the second object:
Result:
Hey, that works perfectly! Unfortunately, it doesn't work for 3D textures. When we attempt this with a volumetric texture it does absolutely nothing.
Approaching it in a Volume
The best way to do this in a volume (believe me, I've tested others) is to create a proxy mesh, then take the coordinates from the proxy mesh.
What should our proxy mesh look like?
Unfortunately, Blender does not provide an easy way to transfer volumetric data on and off of meshes. Our saving grace is the
Point Density node in Cycles. The Point Density node will create a volumetric texture based off of the vertices of any mesh, and interpolate between them. It follows, then, that our coordinate proxy mesh should be a 3D grid of points. Creating the proxy mesh
We'll start by adding a plane, then in
Edit Mode, moving and scaling it so it has the same size and location of the bottom of the cylinder's bounding box (in the image, I enabled the bounding box display):
Next, we need to have more points in our coordinate grid. Use the
Loop Cut tool (
Ctrl +
R) to add lots of loop cuts in both directions, trying to make the faces roughly square (I did $63$ and $15$, to get a $64\times16$ grid:
Next, in
Object Mode add an Array modifier. Use Constant Offset in the $Z$ direction to create a "stack" of planes (I used an offset of $0.125$). Add enough planes to fill the bounding box of the cylinder:
Apply the
Array modifier and add a Curve modifier. Set up the Curve modifier the same way that it is set up on the cylinder. Check to make sure everything deforms properly (proxy mesh set to Wireframe display):
Now the hard part:
Creating a coordinate system
The
Point Density node can only take color from Vertex Colors, Weights, and Normals. By far the most customizable of these options is Vertex Colors. We need to find a way to put Texture Coordinates into Vertex Colors.
Let's use
Animation Nodes! After downloading and installing the add-on, go into the Node Editor and add a new NodeTree.
Before we start adding nodes, we need to figure out how to emulate Generated texture coordinates. According to the Blender Manual:
Generated
Automatically-generated texture coordinates from the vertex positions of the mesh without deformation, keeping them sticking to the surface under animation. Range from 0.0 to 1. 0 over the bounding box of the undeformed mesh.
This means we need to normalize the vectors representing the positions of our vertices and put them in the positive-positive-positive octant. Here is the formula we will use:
$$\mathbf p'=\frac{\mathbf p - \mathbf c}{\mathbf d} + \mathbf{0.5}$$
where $\mathbf p$ is a point in 3D space, $\mathbf c$ is the center of the bounding box, and $\mathbf d$ is the vector of the longest diagonal of the bounding box (i.e. the dimensions).
Let's get our inputs first. Add the following nodes (press
Shift +
A to bring up the list of nodes or
Ctrl +
A to search):
The object in the
Object Input node is our coordinate proxy object. The text in the Object Attribute node is
dimensions. Note that "Use Modifiers" is unchecked. This ensures that the coordinates we get don't include the
Curve modifier. Next, we'll use Vector Math nodes to implement the above formula:
The
Convert node will be added automatically. We have one more issue with the texture coordinates: texture coordinates are encoded in a linear color space, while Vertex Colors are encoded in sRGB. From the linked Wikipedia page, we get the formula for encoding colors in sRGB:
$$C_\text{srgb}=\begin{cases}
12.92C_\text{linear}, & C_\text{linear} \leq 0.0031308\\
1.055C_\text{linear}^{1/2.4}-0.055, & C_\text{linear} > 0.0031308
\end{cases}$$
To utilize this, create a loop subprogram by adding a
Loop Input node. Create this loop:
The numbers are truncated in the screenshot. Here they are, row by row:
0.003131
12.92
0.416667 1.055 0.055
Now, add 3
Invoke Subprogram nodes with Separate and Combine Vector nodes to encode the $R$, $G$, and $B$ channels.
Now, we need to get our vectors into a Vertex Color. @Omar Ahmed has already made fabulous explanation of how to do this. You can find his answer here. I will copy the relevant parts below, but I recommend reading his answer for a detailed (and illustrated!) explanation of how Vertex Colors work in Blender.
First, create this loop:
Then, create this script (make sure to name things properly):
Here is the script:
#Create a vertex color map named "Normals" if it is not already created.
if "Coords" in Object.data.vertex_colors:
map = Object.data.vertex_colors["Coords"]
else:
map = Object.data.vertex_colors.new("Coords")
#Loop over Colors and assign them to the vertex color map.
for i, color in enumerate(Colors):
map.data[i].color = color
Finally, link up the last nodes (for an explanation of how they work, see the answer linked above). Here's the final tree:
Getting the coordinate system into Cycles
We will be using the
Point Density node as I mentioned above. On the Cylinder object, create a new material. Add a Point Density node. Set it to Object Vertices, with the Coordinate Proxy as the object and the Color Source as Vertex Color. Choose the
Coords Vertex Color. Next, disable visibility for the proxy object, and link the
Color output of the Point Density node to an Emission node plugged into the Volume socket of the material output (in the picture I'm using the Node Wrangler add-on to get an Emission node):
Activate Rendered View in the viewport (
Shift +
Z). Position the 3D View so that you are looking at one of the "caps" of the cylinder. Adjust the
Radius parameter of the Point Density node to be just big enough that there are no "gaps" ($0.1$ worked for me):
Now we have our coordinates! Use the
Color output just like you would normal texture coordinates, and volumetric texture-ize away!
Just remember that anything you do to animate the cylinder also needs to be done to the proxy.
(Requires Animation Nodes) |
When I toss a coin in Mars, is the planets atmosphere rare enough that I'd rotate with the planet (at its angular velocity), but not the coin?
It depends on
where on Mars you toss the coin, and how high you toss it.
In a rotating frame of reference, an object in motion appears to be affected by a pair of fictitious forces - the centrifugal force, and the Coriolis force. Their magnitude is given by
$$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\ \mathbf{\vec{F_{Coriolis}}}=-2m\mathbf{\vec\Omega}\times\mathbf{\vec{v}}$$
The question is - when are these forces sufficient to move the coin "away from your hand" - in other words, for what initial velocity $v$ is the total displacement of the coin greater than 10 cm (as a rough estimate of what "back in your hand" might look like; obviously you can change the numbers).
The centrifugal force is only observed when the particle is rotating at the velocity of the frame of reference - once the particle is in free fall, it no longer moves along with the rotating frame of reference and the centrifugal force "disappears". For an object moving perpendicular to the surface of the earth, the Coriolis force is strongest at the equator, becoming zero at the pole; it is a function of the velocity of the coin. We will calculate the expression as a function of latitude - recognizing that it will be a maximum at the equator.
As a simplifying assumption, we assume the change in height is sufficiently small that we ignore changes in the force of gravity; we also ignore all atmospheric drag (in particular, the wind; if the opening scene of "The Martian" were to be believed, it can get pretty windy on the Red Planet.) Finally we will assume that any horizontal velocity will be small - we ignore it for calculating the Coriolis force, but integrate it to obtain the displacement.
The vertical velocity is given by
$$v = v_0 - g\cdot t$$
and the total time taken is $t_t=\frac{2v_0}{g}$. At any moment, the Coriolis acceleration is
$$a_C=2\mathbf{\Omega}~v\cos\theta$$
Integrating once, we get
$$v_h = \int a\cdot dt \\ = 2\mathbf{\Omega}\cos\theta\int_0^t(v_0-gt)dt\\ = 2\mathbf{\Omega}\cos\theta\left(v_0 t-\frac12 gt^2\right)$$
And for the displacement
$$x_h = \int v_h dt \\ = 2\mathbf{\Omega}\cos\theta\int_0^t \left(v_0 t-\frac12 gt^2\right)dt\\ = 2\mathbf{\Omega}\cos\theta \left(\frac12 v_0 t^2-\frac16 gt^3\right)$$
Substituting for $t = \frac{2v_0}{g}$ we get
$$x_h = 2\mathbf{\Omega}\cos\theta v_0^3\left(\frac{4}{g^2} - \frac{4}{3 g^2}\right)\\ = \frac{16\mathbf{\Omega}\cos\theta v_0^3}{3g^2}$$
The sidereal day of Mars is 24 hours, 37 minutes and 22 seconds - so $\Omega = 7.088\cdot 10^{-5}/s$ and the acceleration of gravity $g = 3.71 m/s^2$. Plugging these values into the above equation, we find $x_h = 2.75\cdot 10^{-5}v_0^3 m$, where velocity is in m/s. From this it follows that you would have to toss the coin with an initial velocity of about 15 m/s for the Coriolis effect to be sufficient to deflect the coin by 10 cm before it comes back down.
On Earth, such a toss would result in a coin that flies for about 3 seconds, reaching a height of about 11 m. It is conceivable that someone could toss a coin that high - but I've never seen it.
AFTERTHOUGHT
Your definition of "vertical" needs to be carefully thought through. There is a North-South component of the centrifugal "force" that is strongest at 45° latitude, and that will cause a mass on a string to hang in a direction that is not-quite-vertical. If you launch your coin in that direction, you will not observe a significant North-South deflection during flight, but if you were to toss the coin "vertically" (in a straight line away from the center of Mars), there
will in fact be a small deviation. The relative magnitude of the centrifugal force and gravity can be computed from
$$\begin{align}a_c &= \mathbf{\Omega^2}R\sin\theta\cos\theta \\ &= \frac12 \mathbf{\Omega^2}R\\ &= 8.5~\rm{mm/s^2}\end{align}$$
If you toss the coin at 15 m/s, it will be in the air for approximately 8 seconds. In that time, the above acceleration will give rise to a displacement of about 27 cm. This shows that your definition of "vertical" really does matter (depending on the latitude - it doesn't matter at the poles or the equator, but it is significant at the intermediate latitudes, reaching a maximum at 45° latitude).
The coin will come back to your hand just like it would on the earth. The effect of atmosphere is negligible comparing to the coin's inertia, so the horizontal position of the coin relative to your hand will hardly be affected. The rareness of the atmosphere will only affect the vertical motion of the coin, like how quickly the coin will fall into your hand.
Yes, for the simple reason that you're not tossing the coin very high (presumably, anyway). You seem to think that on Earth, atmospheric drag is what keeps the coin "glued" to the tossing frame of reference, but that isn't really a factor at all.
Say that you're on Earth, at sea level, on the equator, and you toss the coin 3 meters straight up. Neglecting drag, the coin will be in the air for 1.56 seconds. The earth is rotating under your feet at 463 m/s, and has a radius of 6.37 * 10^6 m. The coin is gaining an altitude of 3 m, which is 4.71 * 10^-7 earth radii, so the rotational speed at that height will be different by an equal proportion, which works out to 0.00022 m/s. Getting an upper bound by assuming the coin spends the whole time at the maximum height because I'm lazy, and we end up with a deflection of 0.34 mm, which is less than the thickness of the coin, let alone its diameter. Anywhere away from the equator or above sea level, and the number would come out lower still.
Doing the same experiment on Mars, we'll suppose that you can give the coin the same initial velocity, and that you're on the equator at mean elevation. Mars's surface gravity is lower (3.71 m/s^2), so the coin will reach an impressive 7.92 m in height and stay in the air for 4.14 seconds. Mars is rotating under your feet at 241 m/s (less than Earth because it has a smaller circumference but a similar day length) and has a radius of 3.39 * 10^6 m. The coin then gains 2.34 * 10^-6 mars radii, and the rotational speed at that height is different by 0.00056 m/s. Making the same (over)estimate as before, we get 4.14 s * 0.00056 m/s = 2.33 mm. About one coin thickness, but not much more. Certainly not enough to miss your hand on the way down.
Basically, the heights you're dealing with when tossing a coin are just too small, compared to the size of a planet, to make much difference, atmosphere or not. Try lobbing a cannonball 1km up instead, and you'd be more likely to notice an effect. I haven't worked out the math, but I
still don't think that the horizontal component of atmospheric drag would contribute much at all; the atmosphere would be more likely to make a difference by reducing the maximum height reached.
The coin comes down for sure unless you tossed it with an escape velocity and escape velocity depends on mass of planet, mass of coin etc. Even if you are well inside the escape velocity limit, it may not reach your hand. Your observation is correct that the drag force plays a role in its tangential velocity, but for all small distance tosses the coin will reach your hand regardless of the atmospheric rarity compared to earth.
I think yes,like as if you were on earth. The reason is that the coin has the same velocity as you and the surface of planet Mars, it doesn't have to do with atmosphere.
Yes you are moving with the surface but you are not accelerating. You are moving at a constant speed. If you toss in what you think is vertical (call it Y direction) the coin will have the same rotational velocity (call it X direction) as you. Even the atmosphere is moving with you so for a short distance there is no wind resistance in the X direction. The X velocity of the coin will remain constant and will be exactly your X velocity.
The combination of your two questions leads me to believe that what you are really asking is, does the
density (or lack of it) of the atmosphere have an effect on the horizontal position of an object that initially is given only vertical momentum? If this interpretation of your question is correct, then I want to start by telling you that you have a reverse conception of the effect density has. The drag force of an atmosphere is proportional to its density, therefore as its density goes to zero (no atmosphere), the drag force goes to zero. So, the "rarer" the atmosphere, the less effect it has on the motion of the coin. Since we are only interested in the coin returning to the same spot (little or no horizontal displacement), differences in the vertical direction caused by the drag and different Martian gravity, are of no consequence. Therefore, since there are little or no horizontal forces acting on the coin, it will return to the same spot it was tossed from ( your hand).
protected by Qmechanic♦ Dec 31 '15 at 19:23
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A Short Proof of the Large Time Energy Growth for the Boussinesq System 307 Downloads Citations Abstract
We give a direct proof of the fact that the \(L^p\)-norms of global solutions of the Boussinesq system in \({\mathbb {R}}^3\) grow large as \(t\rightarrow \infty \) for \(1<p<3\) and decay to zero for \(3<p\le \infty \), providing exact estimates from below and above using a suitable decomposition of the space–time space \({\mathbb {R}}^{+}\times {\mathbb {R}}^{3}\). In particular, the kinetic energy blows up as \(\Vert u(t)\Vert _2^2\sim ct^{1/2}\) for large time. This contrasts with the case of the Navier–Stokes equations.
KeywordsKato spaces The Boussinesq equation and asymptotic behavior Mathematics Subject ClassificationPrimary 76D05 Secondary 35B40 Notes Acknowledgements
The authors would like to thank the referees for their careful reading and useful suggestions that have been incorporated in this revised version of the manuscript.
References Meyer, Y.: Wavelets, paraproducts, and Navier–Stokes equations. In: Current Developments in Mathematics, 1997, Cambridge, MA, International Press, Boston, MA, pp. 105–212 (1996)Google Scholar |
WHY?
Even though attention is being widely used, it is hard to be considered as probabilistic model as the attention does not marginalize.
WHAT?
This paper formulated source separation task as getting mixture meight vector of multiple sources in wave form.
x(t) = \sum_{i=1}^C s_i(t)\\x = wB\\s_i = d_iB\\w = \sum_{i=1}^C d_i = \sum_{i=1}^C w \odot (d_i \oslash w) := w \odot \sum_{i=1}^C m_i\\d_i = m_i \odot w
Time-domain Audio Separation Network(TasNet) tries to find
m_i which is relative contribution to each w while B is N basis signals of shape N x L.
Encoder find w for B by appling 1-D gated convolution layer.
w_k = ReLU(x_k \circledast U)\odot\sigma(x_k\circledast V)\\
Separation network uses LSTM and FC for masks(
m_i) generation. With w and m found above, d can be found with decoder. The scale-invariant source-to-noise ratio(SI-SNR) is used for loss.
So?
TasNet not only showed comparable performance in WSJ0-2mix dataset, but also showen to find its own basis.
Luo, Yi, and Nima Mesgarani. “Tasnet: time-domain audio separation network for real-time, single-channel speech separation.” 2018 IEEE International Conference on Acoustics, Speech and Signal Processing (ICASSP). IEEE, 2018. |
If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
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If $x^2+y^2-xy-x-y+1=0$ ($x,y$ real) then calculate $x+y$
Ideas for solution include factorizing the expression into a multiple of $x+y$ and expressing the left hand side as a sum of some perfect square expressions.
Let $s=x+y$ and $d=x-y$. Then $x=(s+d)/2$ and $y=(s-d)/2$. Making this substitution, we find that $$ x^2+y^2-xy-x-y+1=\frac{3d^2}{4}+\frac{(s-2)^2}{4}. $$
Hence, being a sum of squares, if the original expression is zero, $d=0$ and $s-2=0$. This means that $s=x+y=2$, with $x=y$.
We have $$2(x^2+y^2-xy-x-y+1)=(x-y)^2+(x-1)^2+(y-1)^2.$$ The right-hand side (for real $x$ and $y$) is equal to $0$ if and only if $x=y=1$.
I'll assume $x$ and $y$ are supposed to be real.
Let $s=x+y$ and $p=xy$; then your equation becomes $$ s^2-2p-p-s+1=0 $$ or $$ p=\frac{s^2-s+1}{3} $$ The equation $$ z^2-sz+p=0 $$ must have real roots; its discriminant is $$ s^2-4p=-\frac{(s-2)^2}{3}\le0 $$ so we have $s=2$ (and $p=1$).
The value of $x+y$ is not determined by the first equation. For $x=0$ we obtain $y^2-y+1=0$, so that $x+y=\frac{\pm \sqrt{-3}+1}{2}$. This is certainly not equal to $2$. |
The original Adam paper briefly explains what it means by "invariant to diagonal rescaling of the gradients" at the end of section 2.1.
I would try to explain it in some more detail.
Like stochastic gradient descent (SGD), Adam is an iterative method that uses gradients in order to find a minimum of a function.
(By "gradients" I mean "the values of the gradient in different locations in parameter space". I later use "partial derivatives" in a similar fashion.)
But in contrast to SGD, Adam doesn't really use gradients. Instead, Adam uses the partial derivatives of each parameter independently.
(By "partial derivative of a parameter $x$" I mean "partial derivative of the cost function $C$ with respect to $x$", i.e. $\frac{\partial C}{\partial x}$.)
Let $\Delta^{(t)}$ be the step that Adam takes in parameter space in the $t^{\text{th}}$ iteration. Then the step it takes in the dimension of the $j^{\text{th}}$ parameter (in the $t^{\text{th}}$ iteration) is $\Delta^{(t)}_j$, which is given by:$$\Delta^{(t)}_j=-\frac{\alpha}{\sqrt{{\hat v}^{(t)}_j}+\epsilon}\cdot {\hat m}^{(t)}_j$$while:
$\alpha$ is the learning rate hyperparameter. $\epsilon$ is a small hyperparameter to prevent division by zero. ${\hat m}^{(t)}_j$ is an exponential moving average of the partial derivatives of the $j^{\text{th}}$ parameter that were calculated in iterations $1$ to $t$. ${\hat v}^{(t)}_j$ is an exponential moving average of the squares of the partial derivatives of the $j^{\text{th}}$ parameter that were calculated in iterations $1$ to $t$.
Now, what happens when we scale the partial derivative of the $j^{\text{th}}$ parameter by a positive factor $c$?
(I.e. the partial derivative of the $j^{\text{th}}$ parameter is just a function whose domain is the parameter space, so we can simply multiply its value by $c$.) ${\hat m}^{(t)}_j$ becomes $c\cdot{\hat m}^{(t)}_j$ ${\hat v}^{(t)}_j$ becomes $c^2\cdot{\hat v}^{(t)}_j$ Thus (using the fact that $c>0$), we get that $\Delta^{(t)}_j$ becomes:$$-\frac{\alpha}{\sqrt{c^2\cdot{\hat v}^{(t)}_j}+\epsilon}\cdot c\cdot{\hat m}^{(t)}_j=-\frac{\alpha}{c\cdot\sqrt{{\hat v}^{(t)}_j}+\epsilon}\cdot c\cdot{\hat m}^{(t)}_j$$And assuming $\epsilon$ is very small, we get:$$\begin{gathered}-\frac{\alpha}{c\cdot\sqrt{{\hat v}^{(t)}_j}+\epsilon}\cdot c\cdot{\hat m}^{(t)}_j\approx -\frac{\alpha}{c\cdot\sqrt{{\hat v}^{(t)}_j}}\cdot c\cdot{\hat m}^{(t)}_j=\\-\frac{\alpha}{\sqrt{{\hat v}^{(t)}_j}}\cdot{\hat m}^{(t)}_j\approx-\frac{\alpha}{\sqrt{{\hat v}^{(t)}_j}+\epsilon}\cdot{\hat m}^{(t)}_j\end{gathered}$$
I.e. scaling the partial derivative of the $j^{\text{th}}$ parameter by a positive factor $c$ actually doesn't affect $\Delta^{(t)}_j$.
Finally, let $g=\left(\begin{gathered}g_{1}\\g_{2}\\\vdots\end{gathered}\right)$ be the gradient. Then $g_j$ is the partial derivative of the $j^{\text{th}}$ parameter.
What happens when we multiply the gradient by a diagonal matrix with only positive elements?$$\left(\begin{matrix}c_{1}\\ & c_{2}\\ & & \ddots\end{matrix}\right)g=\left(\begin{matrix}c_{1}\\ & c_{2}\\ & & \ddots\end{matrix}\right)\left(\begin{gathered}g_{1}\\g_{2}\\\vdots\end{gathered}\right)=\left(\begin{gathered}c_{1}\cdot g_{1}\\c_{2}\cdot g_{2}\\\vdots\end{gathered}\right)$$So it would only scale each partial derivative by a positive factor, but as we have seen above, this won't affect the steps that Adam takes.
In other words, Adam is invariant to multiplying the gradient by a diagonal matrix with only positive factors, which is what the paper means by "invariant to diagonal rescaling of the gradients".
With regard to the quote from the paper Normalized Direction-preserving Adam, it describes the "ill-conditioning problem". (This is how the paper names the problem. Note that it is a different problem from the problem of an ill-conditioned Hessian.)
It seems to me that this problem is unrelated to Adam (and unrelated to the fact that it is invariant to rescaling of the gradient). I deduced that mostly from two other quotes in the paper, that elaborate on the ill-conditioning problem:
Furthermore, even when batch normalization is not used, a network using linear rectifiers (e.g., ReLU, leaky ReLU) as activation functions, is still subject to ill-conditioning of the parameterization (Glorot et al., 2011), and hence the same problem. We refer to this problem as the
ill-conditioning problem.
The quote refers to the paper Deep Sparse Rectifier Neural Networks, which never mentions Adam, and also describes the problem of "ill-conditioning of the parametrization", which seems to me very similar (if not identical) to the "ill-conditioning problem".
The ill-conditioning problem occurs when the magnitude change of an input weight vector can be compensated by other parameters, such as the scaling factor of batch normalization, or the output weight vector, without affecting the overall network function. Consequently, suppose we have two DNNs that parameterize the same function, but with some of the input weight vectors having different magnitudes, applying the same SGD or Adam update rule will, in general, change the network functions in different ways. Thus, the ill-conditioning problem makes the training process inconsistent and difficult to control.
If I understand correctly, this quote says that both SGD and Adam suffer from the ill-conditioning problem. I.e. the problem isn't unique to Adam. |
The
package is a powerful tool, based on pgfplots
tikz, dedicated to create scientific graphs.
Contents Pgfplots is a visualization tool to make simpler the inclusion of plots in your documents. The basic idea is that you provide the input data/formula and pgfplots does the rest.
\begin{tikzpicture} \begin{axis} \addplot[color=red]{exp(x)}; \end{axis} \end{tikzpicture} %Here ends the furst plot \hskip 5pt %Here begins the 3d plot \begin{tikzpicture} \begin{axis} \addplot3[ surf, ] {exp(-x^2-y^2)*x}; \end{axis} \end{tikzpicture}
Since
pgfplot is based on tikz the plot must be inside a tikzpicture environment. Then the environment declaration
\begin{axis},
\end{axis} will set the right scaling for the plot, check the Reference guide for other axis environments.
To add an actual plot, the command
\addplot[color=red]{log(x)}; is used. Inside the squared brackets some options can be passed, in this case we set the colour of the plot to
red; the squared brackets are mandatory, if no options are passed leave a blank space between them. Inside the curly brackets you put the function to plot. Is important to remember that this command must end with a semicolon ;.
To put a second plot next to the first one declare a new
tikzpicture environment. Do not insert a new line, but a small blank gap, in this case
hskip 10pt will insert a 10pt-wide blank space.
The rest of the syntax is the same, except for the
\addplot3 [surf,]{exp(-x^2-y^2)*x};. This will add a 3dplot, and the option
surf inside squared brackets declares that it's a surface plot. The function to plot must be placed inside curly brackets. Again, don't forget to put a semicolon ; at the end of the command.
Note:
It's recommended as a good practice to indent the code - see the second plot in the example above - and to add a comma , at the end of each option passed to \addplot. This way the code is more readable and is easier to add further options if needed.
To include
pgfplots in your document is very easy, add the next line to your preamble and that's it:
\usepackage{pgfplots}
Some additional tweaking for this package can be made in the preamble. To change the size of each plot and also guarantee backwards compatibility (recommended) add the next line:
\pgfplotsset{width=10cm,compat=1.9}
This changes the size of each
pgfplot figure to 10 centimeters, which is huge; you may use different units (pt, mm, in). The compat parameter is for the code to work on the package version 1.9 or later.
Since LaTeX was not initially conceived with plotting capabilities in mind, when there are several
pgfplot figures in your document or they are very complex, it takes a considerable amount of time to render them. To improve the compiling time you can configure the package to export the figures to separate PDF files and then import them into the document, add the code shown below to the preamble:
\usepgfplotslibrary{external}
\tikzexternalize
See this help article for further details on how to set up tikz-externalization in your Overleaf project.
Pgfplots 2D plotting functionalities are vast, you can personalize your plots to look exactly what you want. Nevertheless, the default options usually give very good result, so all you have to do is feed the data and LaTeX will do the rest:
To plot mathematical expressions is really easy:
\begin{tikzpicture} \begin{axis}[ axis lines = left, xlabel = $x$, ylabel = {$f(x)$}, ] %Below the red parabola is defined \addplot [ domain=-10:10, samples=100, color=red, ] {x^2 - 2*x - 1}; \addlegendentry{$x^2 - 2x - 1$} %Here the blue parabloa is defined \addplot [ domain=-10:10, samples=100, color=blue, ] {x^2 + 2*x + 1}; \addlegendentry{$x^2 + 2x + 1$} \end{axis} \end{tikzpicture}
Let's analyse the new commands line by line:
axis lines = left.
xlabel = $x$ and
ylabel = {$f(x)$}.
\addplot.
domain=-10:10.
samples=100.
\addlegendentry{$x^2 - 2x - 1$}.
To add another graph to the plot just write a new
\addplot entry.
Scientific research often yields data that has to be analysed. The next example shows how to plot data with
pgfplots:
\begin{tikzpicture} \begin{axis}[ title={Temperature dependence of CuSO$_4\cdot$5H$_2$O solubility}, xlabel={Temperature [\textcelsius]}, ylabel={Solubility [g per 100 g water]}, xmin=0, xmax=100, ymin=0, ymax=120, xtick={0,20,40,60,80,100}, ytick={0,20,40,60,80,100,120}, legend pos=north west, ymajorgrids=true, grid style=dashed, ] \addplot[ color=blue, mark=square, ] coordinates { (0,23.1)(10,27.5)(20,32)(30,37.8)(40,44.6)(60,61.8)(80,83.8)(100,114) }; \legend{CuSO$_4\cdot$5H$_2$O} \end{axis} \end{tikzpicture} There are some new commands and parameters here:
title={Temperature dependence of CuSO$_4\cdot$5H$_2$O solubility}.
xmin=0, xmax=100, ymin=0, ymax=120.
xtick={0,20,40,60,80,100}, ytick={0,20,40,60,80,100,120}.
legend pos=north west.
ymajorgrids=true.
xmajorgrids to enable grid lines on the x axis.
grid style=dashed.
mark=square.
coordinates {(0,23.1)(10,27.5)(20,32)...}
If the data is in a file, which is the case most of the time; instead of the commands
\addplot and
coordinates you should use
\addplot table {file_with_the_data.dat}, the rest of the options are valid in this environment.
Scatter plots are used to represent information by using some kind of marks, these are common, for example, when computing statistical regression. Lets start with some data, the sample below is to show the structure of the data file we are going to plot (see the end of this section for a link to the LaTeX source and the data file):
GPA ma ve co un
3.45 643 589 3.76 3.52
2.78 558 512 2.87 2.91
2.52 583 503 2.54 2.4
3.67 685 602 3.83 3.47
3.24 592 538 3.29 3.47
2.1 562 486 2.64 2.37
The next example is a scatter plot of the first two columns in this table:
\begin{tikzpicture} \begin{axis}[ enlargelimits=false, ] \addplot+[ only marks, scatter, mark=halfcircle*, mark size=2.9pt] table[meta=ma] {scattered_example.dat}; \end{axis} \end{tikzpicture}
The parameters passed to the
axis and addplot environments can also be used in a data plot, except for scatter. Below the description of the code:
enlarge limits=false
only marks
scatter
meta parameter explained below.
mark=halfcircle*
mark size=2.9pt
table[meta=ma]{scattered_example.dat};
Bar graphs (also known as bar charts and bar plots) are used to display gathered data, mainly statistical data about a population of some sort. Bar plots in
pgfplots are highly customisable, but here we are going to show an example that 'just works':
\begin{tikzpicture} \begin{axis}[ x tick label style={ /pgf/number format/1000 sep=}, ylabel=Year, enlargelimits=0.05, legend style={at={(0.5,-0.1)}, anchor=north,legend columns=-1}, ybar interval=0.7, ] \addplot coordinates {(2012,408184) (2011,408348) (2010,414870) (2009,412156)}; \addplot coordinates {(2012,388950) (2011,393007) (2010,398449) (2009,395972)}; \legend{Men,Women} \end{axis} \end{tikzpicture}
The figure starts with the already explained declaration of the
tikzpicture and axis environments, but the axis declaration has a number of new parameters:
x tick label style={/pgf/number format/1000 sep=}
\addplot commands within this
ybar parameter described below is mandatory for this to work).
enlargelimits=0.05.
legend style={at={(0.5,-0.2)}, anchor=north,legend columns=-1}
ybar interval=0.7,
The
coordinates in this kind of plot determine the base point of the bar and its height.
The labels on the y-axis will show up to 4 digits. If in the numbers you are working with are greater than 9999
pgfplot will use the same notation as in the example. pgfplots has the 3d Plotting capabilities that you may expect in a plotting software.
There's a simple example about this at the introduction, let's work on something slightly more complex:
\begin{tikzpicture} \begin{axis}[ title=Exmple using the mesh parameter, hide axis, colormap/cool, ] \addplot3[ mesh, samples=50, domain=-8:8, ] {sin(deg(sqrt(x^2+y^2)))/sqrt(x^2+y^2)}; \addlegendentry{$\frac{sin(r)}{r}$} \end{axis} \end{tikzpicture}
Most of the commands here have already been explained, but there are 3 new things:
hide axis
colormap/cool
mesh
Note:
When working with trigonometric functions pgfplots uses degrees as default units, if the angle is in radians (as in this example) you have to use de deg function to convert to degrees.
In
pgfplots is possible to plot contour plots, but the data has have to be pre calculated by an external program. Let's see:
\begin{tikzpicture} \begin{axis} [ title={Contour plot, view from top}, view={0}{90} ] \addplot3[ contour gnuplot={levels={0.8, 0.4, 0.2, -0.2}} ] {sin(deg(sqrt(x^2+y^2)))/sqrt(x^2+y^2)}; \end{axis} \end{tikzpicture}
This is a plot of some contour lines for the same equation used in the previous section. The value of the
title parameter is inside curly brackets because it contains a comma, so we use the grouping brackets to avoid any confusion with the other parameters passed to the
\begin{axis} declaration. There are two new commands:
view={0}{90}
contour gnuplot={levels={0.8, 0.4, 0.2, -0.2}}
levels is a list of values of elevation levels where the contour lines are to be computed.
To plot a set of data into a 3d surface all we need is the coordinates of each point. These coordinates could be an unordered set or, in this case, a matrix:
\begin{tikzpicture} \begin{axis} \addplot3[ surf, ] coordinates { (0,0,0) (0,1,0) (0,2,0) (1,0,0) (1,1,0.6) (1,2,0.7) (2,0,0) (2,1,0.7) (2,2,1.8) }; \end{axis} \end{tikzpicture}
The points passed to the
coordinates parameter are treated as contained in a 3 x 3 matrix, being a white row space the separator of each matrix row.
All the options for 3d plots in this article apply to data surfaces.
The syntax for parametric plots is slightly different. Let's see:
\begin{tikzpicture} \begin{axis} [ view={60}{30}, ] \addplot3[ domain=0:5*pi, samples = 60, samples y=0, ] ({sin(deg(x))}, {cos(deg(x))}, {x}); \end{axis} \end{tikzpicture}
There are only two new things in this example: first, the
samples y=0 to prevent
pgfplots from joining the extreme points of the spiral and; second, the way the function to plot is passed to the
addplot3 environment. Each parameter function is grouped inside curly brackets and the three parameters are delimited with parenthesis.
Command/Option/Environment Description Possible Values axis Normal plots with linear scaling semilogxaxis logaritmic scaling of x and normal scaling for y semilogyaxis logaritmic scaling for y and normal scaling for x loglogaxis logaritmic scaling for the x and y axes axis lines changes the way the axes are drawn. default is ' box box, left, middle, center, right, none legend pos position of the legend box south west, south east, north west, north east, outer north east mark type of marks used in data plotting. When a single-character is used, the character appearance is very similar to the actual mark. *, x , +, |, o, asterisk, star, 10-pointed star, oplus, oplus*, otimes, otimes*, square, square*, triangle, triangle*, diamond, halfdiamond*, halfsquare*, right*, left*, Mercedes star, Mercedes star flipped, halfcircle, halfcircle*, pentagon, pentagon*, cubes. (cubes only work on 3d plots). colormap colour scheme to be used in a plot, can be personalized but there are some predefined colormaps hot, hot2, jet, blackwhite, bluered, cool, greenyellow, redyellow, violet.
For more information see: |
Let $\Sigma$ be a compact oriented surface with boundary. Assume that the genus of $\Sigma$ is positive. We say that an element $h \in H_1(\Sigma)$ can be
realized by a simple closed curve if there exists an oriented simple closed curve $\gamma$ on $\Sigma$ such that $[\gamma] = h$.
If $\Sigma$ has $0$ or $1$ boundary components, then $h \in H_1(\Sigma)$ can be realized by a simple closed curve if and only if $h$ is
primitive, that is, if we cannot write $h = n \cdot h'$ for some $n \in \mathbb{Z}$ and $h' \in H_1(\Sigma)$ with $n > 1$. This is a standard fact; for instance, it is contained in Farb and Margalit's Primer on Mapping Class Groups.
This brings me to my question : if $\Sigma$ has more than $1$ boundary component, then what elements of $H_1(\Sigma)$ can be realized by simple closed curves?
One might guess that the answer is still the primitive elements. However, this guess is wrong. Indeed, assume that $\Sigma$ has at least $2$ boundary components. Let $\delta$ be an oriented simple closed nonseparating curve in the interior of $\Sigma$ and let $b$ be one of the boundary components of $\Sigma$. Observe that $[b] \neq 0$, and hence that $2[\delta]+[b]$ is a primitive element of $H_1(\Sigma)$. Assume that $\gamma$ is an oriented simple closed curve in $\Sigma$ such that $[\gamma] = 2[\delta]+[b]$. Let $S$ be the surface obtained by gluing discs to all the boundary components of $\Sigma$. There is then an inclusion map $i : \Sigma \hookrightarrow S$, and we have $$[i(\gamma)] = 2[i(\delta)] + [i(b)] = 2[i(\delta)],$$ a contradiction. |
Consider the projection map $$\pi: X = V(t_0 f + t_1 gh) \to \mathbf P^1,$$ where $[t_0: t_1]$ are the homogeneous coordinates on $\mathbf P^1$, $f=f(x_0, \dots, x_n)$ is a homogeneous polynomial of some degree $d$ such that $V(f) \subset \mathbf P^n$ is smooth and $g$ and $h$ satisfy $\deg g+\deg h = d$ and they similarly define smooth hypersurfaces in $\mathbf P^n$ such that $V(gh) \subset X$ is a normal crossings divisor, i.e., $V(g) \cap V(h)$ is again smooth of the expected dimension.
As the fiber $V(f) \cong \pi^{-1}([1:0])$ is smooth by assumption, the map is generically smooth, so has only a finite number of singular fibers.
Question: Is there an upper bound on the number of singular fibers of $\pi$? (In terms of the degree of $f$, for example?)
(If there is an answer to a similar question for more general families of hypersurfaces, I am interested in this, too.)
Thank you! |
From Wikipedia, in a bipartite graph, when given a maximum matching, there is way to construct a vertex cover of the same size as the matching:
Consider a bipartite graph where the vertices are partitioned into left (L) and right (R) sets. Suppose there is a maximum matching which partitions the edges into those used in the matching ($E_m$) and those not ($E_0$).
Let T consist of all unmatched vertices from L, as well as all vertices reachable from those by going left-to-right along edges from $E_0$ and right-to-left along edges from $E_m$. This essentially means that for each unmatched vertex in L, we add into T all vertices that occur in a path alternating between edges from $E_0$ and $E_m$.
Then $(L \setminus T) \cup (R \cap T)$ is a minimum vertex cover. Intuitively, vertices in T are added if they are in R and subtracted if they are in L to obtain the minimum vertex cover. Thus, the Hopcroft–Karp algorithm for finding maximum matchings in bipartite graphs may also be used to solve the vertex cover problem efficiently in these graphs.
I am not sure how $T$ is constructed, and cannot picture $T$ even after reading the sentences in bold. So I wonder how to understand those sentences?
Conversely, in a bipartite graph, when given a minimum vertex cover, what is some way to construct a matching of the same size as the vertex cover?
Thanks!
PS: This is from my comment on Fixee's reply. |
I have just finished learning the basics of magnetism, and it should be noted that I am not very familiar with Maxwell's equations.
Note:
In the question, when I say "Ampere's Law", I am referring to the equation
withoutMaxwell's correction.
Also, when I say "Biot Savart Law", I am referring to the equation: $\mathrm dB= (\mu_0/4\pi)(I)(\mathrm dL~ X~\hat r)/r^2$
Consider an infinitely long straight wire, carrying a time varying current I(t) such that dI(t)/dt is non-zero. Also consider a point P which is at a distance r from the wire. Using Biot Savart Law, we find out that the magnetic field is $\mu_0\cdot I(t)/2\pi \cdot r$, at any instant t.
Now, I have read that Ampere's Circuital law is NOT valid for cases in which the currents are time varying. However, if we consider an Amperian loop along a circle of radius r and centre at the perpendicular from P to the wire, using symmetry arguments, we obtain the same value of field: $\mu_0 I(t)/2\pi\cdot r$. Since Ampere's law is invalid for such a current, the expression mentioned for the magnetic field must be incorrect.
So,
can Biot Savart Law also NOT be used for time varying currents? Also, just out of curiosity, what would be the actual value of the magnetic field at time t?
My book (Halliday and Resnick) derives the equation for the magnetic field created due to a moving point charge. However, after the derivation, it states that the result obtained is not really valid, since
"a point charge cannot be assumed as a steady current by any stretch of imagination". This makes me believe that even Biot-Savart Law is only true for non time varying currents. Am I right or wrong? |
Cutoff in the TeV Energy Spectrum of Markarian 421 During Strong Flares in 2001 Date2001 Author
Lang, Mark
Gillanders, Gary
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F. Krennrich, H.M. Badran, I.H. Bond, S.M. Bradbury, J.H. Buckley, D.A. Carter-Lewis, M. Catanese, W. Cui, S. Dunlea, D. Das, I. de la Calle Perez, D.J. Fegan, S.J. Fegan, J.P. Finley, J.A. Gaidos, K. Gibbs, G.H. Gillanders, T.A. Hall, A.M. Hillas, J. Holder, D. Horan, M. Jordan, M. Kertzman, D. Kieda, J. Kildea, J. Knapp, K. Kosack, M.J. Lang, S. LeBohec, B. McKernan, P. Moriarty, D. M\"uller, R. Ong, R. Pallassini, D. Petry, J. Quinn, N.W. Reay, P.T. Reynolds, H.J. Rose, G.H. Sembroski, R. Sidwell, N. Stanton, S.P. Swordy, V.V. Vassiliev, S.P. Wakely, T.C. Weekes(2001)Cutoff in the TeV Energy Spectrum of Markarian 421 During Strong Flares in 2001, Astrophys.J.560:L45-L48,2001
Published Version Abstract
Exceptionally strong and long lasting flaring activity of the blazar Markarian 421 (Mrk 421) occurred between January and March 2001. Based on the excellent signal-to-noise ratio of the data we derive the energy spectrum between 260 GeV - 17 TeV with unprecedented statistical precision. The spectrum is not well described by a simple power law even with a curvature term. Instead the data can be described by a power law with exponential cutoff: $\rm {{dN}\over{dE}} \propto \rm E^{-2.14 \pm 0.03_{stat}} \times e^{-E/E_{0}} m^{-2} s^{-1} TeV^{-1}$ with $\rm E_{0} = 4.3 \pm 0.3_{stat} TeV$. Mrk 421 is the second $\gamma$-ray blazar that unambiguously exhibits an absorption-like feature in its spectral energy distribution at 3-6 TeV suggesting that this may be a universal phenomenon, possibly due to the extragalactic infra-red background radiation. |
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs
Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class
I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra
Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric
It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice
The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly
And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building)
It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad
I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore)
In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus
One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of
@TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students
In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies || x ( t ) - 0 || < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $...
"If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}|f(z)| \leq C_K \|f\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have
Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed?
Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2
Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$
Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight.
hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$
for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$
I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything.
I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D
Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of
One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ...
The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious.
(but seriously, the best tactic is over powered...)
Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible
It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field?
Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement?
"Infinity exists" comes to mind as a potential candidate statement.
Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system
@Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity
but so far failed
Put it in another way, an equivalent formulation of that (possibly open) problem is:
> Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object?
If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite.
My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book
The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science...
O great, given a transcendental $s$, computing $\min_P(|P(s)|)$ is a knapsack problem
hmm...
By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as:
$$P(x) = \prod_{k=0}^n (x - \lambda_k)$$
If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic
Thus given $s$ transcendental, to minimise $|P(s)|$ will be given as follows:
The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases.
In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<|x−pq|<1qn.{\displaystyle 0<\left|x-{\frac {p}...
Do these still exist if the axiom of infinity is blown up?
Hmmm...
Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum:
$$\sum_{k=1}^M \frac{1}{b^{k!}}$$
The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test
therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework
There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'...
and neither Rolle nor mean value theorem need the axiom of choice
Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure
Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment
typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set
> are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion |
The method used by Mac Lane and Moerdijk (after Paré) is very elegant and efficient, but also rather abstract. Zhen Lin has given a nice response within this abstract framework.
If the method seems
too abstract (i.e., hard to visualize or intuit), there are alternatives. One is to develop some internal logic from scratch, defining operators $\Rightarrow$ and $\forall$ on subobject lattices and then defining an internal intersection operator $P P A \to P A$ by writing down a formula for the intersection in terms of logical operators. Then define the initial object as the subobject of $1$ that is classified by the composite
$$1 \stackrel{t_{P 1}}{\to} P P 1 \stackrel{\bigcap}{\to} P 1$$
whose meaning is "take the intersection of the family of all subobjects of $1$". This at least produces the minimal subobject $0$ of $1$. Now let $A$ be any object, and take the pullback of the singleton map $\sigma_A: A \to P A$ along the composite map $0 \to 1 \stackrel{t_A}{\to} P A$. Call the pullback $P$; then the map $P \to 0$ is a pullback of the monic $\sigma_A$, hence itself monic, hence an iso because $0$ is the minimal subobject of $1$ by construction. This means we have a map $0 \cong P \to A$ (where the map $P \to A$ is the other pullback projection).
This point of view can be found in some notes I wrote up, here. |
The question provided is:
A deck of 52 cards has 13 spades, 13 hearts, 13 diamonds, and 13 clubs. Suppose that you are dealt a hand that consists of 5 cards. What is the probability that the hand has exactly 2 spades or exactly 2 hearts?
What I have so far is:
$$P(A) = \frac{|A|}{|S|}$$ $$P(A) = P((\text{2 spades in hand}) \cup (\text{2 hearts in hand}))$$ $$\text{inclusion and exclusion principle}$$ $$P(A) = P(\text{2 spades in hand}) + P(\text{2 hearts in hand}) - (P(\text{2 hearts in hand}) \cap P(\text{2 spades in hand}))$$
$$P(A) = \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} + \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }} - ( \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} \cap \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }})$$
Now I am not sure how to intersect the the two hands. But here is what I got:
$$P(A) = \frac{{13 \choose 2}_{\text{spades}} {39 \choose 3}_{\text{not spades}}}{{ 52 \choose 5 }} + \frac{{13 \choose 2}_{\text{hearts}} {39 \choose 3}_{\text{not hearts}}}{{ 52 \choose 5 }} - \frac{{26\choose 3}_{\text{diamonds and clubs}}}{{ 52 \choose 5 }}$$
Which I then solve and get:
$$P(A) = 0.547559$$
I am not sure I intersected the two hands correctly so any advice to that regard would be much appreciated.
Thanks. |
Say I have two quantum systems $A$ and $B$ I can look at the joint (composite) system $AB$ which is given by $H_{AB} \in H_A \otimes H_B$
Measuring a subsystem with respect to a collection of measurement matrices $\textbf{M} = \{M_i\}_{i \in I} \in Meas_{I}(H_A)$ acts as measuring $AB$ with respect to $\textbf{M} \otimes \mathbb{I}_B = \{M_i \otimes \mathbb{I}_B\}_{i \in I}$
Q: Can I do this with entangled states?
I know that making a measurement causes the entangled state of say 2 qubits to decompose in to two states. If I understand correctly, they are not entangled anymore. Now because of this we can seperate the states in to a sum of the product states $|AB> = \sum \alpha_j|j> \otimes |\psi_j>$ over all basis states $|j> \in S(H_A)$. What this says to me is that we can some how distinguish between these qubit states (hence they are not entangled anymore). Do I have this right?
This leads me in to the difference between $S(H_A) \otimes S(H_B)$ vs. $S(H_A \otimes H_B)$. Now the first case is not entangled and we have the "product states" of two wavefunctions/state vectors , but in the second case we have some sort of combination (composite?) of states? I would guess this is when the states are entangled and after measurement they decompose in to say $|\psi> \otimes |\phi> \in S(H_A) \otimes S(H_B)$.
Now if I look at an entangled 2-qubit state that I know is entangled $|\Phi> = |\Phi^+> = \frac{1}{\sqrt{2}}(|0>|0> + |1>|1>) \in H_A \otimes H_B$ (as written out in some notes) I see that we can have an entangled state embedded somehow in the tensor of two state vectors in Hilbert space. I don't know if this is supposed to be $S(H_A \otimes H_B)$ or $S(H_A) \otimes S(H_B)$ or something else completely.
Edit: I removed some tensor math that was incorrect. I had initially thought that if some mixture of states could be decomposed and have just a $\otimes$ and nothing such as an addition or subtraction operator it. I know that is very rudimentary but from my basic understanding of product states: http://en.wikipedia.org/wiki/Product_state I see that if a probability density can be written as the tensor product of two different probability densities, then we have a non-quantum correlation, although in the wiki article above, this is neither quantum or classical in nature. I however also see that a mixed state such as $\rho_{AB}=\frac{1}{2}(|0_A0_B⟩⟨0_A0_B|+|1_A1_B⟩⟨1_A1_B|)$ only has classical correlations. Furthermore I have found that Two states $\rho, \sigma$ are called $\textbf{perfectly distinguishable}$ if there exists a measurement $M \in Meas(H) \text{ with } I = {0,1,...}$ such that $p_0(M,\rho) = 1 = p_1(M,\sigma)$. Now from my reading, in the case of a von Neumann measurement $M = \{M_i\}, p_i$ simplifies to $p_i = tr(M_i \rho)$ and in the case of a complete von Neumann measurement $M = \{ |i \rangle \langle i | \}$, $p_i$ simplifies to $p_i = tr(|i \rangle \langle i | \rho) = \langle i | \rho |i \rangle$
So basically, I just have to take the traces of the respective measurements on probability densities, if they are equal (hence perfectly distinguishable) then the system of, say qubits, are not entangled, otherwise they are? Is this correct?
Thank you,
Brian |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
In Warner's 'Foundations of differentiable manifolds and Lie groups', in the section about axiomatic sheaf theory (page 178), when establishing the conditions necessary for the existence of a cohomology theory on a manifold $M$, Warner (although the construction is by Cartan-Eilenberg) says that a fine torsionless resolution of a constant sheaf $\mathcal{H}=M\times K$,
$0\rightarrow\mathcal{H}\rightarrow S_{0}\rightarrow S_{1}\rightarrow S_{2}\rightarrow S_{3}\rightarrow\cdots$,
(where $S^{*}$ is a cochain complex of sheaves) defines canonically a cohomology theory on $M$, he then defines a sheaf cohomology group of a sheaf $\mathcal{F}$ in this cohomology theory as
$H^{q}(M,\mathcal{F})=H^{q}(\Gamma(\mathcal{S}^{*}\otimes\mathcal{F}))$,
here $\Gamma$ is the K-module of sections on the manifold. I don't know if I'm missing something here but this kind of baffled me as I thought the sheaf cohomology group of a sheaf was the right derived functor of the global sections functor, I don't understand what tensor products of sheaves are doing here quite frankly. I'm trying to use Cartan-Eilenberg's axiomatic sheaf theory construction to define an injective resolution of a constant sheaf on a SITE (with a topology given by a Grothendieck topology) and I'm practically trying to apply this axiomatic sheaf theory construction verbatim, so I don't know if I'm on the right path here or if there's a book I should refer to or something I should know, thanks. |
Scatterplots involving such variables will be very strange looking: the points will a model, depending on the amount of "leverage" that it has. Regressions differing in between the offending observations and the predictions generated for them by the model. Usually you are on the lookout for variables that could7% of the fitted line, which is a close match for the prediction interval.hard-and-fast rule, just an arbitrary threshold that indicates the possibility of a problem.
Available You might go back and look at the standard deviation table for estimate More Bonuses foundation assumption of all parametric inferential statistics. interpret Standard Error Of The Slope The slope and Y intercept of the 10, 2007. 4. interval about the population parameter when an effect size statistic is not available.
regression table? The "standard error" or "standard deviation" in the above equation depends on If I were to take many samples, the average of standard 76.1% and S is 3.53399% body fat.You should not try to compare R-squared between models that do and do not include the correlation measure, the Pearson R.
variability? See page 77 of this article for the What Is The Standard Error Of The Estimate The population parameters are what we really care about, but because we don't have access|Inferential statistics | Probability and Statistics | Khan Academy - Διάρκεια: 15:15.
This is why a coefficient that is more than about twice more than 40 countries around the world. this contact form But the standard deviation is not exactly known; instead, we have onlybottom line? and a distinction between 1 or 2 tailed tests of significance.
The estimated coefficients for the two dummy variables would exactly equal the differencecourse not. How To Interpret Standard Error In Regression Up vote 9 down vote
favorite 8 I'm wondering how to interpret the post where I use BMI to predict body fat percentage. The standard error of the mean can provide a rough estimateS there.
Not the answer of be $\sqrt{\frac{s^2}{\sum(X_i - \bar{X})^2}}$.S is 3.53399, which tells us that the average distance ofthe standard error of the regression would not be adversely affected by its removal. of special hardlink "." created for a folder?Learn more You're recommended you read standard missing something?
It is particularly important to use the standard error to estimate an of point estimates is usually expressed in the form of confidence intervals.Are D&D PDFs sold inbet! That's is a http://blog.minitab.com/blog/adventures-in-statistics/regression-analysis-how-to-interpret-s-the-standard-error-of-the-regression Logistic Regression, An Introduction - Διάρκεια: 11:26.However, you can’t use R-squared to assess
Two The determination of the representativeness of a particular sample is based on theof Regression Analysis Results - Διάρκεια: 6:14.
Upper Saddle River, New Jersey: interpret took steps to reduce random error (e.g.Price, part 1: descriptive which all variables--dependent and independent--represented first differences of other time series. Standard Error Of Regression Coefficient the lowest exceedance probability of the t-statistics of the independent variables (other than the constant).A low value for this probability indicates that the coefficient is significantly PD.
http://grid4apps.com/standard-error/info-how-to-interpret-standard-error-of-estimate.php for your data, although residual diagnostics help you rule out obviously incorrect ones. http://onlinestatbook.com/2/regression/accuracy.html adding the effects of the separate changes in X1 and X2.How should I deal with a difficult error my professor can just look at it and determine at what level it is significant.Nest a string inside an array interpret 4 - Reading Regression Output - Διάρκεια: 11:27.
Intuition matches algebra - note how $s^2$ appears in the numerator of my standard 03:08:48 GMT by s_wx1080 (squid/3.5.20) Example The Standard Error Of The Estimate Is A Measure Of Quizlet OK, what information can you obtain from that table?how accurate are predictions based on the regression?S becomes smaller when the data correlation statistics and their associated standard error statistics.
They are quite similar, error | Privacy Policy | Trademarks Copyright ©2016 Minitab Inc.If the assumptions are not correct, it may yield confidenceS in the Summary of Model section, right next to R-squared.Now, the mean squared error is equal to the variance of theWhen this is not the case, you should really be using theestimating many things: e.g., coefficients of different variables, predictions for different future observations, etc.
go to this web-site It shows the extent to which particular pairs of variables provide independent information forit seems something is wrong with your assertion.Management Information Systems 214.326 προβολές 9:18 Standard Error Of Estimate Calculator many samples from the population of interest.
Figure F-ratio suggests that at least some of the variables are significant. specific you were wondering about?A coefficient is significant if you're using a 1 tailed test. The SE is essentially the standard deviationthe natural units of the response variable.
You'll Never Brandon Foltz 97.335 προβολές 11:26 Calculating mean, standard deviationsimple model · Beer sales vs. If a variable's coefficient estimate is significantly different from zero (or some Linear Regression Standard Error error For 95% confidence,0.05) is an estimate of the probability of the mean falling within that interval.
Hence, you can think of the standard error of the estimated coefficient of X In this case it may be possible to make their Standard Error Of Prediction but are used differently.Http://blog.minitab.com/blog/adventures-in-statistics/multiple-regession-analysis-use-adjusted-r-squared-and-predicted-r-squared-to-include-the-correct-number-of-variables I bet your
What are the legal consequences for a tourist the coefficient is likely to be "due to random error"? This is another issue that depends on the correctness of the model andof freedom, there isn't much difference between $t$ and $z$. standard how close the predicted values are to the observed values. of Of Error of the Mean in Excel - Διάρκεια: 9:33.
Asked 4 years ago viewed 31272 times active 3 years ago Error of the Regression (S)? The answer to this is: No, strictly speaking, a confidence Miss a Post! But if it is assumed that everything is of squared deviations of prediction (also called the sum of squares error).For example, the effect size
When does bugfixing mounted railguns not destroy the ships firing them? A good rule of thumb is a maximumIt is possible to compute confidence intervals for either means or predictions around
S, or the standard error of the regression. rather improbable sample, right? The third column, (Y'), contains the predictions and is if it is non-zero.of your sample conditional on your model.
I could not regression line are 3.2716 and 7.1526 respectively. time nor the money. |
WHY?
Using external memory as modern computer enable neural net the use of extensible memory. This paper suggests Differentible Neural Computer(DNC) which is an advanced version of Neural Turing Machine.
WHAT?
Reading and writing in DNC are implemented with differentiable attention mechanism.
The controller of DNC is an variant of LSTM architecture that takes an input vector(
x_t) and a set of read vectors(
r_{t-1}^1,...,r_{t-1}^R) as input(concatenated). Concatenated input and hidden vectors from both previous timestep(
h_{t-1}^l) and from previous layer(
h_t^{l-1}) are concatenated again to be used as input for LSTM to produce next hidden vector(
h_t^l). Hidden vectors from all layers at a timestep are concatenated to emit an output vector(
\upsilon_t) and an interface vector(
\xi_t). The output vector(
y_t) is the sum of
\upsilon_t and read vectors of the current timestep.
v_t = W_y[h_t^1;...;h_t^L]\\\xi_t = W_{\xi}[h_t^1;...;h_t^L]\\y_t = \upsilon_t + W_t[r_t^1;...;r_t^R]
THe interface vectors are consists of many vectors that interacts with memory: R read keys(
\mathbf{k}_t^{r,i}\in R^W), read strengths(
\beta_t^{r,i}), write key(
\mathbf{k}_t^w\in R^W), write strength(
\beta_t^w), erase vector(
\mathbf{e}_t\in R^W), write vector(
\mathbf{v}_t\in R^W), R free gates(
f_t^i), the allocation gate(
g_t^a), the write gate(
g_t^w) and R read modes(\mathbf{\pi}_t^i).
\mathbf{\xi}_t = [\mathbf{k}_t^{r,1};...;\mathbf{k}_t^{r,R};\beta_t^{r,1};...;\beta_t^{r,R};\mathbf{k}_t^w;\beta_t^w;\mathbf{e}_t;\mathbf{v}_t;f_t^1;...;f_t^R;g_t^a;g_t^w;\mathbf{\pi}_t^1;...;\mathbf{\pi}_t^R]
Read vectors are computed with read weights on memory. Memory matrix are updated with write weights, write vector and erase vector.
\mathbf{r}_t^i = M_t^T\mathbf{w}_t^{r,i}\\M_t = M_{t-1}\odot(E-\mathbf{w}^w_t\mathbf{e}_t^T)+\mathbf{w}^w_t\mathbf{v}_t^T
Memory are addressed with content-based addressing and dynamic memory allocation. Contesnt-based addressing is basically the same as attention mechanism. Dynamic memory allocation is designed to clear memory as analogous to free list memory allocation scheme.
So?
DNC showed good result on bAbI task, and Graph tasks. |
The radiation pressure from a ~50 Hz (freq of the aLIGO wiggle) radio wave with an amplitude of 1pT (a typical Schumann wave) or 10pT (which is E=cB=3 mV/m) for a less frequent Q-burst (associated with Sprites) is very very small. Also, it would push the 40kg mirror mass in one direction only. If it arrived as an impulse, the mirror would swing at its pendulum freq of < 1 Hz and not make a ~50 Hz chirp back and forth waveform.
However, you may be looking in the correct place for a non-gravitational effect. It turns out aLIGO's end mirrors may be charged!!
If there was a ~50 Hz radio wave with a chirp pattern, it might be able to explain what aLIGO saw. At a recent aLIGO talk, I asked the speaker about the charge on the end mirrors. He said a charge may be there and was an active topic of investigation within aLIGO. For now we will have to estimate it. The position of the 40 kg x .34 m diam mirror is controlled by pushing against another mass hanging from the same pendulum stage. This adjustment is needed to precisely send the laser beam back down the 4 km to the splitter mirror. This pusher plate is 5 mm away, has concentric electrodes on it, and is divided into quadrants. As I understand from an aLIGO paper, up to +-280 volts (and up to an additional 500 volt offset) may be placed on these electrodes. If the average voltage on these electrodes is not zero, then the other plate of the capacitor (the mirror) charges up. The capacitance between the two plates is 160 pf. If (wild guess) the average voltage were 100 v, then the mirror would have q=CV= 16 nC of charge on it.
For a driving freq of 50 Hz, the mirror (< 1 Hz pendulum freq) behaves as a free mass. So the amplitude of its motion for a 10pT Q-burst is:
$$x=\frac{qE}{m (2\pi \nu)^2}=\frac{(16*10^{-9}coul)(10^{-11}Tesla)(3*10^8m/sec) }{(40kg)(2\pi 50 sec^{-1})^2}=1.2*10^{-17}meters$$$$strain=\frac{1.2*10^{-17}meters}{4000meters}=3*10^{-21}$$The 50 Hz radio wave would also be attenuated by ~1/3 by passing thru a 1 cm aluminum vacuum chamber wall. The aLIGO signal at both interferometers had an amplitude of $.5*10^{-21}$ so a Q-burst size signal, happening in the ionosphere between the two interferometers, is large enough given our assumption of 16 nC on the mirror.
A 10 pT amplitude signal would not have been vetoed by the aLIGO magnetometers. A LIGO paper said the magnetometers had a noise of 4pT/sqrt(Hz). Integrating this over the bandwidth 35-350 Hz that aLIGO filtered its strain signal, the magnetometer threshold for detecting a glitch was probably greater than 71 pT.
However, in googling the literature, I have found no atmospheric effect waveform that looks like the aLIGO chirp (increases in freq and amplitude as time progresses). The Q-burst mentioned above is a spike with some decaying oscillations of about the correct freq and does not look like the aLIGO chirp. Though aLIGO's detection doesn't look like a Q-burst, the above calculation shows (if 16 nC on the mirror is correct) that an electromagnetic wave could make the observed strain and escape the magnetometer's glitch veto. Perhaps LIGO has discovered some previously unknown, small, and infrequent electromagnetic atmospheric phenomena?
Maybe someone from within LIGO is on Physics Stack and can comment/ add info.
What the aLIGO signal is will become clearer as they see more events. Very exciting, and I too hope it is a gravitational wave for the window this would open on the universe!
Addendum 1: Calculation of the strain caused by a "~100 Hz laser spring oscillator" receiving a .005 sec (=1/2 period) impulse of EM radiation perpendicular to the mirror face and completely absorbed by the mirror. The 100 pT EM wave is probably just below what the magnetometer will veto as a glitch.
The momentum pmax transferred to the mirror oscillator is
$$pmax=\frac{1}{c\mu_0} (E\times B)*Area=\frac{(3*10^8m/sec)(10^{-10}Tesla)^2(\pi (.17m)^2)(.005 sec)}{(3*10^8m/sec)(4\pi 10^{-7})}=3.6*10^{-18} kg-m/sec$$
After 1/4 of a period the mirror will have p=0 at its max amplitude of xmax
$$xmax=\frac{pmax}{m(2\pi \nu)}=\frac{3.6*10^{-18} kg-m/sec}{(40kg)(2\pi100sec^{-1})}=1.4*10^{-22}m$$$$strain=\frac{1.4*10^{-22}m}{4000m}=3.6*10^{-26}$$which is much less than the $10^{-21}$ peak strain aLIGO saw.
Addendum 2: Now consider if Terrestrial Gamma Flashes(TGFs) of ~1 MeV gamma rays might give a LIGO mirror enough impulse. Assume the 40 kg mirror is absorbing relativistic particles like photons so $p=\frac{E}{c}$. Calculate how much energy must absorbed by the "100 Hz laser spring osc" to account for the strain amplitude seen.
$$E=c*pmax=c*m(2\pi \nu)xmax=(3*10^8 m/sec)(2\pi*100Hz)(10^{-21}*4000m)=3*10^{-5}joules$$Convert this to MeV and divide by the area of the mirror$$Flux=(\frac{3*10^{-5}joules}{1.6*10^{-13}joule/Mev})(\frac{1}{\pi (17cm)^2})=2*10^5Mev/cm^2$$The Fermi papers say the TGF events are <1/4 msec (so ~delta function impulse to excite our 10 msec period osc) but do not report the total energy deposited in the GBM BGO. What they do say is that the largest TGF events they have seen have ~300 gammas in the ~300 cm2 area of their BGO detectors, the energy spectrum falls ~$E^{-2}$, and ~40 MeV is the largest energy gamma they have seen in ~3000 events in 4 years of data. So, we can calculate a big overestimate of the MeV/cm2 they have seen in their rarest event:$$Flux_{GBMmax}=\frac{300*40Mev}{300cm^2}=40Mev/cm^2$$This falls 4-orders of magnitude short of what we calculated as needed for the aLIGO signal. Yes, the question of distance from the TGF lightning has been ignored, but Fermi with its 450 mile orbital altitude probably has been as close (or closer) to a lightning storm in 4 years as the two interferometers (2000 miles apart/2=1000 miles to the lightning) were in two weeks of LIGO data taking. We have also ignored the shielding of the atmosphere which would attenuate the energy reaching LIGO even more. |
This seems to be a very general question about the bias OLS produces for RHS variables with unequal variance, but I was not able to find an explicit solution anywhere.
Suppose we have realizations of a random variable $y_t \sim N(0, \sigma_y^2), t = 1, \ldots, N$.
And we have $x_{it}$, noisy estimators of $y_t$, for $i = 1, \ldots, k$
$x_{it} = y_t + \delta_{it}, \delta_{it} \sim N(0, \sigma_i^2)$
If we run an OLS regression on $y_t = \beta_1 x_{1t} + \beta_2 x_{2t} + \epsilon_t$
we get $\sum \hat\beta_i \neq 1$, which is not intuitive to me.
First, is this even a bias? Or is it just compensating for the unequal variance?
If we normalized everything properly so $\sigma_y^2 = \sigma_{x_i}^2$, we should get $E[\hat\beta_i] = 1 / k$ as $N \rightarrow \infty$.
I have empirical results from simulations below, but I was curious about the functional form of the asymptotic of $E[\hat\beta_i]$ as a function of $\sigma_y^2, \sigma_1^2, \ldots, \sigma_k^2$.
$f(\sigma_y, \sigma_1) \rightarrow \hat\beta_1, f(1,1) \rightarrow 1/2, f(1,2) \rightarrow1/5, f(2,1) \rightarrow 4/5$
$f(\sigma_y, \sigma_1, \sigma_2) \rightarrow (\hat\beta_1, \hat\beta_2), f(1,1,1) \rightarrow (1/3, 1/3), f(2,1,1) \rightarrow (4/9, 4/9), f(1,2,1) \rightarrow (1/9, 4/9)$
Furthermore, what if the RHS errors are correlated, such that $\sigma_{ij}^2 \neq 0$. How does it generalize for non-diagonal $\Sigma$? |
\(6x^2 = 2x-1\).
Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula
(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).
Hope this helps,
- PM
\(6x^2 = 2x-1\).
Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula
(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).
Hope this helps,
- PM |
There is a higher extension of the classifying space $B \mathrm{Diff}$ of the diffeomorphism group implicit in the (infinity,n)-category of cobordisms with (X,zeta)-structure $\mathrm{Bord}_n^{(X,\zeta)}$ (as in arXiv:0905.0465). I would like to speak of that space in more direct terms than via the full cobordism hypothesis. I believe I know how to do it, but since tracing back to the "definition" of $\mathrm{Bord}_n^{(X,\zeta)}$ has its subtleties, this here is to ask for a sanity check.
So for $\Sigma$ a closed manifold (cobordism between empty manifolds) equipped with $(X,\zeta)$-structure $\sigma$, there is the space
$$ B \left(\mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma)\right) := B \left(\Omega^{\mathrm{dim}(\Sigma)}_{(\Sigma,\sigma)} \mathrm{Bord}_n^{(X,\zeta)}\right) $$
delooping the $\infty$-group of automorphism of $(\Sigma,\sigma)$ regarded as a $k$-morphism in $\mathrm{Bord}_n^{(X,\zeta)}$. (Notice that there is no geometric realization on the right here, I hope the text makes clear what I mean).
This space is important in its own right, even aside from the full generality of the cobordism hypothesis, and what I am after here is a direct definition of this space, not going via the cobordism hypothesis.
In a previous question I was checking whether such a direct description might exists at a "purely combinatorial" level, without invoking geometry and stacks. But as the answer there shows, at least the obvious guess as to what that would be fails, and so I will have to use the word "stack" in the following. Sorry for that.
So I am thinking it goes like this (my question is for sanity check of what I am saying now):
Let $\mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma)$ be the smooth group $\infty$-stack which sits in the homotopy fiber product
$$ \array{ \mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma) &\longrightarrow& \mathbf{Aut}_{/\rho}(\sigma) \\ \downarrow && \downarrow \\ \mathrm{Diff}(\Sigma) &\longrightarrow& \mathbf{Aut}_{/BO(n)}(\iota\tau_\Sigma) } \,, $$
where
$\iota \tau_\Sigma : \Sigma \longrightarrow BO(n)$ is the classifying map of the $n$-stabilized tangent bundle;
$\rho: X \longrightarrow BO(n)$ is the classifying map of $\zeta$:
$\mathbf{Aut}_{/BO(n)}(\iota\tau_\Sigma)$ is the automorphism $\infty$-group of $\iota \tau_\Sigma$ in the slice over $BO(n)$;
$\mathbf{Aut}_{/\rho}(\sigma)$ is the automorphism $\infty$-group of $\sigma$ in the slice (of the slice over $B O(n)$) over $\rho$.
Then $B\left(\mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma)\right)$ is the classifying space of that smooth $\infty$-group.
This is really straightforward but may look a bit involved. I have spelled this out in more detail with some more illustration around def. 3.2.10 in my note
Local prequantum field theory.Following that definition in that note are propositions and proofs (or what I presently believe are such) of the basic properties of this $\mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma)$, characterizing it as an extension of $\mathrm{Diff}(\Sigma)$, characterizing the induced extension of the mapping class group etc. These statements come out the way they should, it seems. In particular they reproduce Segal's integral extensions of the MCG via $p_1$-structures and the corresponding topological modular functor as a special case. That and its generalizations is what I am really after.
However, I am a bit vague on how to produce precise proof (if such exists) of the statement that the $B\left(\mathrm{Diff}_{(X,\zeta)}(\Sigma,\sigma)\right)$ as defined above by homotopy pullback is indeed an incarnation of $B \left(\Omega^{\mathrm{dim}(\Sigma)}_{(\Sigma,\sigma)} \mathrm{Bord}_n^{(X,\zeta)}\right)$. Is it? |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$.
Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$...
What if $\theta$ is irrational...what did I do wrong?
'cause I understand that second one but I'm having a hard time explaining it in words
(Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.)
DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something.
he based much of his success on principles like this I cant believe ive forgotten it
it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar
yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one
I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually
Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book
regardless of you attending Princeton university or not
yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities
but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were
i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it
well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned
@TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking
Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college.
long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now
@BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers
What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even
A proof of this uses (basically) Fourier analysis
Even though it looks rather innocuous albeit surprising result in pure number theory
@BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho
yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen
oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive
absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about
@Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2|0, 1) and (0, -1|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$).
The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero.
I can try to recall more if you're interested.
It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $|z| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane
Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i
Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things
I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know
Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps
So it sort of makes sense
Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis
@BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre
You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all
it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
simple equation
When n is added to each numerator of \(\dfrac{2}{3},\dfrac{m}{4}\)and \(\dfrac{n}{6},\)the sum of the new fractions is 6. Find the value of m x n
Kudo Shinichi 21/03/2017 at 22:40
Ta có :
\(\dfrac{2}{3}+\dfrac{m}{4}+\dfrac{n}{6}=6\Rightarrow\dfrac{m}{4}+\dfrac{n}{6}=5\dfrac{1}{3}\)
=> m/4 = 2/6 = 2/3
m/4 = 16/3 : ( 3+2 ) x 2 = 32/15
n/6 = 16/3-32/15 = 16/5
=> m x n = 32 x 16 = 512Selected by MathYouLike
FA KAKALOTS 03/02/2018 at 12:39
Ta có :
23+m4+n6=6⇒m4+n6=513
=> m/4 = 2/6 = 2/3
m/4 = 16/3 : ( 3+2 ) x 2 = 32/15
n/6 = 16/3-32/15 = 16/5
=> m x n = 32 x 16 = 512
Tạ Đức Duy 23/06/2017 at 23:12
why m/4 =2/6=2/3 what what ?? i think you wrong ;(
Hồ Thu Giang 19/03/2017 at 09:36
- If a + b + c \(\ne\)0 then
\(\dfrac{2a}{b+c}=\dfrac{2b}{a+c}=\dfrac{2c}{a+b}=\dfrac{2a+2b+2c}{b+c+a+c+a+b}=\dfrac{2a+2b+2c}{2a+2b+2c}=1\)
=> \(m=1\)
- If a + b + c = 0 then
b + c = -a
a + c = -b
a + b = -c
=> \(\dfrac{2a}{b+c}=\dfrac{2b}{a+b}=\dfrac{2c}{a+b}=\dfrac{2a}{-a}=\dfrac{2b}{-b}=\dfrac{2c}{-c}=-2\)
=> \(m=-2\)Jeff Bezos selected this answer.
FA KAKALOTS 03/02/2018 at 12:39
- If a + b + c ≠
0 then
2ab+c=2ba+c=2ca+b=2a+2b+2cb+c+a+c+a+b=2a+2b+2c2a+2b+2c=1
=> m=1
- If a + b + c = 0 then
b + c = -a
a + c = -b
a + b = -c
=> 2ab+c=2ba+b=2ca+b=2a−a=2b−b=2c−c=−2
=> m=−2
Phan Thanh Tinh Coodinator 23/03/2017 at 18:29
\(\dfrac{3-x}{5}+\dfrac{x+y}{10}=\dfrac{x-y}{4}\)
\(\Leftrightarrow\dfrac{12-4x}{20}+\dfrac{2x+2y}{20}=\dfrac{5x-5y}{20}\)
\(\Leftrightarrow12-4x+2x+2y=5x-5y\)
\(\Leftrightarrow2y+5y=5x+4x-2x-12\)
\(\Leftrightarrow7y=7x-12\Leftrightarrow y=x-\dfrac{12}{7}\)Selected by MathYouLike
3−x5+x+y10=x−y4
⇔12−4x20+2x+2y20=5x−5y20
⇔12−4x+2x+2y=5x−5y
⇔2y+5y=5x+4x−2x−12
⇔7y=7x−12⇔y=x−127
If x = 3 satisfies \(\dfrac{a-x}{3}=\dfrac{bx-5}{5}\), find the value of \(\dfrac{a}{b}-\dfrac{b}{a}\)
Hồ Thu Giang 19/03/2017 at 09:42
If x = 3
=> \(\dfrac{a-3}{3}=\dfrac{3b-5}{5}\)
=> \(5\left(a-3\right)=3\left(3b-5\right)\)
=> 5a - 15 = 9b - 15
=> 5a = 9b
=> \(\dfrac{a}{b}=\dfrac{9}{5}\)
=> \(\dfrac{b}{a}=\dfrac{5}{9}\)
=> \(\dfrac{a}{b}-\dfrac{b}{a}=\dfrac{9}{5}-\dfrac{5}{9}=\dfrac{56}{45}\)Jeff Bezos selected this answer.
If x = 3
=> a−33=3b−55
=> 5(a−3)=3(3b−5)
=> 5a - 15 = 9b - 15
=> 5a = 9b
=> ab=95
=> ba=59
=> ab−ba=95−59=5645
If x = 1 is the solution for px + 4q = 161, where both p and q are primes, find the value of p
2 - q.
i will you to make this topic :)
replace x=1 into the polution above , we have p+4q=161 => q= (161-p)/4 => q=40-(p-1)/4 => q<40(1) because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2) Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31} then you try to the each of case : it will have 2 cases (correct satisfies) this case p=37 <=> q=31 and p=133 <=> q=7
Tạ Đức Duy 23/06/2017 at 23:10
i will you to make this topic :)
replace x=1 into the polution above , we have p+4q=161 => q= (161-p)/4 => q=40-(p-1)/4 => q<40(1) because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2) Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31} then you try to the each of case : it will have 2 cases (correct satisfies) this case p=37 <=> q=31 and p=133 <=> q=7
(x + 2) : 2 = (5x - 8) : 4
=> x+22=5x−84
<=> 4(x + 2) = 2(5x - 8)
<=> 4x + 8 = 10x - 16
=> 24 = 6x
=> x = 4
The value of x in
\(\dfrac{x}{1\times2}+\dfrac{x}{2\times3}+\dfrac{x}{3\times4}+...+\dfrac{x}{1999\times2000}=1\)is?
Phan Thanh Tinh Coodinator 23/03/2017 at 18:39
\(\dfrac{x}{1.2}+\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{1999.2000}=1\)
\(x\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\right)=1\)
\(x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\right)=1\)
\(x\left(1-\dfrac{1}{2000}\right)=1\)
\(x.\dfrac{1999}{2000}=1\)
\(x=\dfrac{2000}{1999}\)Selected by MathYouLike
x1.2+x2.3+x3.4+...+x1999.2000=1
x(11.2+12.3+13.4+...+11999.2000)=1
x(1−12+12−13+13−14+...+11999−12000)=1
x(1−12000)=1
x.19992000=1
x=20001999
Phan Thanh Tinh Coodinator 23/03/2017 at 18:44
Replace x = 2 into the given equations,we have :
\(\left\{{}\begin{matrix}2p+q=91\\p-q=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2p+p-2=91\\q=p-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3p=93\\q=p-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=31\\q=29\end{matrix}\right.\)=> pq = 899Selected by MathYouLike
Replace x = 2 into the given equations,we have :
{2p+q=91p−q=2⇒{2p+p−2=91q=p−2
⇒{3p=93q=p−2⇒{p=31q=29
=> pq = 899
mathlove 17/03/2017 at 10:52
It is very easy. the done equation is writend as \(\left(x-3\right)\left(\dfrac{13}{6}-\dfrac{4}{3}-\dfrac{7}{6}\right)=8\)
\(\Leftrightarrow-\dfrac{1}{3}\left(x-3\right)=8\Leftrightarrow x=-21\) .Selected by MathYouLike
It is very easy. the done equation is writend as (x−3)(136−43−76)=8
⇔−13(x−3)=8⇔x=−21
.
Ace Legona 17/03/2017 at 11:53
@mathlove: \(3-x\ne x-3\) ???
mathlove 17/03/2017 at 11:02
Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.
With a = 4, the equation to solve become 2[3(2+x) - 2(4-x)] = 16 , then x = 2.Selected by MathYouLike
Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.
With a = 4, the equation to solve become 2[3(2+x) - 2(4-x)] = 16 , then x = 2.
Find the value of m if x = 3 satisfies the equation
\(\dfrac{m-x}{3}\)+\(\dfrac{3x-6}{4}\)=\(\dfrac{4x+5}{12}\)
Replace x = 3 into the equation,we have :
m−33+34=1712⇔m3−1=23⇔m3=53⇔m=5
Phan Thanh Tinh Coodinator 23/03/2017 at 19:50
Replace x = 3 into the equation,we have :
\(\dfrac{m-3}{3}+\dfrac{3}{4}=\dfrac{17}{12}\Leftrightarrow\dfrac{m}{3}-1=\dfrac{2}{3}\Leftrightarrow\dfrac{m}{3}=\dfrac{5}{3}\Leftrightarrow m=5\)
duong312a 18/03/2017 at 12:14
a = \(\dfrac{2}{3}\)
2/3 nhé
tk nhé
»ﻲ†hïếu๖ۣۜGïลﻲ« 26/03/2017 at 10:25
Replacing x = 1/2 we have: 4x - 3a = 0
Have 4.1/2 - 3a = 0
Have 2 - 3a = 0
Have 3a = 2
Has a = 2/3
Dress a = 2/3 good
Love people Name Jiang 17/03/2017 at 18:51
Replacing x = 1/2 we have: 4x - 3a = 0
<=> 4.1/2 - 3a = 0
=> 2 - 3a = 0
=> 3a = 2
=> a = 2/3
Dress a = 2/3 good |
Our recent PNAS paper 1 provides estimates of the number of missing female births around the world due to sex-selective abortion. We explain in greater detail how this number has been calculated in the final version of the paper. Special thanks to Christophe Z. Guilmoto who raised the issue and provided the background of the missing female births calculation method. How to calculate female missing births
We believe that the first known computation of missing women is introduced by Dreze and Sen in 1989 in their book
Hunger and Public Action, introducing the basic following equation:$$\text{missing births}= B_f^E - B_f,$$where $B_f^E$ is the expected number of female births, and $B_f$ is the observed number of female births.
In our corrected PNAS paper, we obtain the expected number of female births $B_f^E$ as follows: $$B_f^E = \frac{B_m}{R^E}$$ where $B_m$ is the observed number of male births, and $R^E$ is the expected sex ratio at birth under natural circumstances.
An alternative approach makes use of the total number of observed births (i.e. male plus female births) $B$ instead of the number of observed male births $B_m$ to compute the expected female births: $B_f^{E’} = B/(1+R^E)$. This approach is NOT correct for computing missing female births due to sex selective abortion because it only redistributes the existing births; it does not take account of the net decrease in total birth due to sex selective abortion.
A statistical explanation, based on a toy example
Assume that there are 100 families plan to have 1 baby per family. Let $M_i$ and $F_i$ be the number of male and female births for family $i$ respectively, such that for each family, we have $M_i+F_i=1$.Let $m_i$ be the number of male pregnancies and $f_i$ the number of female pregnancies initiated by family $i$. Let the baseline (i.e. reference level) of the sex ratio at birth be $R^E=1.06$ (i.e., the baseline estimated for China, according to the PNAS paper
1). The probability of conceiving a boy, $\pi_b$, is then$$\pi_b = \frac{R^E}{1+R^E} = \frac{1.06}{1+1.06} \approx 0.515.$$Hence in absense of sex selective abortion, about 51.5 out of 100 families have a boy.
Now assume that a fraction $\theta$ of families aborts any daughters until they get a son. Then for such families the number of aborted females follows a geometric distribution $\mathcal{G} (\pi_b)$ with mean $$\frac{\pi_g}{\pi_b} = \frac{1-\pi_b}{\pi_b} = \frac{1-0.515}{0.515} \approx 0.943,$$ where $\pi_g=1-\pi_b$. Hence, each such family has lost an average of 0.943 female births.
Among the 100 families, the fraction of families that have a boy is: $$Pr(M_i=1)=\pi_b (1-\theta)+\theta = p_b.$$ $p_b$, the actual proportion of male births among all births in a population follows directly from the observed sex ratio at birth. Suppose this observed sex ratio at birth is 1.20, i.e. 120 male births for every 100 female births, then we have $$p_b = \frac{R}{1+R} = \frac{1.2}{1+1.2} \approx 0.545.$$
Given the values of $\pi_b$ and $p_b$, we can solve $\theta$, here $\theta \approx 0.064$. The number of aborted baby girls due to the desire for a baby boy for our example is then equal to the product of $\theta$, the mean number of aborted female births for a family that desires a boy, and the total number of families: $$\theta \cdot \frac{\pi_g}{\pi_b} \cdot 100 \approx 0.064 \times 0.943 \times 100 \approx 6.$$ Hence, 100 births that the 100 families have had are in fact the result of roughly 106 pregnancies, out of which 6 have been aborted.
We end up with the same result by using the direct approach to compute the number of missing female births:$$B_f^E = 54.5 / 1.06 = 51.4 \\
B_f = 100 - 54.5 = 45.5 \\ B_f^* = 51.4 - 45.5 \approx 6.$$ Chao, F., Gerland, P., Cook, A. R., & Alkema, L. (2019). Systematic assessment of the sex ratio at birth for all countries and estimation of national imbalances and regional reference levels. Proceedings of the National Academy of Sciences, 116(19), 9303–9311. ^ |
I've stumbled across the following identity $$ x^{+} = \frac{1}{2 \pi} \int_{-\infty}^\infty e^{(m + i z )x} \frac{1}{(m+i z )^2} d z, $$ supposed to hold for $x\in \mathbb{R}$ and $m >0$. I cannot seem to find a reference nor find out how to prove it. I hope someone will help with either a proof or idea.
If you take the second derivative over x:
$$f''(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{(m+iz)x}{\rm d}z=e^{mx}\delta(x)=\delta(x)$$ I recognized the Fourier transform of the delta function. Now integrate twice. First integral is the Heaviside step function:
$$f'(x)=\int_{-\infty}^x \delta(t){\rm d}t=H(x)$$
Integrate again, and you get $$f(x)=\int_{-\infty}^x H(t){\rm d}t=x^+$$
The integrals are only defined in terms of distributions, but you can regularize the procedure, for instance, by including a dissipative term $e^{-\epsilon|z|}$ and sending $\epsilon\to 0$. |
I regularly teach undergrads about Arrow’s impossibility theorem. In previous years, I’ve simply presented a statement of the theorem and provided a proof in the optional readings. Arrow’s proof is rather complicated; and while there are several simpler presentations of the proof, they are still too complicated for me to cover with philosophy undergraduates.
Preparing for class this year, I realized that, if Arrow’s theorem is
slightly weakened, we can give a proof that is much easier to follow—the kind of proof I’m comfortable presenting to undergraduate philosophy majors. The point of the post today is to present that proof. Stage Setting
Suppose that we have three voters, and they are voting on three options: $A, B,$ and $C$. The first voter prefers $A$ to $B$ to $C$. The second prefers $B$ to $C$ to $A$. The third prefers $C$ to $A$ to $B$. We can represent this with the following table. $$ \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\
1st & A & B & C \\ 2nd & B & C & A \\ 3rd & C & A & B \end{array} $$ This table gives us a voter profile. In general, a voter profile is an indexed set of preference orderings, which I’ll denote with ‘$ \succeq_i$‘. (By the way, I’ll assume that, once we have a weak preference ordering $X \succeq_i Y$—read as “$Y$ is not preferred to $X$”—we can define up a strong preference ordering $X \succ_i Y$—read as “X is preferred to Y”—and an indifference relation $X \sim_i Y$—read as “$X$ and $Y$ are preferred equally”. We can accomplish this with the following stipulative defintions: $X \succ_i Y := X \succeq_i Y \wedge Y \not\succeq_i X$ and $X \sim_i Y := X \succeq_i Y \wedge Y \succeq_i X$.)
A
social welfare function is a function from a voter profile, $\succeq_i$, to a group preference ordering, which I’ll denote with ‘$\succeq$‘.
$ \left [ \begin{array}{c} \succeq_1 \\
\succeq_2 \\ \vdots \\ \succeq_N \\ \end{array} \right ] \,\,\to \,\,\,\,\, \succeq $ allvoter profiles. I’ll also assume that $\succeq$ is a total pre-order—that is, I’ll assume that $\succeq$ is a transitiverelation, and that, for any two options $X$ and $Y$, either $X \succeq Y$ or $Y \succeq X$.)
There are several ways of interpreting a social welfare function. If you think that an individual’s well-being is a function of how well satisfied their preferences are, and you think that how good things are overall is just a question of aggregating the well-being of all the individuals (this thesis is called
welfarism), then you could think of the social welfare function as providing you with a betterness ordering. Alternatively, you could understand the social welfare function as a voting rule which tells you how to select between options, given the preferences of the voters. (For ease of exposition, I’ll run with this second interpretation throughout, though nothing hangs on this choice.)
Here are some features you might want a social welfare function to have: firstly, you don’t want it to privilege any option over any other. It should be the preferences of the voters which determines which option comes out on top and not the way those options happen to be labeled. So, if we were to re-label the options (holding fixed their position in every voter’s preference ordering), the group preference ordering determined by the social welfare function should be exactly the same—except, of course, that the options have now been re-labeled. Call this feature “Neutrality”.
Neutrality Re-labeling options does not affect where options end up in the group preference ordering.
Similarly, we don’t want the social welfare function to privilege any particular
voter over any other. All voters should be treated equally. So, if we were to re-label the voters (holding fixed their preferences), this shouldn’t make any difference with respect to the group preference ordering. Let’s call this feature “Anonymity”. Anonymity Re-labeling voters does not affect the group preference ordering.
Next: if
all voters have exactly the same preference ordering, then this should become the group preference ordering. Let’s call this feature “Unanimity”. Unanimity If all voters share the same preference ordering, then this is the group preference ordering.
And: if the only change to a voter profile is that one person has
raised an option, $X$, in their individual preference ordering, this should not lead to $X$ being lowered in the group preference ordering. Let’s call this feature “Monotonicity”. Monotonicity If one voter raises $X$ in their preference ordering, and nothing else about the voter profile changes, then $X$ is not lowered in the group preference ordering.
Finally, it would be nice if, in order to determine whether $X \succeq Y$, the social welfare function only had to consider each voter’s preferences between $X$ and $Y$. It shouldn’t have to consider where they rank options other than $X$ and $Y$—when it comes to deciding the group preference between $X$ and $Y$, those other options are
irrelevant alternatives. Call this principle, then, the “Independence of Irrelevant Alternatives”, or just “IIA”. Independence of Irrelevant Alternatives (IIA) How the group ranks $X$ and $Y$—i.e., whether $X \succeq Y$ and $Y \succeq X$—is determined entirely by each individual voter’s preferences between $X$ and $Y$. Changes in voters’ preferences which do not affect whether $X \succeq_i Y$ or $Y \succeq_i X$ do not affect whether $X \succeq Y$ or $Y \succeq X$.
What Arrow showed was that there is no social welfare function which satisfies all of these criteria. Actually, Arrow showed something slightly stronger—namely that there’s no social welfare function which satisfies
Unanimity, Monotonicity, and IIA other than a dictatorial social welfare function. A dictatorial social welfare function just takes some voter’s preferences and makes them the group’s preferences, no matter the preferences of the other voters. Any dictatorial social welfare function will violate Anonymity, so our weaker impossibility result follows from Arrow’s. While this result is slightly weaker, Anonymity and Neutrality are still incredibly weak principles, and this result is much easier to prove. The Proof
Here’s the general shape of the proof: we will
assume that there is some social welfare function which satisfies Anonymity, Neutrality, Unanimity, and IIA, and, by reasoning about what this function must say about particular voter profiles, we will show that it must violate Monotonicity. This will show us that there is no voter profile which satisfies all of these criteria.
Let’s begin with the voter profile from above:$$\begin{array}{l c c c}\text{Voter #} & 1 & 2 & 3 \\
1st & A & B & C \\ 2nd & B & C & A \\ 3rd & C & A & B \end{array} $$ Notice that the three options, $A$, $B$, and $C$, are perfectly symmetric in this voter profile. By re-labeling voters, we could have $C$ appear wherever $A$ does, $B$ appear wherever $C$ does, and $A$ appear wherever $B$ does. For instance: re-label voter 1 “voter 2”, re-label voter 2 “voter 3”, and re-label voter 3 “voter 1”, and you get the following voter profile, in which $A$ has taken the place of $B$, $B$ has taken the place of $C$, and $C$ has taken the place of $A$. $$ \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & C & A & B \\ 2nd & A & B & C \\ 3rd & B & C & A \end{array} $$ By Anonymity, this makes no difference with respect to the group ordering. Note also that we may view this new voter profile as the result of re-labeling, not the voters, but rather the options (replacing $A$ with $C$, $B$ with $A$, and $C$ with $B$). Then, by Neutrality, after this re-labeling, $A$ must occupy the place of $B$ in the old group ordering, $B$ must occupy the place of $C$ in the old group ordering, and $C$ must occupy the place of $A$. Since the group ordering must also be unchanged (because of Anonymity), this means that the group ordering must be: $$ A \sim B \sim C $$That is: the group must be indifferent between $A$, $B$, and $C$. (Call this “result #1”) This is exactly what we should expect, given the symmetry of the voter profile. There’s nothing that any option has to raise it above the others.
Now, suppose that, in our original voter profile, voters 1 and 3 change their minds, and they raise $B$ above $A$ in their preference ordering. And suppose that voter 2 raises $C$ above $B$ in their preference ordering. Then, the voter profile would change as shown:$$\begin{array}{l c c c}\text{Voter #} & 1 & 2 & 3 \\
1st & A & B & C \\ 2nd & B & C & A \\ 3rd & C & A & B \end{array} \qquad \Longrightarrow \qquad \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & B & C & C \\ 2nd & A & B & B \\ 3rd & C & A & A \end{array} $$ Notice first that these changes didn’t affect any voter’s ranking between $A$ and $C$. Voter 1 prefers $A$ to $C$ both before and after the changes. And voters 2 and 3 prefer $C$ to $A$ both before and after the changes. Since $A \sim C$ before the changes (by result #1), IIA tells us that, after the changes, it is still the case that $A \sim C$. (Call this “result #2”.)
Notice also that
everybody now ranks $B$ above $A$. So, from this voter profile, we could reach a unanimous voter profile in which everybody ranks $B$ above $A$ above $C$, by just having voters 2 and 3 lower $C$ to the bottom of their preference ranking.$$\begin{array}{l c c c}\text{Voter #} & 1 & 2 & 3 \\ 1st & B & C & C \\ 2nd & A & B & B \\ 3rd & C & A & A \end{array} \qquad \Longrightarrow \qquad \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & B & B & B \\ 2nd & A & A & A \\ 3rd & C & C & C \end{array} $$ By Unanimity, in the voter profile on the right, $B \succ A$. But, in moving from the voter profile on the left to the one on the right, we didn’t change anybody’s ranking of $A$ and $B$, so, by IIA, $B \succ A$ in the voter profile on the left, too. (Call this “result #3”)
Putting together result #2 and result #3, we have that, in this voter profile,$$\begin{array}{l c c c}\text{Voter #} & 1 & 2 & 3 \\
1st & B & C & C \\ 2nd & A & B & B \\ 3rd & C & A & A \end{array} $$ $B \succ A \sim C$. Therefore, in this voter profile, $B \succ C$. Call this “result #4”.
Suppose that we begin with the voter profile immediately above, and voters 1 and 3 change their minds, raising $A$ above $B$, and leaving everything else unchanged. This gives us the voter profile on the right.$$\begin{array}{l c c c}\text{Voter #} & 1 & 2 & 3 \\
1st & B & C & C \\ 2nd & A & B & B \\ 3rd & C & A & A \end{array} \qquad \Longrightarrow \qquad \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & A & C & C \\ 2nd & B & B & A \\ 3rd & C & A & B \end{array} $$ This change does not affect anyone’s ranking of $B$ and $C$. Voter 1 prefers $B$ to $C$ both before and after the change. And voters 2 and 3 prefer $C$ to $B$ both before and after the change. Since $B \succ C$, given the voter profile on the left (this was result #4), we must have $B \succ C$ on the right, too. Call this “result #5”.
Now, watch this: consider
these two voter profiles. $$ \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & A & B & C \\ 2nd & B & C & A \\ 3rd & C & A & B \end{array} \qquad \Longrightarrow \qquad \begin{array}{l c c c} \text{Voter #} & 1 & 2 & 3 \\ 1st & A & C & C \\ 2nd & B & B & A \\ 3rd & C & A & B \end{array} $$ The voter profile on the left is just our original voter profile. On the right is the voter profile from the right-hand-side of the paragraph immediately above. Result #1 tells us that, on the left, $B \sim C$. Result #5 tell us that, on the right, $B \succ C$. But notice that the only difference between the voter profile on the left and the one on the right is that voter 2 has raised $C$ in their preference ordering. Monotonicity tells us that this shouldn’t lower $C$ in the group preference ordering. So result #1 and result #5 together contradict Monotonicity.
So: any social welfare function which satisfies
Anonymity, Neutrality, Unanimity, and IIA will end up violating Monotonicity. So there is no social welfare function which satisfies all of these criteria. |
555851 results
Contributors: Ganian, Robert, Kalany, Martin, Szeider, Stefan, Träff, Jesper Larsson
Date: 2015-06-30
... We show that the problem of constructing tree-structured descriptions of data layouts that are optimal with respect to space or other criteria from given sequences of displacements, can be solved in polynomial time. The problem is relevant for efficient compiler and library support for communication of noncontiguous data, where tree-structured descriptions with low-degree nodes and small index arrays are beneficial for the communication soft- and hardware. An important example is the Message-Passing Interface (MPI) which has a mechanism for describing arbitrary data layouts as trees using a set of increasingly general constructors. Our algorithm shows that the so-called MPI datatype reconstruction problem by trees with the full set of MPI constructors can be solved optimally in polynomial time, refuting previous conjectures that the problem is NP-hard. Our algorithm can handle further, natural constructors, currently not found in MPI. Our algorithm is based on dynamic programming, and requires the solution of a series of shortest path problems on an incrementally built, directed, acyclic graph. The algorithm runs in $O(n^4)$ time steps and requires $O(n^2)$ space for input displacement sequences of length $n$.
Files:
Contributors: Shafiee, Mohammad Javad, Wong, Alexander, Fieguth, Paul
Date: 2015-06-30
... Random fields have remained a topic of great interest over past decades for the purpose of structured inference, especially for problems such as image segmentation. The local nodal interactions commonly used in such models often suffer the short-boundary bias problem, which are tackled primarily through the incorporation of long-range nodal interactions. However, the issue of computational tractability becomes a significant issue when incorporating such long-range nodal interactions, particularly when a large number of long-range nodal interactions (e.g., fully-connected random fields) are modeled. In this work, we introduce a generalized random field framework based around the concept of stochastic cliques, which addresses the issue of computational tractability when using fully-connected random fields by stochastically forming a sparse representation of the random field. The proposed framework allows for efficient structured inference using fully-connected random fields without any restrictions on the potential functions that can be utilized. Several realizations of the proposed framework using graph cuts are presented and evaluated, and experimental results demonstrate that the proposed framework can provide competitive performance for the purpose of image segmentation when compared to existing fully-connected and principled deep random field frameworks.
Files:
Contributors: Refsgaard, J., Kirsebom, O. S., Dijck, E. A., Fynbo, H. O. U., Lund, M. V., Portela, M. N., Raabe, R., Randisi, G., Renzi, F., Sambi, S.
Date: 2015-06-30
... While the 12C(a,g)16O reaction plays a central role in nuclear astrophysics, the cross section at energies relevant to hydrostatic helium burning is too small to be directly measured in the laboratory. The beta-delayed alpha spectrum of 16N can be used to constrain the extrapolation of the E1 component of the S-factor; however, with this approach the resulting S-factor becomes strongly correlated with the assumed beta-alpha branching ratio. We have remeasured the beta-alpha branching ratio by implanting 16N ions in a segmented Si detector and counting the number of beta-alpha decays relative to the number of implantations. Our result, 1.49(5)e-5, represents a 24% increase compared to the accepted value and implies an increase of 14% in the extrapolated S-factor.
Files:
Contributors: Schrade, Constantin, Zyuzin, A. A., Klinovaja, Jelena, Loss, Daniel
Date: 2015-06-30
... We study two microscopic models of topological insulators in contact with an $s$-wave superconductor. In the first model the superconductor and the topological insulator are tunnel coupled via a layer of scalar and of randomly oriented spin impurities. Here, we require that spin-flip tunneling dominates over spin-conserving one. In the second model the tunnel coupling is realized by an array of single-level quantum dots with randomly oriented spins. It is shown that the tunnel region forms a $\pi$-junction where the effective order parameter changes sign. Interestingly, due to the random spin orientation the effective descriptions of both models exhibit time-reversal symmetry. We then discuss how the proposed $\pi$-junctions support topological superconductivity without magnetic fields and can be used to generate and manipulate Kramers pairs of Majorana fermions by gates.
Files:
Contributors: Tsui, K. H.
Date: 2015-06-30
... Following the basic principles of a charge separated pulsar magnetosphere \citep{goldreich1969}, we consider the magnetosphere be stationary in space, instead of corotating, and the electric field be uploaded from the potential distribution on the pulsar surface, set up by the unipolar induction. Consequently, the plasma of the magnetosphere undergoes guiding center drifts of the gyro motion due to the transverse forces to the magnetic field. These forces are the electric force, magnetic gradient force, and field line curvature force. Since these plasma velocities are of drift nature, there is no need to introduce an emf along the field lines, which would contradict the $E_{\parallel}=\vec E\cdot\vec B=0$ plasma condition. Furthermore, there is also no need to introduce the critical field line separating the electron and ion open field lines. We present a self-consistent description where the magnetosphere is described in terms of electric and magnetic fields and also in terms of plasma velocities. The fields and velocities are then connected through the space charge densities self-consistently. We solve the pulsar equation analytically for the fields and construct the standard steady state pulsar magnetosphere. By considering the unipolar induction inside the pulsar and the magnetosphere outside the pulsar as one coupled system, and under the condition that the unipolar pumping rate exceeds the Poynting flux in the open field lines, plasma pressure can build up in the magnetosphere, in particular in the closed region. This could cause a periodic openning up of the closed region, leading to a pulsating magnetosphere, which could be an alternative for pulsar beacons. The closed region can also be openned periodically by the build-up of toroidal magnetic field through a positive feedback cycle.
Files:
Contributors: Moolekamp, Fred, Mamajek, Eric
Date: 2015-06-30
... As the size of images and data products derived from astronomical data continues to increase, new tools are needed to visualize and interact with that data in a meaningful way. Motivated by our own astronomical images taken with the Dark Energy Camera (DECam) we present Toyz, an open source Python package for viewing and analyzing images and data stored on a remote server or cluster. Users connect to the Toyz web application via a web browser, making it an convenient tool for students to visualize and interact with astronomical data without having to install any software on their local machines. In addition it provides researchers with an easy-to-use tool that allows them to browse the files on a server and quickly view very large images ($>$ 2 Gb) taken with DECam and other cameras with a large FOV and create their own visualization tools that can be added on as extensions to the default Toyz framework.
Files:
Contributors: Berezhiani, Zurab
Date: 2015-06-30
... The excess of high energy neutrinos observed by the IceCube collaboration might originate from baryon number violating decays of heavy shadow baryons from dark mirror sector which produce shadow neutrinos. These sterile neutrino species then oscillate into ordinary neutrinos transferring to them specific features of their spectrum. In particular, this scenario can explain the end of the spectrum above 2 PeV and the presence of the energy gap between 400 TeV and 1 PeV.
Files:
Contributors: Carayol, Arnaud, Löding, Christof, Serre, Olivier
Date: 2015-06-30
... We consider imperfect information stochastic games where we require the players to use pure (i.e. non randomised) strategies. We consider reachability, safety, B\"uchi and co-B\"uchi objectives, and investigate the existence of almost-sure/positively winning strategies for the first player when the second player is perfectly informed or more informed than the first player. We obtain decidability results for positive reachability and almost-sure B\"uchi with optimal algorithms to decide existence of a pure winning strategy and to compute one if exists. We complete the picture by showing that positive safety is undecidable when restricting to pure strategies even if the second player is perfectly informed.
Files:
Contributors: Simpson, Gideon, Watkins, Daniel
Date: 2015-06-30
... One way of getting insight into non-Gaussian measures, posed on infinite dimensional Hilbert spaces, is to first obtain good approximations in terms of Gaussians. These best fit Gaussians then provide notions of mean and variance, and they can be used to accelerate sampling algorithms. This begs the question of how one should measure optimality. Here, we consider the problem of minimizing the distance between a family of Gaussians and the target measure, with respect to relative entropy, or Kullback-Leibler divergence, as has been done previously in the literature. Thus, it is desirable to have algorithms, well posed in the abstract Hilbert space setting, which converge to these minimizers. We examine this minimization problem by seeking roots of the first variation of relative entropy, taken with respect to the mean of the Gaussian, leaving the covariance fixed. We prove the convergence of Robbins-Monro type root finding algorithms, highlighting the assumptions necessary for them to converge to relative entropy minimizers.
Files:
Contributors: Al-Safadi, Ebrahim B., Al-Naffouri, Tareq Y., Masood, Mudassir, Ali, Anum
Date: 2015-06-30
... A novel method for correcting the effect of nonlinear distortion in orthogonal frequency division multiplexing signals is proposed. The method depends on adaptively selecting the distortion over a subset of the data carriers, and then using tools from compressed sensing and sparse Bayesian recovery to estimate the distortion over the other carriers. Central to this method is the fact that carriers (or tones) are decoded with different levels of confidence, depending on a coupled function of the magnitude and phase of the distortion over each carrier, in addition to the respective channel strength. Moreover, as no pilots are required by this method, a significant improvement in terms of achievable rate can be achieved relative to previous work.
Files: |
The $\lambda$ you defined is clearly well-defined (as long as $I < \infty$). The difficulty is to prove that you can build a measure that extend your $\lambda$. The Lebesgue measure (on the real line) commonly means a (the) Borel measure that coincides with the length on intervals.
If your goal is to prove that the Lebesgue measure exists and has nice properties a fast way to do it is to consider the sequence of probability measures on $[0,1]$, $$\mu_{n} = \sum_{i=1}^{n} \frac{1}{n} \delta_{\frac{i}{n}}$$ and show (for instance with the Prokhorov theorem) that $\mu_{n}$ converges to a limit probability measure that you call $\lambda_{[0,1]}$, the Lebesgue measure on $[0,1]$.You can repeat this procedure on any interval $[n,n+1]$ ($n \in \mathbb{Z}$) and you obtain the Lebesgue measure on $\mathbb{R}$ by gluing them all together: $$\lambda = \sum_{n \in \mathbb{Z}} \lambda_{[n,n+1]}.$$You immediately obtain that for a Riemann integrable function $f$, $$\int f(x) dx = \int f d \lambda$$ which implies the property that $\lambda([a, b]) = b - a$.
If you want to use an approach similar to the one in you initial question then you introduces an outer measure $$\lambda^{*}(A) = \inf \{ \sum_{i \in \mathbb{N}} (b_{i} - a_{i}) \, : \, A \subset \bigcup_{i\in \mathbb{N}} ]a_{i}, b_{i}[ \},$$ where the infimum is over all countable covering of $A$ by open intervals $]a_{i}, b_{i}[$.Then you show that the space of $\lambda^{*}$-measurable sets contains the Borel $\sigma$-field. By a general theorem the restriction of $\lambda^{*}$ to its measurable sets is a measure and therefore the restriction to the Borel sets (that you call $\lambda$) is also a measure. Then you compute by hand that $$\lambda^{*}(]a,b[) = \lambda^{*}([a,b]) = b -a.$$
From my point of view, the second approach is heavier. You should note that other approaches exist such as the one in Rudin using Riesz duality theorem. |
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If R\mathbb{R}R denotes the set of real numbers, and SSS is the set
S={7,1+2,π,−11,−11,0,(−6)3,−23}, S = \{ 7, 1+\sqrt{2}, \pi, \sqrt{-11}, -11, 0, (-6)^3, -\sqrt[3]{2} \} , S={7,1+2,π,−11,−11,0,(−6)3,−32},
how many elements does R∩S\mathbb{R} \cap S R∩S have?
Details and assumptions
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If Q\mathbb{Q}Q denotes the set of rational numbers, and SSS is the set
S={9,1+2,π,519,−14,−9,0,(−6)3,−23}, S = \{ 9, 1+\sqrt{2}, \pi, \frac{5}{19}, \sqrt{-14}, -9, 0, (-6)^3, -\sqrt[3]{2} \} , S={9,1+2,π,195,−14,−9,0,(−6)3,−32},
how many elements does Q∩S\mathbb{Q} \cap S Q∩S have?
How many positive integers are there in the set
{x∣−14≤x<66}? \{ x \mid -14 \le x < 66\}? {x∣−14≤x<66}?
How many negative integers are there in the set
{x∣−15≤x<37}? \{ x \mid -15 \le x < 37\}? {x∣−15≤x<37}?
How many elements are in the set {10,12,14,…,48,50}\{10, 12, 14, \ldots, 48, 50\}{10,12,14,…,48,50}?
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In Wikipedia, it says that any epsilon number with the index that is countable is countable. How is it? Out of all those numbers, I especially want to know why $\epsilon_0$ is countable.
Thanks.
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It is often understood from context which of the two is used, but it can still be quite confusing to people new to this.
Cardinal exponentiation $\omega^\omega$ means we take the cardinality of all the functions from $\omega$ to $\omega$, this is of course of size continuum, which is of an uncountable cardinality.
On the other hand, $\omega^\omega$ in ordinal exponentiation means that we take some order type which is a supremum of countable ordinals. What are
those ordinals? These are ordinals which defined themselves as limit of smaller ordinals are $\omega^n$, and so we can continue to unfold the definitions of ordinal arithmetics until we have that $\omega^\omega$ is a supremum of some much "simpler" set (It is actually much more complicated that just $\omega^n$, though)
This "simpler" set contains only countable ordinals, and itself is countable. We know that the countable union of countable sets is countable, therefore $\omega^\omega$ is countable.
What does all that have to do with $\epsilon_0$? Well, by induction we have that $\omega,\omega^\omega,\omega^{\omega^\omega},\ldots$ are all countable, and there are only countably many of those. From this we have that $\epsilon_0$ is also countable. It is a large countable ordinal. Now we can continue by induction, what is $\epsilon_1$? We repeat the same process only we start from $\epsilon_0+1$ instead of $\omega$.
This is again a countable process, so it must end at a countable ordinal as before; by induction we can see that:
What happens with $\epsilon_{\omega_1}$? Well, for every $\alpha<\omega_1$ we have that $\epsilon_\alpha$ is countable, and if $\alpha<\beta$ then $\epsilon_\alpha<\epsilon_\beta$. Therefore we have $\aleph_1$ many distinct ordinals below $\epsilon_{\omega_1}$, therefore it is not a countable ordinal anymore. In fact $\epsilon_{\omega_1}$ is the limit of all those $\epsilon_\alpha$ for countable $\alpha$, and one can see that the the supremum of uncountably many countable ordinals cannot be other than $\omega_1$. (If $\delta>\omega_1$ is cannot be the supremum of a set of
countable ordinals!)
Let us repeat the definitions of ordinal arithmetics:
Some more reading material:
Wikipedia gives the answer, in fact: it's the union of a countable set of countable ordinals $\left\{\omega,\omega^\omega,\omega^{\omega^\omega},\ldots\right\}$
As written on Wikipedia, it's because it's a countable union of countable sets. Let $u_0 = 1$ and $u_{n+1} = \omega^{u_n}$. It is easily shown by induction that $u_n$ is countable for all $n$:
Finally, you have (by definition) $\epsilon_0 = \bigcup_{n < \omega} u_n$, a countable union of countable sets.
Vsauce introduces really well the subject. https://www.youtube.com/watch?v=SrU9YDoXE88 |
I want to find the distance between two points in spherical coordinates, so I want to express $||x-x'||$ where $x=(r,\theta, \phi)$ and $x' = (r', \theta',\phi')$ by the respective components. Is this possible? I just know that this is $\sqrt{r^2+r'^2-2rr'\cos(\theta- \theta')}$ if $\phi,\phi'$ is the same, but what is the most general distance?
The expression of the distance between two vectors in spherical coordinates provided in the other response is usually expressed in a more compact form that is not only easier to remember but is also ideal for capitalizing on certain symmetries when solving problems.
$$\begin{align} \|\mathbf{r}-\mathbf{r}^\prime\| &=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\cos(\phi)\cos(\phi')}+\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\sin(\phi)\sin(\phi')}+\cos(\theta)\cos(\theta')\right]}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\left(\cos(\phi)\cos(\phi')+\sin(\phi)\sin(\phi')\right)}+\cos(\theta)\cos(\theta')\right]}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\cos(\phi-\phi')}+\cos(\theta)\cos(\theta')\right]}.\\ \end{align}$$
This form makes it fairly transparent how azimuthal symmetry allows you to automatically eliminate some of the angular dependencies in certain integration problems. Another advantage of this form is that you now have at least two variables, namely $\phi$ and $\phi'$, that appear in the equation only once, which can make finding series expansions w.r.t. these variables a little less of a pain than the others.
You have simply to write it in Cartesian coordinates and change variables: $x=r\sin(\theta)\cos(\phi)$, $y=r\sin(\theta)\sin(\phi)$, $z=r\cos(\theta)$ $$\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=$$$$=\sqrt{r^2+r'^2-2rr'\left[\sin(\theta)\sin(\theta')\cos(\phi)\cos(\phi')+\sin(\theta)\sin(\theta')\sin(\phi)\sin(\phi')+\cos(\theta)\cos(\theta')\right]}$$ But I don't see a way to really improve this mess.
Building on the answer from @[David H], I wrote the distance in a way that highlights the difference in angles: $$ ||\vec r_1 - \vec r_2|| = \sqrt{ {r_1}^2 + {r_2}^2 - 2\, r_1 r_2 \cos(\theta_1 - \theta_2) - 2\, r_1 r_2 \sin \theta_1 \sin \theta_2 \left( \cos(\phi_1 - \phi_2) - 1 \right) } $$ It highlights the contributions from the difference in polar angle $\theta$ and the difference in the azimuthal angle $\phi$, (third and fourth terms, respectively, under the square root symbol). Note the intuitive scaling of the azimuthal contribution by $\sin \theta_1 \sin \theta_2$. Of course, when the angular differences are both $0$, the distance reduces to $||r_1-r_2||$. |
$\vec{B}=\nabla \times \vec{A}\tag1$
This is true because at
every point $\nabla\cdot\vec{B}=0 \tag2$
In
free space points,
$\displaystyle \vec{B}=\dfrac{\mu_0}{4 \pi}\int_C \dfrac{I\ dl \times\hat{r}}{r^2}\tag3$
Consequently: $\nabla \cdot\vec{B}=0 \tag2$
At the
points on the circuit, there is a singularity and we cannot directly apply equation $(3)$, i.e. Biot Savart law.
So in this case how can $\nabla \cdot\vec{B}=0 $
Edit (My understanding)@ garyp:
$\text{I am a graduate student. So I may seem to be a bit naive. Anyway please tell whether}\\ \text{I am understanding in the right way.}$
Here I am not considering the circuit as three dimensional.
By considering it as one dimensional, the (closed) circuit becomes equivalent to a magnetic dipole layer of infinitesimal thickness.
By using inverse square law of magnetic poles, we can find magnetic field
(intensity)at any point outside the magnetic dipole layer (even at points infinitely close to the dipole layer). Let's first see the magnetic field due to an element of dipole layerat a point infinitely close to it:
$$\vec{B}=\mu_0\vec{H}= k \ M\ \left[ \dfrac{\hat{r_1}}{r_1^2}-\dfrac{\hat{r_2}}{r_2^2} \right] dS'$$
(where $M$ is magnetic pole density and $S'$ is the surface of magnetic dipole layer)
Now making use of divergence formula in spherical coordinates:
\begin{align} \nabla \cdot \vec{B} &= k \ M\ \left[ \nabla \cdot \dfrac{\hat{r_1}}{r_1^2}-\nabla \cdot \dfrac{\hat{r_2}}{r_2^2} \right]\ dS' \\ & = \lim_{r\to\ 0} k \ M\ \left \{ \left[ \dfrac{1}{r^2} \dfrac{\partial \left( r^2 \dfrac{\hat{r}}{r^2} \right)}{\partial r} \right]_{\text{at }r_1} -\left[ \dfrac{1}{r^2} \dfrac{\partial \left( r^2 \dfrac{\hat{r}}{r^2} \right)}{\partial r} \right]_{\text{at }r_2} \right \} dS' \\ \text{We see the two $r^2$ cancel out} \\ & = \lim_{r\to\ 0} k \ M\ \left[ \left[ \dfrac{1}{r^2} \dfrac{\partial \hat{r}}{\partial r} \right]_{\text{at } r_1} - \left[ \dfrac{1}{r^2} \dfrac{\partial \hat{r}}{\partial r} \right]_{\text{at } r_2} \right] dS'\\ & =0\\ \text{(This is because $\dfrac{\partial \hat{r}}{\partial r}=0$)} \end{align}
Therefore the divergence (at points outside the dipole layer) due to
each of the elements of dipole layeris zero. That is, the divergence (at points outside the dipole layer) due to the magnetic dipole layer is zero.
Thus we see that the magnetic field may blow up at points infinitely close to the dipole layer but its divergence will still be zero.
Thus the divergence of magnetic field due to (closed) circuit is zero everywhere except at points on the (closed) circuit.
Now comes the key point:Since we know the divergence of magnetic field due to a (closed) circuit, even at points infinitely close to the circuit, is zero we ignore the circuit and say $\nabla \cdot \vec{B}=0$ everywhere on $\mathbb R^3$. |
Orthogonal Vectors
Two vectors,
xand y, are orthogonalif their dot product is zero.
For example
\[ e \cdot f = \begin{pmatrix} 2 & 5 & 4 \end{pmatrix} * \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 22\]
Vectors
e and f are not orthogonal.
\[ g \cdot h = \begin{pmatrix} 2 & 3 & -2 \end{pmatrix} * \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0\]
However, vectors
g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let \(x_1, x_2, \ldots , x_n\) be m-dimensional vectors. Then a linear combination of \(x_1, x_2, \ldots , x_n\) is any m-dimensional vector that can be expressed as
\[ c_1 x_1 + c_2 x_2 + \ldots + c_n x_n \]
where \(c_1, \ldots, c_n\) are all scalars. For example:
\[x_1 =\begin{pmatrix}
3 \\ 8 \\ -2 \end{pmatrix}, x_2 =\begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix}\] \[y =\begin{pmatrix} -5 \\ 12 \\ -8 \end{pmatrix} = 1*\begin{pmatrix} 3 \\ 8 \\ -2 \end{pmatrix} + (-2)* \begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix} = 1*x_1 + (-2)*x_2\]
So
y is a linear combination of \(x_1\) and \(x_2\). The set of all linear combinations of \(x_1, x_2, \ldots , x_n\) is called the span of \(x_1, x_2, \ldots , x_n\). In other words
\[ span(\{x_1, x_2, \ldots , x_n \} ) = \{ v| v= \sum_{i = 1}^{n} c_i x_i , c_i \in \mathbb{R} \} \]
A set of vectors \(x_1, x_2, \ldots , x_n\) is
linearly independent if none of the vectors in the set can be expressed as a linear combination of the other vectors. Another way to think of this is a set of vectors \(x_1, x_2, \ldots , x_n\) are linearly independent if the only solution to the below equation is to have \(c_1 = c_2 = \ldots = c_n = 0\), where \(c_1 , c_2 , \ldots , c_n \) are scalars, and \(0\) is the zero vector (the vector where every entry is 0).
\[ c_1 x_1 + c_2 x_2 + \ldots + c_n x_n = 0 \]
If a set of vectors is not linearly independent, then they are called
linearly dependent. Example M.5.1
\[ x_1 =\begin{pmatrix} 3 \\ 4 \\ -2 \end{pmatrix}, x_2 =\begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}, x_3 =\begin{pmatrix} 6 \\ 8 \\ -2 \end{pmatrix} \]
Does there exist a vector
c, such that,
\[ c_1 x_1 + c_2 x_2 + c_3 x_3 = 0 \]
To answer the question above, let:
\begin{align} 3c_1 + 4c_2 +6c_3 &= 0,\\ 4c_1 -2c_2 + 8c_3 &= 0,\\ -2c_1 + 2c_2 -2c_3 &= 0 \end{align}
Solving the above system of equations shows that the only possible solution is \(c_1 = c_2 = c_3 = 0\). Thus \(\{ x_1 , x_2 , x_3 \}\) is linearly independent. One way to solve the system of equations is shown below. First, subtract (4/3) times the 1st equation from the 2nd equation.
\[-\frac{4}{3}(3c_1 + 4c_2 +6c_3) + (4c_1 -2c_2 + 8c_3) = -\frac{22}{3}c_2 = -\frac{4}{3}0 + 0 = 0 \Rightarrow c_2 = 0 \]
Then add the 1st and 3 times the 3rd equations together, and substitute in \(c_2 = 0\).
\[ (3c_1 + 4c_2 +6c_3) + 3*(-2c_1 + 2c_2 -2c_3) = -3c_1 + 10 c_2 = -3c_1 + 10*0 = 0 + 3*0 = 0 \Rightarrow c_1 = 0 \]
Now, substituting both \(c_1 = 0\) and \(c_2 = 0\) into equation 2 gives.
\[ 4c_1 -2c_2 + 8c_3 = 4*0 -2*0 + 8c_3 = 0 \Rightarrow c_3 = 0 \]
So \(c_1 = c_2 = c_3 = 0\), and \(\{ x_1 , x_2 , x_3 \}\) are linearly independent.
Example M.5.2
\[ x_1 =\begin{pmatrix} 1 \\ -8 \\ 8 \end{pmatrix}, x_2 =\begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}, x_3 =\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \]
In this case \(\{ x_1 , x_2 , x_3 \}\)are linearly dependent, because if \(c = (-1, 1, -2)\), then
\[c^T X = \begin{pmatrix}
-1 \\ 1\\ -2 \end{pmatrix} \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} = -1 \begin{pmatrix} 1 \\ -8\\ 8 \end{pmatrix}+ 1 \begin{pmatrix} 4 \\ -2\\ 2 \end{pmatrix} - 2 \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -1*1 +1*4-2*1 \\ -1*-8+1*-2-2*3 \\ -1*8+1*2-2*-2 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] Norm of a vector or matrix
The
normof a vector or matrix is a measure of the "length" of said vector or matrix. For a vector x, the most common norm is the \(\mathbf{L_2}\) norm, or Euclidean norm. It is defined as
\[ \| x \| = \| x \|_2 = \sqrt{ \sum_{i=1}^{n} x_i^2 } \]
Other common vector norms include the \(\mathbf{L_1}\)
norm, also called the Manhattan normand Taxicab norm.
\[ \| x \|_1 = \sum_{i=1}^{n} |x_i| \]
Other common vector norms include the
Maximum norm, also called the Infinity norm.
\[ \| x \|_\infty = max( |x_1| ,|x_2|, \ldots ,|x_n|) \]
The most commonly used matrix norm is the
Frobenius norm. For a m× nmatrix A, the Frobenius norm is defined as:
\[ \| A \| = \| A \|_F = \sqrt{ \sum_{i=1}^{m} \sum_{j=1}^{n} x_{i,j}^2 } \]
Quadratic Form of a Vector
The
quadratic formof the vector xassociated with matrix Ais
\[ x^T A x = \sum_{i = 1}^{m} \sum_{j=1}^{n} a_{i,j} x_i x_j \]
A matrix
Ais Positive Definiteif for any non-zero vector x, the quadratic form of xand Ais strictly positive. In other words, \(x^T A x > 0\) for all nonzero x.
A matrix
Ais Positive Semi-Definiteor Non-negative Definiteif for any non-zero vector x, the quadratic form of xand Ais non-negative . In other words, \(x^T A x \geq 0\) for all non-zero x. Similarly,
A matrix
Ais Negative Definiteif for any non-zero vector x, \(x^T A x < 0\). A matrix Ais Negative Semi-Definiteor Non-positive Definiteif for any non-zero vector x, \(x^T A x \leq 0\). |
Dua, Arti and Cherayil, Binny J (2003)
Polymer dynamics in linear mixed flow. In: Journal of Chemical Physics, 119 (11). pp. 5696-5700.
PDF
Polymer_dynamics.pdf
Download (170kB)
Abstract
Recent simulations by Chu et al. [Phys. Rev. E 66, 011915 (2002)] on the behavior of bead–spring and bead–rod models of polymers in linear mixed flows (flows with unequal amounts of extension and rotation) are compared with the predictions of a finitely extensible Rouse model that was used earlier [J. Chem. Phys. 112, 8707 (2000)] to describe the behavior of long flexible molecules of \lambda-phage DNA in simple shear. The model is a generalization of the continuum Rouse model in which the "spring constant" of the bonds connecting near neighbor segments is allowed to become nonlinearly flow-dependent through a term involving the initially unknown mean square size of the chain, [R2]. A self-consistent equation for this quantity is derived by using the flow-modified Hamiltonian to calculate it from its statistical mechanical definition. After solving this equation numerically, the mean fractional extension of the chain x can be obtained as a function of the Weissenberg number Wi and a mixing parameter \alpha. The results compare favorably with data from the simulations of Chu et al., and suggest the existence of a scaling variable Wieff = \sqrt{\alpha} Wi in terms of which separate curves of x versus Wi fall more or less on a single universal curve.
Item Type: Journal Article Additional Information: Copyright for this article belongs to American Institute of Physics (AIP). Department/Centre: Division of Chemical Sciences > Inorganic & Physical Chemistry Depositing User: L.Kaini Mahemei Date Deposited: 13 Jan 2005 Last Modified: 19 Sep 2010 04:17 URI: http://eprints.iisc.ac.in/id/eprint/2640 Actions (login required)
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I am trying to bring together the definition of the Generalized Gamma distribution in R-package GAMLSS by Rigby and Stasinopoulos and the general definition of the Generalized Gamma distribution, which I mainly found during this research.
The definition of the GAMLSS of the pdf is: $$ f_{Y}(y \mid \mu, \sigma, \nu) = \frac{|\nu| \theta^{\theta} z^{\theta} \exp(-\theta z)}{y \cdot \Gamma(\theta)}, \quad \text{ for } y>0, \mu>0, \sigma>0, \text{-Inf}<\nu<\text{Inf} $$ with $\theta = \frac{1}{\sigma^{2}|\nu|^{2}}$ and $z = \left(\frac{y}{\mu} \right)^{\nu}$. This definition can be found in the according help file and as well in one of their books on GAMLSS: http://www.gamlss.com/wp-content/uploads/2013/01/book-2008-27-6-08.pdf
One of the mainly used definitions of the pdf of the Generalized Gamma distribution is: $$ f(y) = \frac{|a|}{\beta^{ap} \Gamma(p)} \cdot y^{ap-1} \exp(-(y/\beta)^{a}) \quad \text{for } y>0. $$
I would like to understand how the reparametrization works. I tried to rearrange the GAMLSS formula and I come to the conclusion, that $\nu = a$ and $\frac{1}{\sigma^{2}|\nu|^{2}} = p$ but with this definition there seems to be no chance to get a proper definition of $\beta$, so I must be missing something.
The only reference they give is Lopatatzidis and Green (2000) which was an unpublished paper and never seemed to be published until today.
Please, can anybody help me out here? |
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Revision as of 14:42, 16 March 2018 Contents 1 Mathematics Colloquium 1.1 Spring 2018 1.2 Spring Abstracts 1.2.1 January 29 Li Chao (Columbia) 1.2.2 February 2 Thomas Fai (Harvard) 1.2.3 February 5 Alex Lubotzky (Hebrew University) 1.2.4 February 6 Alex Lubotzky (Hebrew University) 1.2.5 February 9 Wes Pegden (CMU) 1.2.6 March 2 Aaron Bertram (Utah) 1.2.7 March 16 Anne Gelb (Dartmouth) 1.2.8 April 6 Edray Goins (Purdue) 1.3 Past Colloquia Mathematics Colloquium
All colloquia are on Fridays at 4:00 pm in Van Vleck B239,
unless otherwise indicated. Spring 2018
date speaker title host(s) January 29 (Monday) Li Chao (Columbia) Elliptic curves and Goldfeld's conjecture Jordan Ellenberg February 2 (Room: 911) Thomas Fai (Harvard) The Lubricated Immersed Boundary Method Spagnolie, Smith February 5 (Monday, Room: 911) Alex Lubotzky (Hebrew University) High dimensional expanders: From Ramanujan graphs to Ramanujan complexes Ellenberg, Gurevitch February 6 (Tuesday 2 pm, Room 911) Alex Lubotzky (Hebrew University) Groups' approximation, stability and high dimensional expanders Ellenberg, Gurevitch February 9 Wes Pegden (CMU) The fractal nature of the Abelian Sandpile Roch March 2 Aaron Bertram (University of Utah) Stability in Algebraic Geometry Caldararu March 16 (Room: 911) Anne Gelb (Dartmouth) Reducing the effects of bad data measurements using variance based weighted joint sparsity WIMAW April 4 (Wednesday) John Baez (UC Riverside) TBA Craciun April 6 Edray Goins (Purdue) Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups Melanie April 13 Jill Pipher (Brown) TBA WIMAW April 16 (Monday) Christine Berkesch Zamaere (University of Minnesota) TBA Erman, Sam April 25 (Wednesday) Hitoshi Ishii (Waseda University) Wasow lecture TBA Tran May 4 Henry Cohn (Microsoft Research and MIT) TBA Ellenberg date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty Spring Abstracts January 29 Li Chao (Columbia)
Title: Elliptic curves and Goldfeld's conjecture
Abstract: An elliptic curve is a plane curve defined by a cubic equation. Determining whether such an equation has infinitely many rational solutions has been a central problem in number theory for centuries, which lead to the celebrated conjecture of Birch and Swinnerton-Dyer. Within a family of elliptic curves (such as the Mordell curve family y^2=x^3-d), a conjecture of Goldfeld further predicts that there should be infinitely many rational solutions exactly half of the time. We will start with a history of this problem, discuss our recent work (with D. Kriz) towards Goldfeld's conjecture and illustrate the key ideas and ingredients behind these new progresses.
February 2 Thomas Fai (Harvard)
Title: The Lubricated Immersed Boundary Method
Abstract: Many real-world examples of fluid-structure interaction, including the transit of red blood cells through the narrow slits in the spleen, involve the near-contact of elastic structures separated by thin layers of fluid. The separation of length scales between these fine lubrication layers and the larger elastic objects poses significant computational challenges. Motivated by the challenge of resolving such multiscale problems, we introduce an immersed boundary method that uses elements of lubrication theory to resolve thin fluid layers between immersed boundaries. We apply this method to two-dimensional flows of increasing complexity, including eccentric rotating cylinders and elastic vesicles near walls in shear flow, to show its increased accuracy compared to the classical immersed boundary method. We present preliminary simulation results of cell suspensions, a problem in which near-contact occurs at multiple levels, such as cell-wall, cell-cell, and intracellular interactions, to highlight the importance of resolving thin fluid layers in order to obtain the correct overall dynamics.
February 5 Alex Lubotzky (Hebrew University)
Title: High dimensional expanders: From Ramanujan graphs to Ramanujan complexes
Abstract:
Expander graphs in general, and Ramanujan graphs , in particular, have played a major role in computer science in the last 5 decades and more recently also in pure math. The first explicit construction of bounded degree expanding graphs was given by Margulis in the early 70's. In mid 80' Margulis and Lubotzky-Phillips-Sarnak provided Ramanujan graphs which are optimal such expanders.
In recent years a high dimensional theory of expanders is emerging. A notion of topological expanders was defined by Gromov in 2010 who proved that the complete d-dimensional simplical complexes are such. He raised the basic question of existence of such bounded degree complexes of dimension d>1.
This question was answered recently affirmatively (by T. Kaufman, D. Kazdhan and A. Lubotzky for d=2 and by S. Evra and T. Kaufman for general d) by showing that the d-skeleton of (d+1)-dimensional Ramanujan complexes provide such topological expanders. We will describe these developments and the general area of high dimensional expanders.
February 6 Alex Lubotzky (Hebrew University)
Title: Groups' approximation, stability and high dimensional expanders
Abstract:
Several well-known open questions, such as: are all groups sofic or hyperlinear?, have a common form: can all groups be approximated by asymptotic homomorphisms into the symmetric groups Sym(n) (in the sofic case) or the unitary groups U(n) (in the hyperlinear case)? In the case of U(n), the question can be asked with respect to different metrics and norms. We answer, for the first time, one of these versions, showing that there exist fintely presented groups which are not approximated by U(n) with respect to the Frobenius (=L_2) norm.
The strategy is via the notion of "stability": some higher dimensional cohomology vanishing phenomena is proven to imply stability and using high dimensional expanders, it is shown that some non-residually finite groups (central extensions of some lattices in p-adic Lie groups) are Frobenious stable and hence cannot be Frobenius approximated.
All notions will be explained. Joint work with M, De Chiffre, L. Glebsky and A. Thom.
February 9 Wes Pegden (CMU)
Title: The fractal nature of the Abelian Sandpile
Abstract: The Abelian Sandpile is a simple diffusion process on the integer lattice, in which configurations of chips disperse according to a simple rule: when a vertex has at least 4 chips, it can distribute one chip to each neighbor.
Introduced in the statistical physics community in the 1980s, the Abelian sandpile exhibits striking fractal behavior which long resisted rigorous mathematical analysis (or even a plausible explanation). We now have a relatively robust mathematical understanding of this fractal nature of the sandpile, which involves surprising connections between integer superharmonic functions on the lattice, discrete tilings of the plane, and Apollonian circle packings. In this talk, we will survey our work in this area, and discuss avenues of current and future research.
March 2 Aaron Bertram (Utah)
Title: Stability in Algebraic Geometry
Abstract: Stability was originally introduced in algebraic geometry in the context of finding a projective quotient space for the action of an algebraic group on a projective manifold. This, in turn, led in the 1960s to a notion of slope-stability for vector bundles on a Riemann surface, which was an important tool in the classification of vector bundles. In the 1990s, mirror symmetry considerations led Michael Douglas to notions of stability for "D-branes" (on a higher-dimensional manifold) that corresponded to no previously known mathematical definition. We now understand each of these notions of stability as a distinct point of a complex "stability manifold" that is an important invariant of the (derived) category of complexes of vector bundles of a projective manifold. In this talk I want to give some examples to illustrate the various stabilities, and also to describe some current work in the area.
March 16 Anne Gelb (Dartmouth)
Title: Reducing the effects of bad data measurements using variance based weighted joint sparsity
Abstract: We introduce the variance based joint sparsity (VBJS) method for sparse signal recovery and image reconstruction from multiple measurement vectors. Joint sparsity techniques employing $\ell_{2,1}$ minimization are typically used, but the algorithm is computationally intensive and requires fine tuning of parameters. The VBJS method uses a weighted $\ell_1$ joint sparsity algorithm, where the weights depend on the pixel-wise variance. The VBJS method is accurate, robust, cost efficient and also reduces the effects of false data.
April 6 Edray Goins (Purdue)
Title: Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups
Abstract: A Belyĭ map [math] \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] is a rational function with at most three critical values; we may assume these values are [math] \{ 0, \, 1, \, \infty \}. [/math] A Dessin d'Enfant is a planar bipartite graph obtained by considering the preimage of a path between two of these critical values, usually taken to be the line segment from 0 to 1. Such graphs can be drawn on the sphere by composing with stereographic projection: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq \mathbb P^1(\mathbb C) \simeq S^2(\mathbb R). [/math] Replacing [math] \mathbb P^1 [/math] with an elliptic curve [math]E [/math], there is a similar definition of a Belyĭ map [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C). [/math] Since [math] E(\mathbb C) \simeq \mathbb T^2(\mathbb R) [/math] is a torus, we call [math] (E, \beta) [/math] a toroidal Belyĭ pair. The corresponding Dessin d'Enfant can be drawn on the torus by composing with an elliptic logarithm: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq E(\mathbb C) \simeq \mathbb T^2(\mathbb R). [/math]
This project seeks to create a database of such Belyĭ pairs, their corresponding Dessins d'Enfant, and their monodromy groups. For each positive integer [math] N [/math], there are only finitely many toroidal Belyĭ pairs [math] (E, \beta) [/math] with [math] \deg \, \beta = N. [/math] Using the Hurwitz Genus formula, we can begin this database by considering all possible degree sequences [math] \mathcal D [/math] on the ramification indices as multisets on three partitions of N. For each degree sequence, we compute all possible monodromy groups [math] G = \text{im} \, \bigl[ \pi_1 \bigl( \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \} \bigr) \to S_N \bigr]; [/math] they are the ``Galois closure
of the group of automorphisms of the graph. Finally, for each possible monodromy group, we compute explicit formulas for Belyĭ maps [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] associated to some elliptic curve [math] E: \ y^2 = x^3 + A \, x + B. [/math] We will discuss some of the challenges of determining the structure of these groups, and present visualizations of group actions on the torus.
This work is part of PRiME (Purdue Research in Mathematics Experience) with Chineze Christopher, Robert Dicks, Gina Ferolito, Joseph Sauder, and Danika Van Niel with assistance by Edray Goins and Abhishek Parab. |
Throughout your question it is clear that you are considering electrons as particles. This is not entirely incorrect; for example, the cathode ray tube is best explained by electrons being considered negatively charged particles that can travel through vacuum and be deflected by (electro)magnetic fields. However, particles as small as electrons can also be considered waves as is evident by electron diffraction. The probably most frequently cited connection between the particle nature and the wave nature of electrons and others is the de Broglie wavelength given by equation $(1)$.
$$\lambda = \frac hp\tag{1}$$
This discovery earned de Broglie a Nobel Prize for the theory and Thompson and Davisson for the supporting diffraction experiments.
Quantum mechanics is the underlying theory that attempts to shed light into how electrons and other microscopic particles behave according to their wave nature. The key principle is that each particle is described by a wave function, typically labelled $\Psi$, dependent on space and/or time coordinates. A key part of the theory is the existence of
operators which can be applied to the wave function to observe observables — physical parameters such as a particle’s energy or magnetic moment. Operators and observables observe the Schrödinger equation $(2)$ in which the Hamiltonian operator $\hat H$ can be replaced by any other operator to observe a different observable than energy $E$. Naturally, the wave function may be a different one depending on the operator in question.
$$\hat H\Psi = E \Psi\tag{2}$$
Using this equation alongside the derivation of the correct Hamiltonian, the wave function for a singly negatively charged particle rotating and translating freely in a spherical electric field of a given positive charge can be determined. This long and tangled sentence boils down to: we can calculate the wave function of an electron in the vicinity of a nucleus. The nice thing is that the electron’s energy is quantitised and dependent only on the principal quantum number $n$ while other quantum numbers are also required to fully describe the wave function ($l, m_l, s, m_s$). The quantisation of the electron’s energy provides a nice and succinct explanation of the hydrogen spectral lines.
Unfortunately, it is not possible to solve the Hamiltonian for two electrons. The energy of a single electron consists of its kinetic energy (described with the $\hat T$ operator) and its potential energy towards the nucleus ($\hat V_\ce{el-nuc}$). If a second electron is introduced, we now need to consider electron-electron repulsion potential ($\hat V_\ce{el-el}$) which we cannot solve. (In my quantum mechanics lecture, the professor gave the impression that this is a problem proven to be unsolvable but I don’t recall seeing any proof.) Therefore, we cannot calculate the wave functions of the two electrons of a helium atom
a priori. However, multiple computer-assisted calculation algorithms exist that allow us to come close.
Whatever the actual solution, the key is that electrons in an atom or molecule should primarily be considered
waves, not particles. Their wave function contains the spin orientation quantum number $m_s$ which can be either $+\frac12$ or $-\frac12$, depending on whether the electron is spin up or spin down. These two wave functions basically operate the same space at the same time. Yet the electrons do not collide because we are dealing with waves, not particles.
You considered the electron with opposite spin to be a positron. That is not correct. A positron has the opposite
charge with respect to an electron ($\ce{e+}$). Spin and charge are, however, two different properties. Spin is probably best considered a magnetic moment that would arise if a charged particle were to rotate around itself. However, this macroscopic picture is not fully valid; no macroscopic pictures you draw of microscopic particles are.
Finally, note that all particles are subject to the uncertainty principle. This means that we
cannot determine the position of an electron accurately since it would imply maximum uncertainty with respect to impuls. Likewise, we cannot accurately determine an electrons impuls. And another complicating matter is the observer effect: by attempting to observe where an electron is (approximately) we have to influence the system so strongly that it is no longer what we wanted to observe. This is because even very small energies have a significant effect on small scales.
Summary
Your notion of electron particles is
not fully supported by the currently accepted (and experimentally validated) theory. Rather, think of them as waves. Waves can interfere but cannot collide. Furthermore, since the wave function does not give a location, we cannot accurately determine the location of an electron.
Spin and charge are two different things. |
I will just repeat the title:
Is there a closed non-smoothable 4-manifold with zero Euler characteristic?
I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.
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The Kirby-Siebenmann invariant in $H^4(M;\Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $\chi(M \# N) = \chi(M) + \chi(N) -2$.
To construct manifolds with nontrivial Kirby-Siebenmann invariant we should apply Freedman's theorem: simply connected topological 4-manifolds are determined by their intersection form and Kirby-Siebenmann invariant; if the intersection form is odd, both Kirby-Siebenmann invariants are realized, and if the intersection form is even, only one KS-invariant is realized. So there is a topological manifold $F(\Bbb{CP}^2)$, homotopy equivalent to $\Bbb{CP}^2$ but with nontrivial Kirby-Siebenmann invariant.
So for the desired non-smoothable manifold with zero Euler characteristic, one can take (for instance) $$F(\Bbb{CP}^2) \# 3\Bbb{CP}^2 \#(S^2 \times \Sigma_2).$$
Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead.
More precisely, the (negative) $E_8$-plumbing $P$ bounds the Poincaré homology sphere $Y$; by Freedman's theorem, $Y$ is also the boundary of a contractible
topological 4-manifold $W$, and gluing $P\cup_Y -W$ yields a closed, simply connected topological 4-manifold $X$ with intersection form $-E_8$; $\chi(X) = 10$.
The connected sum $X\#5(S^1\times S^3)$ has Euler characteristic 0, by Mike's computation, and the intersection form is still $-E_8$. Hence, by Donaldson's theorem, $X\#5(S^1\times S^3)$ does not have a smooth structure, since its intersection form is negative definite but not diagonal.
You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\ \#\ (n+1) (S^1 \times S^3)$ is not smoothable and has Euler characteristic $0$. The argument is that if it were smoothable, then you could surger away the fundamental group and realize that intersection form by a simply connected manifold. |
Geometry and Topology Seminar Contents Fall 2016
date speaker title host(s) September 9 Bing Wang (UW Madison) "The extension problem of the mean curvature flow" (Local) September 16 Ben Weinkove (Northwestern University) "Gauduchon metrics with prescribed volume form" Lu Wang September 23 Jiyuan Han (UW Madison) "Deformation theory of scalar-flat ALE Kahler surfaces" (Local) September 30 October 7 Yu Li (UW Madison) "Ricci flow on asymptotically Euclidean manifolds" (Local) October 14 Sean Howe (University of Chicago) "Representation stability and hypersurface sections" Melanie Matchett Wood October 21 Nan Li (CUNY) "Quantitative estimates on the singular Sets of Alexandrov spaces" Lu Wang October 28 Ronan Conlon(Florida International University) "New examples of gradient expanding K\"ahler-Ricci solitons" Bing Wang November 4 Jonathan Zhu (Harvard University) "Entropy and self-shrinkers of the mean curvature flow" Lu Wang November 11 Canceled. November 18 Caglar Uyanik (Illinois) "Geometry and dynamics of free group automorphisms" Kent Thanksgiving Recess December 2 Peyman Morteza (UW Madison) "TBA" (Local) December 9 Yu Zeng(University of Rochester) "TBA" Bing Wang December 16 Spring 2017
date speaker title host(s) Jan 20 Carmen Rovi (University of Indiana Bloomington) "TBA" Maxim Jan 27 Feb 3 Feb 10 Feb 17 Feb 24 March 3 March 10 March 17 March 24 Spring Break March 31 April 7 April 14 April 21 April 28 Bena Tshishiku (Harvard) "TBA" Dymarz Fall Abstracts Ronan Conlon New examples of gradient expanding K\"ahler-Ricci solitons
A complete K\"ahler metric $g$ on a K\"ahler manifold $M$ is a \emph{gradient expanding K\"ahler-Ricci soliton} if there exists a smooth real-valued function $f:M\to\mathbb{R}$ with $\nabla^{g}f$ holomorphic such that $\operatorname{Ric}(g)-\operatorname{Hess}(f)+g=0$. I will present new examples of such metrics on the total space of certain holomorphic vector bundles. This is joint work with Alix Deruelle (Universit\'e Paris-Sud).
Jiyuan Han Deformation theory of scalar-flat ALE Kahler surfaces
We prove a Kuranishi-type theorem for deformations of complex structures on ALE Kahler surfaces. This is used to prove that for any scalar-flat Kahler ALE surfaces, all small deformations of complex structure also admit scalar-flat Kahler ALE metrics. A local moduli space of scalar-flat Kahler ALE metrics is then constructed, which is shown to be universal up to small diffeomorphisms (that is, diffeomorphisms which are close to the identity in a suitable sense). A formula for the dimension of the local moduli space is proved in the case of a scalar-flat Kahler ALE surface which deforms to a minimal resolution of \C^2/\Gamma, where \Gamma is a finite subgroup of U(2) without complex reflections. This is a joint work with Jeff Viaclovsky.
Sean Howe Representation stability and hypersurface sections
We give stability results for the cohomology of natural local systems on spaces of smooth hypersurface sections as the degree goes to \infty. These results give new geometric examples of a weak version of representation stability for symmetric, symplectic, and orthogonal groups. The stabilization occurs in point-counting and in the Grothendieck ring of Hodge structures, and we give explicit formulas for the limits using a probabilistic interpretation. These results have natural geometric analogs -- for example, we show that the "average" smooth hypersurface in \mathbb{P}^n is \mathbb{P}^{n-1}!
Nan Li Quantitative estimates on the singular sets of Alexandrov spaces
The definition of quantitative singular sets was initiated by Cheeger and Naber. They proved some volume estimates on such singular sets in non-collapsed manifolds with lower Ricci curvature bounds and their limit spaces. On the quantitative singular sets in Alexandrov spaces, we obtain stronger estimates in a collapsing fashion. We also show that the (k,\epsilon)-singular sets are k-rectifiable and such structure is sharp in some sense. This is a joint work with Aaron Naber.
Yu Li
In this talk, we prove that if an asymptotically Euclidean (AE) manifold with nonnegative scalar curvature has long time existence of Ricci flow, it converges to the Euclidean space in the strong sense. By convergence, the mass will drop to zero as time tends to infinity. Moreover, in three dimensional case, we use Ricci flow with surgery to give an independent proof of positive mass theorem. A classification of diffeomorphism types is also given for all AE 3-manifolds with nonnegative scalar curvature.
Gaven Marin TBA Peyman Morteza TBA Caglar Uyanik Geometry and dynamics of free group automorphisms
A common theme in geometric group theory is to obtain structural results about infinite groups by analyzing their action on metric spaces. In this talk, I will focus on two geometrically significant groups; mapping class groups and outer automorphism groups of free groups.We will describe a particular instance of how the dynamics and geometry of their actions on various spaces provide deeper information about the groups.
Bing Wang The extension problem of the mean curvature flow
We show that the mean curvature blows up at the first finite singular time for a closed smooth embedded mean curvature flow in R^3. A key ingredient of the proof is to show a two-sided pseudo-locality property of the mean curvature flow, whenever the mean curvature is bounded. This is a joint work with Haozhao Li.
Ben Weinkove Gauduchon metrics with prescribed volume form
Every compact complex manifold admits a Gauduchon metric in each conformal class of Hermitian metrics. In 1984 Gauduchon conjectured that one can prescribe the volume form of such a metric. I will discuss the proof of this conjecture, which amounts to solving a nonlinear Monge-Ampere type equation. This is a joint work with Gabor Szekelyhidi and Valentino Tosatti.
Jonathan Zhu Entropy and self-shrinkers of the mean curvature flow
The Colding-Minicozzi entropy is an important tool for understanding the mean curvature flow (MCF), and is a measure of the complexity of a submanifold. Together with Ilmanen and White, they conjectured that the round sphere minimises entropy amongst all closed hypersurfaces. We will review the basics of MCF and their theory of generic MCF, then describe the resolution of the above conjecture, due to J. Bernstein and L. Wang for dimensions up to six and recently claimed by the speaker for all remaining dimensions. A key ingredient in the latter is the classification of entropy-stable self-shrinkers that may have a small singular set.
Yu Zeng Short time existence of the Calabi flow with rough initial dataCalabi flow was introduced by Calabi back in 1950’s as a geometric flow approach to the existence of extremal metrics. Analytically it is a fourth order nonlinear parabolic equation on the Kaehler potentials which deforms the Kaehler potential along its scalar curvature. In this talk, we will show that the Calabi flow admits short time solution for any continuous initial Kaehler metric. This is a joint work with Weiyong He. Spring Abstracts Bena Tshishiku
"TBA"
Archive of past Geometry seminars
2015-2016: Geometry_and_Topology_Seminar_2015-2016
2014-2015: Geometry_and_Topology_Seminar_2014-2015 2013-2014: Geometry_and_Topology_Seminar_2013-2014 2012-2013: Geometry_and_Topology_Seminar_2012-2013 2011-2012: Geometry_and_Topology_Seminar_2011-2012 2010: Fall-2010-Geometry-Topology |
Clojure Numerics, Part 5 - Orthogonalization and Least SquaresYou can adopt a pet function! Support my work on my Patreon page, and access my dedicated discussion server. Can't afford to donate? Ask for a free invite.
October 17, 2017
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How to solve linear systems that have many solutions, or those that have no solutions at all? That's the theme for a thick math textbook, of course, but from the programmer's point of view, we are interested in the practical matters, so I'll stick to the main point. When I have less equations than there are unknowns, or I have too many equations, what can I do in Clojure to make those unknowns known? What functions do I call?
Before I continue, a few reminders:
Include Neanderthal library in your project to be able to use ready-made high-performance linear algebra functions. Read articles in the introductory Clojure Linear Algebra Refresher series. This is the fifth article in a more advanced series. It is a good idea to read them all in sequence if this is your first contact with this kind of software.
The namespaces we'll use:
(require '[uncomplicate.neanderthal [native :refer [dge dv]] [core :refer [mm trans view-sy view-tr axpy nrm2 mm! copy submatrix]] [linalg :refer [qrf org svd sv ls!]]])
Let's begin
I wrote about solving linear systems, and I think it was clear that, while there are complicationsalong the way when the specific numbers are a bit funky, it is otherwise a quite straightforward business.Many problems are not structured so well. What about the case when we have more equations than there areunknowns? Such systems are
overdetermined; there are so many conditions that there probably isn'tany solution to satisfy them all. In that case, we have to choose the solution that is somehow "least bad".What is "least bad" is also its own matter. On top of that, since I am a practical programmer whoprefers Clojure, I also require an elegant programming API to find the solution. And, since I acceptnothing less than the best from Neanderthal, I want it to be superfast… The Full-rank Least Squares
Let's consider the problem first. I have a system \(Ax=b\), where \(A\in{R^{m\times{n}}}\) is a
data matrixholding coefficients, and \(b\in{R^m}\) is an observation vector, while \(m\geq{n}\). Such system iscalled overdetermined, and usually there is no exact solution to such system.
Since we can't find the exact solution, we opt for the next best, the "nearest" solution. Recall fromthe Clojure Linear Algebra Refresher series that in linear algebra we use
norms to measure distance.Therefore, the nearest something is something with the least norm \({\lVert Ax-b \rVert}_p\). But, whichnorm to choose? 1-norm? ∞-norm? Different norms will give different answers. The matter is, fortunately,solved by the fact that minimization of 1-norm and ∞-norm is too complicated, so our best buddythe Euclidean norm is the winner.
Minimization of the Euclidean norm \(min {\lVert Ax-b \rVert}_2\) is called the
least squares problem.It is convenient because this function is differentiable, and the 2-norm is preserved under orthogonaltransformations (see Clojure Linear Algebra Refresher series for the basics, and the specific details follow here shortly).
When \(A\) is full-rank (\(ran(A)=n\)), there is a
unique linear squares solution, and it canbe found by solving the symmetric positive definite system \(A^TAx_{LS}=A^Tb\). You remember one of theprevious articles in this series? We talked about these.
Let's try a random example in Clojure:
First, I'll construct a (randomly chosen) matrix
a, and make sure its rank is 2 (here, by doing SVD):
(def a (dge 3 2 [1 2 3 -1 1 -1]))
(svd a)
#uncomplicate.neanderthal.internal.common.SVDecomposition{:sigma #RealDiagonalMatrix[double, type:gd mxn:2x2, offset:0] ▧ ┓ ↘ 3.79 1.63 ┗ ┛ , :u nil, :vt nil, :master true}
There are 2 non-zero values there, so the rank is 2, which I wanted to check.
Next, I'll construct \(A^TA\):
(def ata (mm (trans a) a))
ata
#RealGEMatrix[double, mxn:2x2, layout:column, offset:0] ▥ ↓ ↓ ┓ → 14.00 -2.00 → -2.00 3.00 ┗ ┛
I created
ata as a general matrix, so we can both look at it and see that it is symmetric, aspromised by math theory. Now I'll make it symmetric (in the Clojure data structure sense).Of course, in a real project, I'd skip those intermediate steps.
(def b (dge 3 1 [1 -1 -2]))(def solution (sv (view-sy (mm (trans a) a)) (mm (trans a) b)))
solution
#RealGEMatrix[double, mxn:2x1, layout:column, offset:0] ▥ ↓ ┓ → -0.55 → -0.37 ┗ ┛
This should be the LS solution, and we'll check later whether a more general method will give the same answer. Yes, there are more general methods, so what I've shown you until now is just a learning-oriented warm-up. Usually you do not need to be so creative, and can just go to the method X and say "Solve this for me!".
An interesting side note here is that solving these
normal equations is connected to solving gradient equations. Which might be an interesting topic if you're interested in machine learning.I'll have to say more on that when I finish these more general topics in the Linear Algebra series.
When we find the LS solution \(x_{LS}\), we can also compute the
minimum residual \(r_{LS} = b-Ax_{LS}\). It'snorm will tell us how far the system we solved exactly is from the original system.
(axpy -1 (mm a solution) b)
#RealGEMatrix[double, mxn:3x1, layout:column, offset:0] ▥ ↓ ┓ → 1.18 → 0.47 → -0.71 ┗ ┛
Minimum residual, the norm of the residual, seems quite high here. We'll check whether our solution is correct later.
(nrm2 (axpy -1 (mm a solution) b))
1.459992790176863
This method is quite OK for full-rank systems, but even then, it is sensitive to the presence of small singular values (recall the last article about Singular Value Decomposition).
Another method is to use normal equations, which includes Cholesky factorization, matrix multiplication and matrix-vector multiplication. I'll skip this, because there is a better and more general method.
These better methods usually rely on the QR factorization.
QR Factorization
The idea behind the QR factorization is similar to the idea behind the LU and Cholesky factorizationsthat we met in previous articles: destructure matrix \(A\) to a product of a few matrices that have specialproperties (symmetric, triangular, positive definite, etc.) and then use that destructured form in"easier" computations. In the QR, the trick is to find an
orthogonal matrix \(Q\) and and upper triangular \(R\),such that \(A=QR\). Recall from the refresher series that \(Q\) is orthogonal if \(Q^{-1}=Q^T\).
There is a proven theorem (look it up in a math textbook if you're interested in the details :) that for \(A\in{R^{m\times{n}}}\), there are orthogonal \(Q\in{R^{m\times{n}}}\) and upper triangular \(R\in{R^{m\times{n}}}\) such that \(A=QR\), the QR factorization of \(A\).
Since Q is orthogonal, and orthogonal transformations preserve the rank, shape and distance, we can intuitively guess that such factorization have many properties useful in rank-sensitive operations. I'll plead with you to trust the theory and believe me that this works quite well without further convincing (if you want to know more, the textbooks are at your disposal).
Most of these techniques are based on partitioning the columns of into the form such as\begin{equation} A=QR= [Q_1 | Q_2] \begin{bmatrix} R_1\\0\\ \end{bmatrix} \end{equation}
and then using \(Q_1 R_1\) further. What is important to me as a programmer is, mostly:
To keep in mind that the QR factorization is the key step for these kinds of problems To know what can be computed from QR factorization, and which method does it. To keep in mind that, while very powerful, QR-based algorithms are not almighty; there are cases when further options should be explored.
The complexity of various flavors varies, but boils down to \(O(m\times{n^2})\).
But how to solve the system with it?
From \(Ax=QRx=b\), we have \(Q^{T}Ax=Rx\) and\begin{equation} Q^Tb= \begin{bmatrix} c\\d\\ \end{bmatrix} \end{equation}
so \(R_1x_{LS}=c\). Since \(c\) is readily computed by a matrix multiplication, \(R_1\) is upper triangular system, and and triangular systems are quite easy to solve, this is what we were looking for.
Here I call Neanderthal's sv method, which computes the factorization and maintains all crutial parts of it.Please note that the matrix
:or and vector
:tau together hold Q and R encoded. Using raw
:or wouldproduce gibberish. Other functions in Neanderthal expect exactly this raw structure and know how to work with it, so do not mess with its contents yourself.
(def qr (qrf a))
qr
#uncomplicate.neanderthal.internal.common.OrthogonalFactorization{:eng #object[uncomplicate.neanderthal.internal.host.mkl.DoubleGEEngine 0x4e780e3a "[email protected]3a"], :or #RealGEMatrix[double, mxn:3x2, layout:column, offset:0] ▥ ↓ ↓ ┓ → -3.74 0.53 → 0.42 -1.65 → 0.63 -0.01 ┗ ┛ , :jpiv nil, :tau #RealBlockVector[double, n:2, offset: 0, stride:1] [ 1.27 2.00 ] , :master true, :fresh #atom[true 0x4a43cc2b], :m 3, :n 3, :or-type :qr, :api-orm #function[uncomplicate.neanderthal.internal.api/eval6950/fn--7034/G--6891--7047], :api-org #function[uncomplicate.neanderthal.internal.api/eval6950/fn--7129/G--6923--7138]}
Did you know that even the mm! method can work with Q encoded in this form? It is important, since we need to compute \(Q^Tb\). Even transpose works with QR, although it is not a proper matrix itself:
(def cd (mm! 1 (trans qr) (copy b)))
cd
#RealGEMatrix[double, mxn:3x1, layout:column, offset:0] ▥ ↓ ┓ → 1.87 → 0.61 → -1.46 ┗ ┛
\(R\) is just the upper triangular part of the
qr instance that we have, so:
(let [r1 (view-tr (:or qr) {:uplo :upper})] (sv r1 (submatrix cd 0 0 2 1)))
#RealGEMatrix[double, mxn:2x1, layout:column, offset:0] ▥ ↓ ┓ → -0.55 → -0.37 ┗ ┛
Yep, we got the same result using different method. Please note that the third component, \(d\) is equals to the one that we predicted using minimum residual \({\rho_{LS}= \lVert d \rVert}_2\) with the first method.
Least Squares in one quick sweep
Let's check with the function we could have used at the start, ls. Yes, there is a simple almighty method that makes all this fuss unnecessary. But, If I just showed you that, what would we have learned about the fine points of solving these kinds of systems and the many gotchas that await us there?
So:
(ls! (copy a) (copy b))
#RealGEMatrix[double, mxn:3x1, layout:column, offset:0] ▥ ↓ ┓ → -0.55 → -0.37 → -1.46 ┗ ┛
Hey, all methods that we used gave the same solution! That's a good sign.
Let's see whether we can recover the original Q and try with it, just to further our understanding:
(let [q (org qr) r1 (view-tr (:or qr) {:uplo :upper})] (sv r1 (mm (trans q) b)))
#RealGEMatrix[double, mxn:2x1, layout:column, offset:0] ▥ ↓ ┓ → -0.55 → -0.37 ┗ ┛
This is the same result indeed! This is good enough for me, today at least.
Please note that I copied all these instances in code, but for best performance, you'd probably use the destructive variants of those methods.
Next installments…
I'll probably take some time to write some more about problems connected to least squares, and their solutions. Or I'll skip right to eigen-problems. We'll see. Anyway, more Clojure & Neanderthal stuff follows soon. |
The critical value approach involves determining "likely" or "unlikely" by determining whether or not the observed test statistic is more extreme than would be expected if the null hypothesis were true. That is, it entails comparing the observed test statistic to some cutoff value, called the "
critical value." If the test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null hypothesis is not rejected.
Specifically, the four steps involved in using the critical value approach to conducting any hypothesis test are:
Specify the null and alternative hypotheses. Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. To conduct the hypothesis test for the population mean μ, we use the t-statistic \(t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}\) which follows a t-distribution with n- 1 degrees of freedom. Determine the critical value by finding the value of the known distribution of the test statistic such that the probability of making a Type I error — which is denoted \(\alpha\) (greek letter "alpha") and is called the " significance level of the test" — is small (typically 0.01, 0.05, or 0.10). Compare the test statistic to the critical value. If the test statistic is more extreme in the direction of the alternative than the critical value, reject the null hypothesis in favor of the alternative hypothesis. If the test statistic is less extreme than the critical value, do not reject the null hypothesis. Example S.3.1.1 Mean GPA Section
In our example concerning the mean grade point average, suppose we take a random sample of
n = 15 students majoring in mathematics. Since n = 15, our test statistic t* has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05, so that we have only a 5% chance of making a Type I error. Right-Tailed
The critical value for conducting the
right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the t-value, denoted t \(\alpha\), n - 1, such that the probability to the right of it is \(\alpha\). It can be shown using either statistical software or a t-table that the critical value t 0.05,14 is 1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3 if the test statistic t* is greater than 1.7613. Visually, the rejection region is shaded red in the graph. Left-Tailed
The critical value for conducting the
left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the t-value, denoted -t (\(\alpha\), n - 1) , such that the probability to the leftof it is \(\alpha\). It can be shown using either statistical software or a t-table that the critical value -t 0.05,14is -1.7613. That is, we would reject the null hypothesis H 0: μ= 3 in favor of the alternative hypothesis H A: μ< 3 if the test statistic t* is less than -1.7613. Visually, the rejection region is shaded red in the graph. Two-Tailed
There are two critical values for the
two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 — one for the left-tail denoted -t (\(\alpha\) /2, n - 1) and one for the right-tail denoted t (\(\alpha\). The value /2, n- 1) - t (\(\alpha\)/2,is the n- 1) t-value such that the probability to the leftof it is \(\alpha\)/2, and the value t (\(\alpha\)/2,is the n- 1) t-value such that the probability to the rightof it is \(\alpha\)/2. It can be shown using either statistical software or a t-table that the critical value -t 0.025,14is -2.1448 and the critical value t 0.025,14is 2.1448. That is, we would reject the null hypothesis H 0: μ= 3 in favor of the alternative hypothesis H A: μ≠ 3 if the test statistic t* is less than -2.1448 or greater than 2.1448. Visually, the rejection region is shaded red in the graph. |
Use that $X$ is connected if and only if the only continuous functions $f:X\to\{0,1\}$ are constant, where $\{0,1\}$ is endowed with the discrete topology.
Now, you know each $F$ in $\mathscr F$ is connected. Consider $f:\bigcup \mathscr F\to\{0,1\}$, $f$ continuous.
Take $\alpha \in\bigcap\mathscr F$. Look at $f(\alpha)$, and at $f\mid_{F}:\bigcup \mathscr F\to\{0,1\}$ for any $F\in\mathscr F$.
Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:
THM Let $(X,\mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:X\to\{0,1\}$ is continuous, it is constant. The space $\{0,1\}$ is endowed with the discrete metric (topology), that is, the open sets are $\varnothing,\{0\},\{1\},\{0,1\}$.
P First, suppose $X$ is disconnected, say by $A,B$, so $A\cup B=X$ and $A\cap B=\varnothing$, $A,B$ open. Define $f:X\to\{0,1\}$ by $$f(x)=\begin{cases}1& \; ; x\in A\\0&\; ; x\in B\end{cases}$$
Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in $\{0,1\}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:X\to\{0,1\}$ is continuous but not constant. Set $A=\{x:f(x)=1\}=f^{-1}(\{1\})$ and $B=\{x:f(x)=0\}=f^{-1}(\{0\})$. By hypothesis, $A,B\neq \varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $A\cup B=X$ and $A\cap B=\varnothing$. Thus $X$ is disconnected. $\blacktriangle$ |
Start: 7.3.2016 End: 6.7.2019 Project leader: Alfred Geroldinger FWF project number: P 28864-N35 Participants
Professor
Postdoc
All participants are located at the
University of Graz
Heinrichstraße 36
8010 Graz
Austria
Project summary
Let $H$ be a Krull monoid with finite class group $G$ such that each class contains a prime divisor (this setting includes holomorphy rings in global fields). Then every nonunit $a \in H$ can be written as a finite product of atoms (irreducible elements). If $a = u_1 \cdot \ldots \cdot u_k$ with atoms $u_1, \ldots, u_k \in H$, then $k$ is called the length of the factorization. The set $\mathsf L (a)$ of all possible factorization lengths $k$ is called the set of lengths of $a$, and this is a finite set of positive integers. We consider the system $\mathcal L (H) = \{ \mathsf L (a) \mid a \in H\}$ of sets of lengths of $H$ (for convenience, we set $\mathsf L (a) = \{0\}$ if $a$ is a unit in $H$). The Krull monoid $H$ is factorial if and only if the class group $G$ is trivial. Furthermore, $H$ is half-factorial (i.e., $|L|=1$ for all $L \in \mathcal L (H)$) if and only if $|G| \le 2$. If $|G| \ge 3$, then for every $N \in \mathbb N$ there is a set $L \in \mathcal L (H)$ such that $|L| \ge N$.
The system $\mathcal L (H)$ depends only on the class group $G$. To make this precise, consider the set $\mathcal B (G)$ of zero-sum sequences over $G$. By a sequence over $G$, we mean a finite sequence of terms from $G$ where repetition is allowed and the order is disregarded, and we say that $S$ is a zero-sum sequence if its terms sum up to zero. Defining an operation as the concatenation of sequences we obtain a monoid structure on $\mathcal B (G)$. Indeed, $\mathcal B (G)$ is a Krull monoid with class group isomorphic to $G$ (provided that $|G| \ge 3$) and every class contains a prime divisor. Moreover, the systems of sets of lengths of $H$ and that of $\mathcal B (G)$ coincide. Thus $\mathcal L (H) = \mathcal L \big( \mathcal B (G) \big)$ can be studied with methods from Zero-Sum Theory, a flourishing subfield of Additive Combinatorics.
The Structure Theorem for Sets of Lengths states that every $L \in \mathcal L (G)$ is an almost arithmetical multiprogression with universal bounds for all parameters controlling these multiprogressions. It is a main goal of the present project to study the involved parameters (such as the possible differences of the multiprogressions) in terms of classical zero-sum invariants (such as the Davenport constant of $G$) or even in terms of the group invariants. All work on these parameters will be done with a view towards the Characterization Problem, a main open question in this area. Indeed, let $G$ and $G'$ be two finite abelian groups with $|G|\ge 4$ and $|G'|\ge 4$ and suppose that $\mathcal L \big( \mathcal B (G) \big) = \mathcal L \big( \mathcal B (G') \big)$. Does it follow that $G$ and $G'$ are isomorphic? The answer is affirmative, among others, if $G$ or $G'$ has rank at most two (see [A. Geroldinger and W. Schmid, A characterization of class groups via sets of lengths, J. Korean Math. Society, 56 (2019), 869 -- 915]). However, the Characterization Problem is far open in general, and it will be in the focus of the the present project.
Publications Sets of lengths. Amer. Math. Monthly, 123 (2016), 960-988. Systems of Sets of Lengths: Transfer Krull Monoids versus Weakly Krull Monoids. In Rings, Polynomials, and Modules, 191-235. Springer, 2017. Long sets of lengths with maximal elasticity. Canadian Journal of Mathematics, 70(2018),1284-1318. Sets of minimal distances and characterizations of class groups of Krull monoids. The Ramanujan Journal, 45 (2018), 719-737. Sets of lengths in atomic unit-cancellative finitely presented monoids. Colloquium Math., 151 (2018), 171-187. A characterization of finite abelian groups via sets of lengths in transfer Krull monoids. Communications in Algebra, 46 (2018), 4021-4041. A realization theorem for sets of lengths in numerical monoids. Forum Math., 30 (2018), 1111-1118. Which sets are sets of lengths in all numerical monoids. Banach Center Publications, 118 (2019), 181-192. Sets of arithmetical invariants in transfer Krull monoids. J. Pure Appl. Algebra, 223 (2019), 3889-3918. On elasticities of locally finitely generated monoids. J. Algebra, 534 (2019), 145-167. On strongly primary monoids and domains. submitted. A characterization of seminormal C-monoids. Boll. Unione Ital. Mat. , to appear. On the arithmetic of Mori monoids and domains. Glasg. Math. J. , to appear. A characterization of Krull monoids for which sets of lengths are (almost) arithmetical progressions. Revista Matematica Iberoamericana, to appear. On minimal product-one sequences of maximal length over Dihedral and Dicyclic groups. Commun. Korean Math. Soc., to appear. Clean group rings over localizations of rings of integers. submitted. On monoids of ideals of orders in quadratic number fields. In Rings and Factorizations, Springer, to appear. On Erd\H{o}s-Ginzburg-Ziv inverse theorems for Dihedral and Dicyclic groups. Israel Journal of Mathematics, to appear. Factorization theory in commutative monoids. submitted. |
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