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Tag Info

New answers tagged angular-momentum

1

I think (but I'm not sure) that if you do it like if the photon has three physical polarization states then you get, as you said $S=0,1,2$. You can compute easily all of these states. But remember, you can express the states of the coupled basis in the uncoupled basis (via a Clebsch-Gordan decomposition). Then, you can delete the terms containing a non ...

1

So the key point to understanding this problem is to understand that it is the modes that contain information about the physical parameters of your photons (such as the momentum or angular momentum), and quantization is just a description of excitation of those modes. For instance the canonical quantization of the plane-wave expansion which you've ...

2

See also Simple Harmonic Motion - What are the units for $\omega_0$ ? and https://en.wikipedia.org/wiki/Joule#Confusion_with_newton-metre Here's a somewhat shorter explanation reflecting my own (possibly incorrect) intuition: Radians aren't "real" units; they're just a trick to keep track of which quantities involve angles and which don't, since it's ...

3

In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by: $$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$ If you do the same for two solar mass neutron stars ...

0

Check ✓ except the the speed depends on the center of rotation, not the choice of the origin. The general rule is that if the origin as velocity $\vec{v}_O$ then a point A located at $\vec{r}_A$ has speed $\vec{v}_A = \vec{v}_O + \vec\omega \times \vec{r}_A$. It happenstance that your origin does not move. Check ✓. In general, torques and angular momental ...

21

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...

3

Your mistake is your definition of the angular velocity $$\omega=rv \tag{not correct}$$ is incorrect. We know that $\omega$ has units of $1/s$, but your assertion gives it units of $m^2/s$. The correct definition is $$\omega=\frac vr \tag{correct}$$ which gives the correct units. Using this: $$\left[L\right] = [m]\left[r^2\right]\cdot\left[v\cdot ... 3 The easiest thing for this exercise is to use Levi-Civita symbol for the vector product:$$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$where I denote by e_i the canonical basis of \mathbb{R}^3. Using this notation, we have:$$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$and$$[L^2,\vec{p}]=e_i[L_j ...

0

It is best to read a book on dynamics to get all the details you need, but here is an overview. I assume all the rotation axes intersect at the origin, and the payload center of mass is located at a distance $\vec{c}$ from the last gimbal (in local coordinates). The mass moment of inertia of the payload is ...

0

As pointed out in the comments, the 'machine' in the movie Contact is just the gimbal part without the gyroscope. The point of this gimbal is to allow the thing being held at the center to be completely free to rotate in all directions. So, if they are working well they have no effect on the system except in the translational degrees of freedom, and they ...

1

It seems like you know what the answer is, but you just don't know how to prove it. You are right though. To make things simpler, just view things in the center of mass frame. Then the total momentum is zero, and, like you said, the total angular momentum is just the sum of the orbital momentum of each planet plus the sum of the spin angular momentum of ...

-5

Black holes pull matter in thus creating a pulling of matter or rotation (gravity) force that matter rotates or spirals around thus creating our shape form and rotational direction of our galaxy. If our sun doesn't die first then yes one day we will be pulled to the center of our galaxy. Every galactic year our solar system speeds up and grows closer to the ...

0

Case a) Body with uniform motion (no rotation), with C the center of the disk and A a point on the edge (for example below, at a distance $R$). \begin{aligned} \vec{v}_A & = (v,0,0) \\ \vec{\omega} & = (0,0,0) \\ \vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (v,0,0) \\ \vec{L} & = m \vec{v}_C = (m v,0,0) \\ \vec{H}_A & = ... 0 From your description, I assume the disk only only translating, not rotating. Is this correct? If so, read on. If not, I'll delete. I'm uncomfortable with the first method that uses L=I\omega. In this equation, it is assumed that every point on the rigid body can be characterized by the same angular velocity \omega. From your description of the motion ... 0 The first method is correct. The second method is wrong because the equation you use only applies to point particles, not continuous masses with volume (such as a disk). You're incorrectly treating the disk as a point particle located at the disk center. If you want to use the second method, you'll need to use this equation for angular momentum of ... 1 The question is essentially just asking you to perform Clebsch-Gordan (CG) decomposition, namely a change of basis on the Hilbert space \mathcal H_{j_1}\otimes\mathcal H_{j_2} of the composite system of spins. Recall that for each of the Hilbert spaces \mathcal H_{j_1} and \mathcal H_{j_1} there exist orthonormal bases of eigenvectors of \mathbf ... 3 Consider two bodies A and B. With respect to an inertial coordinate system with origin at point O, the coords of the particles in A are vectors x_{a}\in V_{3} with a=1,2,\ldots, N_{A} and similarly the coords of the particles of B are x_{b}\in V_{3} with b=1,2,\ldots,N_{B}. The momenta wrt the inertial frame with origin at point O of the particles ... 0 The problem here was when doing the integration by parts. One has to be very careful when remembering which operators act on which expressions. In my case, it turned out that the derivative was acting on the wrong terms when doing the integration by parts, which indeed did add an extra minus sign. 0 I've been wondering a lot about this too. So far I haven't got any straight answers, and I've asked around a lot. Angular acceleration is easy - if you have either a moment of inertia (2D) or an inertia tensor (3D+) I, then the angular acceleration due to a force is simply \alpha=I^{-1}\tau for torque \tau=\vec{F}\textrm{x}\vec{r}, with a force ... 1 Hints: Notice that if we define \mathbf S_{123} = \mathbf S_1 + \mathbf S_2 + \mathbf S_3 and \mathbf S_{12} = \mathbf S_1 + \mathbf S_2, then we have \begin{align} \mathbf S_{123}^2 = \mathbf S_{12}^2 + \mathbf S_3^2 + 2\mathbf S_{12}\cdot\mathbf S_3 \end{align} Notice that your hamiltonian can be written as follows: \begin{align} H = ... 0 It looks like you've missed a sign: \begin{align} Q_i &= \epsilon_{ijk} \int_{p,q} \sqrt{\frac{E_q}{E_p}}p^k \int d^3x (a_qe^{iq\cdot x}-a_q^\dagger e^{-iq\cdot x}) \left(a_p (-i\frac{\partial}{\partial p^j})e^{ip\cdot x} -a_p^\dagger (i\frac{\partial}{\partial p^j})e^{-ip\cdot x}\right). \end{align} Then it looks like the term that goes like ... 0 To the original poster: You appear to be operating under the "hollywood" misconception that a black hole somehow "sucks harder" than the same amount of mass in a non-black-hole form. However, this false "black holes produce an enormous sucking" misconception is one of the many, many concepts of physics that "hollywood" gets totally wrong; a black hole of a ... 0 I'm not totally understanding the question, but I think the critical aspect here is that in atoms, the energy gap between successive levels is much greater than the "thermal" energy of the electrons. So the energy levels really do have to fill up pairwise, with each pair of electrons forming a singlet state with spin zero. I think it might be otherwise in a ... 4 Such an ordering arises from the fact that are arranged chronologically, i.e., according to the dates they were "discovered". The principle quantum number n entered the picture with Bohr's theory of the Hydrogen atom in 1913.Bohr introduced n in his quantization of angular momentum postulate where n is the allowed orbit. Mathematically, L = n{h ... 2 Use Lagrangian mechanics method to answer this problem because it is easier than Newtonian mechanics (IMHO). Let T be the kinetic energy, V be the potential energy then the Lagrangian L is given by L=T-V $$and the Lagrangian equation is$$ \frac{d}{dt}\left(\frac{dL}{d\dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, $$where it is assumed that ... 1 The center of mass is not (0, 0). If you take the average of the two positions, you will get (c, d). The angular momentum \vec L is \vec r \times \vec p. With mass m, this will be:$$ \vec L = m \vec r \times \vec v $$\vec v_\pm is just \dot{\vec r}_\pm = \frac{\mathrm d}{\mathrm dt} \vec r_\pm. So you have to to a time differentiation of ... 0 To prove the conservation of quantities, you need to be able to compute the motion of the system so that you can directly compute these quantities from the time dependent coordinates and verify that they do not change with time. However I don't think that the motion of this system is integrable: it looks like a multiple oscillator and it is very prone to ... 2 I can get you thinking along the right path, but I'll leave most of the doing for you. I suspect that for much of this to make sense, you'll need to actually do it. The idea is to start at the "top" of the ladder:$$|2, 2\rangle = |3/2, 3/2\rangle_1\otimes|1/2, 1/2\rangle_2 =|3/2, 3/2\rangle_1|1/2, 1/2\rangle_2. With this statement (and choice of ...

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### JEE Main 2017 (Online) 8th April Morning Slot

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$-$ 1 and it leaves remainder 6 when divided by x + 1; then :
A
p(2) = 11
B
p(2) = 19
C
p($-$ 2) = 19
D
p($-$ 2) = 11

## Explanation

Let, P(x) = ax2 + bx + c

As, P(0) = 1,

$\therefore\,\,\,$ a(0)2 + b(0) + c = 1

$\Rightarrow $$\,\,\, c = 1 \therefore\,\,\, P(x) = ax2 + bx + 1 If P(x) is divided by x - 1, remainder = 4 \Rightarrow$$\,\,\,$ P$\left( 1 \right) = 4$

$\therefore\,\,\,$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$\Rightarrow $$\,\,\, P(- 1) = 6 \therefore\,\,\, a - b + 1 = 6 . . . .(2) By solving (1) and (2) we get, a = 4, and b = -1 \therefore\,\,\, P(x) = 4x2 - x + 1 P(2) = 4(2)2 - 2 + 1 = 15 P(- 2) = 4 (-2)2 - (- 2) + 1 = 19 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot The sum of all the real values of x satisfying the equation 2(x-1)(x2 + 5x - 50) = 1 is : A 16 B 14 C -4 D - 5 ## Explanation We know, 2x = 1 only when x = 0. Similarly, 2(x-1)(x2 + 5x - 50) = 1 when (x-1)(x2 + 5x - 50) = 0 \Rightarrow (x - 1)(x + 10)(x - 5) = 0 \therefore x = 1, -10, 5 Sum of real values of x = 1 + (-10) + 5 = -4 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) Let S = { x \in R : x \ge 0 and 2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0}. Then S A contains exactly four elements B is an empty set C contains exactly one element D contains exactly two elements ## Explanation Given, 2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 Case 1 : When \sqrt x - 3 \ge 0, then equation becomes 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0 \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0 \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0 \therefore\,\,\, \sqrt x = 0,4 but as \sqrt x - 3 \ge 0 or \sqrt x \ge 3 then \sqrt x \ne 0 \therefore\,\,\, \sqrt x = 4 value is possible. Case 2 : When \sqrt x - 3 < 0 or \sqrt x < 3. There equation becomes - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0 \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0 \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0 \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0 \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0 \therefore\,\,\, \sqrt x = 2,6 as \sqrt x < 3 so \sqrt x \ne 6 \therefore\,\,\, \sqrt x = 2 is possible. So, total possible value of \sqrt x = 2,4 or for x possible values are 4, 16. \therefore\,\,\, Set S contains exactly two elements 4 and 16. 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot If \lambda \in R is such that the sum of the cubes of the roots of the equation, x2 + (2 - \lambda ) x + (10 - \lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is : A 4\sqrt 2 B 2\sqrt 5 C 2\sqrt 7 D 20 ## Explanation Let \alpha , \beta are the roots of the equation, \therefore \alpha + \beta = \lambda - 2 and \alpha$$\beta$ = 10 $-$ $\lambda$

${\alpha ^3} + {\beta ^3}$ = ($\alpha$ + $\beta$)3 $-$ 3$\alpha$$\beta$ ($\alpha$ + $\beta$)

= ($\lambda$ $-$ 2)3 $-$ 3(10 $-$ $\lambda$)($\lambda$ $-$ 2)

= $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

Let $f(\lambda$) = $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

$\therefore$ ${{df(\lambda )} \over {d\lambda }}$ = 3$\lambda ^2$ $-$ 6$\lambda$ $-$ 24

$\therefore$ at maximum of minimum ${{df(\lambda )} \over {d\lambda }}$ = 0

$\therefore$ $\lambda ^2$ $-$ 2$\lambda$ $-$ 8 = 0

$\Rightarrow$ ($\lambda$ + 2) ($\lambda$ $-$ 4) = 0

$\Rightarrow$ $\lambda$ = $-$2, 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 2$\lambda$ $-$ 2

When $\lambda$ = $-$2

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = $-$ 6 < 0

$\therefore$ at $\lambda$ = $-$2, f($\lambda$) has maximum value.

When $\lambda$ = 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 6 > 0

$\therefore$ at $\lambda$ = 4, f($\lambda$) has minimum value.

$\therefore$ When $\lambda$ = 4 equation is,

x2 $-$ 2x + 6 = 0

$\therefore$ ($\alpha$ $-$ $\beta$)2 = ($\alpha$ + $\beta$)2 $-$ 4$\alpha \beta$

$\Rightarrow$ x2 $-$ 4 $\times$ 6

= $-$ 20

$\Rightarrow$ ($\alpha$ $-$ $\beta$) = $2\sqrt 5 i$

$\Rightarrow$ $\left| {\alpha - \beta } \right|$ = $2\sqrt 5$ (ans)

format: plain


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AtCoder Grand Contest 013

A - Sorted Arrays

Time limit時間制限 : 2sec / Memory limitメモリ制限 : 256MB

制約

• 1 \leq N \leq 10^5
• 1 \leq A_i \leq 10^9
• A_i は全て整数である

入力

N
A_1 A_2 ... A_N

入力例 1

6
1 2 3 2 2 1

出力例 1

2

[1,2,3][2,2,1] に分ければよいです。

入力例 2

9
1 2 1 2 1 2 1 2 1

出力例 2

5

入力例 3

7
1 2 3 2 1 999999999 1000000000

出力例 3

3

Score : 300 points

Problem Statement

You are given an array A of length N. Your task is to divide it into several contiguous subarrays. Here, all subarrays obtained must be sorted in either non-decreasing or non-increasing order. At least how many subarrays do you need to divide A into?

Constraints

• 1 \leq N \leq 10^5
• 1 \leq A_i \leq 10^9
• Each A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible number of subarrays after division of A.

Sample Input 1

6
1 2 3 2 2 1

Sample Output 1

2

One optimal solution is to divide the array into [1,2,3] and [2,2,1].

Sample Input 2

9
1 2 1 2 1 2 1 2 1

Sample Output 2

5

Sample Input 3

7
1 2 3 2 1 999999999 1000000000

Sample Output 3

3


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# MATLAB: How to calculate the area of certain portion of an image

how to calculate the area of certain portion of an image ?Image Processing Toolbox

how to calculate the area of certain portion of an image ?

binaryImage = grayImage > 90;measurements = regionprops(binaryImage, 'Area');allAreas = [measurements.Area];


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# Mean squared prediction error

Jump to: navigation, search

In statistics the mean squared prediction error of a smoothing or curve fitting procedure is the expected value of the squared difference between the fitted values $\widehat{g}$ and the (unobservable) function g. If the smoothing procedure has operator matrix L, then

$\operatorname{MSPE}(L)=\operatorname{E}\left[\left( g(x_i)-\widehat{g}(x_i)\right)^2\right].$

The MSPE can be decomposed into two terms (just like mean squared error is decomposed into bias and variance); however for MSPE one term is the sum of squared biases of the fitted values and another the sum of variances of the fitted values:

$\operatorname{MSPE}(L)=\sum_{i=1}^n\left(\operatorname{E}\left[\widehat{g}(x_i)\right]-g(x_i)\right)^2+\sum_{i=1}^n\operatorname{var}\left[\widehat{g}(x_i)\right].$

Note that knowledge of g is required in order to calculate MSPE exactly.

## Estimation of MSPE

For the model $y_i=g(x_i)+\sigma\varepsilon_i$ where $\varepsilon_i\sim\mathcal{N}(0,1)$, one may write

$\operatorname{MSPE}(L)=g'(I-L)'(I-L)g+\sigma^2\operatorname{tr}\left[L'L\right].$

The first term is equivalent to

$\sum_{i=1}^n\left(\operatorname{E}\left[\widehat{g}(x_i)\right]-g(x_i)\right)^2 =\operatorname{E}\left[\sum_{i=1}^n\left(y_i-\widehat{g}(x_i)\right)^2\right]-\sigma^2\operatorname{tr}\left[\left(I-L\right)'\left(I-L\right)\right].$

Thus,

$\operatorname{MSPE}(L)=\operatorname{E}\left[\sum_{i=1}^n\left(y_i-\widehat{g}(x_i)\right)^2\right]-\sigma^2\left(n-2\operatorname{tr}\left[L\right]\right).$

If $\sigma^2$ is known or well-estimated by $\widehat{\sigma}^2$, it becomes possible to estimate MSPE by

$\operatorname{\widehat{MSPE}}(L)=\sum_{i=1}^n\left(y_i-\widehat{g}(x_i)\right)^2-\widehat{\sigma}^2\left(n-2\operatorname{tr}\left[L\right]\right).$

Colin Mallows advocated this method in the construction of his model selection statistic Cp, which is a normalized version of the estimated MSPE:

$C_p=\frac{\sum_{i=1}^n\left(y_i-\widehat{g}(x_i)\right)^2}{\widehat{\sigma}^2}-n+2\operatorname{tr}\left[L\right].$

where p comes from that fact that the number of parameters p estimated for a parametric smoother is given by $p=\operatorname{tr}\left[L\right]$, and C is in honor of Cuthbert Daniel.[citation needed]


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# Don’t waste $1 million on devops infrastructure that you’ll never need (written by lawrence krubner, however indented passages are often quotes). You can contact lawrence at: lawrence@krubner.com, or follow me on Twitter. Samy Dindane asked me a question: How would you deploy a simple React/NodeJS app? I responded on Twitter: Deployment is a big question but I’ll try to summarize. First, ask yourself, do you need to High Availability? When I ask CEOs of early stage startups, they always say “Yes” yet I think only 10% really need High Availability. I think of High Availability as implying multiple servers, multiple load balancers, multiple failover databases, multiple data centers, multiple regions. Lots of redundancy, probably configured for failover. Beware of perfectionism and maximalism. A recent client of mine had raised$5 million then hired Robots And Pencils and spent $1 million to build their MVP, and the CEO insisted on High Availability. He felt certain that his product was going to be the next Pokemon Go: Beyond being a global phenomenon, Pokémon GO is one of the most exciting examples of container-based development in the wild. The application logic for the game runs on Google Container Engine (GKE) powered by the open source Kubernetes project. Niantic chose GKE for its ability to orchestrate their container cluster at planetary-scale, freeing its team to focus on deploying live changes for their players. In this way, Niantic used Google Cloud to turn Pokémon GO into a service for millions of players, continuously adapting and improving. First of all, that’s rare, and second of all, does it actually matter if you lose a few customers because of a few dropped connections? Most of my clients seem to believe they are in markets where they must support uptime of 99.9999999%, even though most of my clients could have 98% uptime during a spike, and it would be fine in the long-run. In the case of my client, the public didn’t like the MVP. Which is a common experience. And the “M” was a lie. If you go for High Availability, then “M” is a lie, unless it’s redefined as “Maximum”. They built a “maximal viable product” instead of a “minimal viable product.” High Availability is excessive; there is nothing “minimum” about it. For a new startup, fast iteration is important. As Steve Blank says, you’re looking for product/market fit. Every variation of your product is a new experiment. The whole point of a Minimal Viable Product is that the public might not like your first idea, that’s why you keep it minimal — don’t waste much money on any one idea, till you have reason to believe that the public likes your idea. You want fast iteration of your business idea, not fast iteration of a devops setup. You want agile product development, not agile devops resource use. Once you have product/market fit, it is nice if you have fine-grained control of devops resources, because you can save a few dollars that way, but it’s not the kind of money you should worry about when you’re still trying to find an idea that the public likes. Small optimizations should wait till you are a stable and mature organization. Fast iteration of extremely minimal products allows you to test many ideas — you thus decrease your risk of failure. High Availability is a symptom of perfectionism, and as such it is the absolute opposite of an MVP. Which startups need to be Highly Available, right from the beginning? Startups doing finance, actually handling money, or startups doing security. That’s it. Unless you’re handling money, or protecting money, you don’t need to start with High Availability. Get there slowly. It’s a nice ideal: when you are big and successful then you should eventually implement it. But don’t start with it. I’ve previously been critical of certain devops strategies, especially Docker/Kubernetes. People have pushed back against my criticism, they ask, “If it’s good enough for Google then surely it is good enough for us?” But Google is a stable and mature company, so it is appropriate for Google to optimize its devops setup, to increase efficiency and lower costs at scale. Are you running a large, mature, stable company? If yes, this essay is not written for you. I’m writing for small startups launching an MVP. That means you don’t need massive redundancy in some cloud strategy. You can consider bare metal servers, which can be very, very cheap. Have you priced a co-lo center? It’s surprising how cheap they can be. You have to balance the risks, based on what your company really needs. I’ve a good friend at a startup who found a co-lo data center in a small college town in Virginia, where they were able to build out a massive MongoDB database, sharded across 4 massive servers, each server holding 256 gigabytes of RAM, 1 terabyte of RAM in total. They also had a Cassandra database and several servers for their front-end. They were paying$7,000 a month for this. The co-lo center was conveniently close to where their devops guy lived. Then the devops guy quit. Because the co-lo center was in this small town, they had to hire another devops person who was also in this small town, which was a limited market to hire in. The company decided that they could lower their risks by moving to AWS, because then they would be able to hire a devops person who lived anywhere. So they re-created their setup on AWS. Suddenly they were paying $35,000 a month. Yes, their bill went up 500% for the same setup. Co-lo is often much cheaper than the cloud, but you do face the risk of being location dependent. This can be a good risk for some companies, and a bad risk for other companies. You need to think carefully about which risks you really need to mitigate, and remember that it is impossible to mitigate all risks. If you avoid High Availability, your setup can be very simple. Bare metal servers, and then deploy an uberjar, or a fat Go binary, or some other easy system. If you have multiple microservices with internal dependencies, you can write a deployment script in Jenkins, and this will still be simpler than a Docker/Kubernetes setup. Not that you have to go with bare metal, I only mention it because it’s faster and cheaper than many of my clients seem to realize. If you want to be in the cloud, you can. And Terraform/Packer gives you an easy way to setup a standard virtual machine (VM) that your developers can use. Terraform/Packer is an easier way to get what developers typically use Docker for. If you want an easy way to scale, you want isolation, easy deployment, security, a standard development environment, plus an AMI that can be spun up many times in AWS, you can use Terraform/Packer. Your resource use will end up being course-grained, compared to the fine-grained control you get with Docker/Kubernetes, but most startups do not need the fine-grained resource control that Docker/Kubernetes makes possible. Terraform means you’re still working with a standard Virtual Machine, which you can run in VirtualBox or Vagrant, as well as run on AWS as an AMI. You can work with something standard, that you’ve probably been using for the last 10 years. You can setup a standard Virtual Machine that runs Linux or Windows, and that means you can stay with the environment that you have many years experience with, using tools that you’ve used for years. By contrast, if you instead use Docker/Kubernetes, you are launching into an adventure that involves a still new eco-system which is still somewhat immature and is still undergoing rapid evolution. If you really want to be loyal to the spirit of MVP, building something really minimal, then avoid Docker and Kubernetes. They are an advanced optimization, and therefore they are not MVP. I mentioned the client who paid Robots And Pencils$1 million to build their MVP? The CEO at this client said, “The public is going to go wild over our product, once they see it, so we need to be ready to scale quickly. I expect we’ll have 10 million users within the first 6 months.” But in fact, after a year, they had 40,000 users, and that was after spending another \$1 million on marketing. But hey, their app was Dockerized and set up in AWS Fargate, so they were ready for 10 million users, whenever those 10 million users decided to show up. None of this followed the real spirit of an MVP: there was nothing minimal here, instead, the project was contaminated with maximalism and perfectionism — the two biggest enemies of an early stage startup.

I wish I could say that this problem only happened at one of my clients, but in fact, nearly every client I’ve had these last few years have been swept up in a kind of perfectionist mania, chasing after expensive (but flawless!) implementations of ideas that had not yet been tested against real customers. This is not the correct way to build a company.

Beware hosted solutions that promise to remove all the devops pain from your life. The CEO felt certain that if Robots And Pencils could Dockerize the app, then it would be easy to achieve infinite scalability via AWS’s Fargate service. While that service is very good, and scales very well, it also turns out that you still need to write auto-scaling rules! Which they never got around to, because by the time they got to that point, they realized their app was not popular with the public. So they spent considerable money trying to achieve “easy scaling” and they never quite got there. Even more sad, or futile, or wasted, is the fact that their server app is a completely self-contained Go binary, with all dependencies included. So they could have easily achieved infinite scaling without Docker.

My basic point is, Docker/Kubenetes brings in a whole new ecosystem, with a lot of new worries, but you can get all the benefits via older Virtual Machine tech, and then script those older Virtual Machines with some mix of Ansible and Terraform (assuming you want to be in the cloud).

Whether you go bare metal or the cloud, MVP should be about a simple approach to devops.

High Availability is a beautiful ideal and every company should work toward that. Eventually, it is something you should achieve. Once you have proven product-market fit, then you can focus on reliability. But should a small startup begin with that, before you even know if the public likes your product? I say, leave High Availability till later. Don’t start with that. High Availability is not MVP, because MVP is about fast iteration of the main idea, not fast iteration of some devops strategy.

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# NE

(Redirected from Ne)

NE, Ne or ne may refer to:

## Science and technology

### Other uses in science and technology

• Electron density (${\displaystyle n_{e}}$)
• Ne, for Number English, a unit of measure for cotton fiber
• Neon, a chemical element with symbol Ne
• Inequality operator, a term used in programming languages as a "not equal to" operator
• Nash equilibrium, a solution concept in game theory
• Norepinephrine, a neurotransmitter


text:
## Sparsity Winning Twice: Better Robust Generalization from More Efficient Training

### Tianlong Chen · Zhenyu Zhang · pengjun wang · Santosh Balachandra · Haoyu Ma · Zehao Wang · Zhangyang Wang

[ Abstract ]
Thu 28 Apr 2:30 a.m. PDT — 4:30 a.m. PDT

Abstract: Recent studies demonstrate the deep networks, even robustified by the state-of-the-art adversarial training (AT), still suffer from large robust generalization gaps, in addition to the much more expensive training costs than standard training. In this paper, we investigate this intriguing problem from a new perspective, i.e., $\textit{injecting appropriate forms of sparsity}$ during adversarial training. We introduce two alternatives for sparse adversarial training: (i) $\textit{static sparsity}$, by leveraging recent results from the lottery ticket hypothesis to identify critical sparse subnetworks arising from the early training; (ii) $\textit{dynamic sparsity}$, by allowing the sparse subnetwork to adaptively adjust its connectivity pattern (while sticking to the same sparsity ratio) throughout training. We find both static and dynamic sparse methods to yield win-win: substantially shrinking the robust generalization gap and alleviating the robust overfitting, meanwhile significantly saving training and inference FLOPs. Extensive experiments validate our proposals with multiple network architectures on diverse datasets, including CIFAR-10/100 and Tiny-ImageNet. For example, our methods reduce robust generalization gap and overfitting by $34.44\%$ and $4.02\%$, with comparable robust/standard accuracy boosts and $87.83\%$/$87.82\%$ training/inference FLOPs savings on CIFAR-100 with ResNet-18. Besides, our approaches can be organically combined with existing regularizers, establishing new state-of-the-art results in AT. All codes are included.

Chat is not available.


text:
# Thread: Prove prime p divides infinitely many members of 1, 11, 111, 1111...

1. ## Prove prime p divides infinitely many members of 1, 11, 111, 1111...

Prove that if p is prime and p > 5, then infinitely many members of the sequence 1, 11, 111, 1111, ... are divisible by p.

I think I can do most of it. Since p > 5, I used Fermat's little theorem to show 10^(p-1) - 1 ≡ 0 mod p. Then 99...99 ≡ 0 mod p (p-1 nines), which means 11...11 ≡ 0 mod p (p-1 ones) since p > 5 so gcd(p,9) = 1.

That shows one member of the sequence is divisible by p, but how do I go from here to show there are infinitely many?

2. I think the rest is indeed easy. Let m = 11...11 (p-1 ones). Suppose n = 11...11 ≡ 0 (mod p). Then 10^(p-1) * n + m ≡ 0 (mod p).

3. Originally Posted by uberbandgeek6
Prove that if p is prime and p > 5, then infinitely many members of the sequence 1, 11, 111, 1111, ... are divisible by p.

I think I can do most of it. Since p > 5, I used Fermat's little theorem to show 10^(p-1) - 1 ≡ 0 mod p. Then 99...99 ≡ 0 mod p (p-1 nines), which means 11...11 ≡ 0 mod p (p-1 ones) since p > 5 so gcd(p,9) = 1.

That shows one member of the sequence is divisible by p, but how do I go from here to show there are infinitely many?
You actually did most of it...

Note that the numbers $9, 99, 999, ...$ are exactly the numbers of the form $10^k-1$, where $k$ is a positive integer.
As you've pointed out the $(p-1)th$ member in the sequence is divisible by $p$.

But what about any $k$th member, where $k$ is a multiple of $p-1$?

We see that $10^k\equiv1\pmod p$ since $p-1$ divides $k$ and so $10^k-1\equiv0\pmod p$. Lastly, since $(9,p)=1$ we may divide by $9$ to get $(10^k-1)/9\equiv0\pmod p$.

It works for every prime $p$ not equal to $2$ or $5$.


text:
CGAL 4.12 - 3D Boolean Operations on Nef Polyhedra

## Functions

template<class Nef_polyhedron , class Polygon_mesh >
void CGAL::convert_nef_polyhedron_to_polygon_mesh (const Nef_polyhedron &nef, Polygon_mesh &pm, bool triangulate_all_faces=false)
Converts an objet of type Nef_polyhedron_3 into a polygon mesh model of MutableFaceGraph. More...

template<class Traits >
istream & CGAL::operator>> (std::istream &in, CGAL::Nef_polyhedron_3< Traits > &N)
This operator reads a Nef polyhedron, which is given in the proprietary file format written by the input operator in and assigns it to N. More...

template<class Traits >
ostream & CGAL::operator<< (std::ostream &out, CGAL::Nef_polyhedron_3< Traits > &N)
This operator writes the Nef polyhedron N to the output stream out using a proprietary file format. More...

template<class Nef_polyhedron_3 >
std::size_t CGAL::OFF_to_nef_3 (std::istream &in, Nef_polyhedron_3 &N)
This function creates a 3D Nef polyhedron from an OFF file which is read from input stream in. More...

## ◆ convert_nef_polyhedron_to_polygon_mesh()

template<class Nef_polyhedron , class Polygon_mesh >
 void CGAL::convert_nef_polyhedron_to_polygon_mesh ( const Nef_polyhedron & nef, Polygon_mesh & pm, bool triangulate_all_faces = false )

#include <CGAL/boost/graph/convert_nef_polyhedron_to_polygon_mesh.h>

Converts an objet of type Nef_polyhedron_3 into a polygon mesh model of MutableFaceGraph.

Note that contrary to Nef_polyhedron_3::convert_to_polyhedron(), the output is not triangulated (but faces with more than one connected component of the boundary). The polygon mesh can be triangulated by setting triangulate_all_faces to true or by calling the function triangulate_faces().

Precondition
Polygon_mesh must have an internal point property map with value type being Nef_polyhedron_3::Point_3.
nef.simple()
Examples:
Nef_3/nef_3_to_surface_mesh.cpp.

## ◆ OFF_to_nef_3()

template<class Nef_polyhedron_3 >
 std::size_t CGAL::OFF_to_nef_3 ( std::istream & in, Nef_polyhedron_3 & N )

#include <CGAL/OFF_to_nef_3.h>

This function creates a 3D Nef polyhedron from an OFF file which is read from input stream in.

The purpose of OFF_to_nef_3 is to create a Nef polyhedron from an OFF file that cannot be handled by the Nef_polyhedron_3 constructors. It handles double coordinates while using a homogenous kernel, non-coplanar facets, surfaces with boundaries, self-intersecting surfaces, and single facets. Every closed volume gets marked. The function returns the number of facets it could not handle.

CGAL::Nef_polyhedron_3<Traits>
Examples:
Nef_3/complex_construction.cpp.

## ◆ operator>>()

template<class Traits >
 istream& CGAL::operator>> ( std::istream & in, CGAL::Nef_polyhedron_3< Traits > & N )

#include <CGAL/IO/Nef_polyhedron_iostream_3.h>

This operator reads a Nef polyhedron, which is given in the proprietary file format written by the input operator in and assigns it to N.

It includes the complete incidence structure, the geometric data, and the marks of each item.

It is recommended to use the CGAL kernels Homogeneous, Simple_homogeneous, or Extended_homogeneous parametrized with any exact number type that models $$\mathbb{Z}$$ (e.g.Gmpz or leda_integer). The input and output iterators of Nef polyhedra parametrized with either of these kernels are compatible as long as the Nef polyhedron is bounded. An unbounded Nef polyhedron can only be read by a Nef polyhedron parametrized with an extended kernel. It is also recommended to use the CGAL stream modifier set_ascii_mode().

CGAL::Nef_polyhedron_3<Traits>


text:
## Electronic Journal of Probability

### Subgaussian concentration and rates of convergence in directed polymers

#### Abstract

We consider directed random polymers in $(d+1)$ dimensions with nearly gamma i.i.d. disorder.  We study the partition function $Z_{N,\omega}$ and establish exponential concentration of $\log Z_{N,\omega}$ about its mean on the subgaussian scale $\sqrt{N/\log N}$ . This is used to show that $\mathbb{E}[ \log Z_{N,\omega}]$ differs from $N$ times the free energy by an amount which is also subgaussian (i.e. $o(\sqrt{N})$), specifically $O( \sqrt{\frac{N}{\log N}}\log \log N)$

#### Article information

Source
Electron. J. Probab., Volume 18 (2013), paper no. 5, 28 pp.

Dates
Accepted: 11 January 2013
First available in Project Euclid: 4 June 2016

https://projecteuclid.org/euclid.ejp/1465064230

Digital Object Identifier
doi:10.1214/EJP.v18-2005

Mathematical Reviews number (MathSciNet)
MR3024099

Zentralblatt MATH identifier
1281.82036

Rights

#### Citation

Alexander, Kenneth; Zygouras, Nikolaos. Subgaussian concentration and rates of convergence in directed polymers. Electron. J. Probab. 18 (2013), paper no. 5, 28 pp. doi:10.1214/EJP.v18-2005. https://projecteuclid.org/euclid.ejp/1465064230

#### References

• Alexander, Kenneth S. Lower bounds on the connectivity function in all directions for Bernoulli percolation in two and three dimensions. Ann. Probab. 18 (1990), no. 4, 1547–1562.
• Alexander, Kenneth S. A note on some rates of convergence in first-passage percolation. Ann. Appl. Probab. 3 (1993), no. 1, 81–90.
• Alexander, Kenneth S. Approximation of subadditive functions and convergence rates in limiting-shape results. Ann. Probab. 25 (1997), no. 1, 30–55.
• Alexander, Kenneth S. Power-law corrections to exponential decay of connectivities and correlations in lattice models. Ann. Probab. 29 (2001), no. 1, 92–122.
• Alexander, K. S. Subgaussian rates of convergence of means in directed first passage percolation. (2011) arXiv:1101.1549
• Borodin, A., Corwin, I., Remenik, D. Log-Gamma polymer free energy fluctuations via a Fredholm determinant identity (2012) arXiv:1206.4573
• Benaïm, Michel; Rossignol, Raphael. Exponential concentration for first passage percolation through modified Poincaré inequalities. Ann. Inst. Henri Poincaré Probab. Stat. 44 (2008), no. 3, 544–573.
• Benjamini, Itai; Kalai, Gil; Schramm, Oded. First passage percolation has sublinear distance variance. Ann. Probab. 31 (2003), no. 4, 1970–1978.
• Chatterjee, S., The universal relation between scaling exponents in first-passage percolation, arXiv:1105.4566
• Chatterjee, S., Disorder chaos and multiple valleys in spin glasses, (2009) arXiv:0907.3381
• Comets, Francis; Shiga, Tokuzo; Yoshida, Nobuo. Directed polymers in a random environment: path localization and strong disorder. Bernoulli 9 (2003), no. 4, 705–723.
• Comets, Francis; Yoshida, Nobuo. Directed polymers in random environment are diffusive at weak disorder. Ann. Probab. 34 (2006), no. 5, 1746–1770.
• Corwin, I., O'Connel, N.M., Seppäläinen, T., Zygouras, N., Tropical combinatorics and Whitaker functions, (2011) arXiv:1110.3489
• Graham, B. Sublinear variance for directed last passage percolation. Journal Th. Prob. (2010)
• Johansson, Kurt. Shape fluctuations and random matrices. Comm. Math. Phys. 209 (2000), no. 2, 437–476.
• Kardar, M., Parisi, G., Zhang, Y.C., Dynamic scaling of growing interfaces phPhys. Rev. Lett. 56 (1986), 889–892
• Kesten, Harry. On the speed of convergence in first-passage percolation. Ann. Appl. Probab. 3 (1993), no. 2, 296–338.
• Krug, J., Spohn, H., Kinetic roughening of growing surfaces ph In Solids Far From Equilibrium: Growth, Morphology and Defects (C. Godreche, ed.). Cambridge Univ. Press. (1991)
• Ledoux, Michel. The concentration of measure phenomenon. Mathematical Surveys and Monographs, 89. American Mathematical Society, Providence, RI, 2001. x+181 pp. ISBN: 0-8218-2864-9
• Wüthrich, Mario V. Scaling identity for crossing Brownian motion in a Poissonian potential. Probab. Theory Related Fields 112 (1998), no. 3, 299–319.
• Seppäläinen, Timo. Scaling for a one-dimensional directed polymer with boundary conditions. Ann. Probab. 40 (2012), no. 1, 19–73.
• Piza, M. S. T. Directed polymers in a random environment: some results on fluctuations. J. Statist. Phys. 89 (1997), no. 3-4, 581–603.


text:
# Find points along a Bézier curve that are equal distance from one another

I'm trying to figure out a generic way of determining a series of points on a Bézier curve where all the points are the same distance from their neighboring points. By distance I mean direct distance between the points not distance along the curve. See the image below

I've written a program that will solve this in an iterative fashion, but I'm wondering if there is a direct solution I could use.

My program starts by defining a circle of some initial radius (R) centered on the start point of the curve. It then intersects this circle with the curve to find the second point , which is R distance away from the start point). It then continues along the curve in this way finding points until it reaches the end of the curve. In most cases the distance between the last intersection point and the end point of the curve will not equal to R. The program then uses that difference to calculate a new value of R to try, and repeats the process.

-
Surely your routine will encounter an incommensurate Bézier (that is, given a particular Bézier curve and a distance, there will be a section of the curve that no longer has a point at the given distance). How would you want that handled? Otherwise, the use of crossing circles is indeed the most straightforward approach. –  Ｊ. Ｍ. Dec 30 '10 at 6:16
@J.M.: I guess that would fall under "reaching the end of the curve". –  Rahul Dec 30 '10 at 7:00
After re-reading (prompted by @Rahul's comment)... so you're actually looking for the distance for the Bézier to be commensurate? –  Ｊ. Ｍ. Dec 30 '10 at 7:05
Yes for a given bezier curve I'm looking for the distance that would hit the end of the curve. Or in other words I want to find the X number of points on the curve that include the start point and end point and where all the points are the same direct distance from their neighbors. –  Eric Anastas Dec 30 '10 at 20:27
This is an interesting question. I'm curious what prompted it. –  I. J. Kennedy Feb 5 '11 at 20:29

So you want the algorithm to "evenly divide" some given Bézier curve into a sequence of n points, such that traveling along the Bézier curve you hit each point in the same order, and such that the direct Euclidean distance from each point to the next point (the chord distance, not the arc distance along the Bézier curve) is the same?

I suspect that you might converge on the desired points a little faster if you pick n "t" values along the curve, and nudge all of them on every iteration -- rather than trying to fix one and going on to the next, and then throwing them all away and starting over every time you pick a new R value.

Say, for example, you want to divide the curve up into n = 100 points. The first point is going to stay fixed at $t_0$=0 and the last point is going to stay fixed at $t_{100}$=1. I'd probably start out at equally spaced t values, $t_0=0; t_1=1/100; t_2=2/100; ...; t_{99}=99/100; t_{100} = 1.$

At each iteration, calculate the distances between neighboring points, then nudge 98 of the t values (being careful not to cross them) to make the distances more equal, sooner or later converging on practically equal distances.

Off the top of my head, I seem to recall 2 methods for nudging those intermediate t values. Let me call the easier-to-understand method "the Babylonian method", and the faster-converging method "the false-position method". (Is there a better name for these techniques?)

## the Babylonian method

Nudge each point closer to halfway between its neighbors.

$d_{7,8}$ = distance( B($t_7$), B($t_8$) )

$r_8$ = (1/2) ( $d_{8,9} - d_{7,8}$ ) / ( $d_{7,8} + d_{8,9}$ )

Because distances are always positive, the ratio values $r$ are always in the range of -1/2 to +1/2 -- zero if $B(t_8)$ has equal distances from $B(t_7)$ and $B(t_9)$, -1/2 if $t_8$ needs to move "halfway" back to $t_7$, +1/2 if $t_8$ needs to be moved "halfway" to $t_9$.

If $r_8$ is positive, $newt_8 = t_8 + r_8(t_9 - t_8)$

If $r_8$ is negative, $newt_8 = t_8 + r_8(t_8 - t_7)$

## the false position method

Each iteration, calculate the accumulated straight-lines distance from the start point to the current location of each point B($t_0$) ... B($t_{100}$).

Nudge each point closer to its "proper" fraction of the distance along that entire accumulated distance.

For example, say (after a few iterations) we find that the accumulated distance from the start point $B(t_0)$ to point $B(t_8)$ is $c_8$ = 6.9, while the accumulated distance to the next point $B(t_9)$ is $c_9$ = 11.5, and the accumulated straight-lines distance from the start to the end is $c_{100}$ = 101.4. We need to make $t_8$ larger and $t_9$ smaller.

Using the false position method to estimate new values for all 98 "t" values, we get

error8L = 8/100 - $c_8 / c_{100}$

error8R = $c_9 / c_{100}$ - 8/100

If error8R is zero, it may help to push t8 all the way to t9. If error8L and error8R are both positive and approximately equal, it may help to push t8 halfway to t9.

$r_8$ = error8L / ( error8R + error8L )

The r value is the proportion of the distance to travel from t8 to t9 -- zero if t8 doesn't need to change this iteration, +1 if t8 needs to be moved all the way to t9.

$newt_8$ = $t_8$ + $r_8$($t_9$ - $t_8$).

If the r value along the segment from the current point to the next point is negative, we need to re-calculate a positive r value along the segment from the previous point to the current point, as we do for $t_9$ in this example:

error9L = 9/100 - $c_8 / c_{100}$

error9R = $c_9 / c_{100}$ - 9/100

$r_9$ = error9R / ( error9R + error9L )

$newt_9$ = $t_9$ - $r_9$($t_9$ - $t_8$).

One way to ensure we never cross the points is to clip each r value (whether in the forward or backward direction) to some maximum value slightly less than 1/2.

In our example, we need to slide $t_8$ a fraction we estimate as $r_8$ = 0.26 of the way "forward" from $t_8$ to $t_9$, while we need to slide $t_9$ a fraction we estimate as $r_9$ = 0.51 (perhaps clipped to 0.499) of the way "backward" from $t_9$ to $t_8$.

Alas, I've pulled these progressive refinement methods out of some dim memories. Perhaps something to do with building sine/cosine tables in one-degree increments?

I hope some kind commenter will remind me of the actual name of these methods, if (as is likely) I've mis-remembered the name or some crucial detail.

-
Thanks a lot! That does make a lot more sense then what I was trying. –  Eric Anastas Feb 28 '13 at 20:43
that helped.. can you please tell the way to find the coordinates of these equidistant points in the curve?? –  Kenpachi Mar 1 '13 at 9:27
@Kenpachi: The Babylonian method gives you a bunch of "t" values. The x,y coordinates on the curve at any t is given by plugging that t value into the Bezier function B(t). For example, if we want the x coordinate at t=0.1 of a typical cubic Bezier curve with control points P0, P1, P2, P3, we calculate B_x(0.1) = (1-0.1)^3*P0_x + 3*(1-0.1)^2*(0.1)*P1_x + 3*(1-0.1)*(0.1)^2*P2_x + (0.1)^3*P3_x. –  David Cary Mar 2 '13 at 1:03
thanks David :D –  Kenpachi Mar 2 '13 at 12:43
@DavidCary since I have used 2 control points to draw the curve with a start point and end point, and i dont see the use of those 4 values to calculate the equidistant points, how shall i do it..ive posted ques .. pls help math.stackexchange.com/questions/321293/… –  Kenpachi Mar 5 '13 at 10:04


text:
In an effort to gain a better understanding of vortex-induced vibrations (VIV), we present three-dimensional numerical simulations of VIV of circular cylinders. We consider operating conditions that correspond to a Reynolds number of $105$, low structural mass and damping ($m*=1.0$, $ζ*=0.005$), a reduced velocity of $U*=6.0$, and allow for two degree-of-freedom ($X$ and $Y$) motion. The numerical implementation makes use of overset (Chimera) grids, in a multiple block environment where the workload associated with the blocks is distributed among multiple processors working in parallel. The three-dimensional grid around the cylinder is allowed to undergo arbitrary motions with respect to fixed background grids, eliminating the need for grid regeneration as the structure moves on the fluid mesh.

1.
Pontaza
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J. P.
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H. C.
, 2004, “
Chimera Reynolds-Averaged Navier-Stokes Simulations of Vortex-Induced Vibration of Circular Cylinders
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text:
2.9

# Finding the meet of two lines

Finding the intersection, or meet, of two lines is simple geometrically–since the point where they cross is usually clear. However the same problem algebraically has an interesting history going back to ancient Chinese mathematicians.

In this step, you will see a clear cut algebraic strategy for solving a pair of general linear equations in $\normalsize{x}$ and $\normalsize{y}$.

## Solving a pair of equations

Suppose we want to solve the system

Note that we have numbered the equations so we can refer to them. Since each equation represents a line, any point $\normalsize{[x, y]}$ satisfying both equations must lie on both lines. Thus solving this pair of equations is the same as geometrically finding the intersection of two lines. We look for an equation where we can conveniently solve for one of the variables. Conveniently here means with a minimum use of fractions! In $\normalsize{(\ref{l2})}$ we see that

Now we substitute that value of y into the other equation:

and simplify

Having the value for $\normalsize{x}$, we go back to $\normalsize{(\ref{l3})}$ to find that

So our solution is $\normalsize{x=1}$ and $\normalsize{y=-2}$, or geometrically the point $\normalsize{[1,-2]}$. As pictured!

This is an important skill, so make sure you know how to do it by working through the following, if this is not already familiar to you.

Q1 (E): Solve $\normalsize{5x-2y=7}$ and $\normalsize{x+7y=31}$.

Q2 (E): Solve $\normalsize{4x+y=39}$ and $\normalsize{8x+7y=53}$.

Here are some questions that amount to the same thing, but are recast in different ways.

Q3 (E): Where do the lines $\normalsize{y=4x-6}$ and $\normalsize{y=-x+9}$ meet?

Q4 (M): I am thinking of two numbers, which add up to $\normalsize{99}$ and differ by $\normalsize{31}$. What are they?

## Answers

A1. The solution is $\normalsize{x=3}$ and $\normalsize{y=4}$.

A2. The solution is $\normalsize{x=11}$ and $\normalsize{y=-5}$.

A3. The lines meet at the point $\normalsize{[3,6]}$.

A4. This question should first be rephrased as two linear equations. Let the two numbers be $x$ and $y$. Then $x+y=99$ and $x-y=31$. These can be solved together to deduce that $x = \normalsize{65}$ and $y = \normalsize{34}$.

## Get a taste of this course

Find out what this course is like by previewing some of the course steps before you join:


text:
Gudhi Namespace Reference

## Classes

class  Euclidean_distance
Compute the Euclidean distance between two Points given by a range of coordinates. The points are assumed to have the same dimension. More...

struct  linear_indexing_tag
Tag for a linear ordering of simplices. More...

struct  Null_output_iterator

OFF file reader top class visitor. More...

OFF file visitor implementation according to Off_reader in order to read points from an OFF file. More...

OFF file reader implementation in order to read points from an OFF file. More...

OFF file visitor implementation according to Off_reader in order to read points from an OFF file. More...

class  Simplex_tree
Simplex Tree data structure for representing simplicial complexes. More...

struct  Simplex_tree_options_fast_persistence

struct  Simplex_tree_options_full_featured

## Functions

template<typename SimplicialComplexForProximityGraph , typename ForwardPointRange , typename Distance >
Proximity_graph< SimplicialComplexForProximityGraph > compute_proximity_graph (const ForwardPointRange &points, typename SimplicialComplexForProximityGraph::Filtration_value threshold, Distance distance)
Computes the proximity graph of the points. More...

void read_points (std::string file_name, std::vector< std::vector< double >> &points)
Read a set of points to turn it into a vector< vector<double> > by filling points. More...

template<typename Graph_t , typename Filtration_value , typename Vertex_handle >
Read a graph from a file. More...

template<typename Vertex_handle , typename Filtration_value >
bool read_simplex (std::istream &in_, std::vector< Vertex_handle > &simplex, Filtration_value &fil)
Read a face from a file. More...

template<typename Simplex_key , typename Filtration_value >
bool read_hasse_simplex (std::istream &in_, std::vector< Simplex_key > &boundary, Filtration_value &fil)
Read a hasse simplex from a file. More...

template<typename Filtration_value >
std::vector< std::vector< Filtration_value > > read_lower_triangular_matrix_from_csv_file (const std::string &filename, const char separator= ';')
Read a lower triangular distance matrix from a csv file. We assume that the .csv store the whole (square) matrix. More...

template<typename OutputIterator >
void read_persistence_intervals_and_dimension (std::string const &filename, OutputIterator out)

std::map< int, std::vector< std::pair< double, double > > > read_persistence_intervals_grouped_by_dimension (std::string const &filename)

std::vector< std::pair< double, double > > read_persistence_intervals_in_dimension (std::string const &filename, int only_this_dim=-1)

## Detailed Description

This file is part of the Gudhi Library. The Gudhi library (Geometric Understanding in Higher Dimensions) is a generic C++ library for computational topology.

Author(s): Pawel Dlotko

This program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.

This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.

You should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/.

## Function Documentation

template<typename SimplicialComplexForProximityGraph , typename ForwardPointRange , typename Distance >
 Proximity_graph Gudhi::compute_proximity_graph ( const ForwardPointRange & points, typename SimplicialComplexForProximityGraph::Filtration_value threshold, Distance distance )

Computes the proximity graph of the points.

If points contains n elements, the proximity graph is the graph with n vertices, and an edge [u,v] iff the distance function between points u and v is smaller than threshold.

Template Parameters
 ForwardPointRange furnishes .begin() and .end() methods. Distance furnishes operator()(const Point& p1, const Point& p2), where Point is a point from the ForwardPointRange, and that returns a Filtration_value.
template<typename Graph_t , typename Filtration_value , typename Vertex_handle >
 Graph_t Gudhi::read_graph ( std::string file_name )

Read a graph from a file.

Template Parameters
 Graph_t Type for the return graph. Must be constructible from iterators on pairs of Vertex_handle Filtration_value Type for the value of the read filtration Vertex_handle Type for the value of the read vertices

File format: 1 simplex per line
Dim1 X11 X12 ... X1d Fil1
Dim2 X21 X22 ... X2d Fil2
etc

The vertices must be labeled from 0 to n-1. Every simplex must appear exactly once. Simplices of dimension more than 1 are ignored.

template<typename Simplex_key , typename Filtration_value >
 bool Gudhi::read_hasse_simplex ( std::istream & in_, std::vector< Simplex_key > & boundary, Filtration_value & fil )

Read a hasse simplex from a file.

File format: 1 simplex per line
Dim1 k11 k12 ... k1Dim1 Fil1
Dim2 k21 k22 ... k2Dim2 Fil2
etc

The key of a simplex is its position in the filtration order and also the number of its row in the file. Dimi ki1 ki2 ... kiDimi Fili means that the ith simplex in the filtration has dimension Dimi, filtration value fil1 and simplices with key ki1 ... kiDimi in its boundary.

template<typename Filtration_value >
 std::vector > Gudhi::read_lower_triangular_matrix_from_csv_file ( const std::string & filename, const char separator = ';' )

Read a lower triangular distance matrix from a csv file. We assume that the .csv store the whole (square) matrix.

Square matrix file format:
0;D12;...;D1j
D21;0;...;D2j
...
Dj1;Dj2;...;0

lower matrix file format:
0
D21;
D31;D32;
...
Dj1;Dj2;...;Dj(j-1);

template<typename OutputIterator >
 void Gudhi::read_persistence_intervals_and_dimension ( std::string const & filename, OutputIterator out )

Reads a file containing persistence intervals. Each line might contain 2, 3 or 4 values: [[field] dimension] birth death The output iterator out is used this way: *out++ = std::make_tuple(dim, birth, death); where dim is an int, birth a double, and death a double. Note: the function does not check that birth <= death.

 std::map > > Gudhi::read_persistence_intervals_grouped_by_dimension ( std::string const & filename )
inline

Reads a file containing persistence intervals. Each line might contain 2, 3 or 4 values: [[field] dimension] birth death The return value is an std::map<dim, std::vector<std::pair<birth, death>>> where dim is an int, birth a double, and death a double. Note: the function does not check that birth <= death.

 std::vector > Gudhi::read_persistence_intervals_in_dimension ( std::string const & filename, int only_this_dim = -1 )
inline

Reads a file containing persistence intervals. Each line might contain 2, 3 or 4 values: [[field] dimension] birth death If only_this_dim = -1, dimension is ignored and all lines are returned. If only_this_dim is >= 0, only the lines where dimension = only_this_dim (or where dimension is not specified) are returned. The return value is an std::vector<std::pair<birth, death>> where dim is an int, birth a double, and death a double. Note: the function does not check that birth <= death.

Examples:
Bottleneck_distance/bottleneck_distance.cpp.
 void Gudhi::read_points ( std::string file_name, std::vector< std::vector< double >> & points )
inline

Read a set of points to turn it into a vector< vector<double> > by filling points.

File format: 1 point per line
X11 X12 ... X1d
X21 X22 ... X2d
etc

template<typename Vertex_handle , typename Filtration_value >
 bool Gudhi::read_simplex ( std::istream & in_, std::vector< Vertex_handle > & simplex, Filtration_value & fil )

Read a face from a file.

File format: 1 simplex per line
Dim1 X11 X12 ... X1d Fil1
Dim2 X21 X22 ... X2d Fil2
etc

The vertices must be labeled from 0 to n-1. Every simplex must appear exactly once. Simplices of dimension more than 1 are ignored.

 GUDHI  Version 2.1.0  - C++ library for Topological Data Analysis (TDA) and Higher Dimensional Geometry Understanding.  - Copyright : GPL v3 Generated on Fri Feb 2 2018 14:51:32 for GUDHI by Doxygen 1.8.11


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## Studia Mathematica

2004 | 160 | 2 | 129-156
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### Duality of matrix-weighted Besov spaces

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We determine the duals of the homogeneous matrix-weighted Besov spaces $Ḃ^{αq}_{p}(W)$ and $ḃ^{αq}_{p}(W)$ which were previously defined in [5]. If W is a matrix $A_{p}$ weight, then the dual of $Ḃ^{αq}_{p}(W)$ can be identified with $Ḃ^{-αq'}_{p'}(W^{-p'/p})$ and, similarly, $[ḃ^{αq}_{p}(W)]* ≈ ḃ^{-αq'}_{p'}(W^{-p'/p})$. Moreover, for certain W which may not be in the $A_{p}$ class, the duals of $Ḃ^{αq}_{p}(W)$ and $ḃ^{αq}_{p}(W)$ are determined and expressed in terms of the Besov spaces $Ḃ^{-αq'}_{p'}({A^{-1}_{Q}})$ and $ḃ^{-αq'}_{p'}({A_{Q}^{-1}})$, which we define in terms of reducing operators ${A_{Q}}_{Q}$ associated with W. We also develop the basic theory of these reducing operator Besov spaces. Similar results are shown for inhomogeneous spaces.
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129-156
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2004
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• Mathematics Department, Duke University, Durham, NC 27708, U.S.A.
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text:
# What value does the sum of the square of the reciprocals of the prime counting function converge to?

Using my only mathematical tool Wolfram Alpha, I noticed that $\sum_{n=2}^\infty \frac{1}{\pi(x)}$ seems to diverge. Naturally, the entered the following sum and, just like the zeta function, saw that it converges when each term is squared! The partial sums from WA suggest that it converges to a number greater than $e$ and would like to know if a closed form is known.

$$\sum_{x=2}^\infty \frac{1}{{\pi(x)}^2} = ?$$

I don't know if there is a more elementary way to do it but I am going to risk total overkill by using (part of) the Prime Number Theorem. There is a constant $c>0$, in fact any $c<1$ will do, such that $$\pi(x)>\frac{cx}{\log x}$$ for all sufficiently large $x$. Therefore $$\frac{1}{\pi(n)^2}<\frac{1}{c^2}\frac{(\log n)^2}{n^2}\ ,$$ for all large $n$, and the series converges by comparison with $$\sum_{n=2}^\infty\frac{1}{n^{2-\varepsilon}}\ .$$
In fact a similar argument will show that for any $a>1$, the series $$\sum_{n=2}^\infty\frac{1}{\pi(n)^a}$$ converges. For $a=1$, the first series you looked at, it does indeed diverge because the terms include all of the numbers $$\frac{1}{1}\,,\ \frac{1}{2}\,,\ \frac{1}{3}\,,\ldots\ ,$$ with many of them repeated.
I doubt that there's a closed form, but you might notice that $$\sum_{n=2}^\infty \dfrac{1}{\pi(n)^2} = -2 + \sum_{j=2}^\infty \left( \dfrac{1}{(j-1)^2} - \dfrac{1}{j^2} \right) p_j$$ where $p_j$ is the $j$'th prime.