text stringlengths 1 1.11k | source dict |
|---|---|
one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Latest Blog Post. The area of a parallelogram is equal to the magnitude of cross-vector products for two adjacent sides. We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Area of a parallelogram. The diagonal of the parallelogram OC will give the resultant vector. We consider area of a parallelogram and volume of a parallelepiped and the notion of determinant in two and three dimensions, whose magnitudes are these for figures with their column vectors as edges. 300+ VIEWS. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is asked Oct 9, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra The | {
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Instead of adding a small semicircle, we can move the contour so that it avoids the singularity.
Since $\frac{\sin(a\pi)}a\cos(at)$ is even in $a$, \begin{align} \int_0^\infty\frac{\sin(a\pi)}a\cos(at)\,\mathrm{d}a &=\frac12\int_{-\infty}^\infty\frac{\sin(a\pi)}a\cos(at)\,\mathrm{d}a\\ &=\frac14\int_{-\infty}^\infty\frac1a\left[\vphantom{\frac12}\sin(a(\pi+t))+\sin(a(\pi-t))\right]\,\mathrm{d}a\tag{1} \end{align} Since $\frac{\sin(a\lambda)}a$ is odd in $\lambda$, we can assume that $\lambda\gt0$ and account for sign later. | {
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cell-biology, neuroscience, neurophysiology
EDIT (Based on comments to Anne's answer)
Time is of course a valid advantage for having a single axon as an information conduction channel. However, the energy or the maintenance cost is the factor that I am actually interested in. Maintenance cost would increase with cell volume (in fact surface area) and having serial neurons would demand more energy (cumulative) than a single axon (including costs of maintaining a nucleus). However a long cell would need much higher number of molecular motors to maintain the traffic flow rate. All responses that require transcriptional control would be slow (such as response to injury). Moreover a single soma also imposes a limit on the number of mitochondria. There should be a limit on how long an axon can be. I am especially interested in case of big animals (with long hindlimbs) that also have a good reflex (perhaps camels, even horses). | {
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oxidation-state
Title: Iodate ion and Octet rule 'm trying to figure out why the oxidation state of the iodate ion is +5. The iodine atom has 7 electrons in its outermost shell (comprised of s and p sub-shells). Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom. This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?
And how is the charge on the iodate ion negative when the third oxygen requires an electron? First of all I strongly advise you to go through the following link as it will make my answer more clear to you .
Iodate ion
Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom.
The bond between the iodine atom and oxygen atom is a covalent bond . The electrons from the iodine atom are not given but shared .
This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule? | {
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homework-and-exercises, thermodynamics, temperature
My attempt :By the formula $$T(t)-T(s)=(T(0)-T(s))e^{-kt}$$
$\color{orange}{\text{Given condition}}$: It takes $t=60\ minutes $ for temperature fall $\color{red}{110^\circ \ C \to 10^\circ \ C}$
Setting the corresponding values, final temperature $T(t)=10^\circ\ C$ , initial temperature $T(0)=110^\circ\ C$ & surrounding body temperature $T(s)=60^\circ\ C$ we get $$10-60=(110-60)e^{-k(60)}$$
$$e^{-60k}=-1$$
$$-60k=\log(-1)$$
$$k=\frac{1}{-60}\log(-1)\tag 1$$
Now, for temperature fall $\color{red}{110^\circ \ C \to 30^\circ \ C}$ , setting $T(t)=30^\circ \ C$, $T(0)=110^\circ\ C$ & $T(s)=10^\circ\ C$ we get
$$30-10=(110-10)e^{-kt}$$
$$e^{-kt}=\frac{20}{100}=\frac{1}{5}$$
$$\implies -kt=\ln\left(\frac{1}{5}\right)$$
Setting the value of $k$ from (1), we get time $t$ as follows
$$t\frac{1}{-60}\log(-1)=\log \frac{1}{5}$$
that is $$t=\log \frac{1}{5} \frac{-60}{\log(-1)}$$ | {
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magnetic-fields
Title: A question about the magnetic field and magnetic poles of a solenoid Q: a solenoid with ends marked a and b is suspended so that the core can rotate in the horizontal plane. A current is maintained in the coil so that the electrons move clockwise when viewed from end A towards end B. How will the coil align itself in earth's magnetic field?
Now An answer I have (from an answer bank our teacher gave us) is "end A will point towards the geographic south pole."
Meaning that end A becomes the south pole of the magnet. Now I don't understand how. | {
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python
It's usually best to put your main code in a main function that you call from here.
ROOT_PATH = os.path.dirname(__file__)
config = ConfigParser.ConfigParser()
config.readfp(open(os.path.join(ROOT_PATH + '/settings.cfg')))
filename = config.get('defaults', 'file_location')
LOG_FILENAME = config.get('defaults', 'log_location')
LOG_FORMAT = '%(asctime)s - %(levelname)s %(message)s'
LOG_LEVEL = config.get('defaults', 'logging_level')
Convention says that ALL_CAPS is for constants. These aren't constants
#TODO Change datefmt to be consistent with Apache logfile
logging.basicConfig(filename=LOG_FILENAME, format=LOG_FORMAT, datefmt='%m/%d %I:%M:%S', level=eval(LOG_LEVEL))
sample_strings = []
sample_strings.append('''Jan 2 15:32:49 cam-1/10.0.0.1 Authentication: [01:26:b2:F8:39:27 ## 172.16.197.23] Anonymous - Successfully logged in, Provider: anon, L2 MAC address: 00:26:B2:F2:39:87, Role: User, OS: Mac OS 10.6''') | {
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function if there is an a > 0 such that f(x+a) = f(x), ∀x ∈ R. Even and odd functions. When integrating even or odd functions, it is useful to use the following A function's being odd or even does not imply differentiability, or even continuity. Theorem (Fourier Convergence Theorem) If f is a periodic func-tion with period 2π and f and f0 are piecewise continuous on [−π,π], then the Fourier series is convergent. • Even/Odd Symmetry. Now in this video I will briefly explained Fourier series in hindi #3 how to compute even and odd functions in fourier series examples in hindi |AEM Fourier claimed (without proof) in 1822 that any function f (x) can be expanded in terms of sines in this way, even discontinuous function! That is, these sine functions form an orthogonal basis for “all” functions! This turned out to be false for various badly behaved f (x), and controversy over the exact conditions for convergence of the The Fourier series is a tool for solving partial differential equations. | {
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# If x and y are positive integers and r is the remainder when (7^(4x+3)
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Math Expert
Joined: 02 Sep 2009
Posts: 58991
If x and y are positive integers and r is the remainder when (7^(4x+3) [#permalink]
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01 Oct 2018, 04:47
1
2
00:00
Difficulty:
55% (hard)
Question Stats:
54% (01:10) correct 46% (01:29) wrong based on 74 sessions | {
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• Calculating one entry in the matrix for $$A\odot B$$ requires computing $$(a_{i1}\wedge a_{1j})\vee (a_{i2}\wedge a_{2j})\vee\cdots \vee (a_{in}\wedge a_{nj})$$.
• So computing one entry takes $$\Theta(n)$$ steps, and calculating the whole matrix for a composition takes $$\Theta(n^3)$$.
• Our algorithm is:
procedure transitive_closure(R)
C = R
P = R
for k from 2 to n:
P = P ⊙ R # P = R^k
C = C ∨ P # C = R ∪ R^2 ∪ ... ∪ R^k
return C
• Since we do the $$\odot$$ operation $$n$$ times here, this algorithm has running time $$\Theta(n^4)$$.
• That's pretty bad, but the best I could do.
• Fortunately, there are people smarter than me. | {
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geology, petrology, igneous, weathering
Title: Are these igneous Liesegang rings on a popular Oahu, Hawaii hike? I hiked the Lanikai Pillbox trail (Oahu, HI). I noticed eroded concentric rings in the igneous rock with a hard center. I'm unable to find the geologic cause of these rings. Are they Liesegang rings? There were several instances (~20) I observed on the hike. Not knowing the setting and judging from the image alone in this trampled environment, I think we're looking at a form of chemical weathering called "spheroidal weathering".
The gist is this (from the linked wiki page):
Penetrating water alters the bed-rock along cracks or joints, causing volume changes between altered and unaltered parts. These differences result in the formation of spheroidal layers that are removed by erosion (or trampling), leaving a central relatively unaltered boulder/pebble.
Liesegang rings are somewhat enigmatic, not sure if the term is applicable here. | {
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F(r - 1) = D[r^(n + 1) - r]
D = F(r - 1) / [r^(n + 1) - r]
D = F(r - 1) / [r(r^n - 1)]
SO: D = 50000(1.0275^4 - 1) / [1.0275^4(1.0275^(4n) - 1] = 13362.60 | {
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java, object-oriented, game
public class Play extends BasicGameState{
private Image backgroundImage;
private UserInfo userInfo;
private Paddle paddle;
private Ball ball;
private Brick[] bricks;
private int numBricks = 3;
public Play(UserInfo userInfo){
this.userInfo = userInfo;
}
@Override
public int getID(){
return Game.play;
}
@Override
public void init(GameContainer gc, StateBasedGame sbg){
bricks = new Brick[numBricks];
try{
backgroundImage = new Image("res/background.jpg");
paddle = new Paddle("res/bat_yellow.png");
ball = new Ball("res/ball_red.png"); | {
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ros, rosbag, ros-kinetic, download
Originally posted by ahendrix with karma: 47576 on 2018-07-31
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by engyasin on 2018-07-31:
ok , that's interesting, I like the recorder overloading theory, because the camera always send images (why stop in the middle?), I will contact the the people that set these rosbag to make sure, .. if the error from rosbag downlaod , I will leave the question unanswered | {
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thermodynamics, kinetic-theory, reversibility, heat-engine, carnot-cycle
state was at $T_\mathrm{high}$, corresponding to the initial temperature of a high-temperature finite body—can the engine return to this original state after a cycle that removes heat from that body, bringing it to $T_\mathrm{high}-\delta T$. Is this correct? | {
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quantum-interpretations, bohmian-mechanics
Title: In pilot wave theory where is the wave? As a non specialist, for a single particle system it's easy to appreciate the concept of a pilot wave extending through all Euclidean space, guiding a particle which ends up at a location determined by the pilot wave and its initial location.
For multiple particles however the wave would presumably need more dimensions to reflect the configuration space of the system.
Is this correct, and if so where does the pilot wave reside?
A related question may be, if quantum computers give an exponential speedup for factorization, then according to pilot wave theory where does the computation take place? The "pilot wave" is just the usual wave function of quantum mechanics. | {
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general-relativity, differential-geometry, vectors, coordinate-systems
Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to $\mathbb{R}^n$, and nonlinear maps do not preserve the linear structure.
On a manifold, there is no sense in attempting to "vectorize" points. A point is a point, an element of the manifold, a vector is a vector, element of a tangent space at a point.
Of course you can map points into $n$-tuples, that is part of the definition of a topological manifold, but there is no reason why the inverse of this map should carry the linear structure over to the manifold. | {
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differential-geometry, tensor-calculus, differentiation, covariance
$$\mathfrak{r}=x^{i}\hat{\mathfrak{e}}_{i}$$
$$\frac{\partial^{2}\mathfrak{r}}{\partial q^{j}\partial q^{k}}=\mathfrak{\hat{e}}_{i}\frac{\partial^{2}x^{i}}{\partial q^{j}\partial q^{k}}$$ | {
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statistical-mechanics, entropy, topological-entropy
(Q3) What is the difference between Gibb's entropy and Shannon's entropy since the formula http://en.wikipedia.org/wiki/Entropy_%28statistical_views%29 is the same. Q1: The important thing to know is that there are several distinct concepts of entropy in statistical physics and mathematics and there is no "the entropy" (life is hard). Only in thermodynamics, the word entropy has clear meaning by itself, because there it is the Clausius entropy. To answer your question, in short: it can be shown that for macroscopic systems (large number of particles), in quasi-static processes that isolated system undergoes, the quantity $\ln V$ behaves as the Clausius entropy in thermodynamics, that is, it does not change as the process proceeds. So it is sometimes called entropy too (I do not think Boltzmann entropy is a good name for it, since it is not clear whether Boltzmann thought this to be "the entropy"; it is said he never wrote in his papers and was first written down by Max Planck). It | {
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"tags": "statistical-mechanics, entropy, topological-entropy",
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wavefunction, atomic-physics, orbitals
We thus modify the hydrogen Hamiltonian by introducing two new operators. One is the average Coulombic repulsion between electrons, and the other is the exchange interaction. However, because we're using the average position of the electrons, then for our spherical atom these operators don't have an angular dependence. Thus the spherical harmonics are still separable as in the hydrogen case, so roughly the shape of the orbitals must remain the same. The only part that can change is the radial part of the wavefunction. Doing the calculations, you'll see that the radial part of the wavefunctions are squeezed or stretched a little bit due to Coulombic repulsion and the exchange interaction between electrons, and the increased Coulombic attraction to the nucleus. But as Wikipedia says, qualitatively they don't change much until you introduce multiple atoms.
Without the mean-field approximation, I suppose even the angular shape would change, but that's beyond me. | {
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acid-base, experimental-chemistry, ph
So, obviously there is something wrong with my reasoning, since nobody calls those areas buffer zones. Is it because it is not useful to have such a highly acid/basic solution? Or is it just because technically, there is no buffering, just small $\rm pH$ changes? Or maybe something else entirely... I think the issue with buffer systems is that they are resistant to changes from strong acids/bases in their normal state. For example, biological systems often are buffered so that they have some resistance to changes in pH, which can wreck havoc on biochemistry.
Recall the basic buffer equation:
$\ce{HA <-> H+ + A-}$ | {
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telescope, telescope-lens
Astro friends: anyone know the largest telescope in the world which has [can have] a real mountable eyepiece through which you can look? Palomar 200-inch is the biggest I know. Others? I know Subaru can project images to a screen, but I mean real look-through-it eyepiece.
... and here's the reply:
In 2003, I had the privilege to see Mars through an eyepiece installed on UT3, almost at its opposition (25" apparent size), with a 0.5" seeing. It was marvelous!
(Another reply cited a blog post about an eyepiece on a 6.5 meter telescope.)
And, a picture of someone looking through one of the 6.5-meter Magellan Telescopes. Which leads me to wonder; it must have been incredibly impressive, but I wonder how much of that giant mirror's light can actually get routed through a teeny eyepiece? The focal plane is clearly much larger than a human pupil, otherwise it wouldn't be able to use those enormous collections of CCDs... | {
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quantum-mechanics, optics, photons, electrons, reflection
Title: Differences between absorption, transparency, reflection, and emission Can someone help me conceptualize the differences between a photon's involvement with absorption, transparency, reflection, and emission?
To be more specific, my current understanding of the matter is that when a photon interacts with an atom containing electrons (not a free electron), if the frequency is high enough, it can be absorbed, and the electron moves to a less-stable, yet higher energy level. In this sense, all other photons that were not absorbed are then reflected back outwards (giving the object its corresponding color). But if the frequency is too low to make the energy gap, the photon passes through the electron cloud and the atoms are transparent in the visual spectrum (as in glass, or air)... | {
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"url": null
} |
organic-chemistry, aromatic-compounds
Boron possesses a vacant $\mathrm{p}_z$ orbital, which can interact with the $\pi$-cloud of the benzene ring.
Considering the fact that boron is treated as a metalloid, it will definitely be less electronegative than carbon, so the dimethylborane group should offer a good +I effect, which could possibly outweigh the -M effects.
Can anyone shed some light on this issue? Are there any other examples of activating meta directors? Note: This requires an extremely high energy scenario to occur.
This is another example that looks promising, it is essentially an ylide that has enough EWing power to meta direct while giving some density to the ring in another resonance structure. []
The phosphonium ylide derivative (phenylphosphonic acid) may provide the same effect if the hypervalent nature of nitrogen is challenged, or if this substance is unobtainable under any circumstances. | {
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quantum-mechanics, condensed-matter, differential-geometry, gauge-theory, berry-pancharatnam-phase
The statement of the adiabatic theorem, ignoring the dynamical phase, is that the wave function at time $t$ is the result of parallel transportation of the wave-function at the initial time (initial vector, with respect to the Berry connection, along the curve of states induced by the curve in parameter space. | {
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# Finding limits using theorems
1. May 27, 2013
### Numnum
1. The problem statement, all variables and given/known data
Use theorems to find the limit:
$$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$
2. Relevant equations
Theorems like
$$f(x)=c$$ is continuous
$$f(x)=x$$ is continuous
$$\lim_{x\rightarrow 0} \cos(x)=1$$
$$\lim_{x\rightarrow 0} \sin(x)=0$$
$$\lim_{x\rightarrow a} \sin(x)=sin(a)$$
$$\lim_{x\rightarrow 0} \sin(x-a)=0$$
3. The attempt at a solution
I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?
2. May 27, 2013
### Dick
Good idea! Let u=x-1. You should also have a theorem about the limit of sin(u)/u as u->0.
Last edited: May 27, 2013
3. May 27, 2013
### Zondrina | {
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quantum-mechanics, hilbert-space, perturbation-theory, approximations
How did we introduce the $e^{ia\lambda}$ term in 11.21? I don't follow the explanation cited, especially due to the repetition of the second order terms.
What does it mean that its "phase can be chosen arbitrarily"? If I multiply the left side of 11.21 with $e^{ib}$, I would have to do the same on the right. Granted I can select $b$ such that the first term on the right has a net phase of zero, but what about the other terms on the right? | {
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experimental-physics, physical-chemistry, models
$$
which again fits the data very nicely. But having two different amplitudes is not a physical solution in the current model, so I'm thinking there's some effect that is still missing in the model.
So it would be great if someone could point me in the right direction here. I don't want a solution, I want ideas how to solve this type of problem. I want advice how to proceed methodically in order to get to the correct result. I would really appreciate that. It's very hard to say anything useful without being deeply steeped in the particulars of your experiment.
However, it sounds like a huge fraction of your conceptual space is revolving around exponential decay, but you're not really using one of the main tools used for understanding those trends - namely, logarithmic scales in your plots.
If the decay section of your plot is really exponential, then if you plot your data in a log-linear plot you should be able to get a nice clean linear asymptote in that plot. | {
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c++, file, console, windows, quiz
switch (choice) {
case Menu_settings_choice::change_color: menu_change_color(); break;
case Menu_settings_choice::set_learned: menu_set_learned(); break;
case Menu_settings_choice::reset_learned: menu_reset_learned(); break;
case Menu_settings_choice::set_languges: menu_set_languages(); break;
case Menu_settings_choice::exit: return;
}
}
}
void Menu_settings::menu_change_color()
{
enum class Menu_change_color_choice {
change_background = 1,
change_letters = 2,
return_to_settings = 3
}; | {
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object-oriented, interview-questions, swift
struct Requirement{
let neededTime : TimeInterval
let skillLevel : Employees
let device : Devices
let material : [ResourceAllocation]
}
struct Employees : OptionSet{
let rawValue: Int
static let junior = Employees(rawValue: 1 << 0)
static let senior = Employees(rawValue: 1 << 1)
static let manager = Employees(rawValue: 1 << 2)
static let all : Employees = [.junior, .senior, .manager]
}
struct Devices : OptionSet{
let rawValue: Int
static let expressoMaker = Devices(rawValue: 1 << 0)
static let coffeeMaker = Devices(rawValue: 1 << 1)
static let mixer = Devices(rawValue: 1 << 2)
}
enum Resource : String{
case milk
case sugar
case cream
case coffeeBag
case teaBag
}
struct ResourceAllocation{
let resource : Resource
let amount : Int
}
struct Snack{
} | {
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php, validation, mvc
$user->setName($username);
$user->setPassword($password);
$this->service->add($user);
$this->view->viewRegisterComplete($user);
return;
}
}
$this->view->viewRegisterForm();
}
Do you think it's a good idea to put the validation in the controller? My problem with the validation in the controller is that if the Model is (for some reason, so don't ask why) modified directly without using the Controller's function then the data wouldn't be validated. General thoughts | {
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hash
Title: Universal hash function for strings of unbounded length Is there a (weakly) universal hash function for strings without any assumption of the string length?
I did not find one on Google / Wikipedia. Any such hash function must have an infinite key.
Suppose that your hash function maps binary strings to some alphabet $\Sigma$ of size $m > 1$. We can identify binary strings with natural numbers, and so can think of the hash function as a function $h\colon K \times \mathbb{N} \to \Sigma$, where $K$ is the (finite) set of keys.
For each $n \in \mathbb{N}$, we can look at the function $h_n\colon K \to \Sigma$ given by $h_n(k) = h(k,n)$. There are finitely many possible functions $h_n$ (at most $m^{|K|}$), and so there must exist $n_1 < n_2$ such that $h_{n_1} = h_{n_2}$. In particular, $\Pr_k[h(k,n_1) = h(k,n_2)] = 1$, and so $h$ is not even a weakly universal function. | {
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cc.complexity-theory, np-complete, planar-graphs
A. Wigderson
The Complexity of the Hamiltonian Circuit Problem for Maximal Planar Graphs
Technical Report #298, Department of EECS, Princeton University, February 1982.
https://www.math.ias.edu/avi/node/820
In a maximal planar graph, every face is a triangle; hence the dual of a maximal planar graph is cubic. It is easy to see that the dual of a maximal planar graph is triply connected.
Therefore your problem is NP-complete. | {
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• Thanks for such a lucid explanation, the stars and bars approach didn't strike for me.. – Csj Apr 10 at 15:22
HINT (in case this is homework)
First of all, your method over-counts. E.g. if a segment is Amy - Bob - Charles - David - Emma, then you would have counted this $$3$$ times: once when you place B & C between A & E and then inserted D later, once when you place B & D between A & E and then inserted C later, and once when you place C & D between A & E and then inserted B later. If a segment has $$4$$ boys you would have over-counted even more. Unfortunately the "degree of over-counting" is not uniform so you cannot just divide by a factor.
It might be easier to count the case of indistinguishable girls and boys first. Since there are only $$3$$ boys leftover after assigning $$2$$ per "gap", you can probably do this by hand. You can also do this formally via stars and bars. Then you can permute the girls $$5!$$ ways the boys $$15!$$ ways. | {
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thermodynamics, kinetic-theory
Is this a correct approach, as far as an explanation is concerned?
P.S. also provide me with an alternative simpler explanation, which I am sure there is... The equation
$$ T = \frac{m\langle v^2 \rangle}{3 k_B} $$
for $\langle v^2 \rangle$ the average speed of a particle in the gas does not hold in frames other than the rest frame of the gas, where the rest frame is the one in which $\langle v \rangle = 0$ for all gas particles. This is because the derivation of this law assumes that the movement of the particles is "random", in particular, that there is no preferred direction, and "no preferred direction" is equivalent to $\langle v \rangle = 0$.
Hence, there is nothing to explain here, your statement | {
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c, graphics, sdl
I would break out of the loop here, since you're going to exit soon anyway.
for ( float theta = 0; theta < (PI * 2); theta += circle.step ) {
circle.new_x = circle.h + (circle.radius * cos ( theta ));
circle.new_y = circle.k - (circle.radius * sin ( theta ));
SDL_RenderDrawLine ( renderer, circle.old_x, circle.old_y,
circle.new_x, circle.new_y ); | {
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type-theory, functional-programming, dependent-types
Title: Defining an HTML Template as an Algebraic Type Wondering if/how you could define a highly nested structure as a Dependent Type (or an Algebraic or Parameterized type). Specifically, an HTML template. Not that they work like this (HTML templates don't have variables to plug in), but imagine a template like this:
<template id="MyTemplate">
<section>
<header>
<h1>{title}</h1>
<h2>{subtitle}</h2>
</header>
<div>{content}</div>
<footer>
<cite>{author}</cite>
<time>{year}</time>
</footer>
</section>
</template>
This is a template, so it basically acts like a class (or a type). So you would instantiate the type like this:
var node = new MyTemplate({
title: 'A title',
subtitle: 'A subtitle',
content: 'Foo bar ...',
author: 'foo@example',
year: 2018
}) | {
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graphs
/** Prints a path from the source to the specified vertex */
public void printPath(String endName) {
if (!graph.containsKey(endName)) {
System.err.printf("Graph doesn't contain end vertex \"%s\"\n", endName);
return;
}
graph.get(endName).printPath();
System.out.println();
}
/** Prints the path from the source to every vertex (output order is not guaranteed) */
public void printAllPaths() {
for (Vertex v : graph.values()) {
v.printPath();
System.out.println();
}
} | {
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java, beginner
Using a special value as to mark "uninitialized" fields robs you from a useful use case: if a user wants to use the application as a part of a script, they cannot just use it to encrypt any string privided by some other program, they have to check if the value is empty first. Java has a universal value for uninitialized fields: null.
The field names are unnecessarily short. A field containing a file name should be named as such. So instead write your field declarations like this:
String inputFileName = null;
String outputFileName = null;
String inputData = null; | {
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> 0then the functions cos nˇ L t and sin nˇ L t, n =1, 2, 3, are periodic with fundamental. Is there any way to solve that? Perhaps an alternative? Many thanks. Introduction to Fourier sine series and Fourier cosine series - Duration: 17:54. Mathematica for Fourier Series and Transforms Fourier Series Periodic odd step function Use built-in function "UnitStep" to define. Do exponential fourier series also have fourier coefficients to be evaluated. For a distribution in a continuous variable x the Fourier transform of the probability density. Then f has. There are countless types of symmetry, but the ones we want to focus on are. < tn ≤ 2L where f (t) is not differentiable, and if at each of these points the left and right-hand limits lim f (t) and lim f (t) exist (although they might not be equal). (Reversibility of Fourier transform for continuous functions) Let f and g be real- or complex-valued functions which are continuous and piecewise smooth on the real line, and suppose that | {
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c#, object-oriented, winforms
LoadProblems
GetDerivedTypesFor
FormatName
CreateInstanceByName
Though if for example FormatName calls some function and especially if it's only used by FormatName, then that function should be right after FormatName and before CreateInstanceByName. Basically for the most part (it's not a hard and fast rule but should be the case usually) a function/method should be defined right after its first use.
private static T CreateInstanceByName<T>(string className)
{
Assembly assembly = Assembly.GetExecutingAssembly();
Type type = assembly.GetTypes().First(t => t.Name == className);
return (T)Activator.CreateInstance(type);
}
I don't really do much reflection, but this seems unnecessary - for every name, you're going to re-get the types and reiterate through them. You already got all the types you wanted in GetDerivedTypesFor, so can't you just create an instance from those types directly?
ProblemExecutionTime.Text = problemOutput.ExecutionTime/1000d + @" seconds"; | {
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algorithms
The cost of getting an $x$ from the first container is \$20. This would satisfy one out of two of our list's requested items. If our strategy for satisfying the list was to brute-force the first container, then the second, the total cost of the list would be \$40.
What gives me trouble is that third choice. There is a 1/9 chance of spending only \$20 and getting both of our $x$ and $y$ in one move. There is a 4/9 chance of spending \$20 and getting either our $x$ or $y$ -- we could then get the remaining one for \$20 and be done for \$40. Finally, there is another 4/9 chance of getting nothing. | {
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ros, gazebo, python, state, ros-indigo
If you still didn't understand very well, I have created a video (https://youtu.be/WqK2IY5_9OQ) that will help you understand it.
Hope it helps.
Originally posted by Ruben Alves with karma: 1038 on 2017-11-08
This answer was ACCEPTED on the original site
Post score: 7 | {
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A' = D * V' /sqrt(N-1)
-.9017 -.3607 -.2843 .1976 -.1637 .0000 .0000 .0000 .0000 .0000
.7000 -.0662 .2108 -.0349 -.2932 .0000 .0000 .0000 .0000 .0000
-.2655 -.0360 .6329 .2515 .0407 .0000 .0000 .0000 .0000 .0000
-.2775 -.8679 .2104 -.2430 .0305 .0000 .0000 .0000 .0000 .0000
-.8970 .5902 .1978 -.2251 -.0856 .0000 .0000 .0000 .0000 .0000
Data are restored as X = S * A'
.8585 .0568 1.2111 1.1736 .3553
-1.2951 .9300 .0384 -.1677 -.9702
-1.0160 1.2390 .1478 1.0005 -1.1788
-.4922 -.1105 -1.0723 -.4102 -.5783
-1.1354 .9282 -.2975 -2.2278 -.1559
.8969 -1.0482 .4764 -.1490 1.9451
1.7236 -.7944 .1691 .3063 -.0586
-.6571 -.6302 .8333 -.3861 .5726
.5776 -.4670 -.6474 .0858 1.5036
.5391 -.1038 -.8589 .7746 -1.4349 | {
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python, mathematics
Title: Probability that Three Pieces Form a Triangle I wrote some code to estimate the solution to the following problem:
If you break a line segment at two random points, what is the probability that the new line segments can form a triangle?
The code is relatively simple, and it works. However, I can't shake the feeling that each of the three functions I've created could be written more "Pythonically." For example, the first function (can_form_triangle) explicitly writes out the combinations of pairs from (a, b, c) and then compares them to the element of (a, b, c) which is not in the pair. There must be a way to express this more succinctly, more elegantly — more Pythonically.
To be clear, I'd just like advice on refactoring the code to be more elegant/Pythonic. I am not looking for advice on how to change the actual functionality to solve the problem more efficiently.
Here's the code:
#!/usr/bin/python3
from random import random | {
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You could use Stirling approximation for $$x \to \infty$$:
$$x!\approx \sqrt{2\pi}\frac{x^{x+\frac{1}{2}}}{e^x}$$
So the limit becomes:
$$\lim_{x \to \infty} \frac{\sqrt{2\pi}\frac{(100x)^{100x+\frac{1}{2}}}{e^x}}{\sqrt{2\pi}\frac{(99x)^{100x+\frac{1}{2}}}{e^x}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{100x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{99x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}}=$$ $$=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(100x)^{99x}}{(99x)^{100x}}=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}$$
For infinites hierarchy $$x^x>>a^x$$ :
$$\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}=0$$
Note that: | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/3160931/find-the-limit-lim-n-rightarrow-infty-frac100n99n100nn"
} |
quantum-mechanics
Title: Why is Heisenberg picture used if Schrödinger picture is simpler? In Heisenberg picture the operators evolve in time instead of states, which evolve in time in Schrödinger picture. But this also means that to solve the equations (at least numerically) one has to work with $N^2$ numbers instead of Schrödinger's $N$ (i.e. Hamiltonian matrix vs state vector).
So, why, despite being dramatically more inefficient for calculations, is Heisenberg picture still used? There are a couple of types of situation in which the Schrodinger picture can be problematic. If you are working in a relativistic context you might not want to be tied down to one particular reference frame in which the states are a function of time and extend across all of space. | {
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inorganic-chemistry, everyday-chemistry, safety
My question is:
What hazards of desiccant silica gel are they concerned about that it is necessary to post the warning "THROW AWAY DO NOT EAT" on the packaging? Essentially you need to understand that the silica gel packet is just $\ce{SiO2}$. These packets are used to absorb moisture (water vapor) to protect goods from spoiling.
To be honest with you here, nothing ever happens when you eat $\ce{SiO2}$. You consume it all the time from food and water. But, the reason these packets are so dangerous is because they might contain impurities such as $\ce{CoCl2}$ (blue color) and $methyl~violet$. These colored products are used to indicate moisture. These products are carcinogenic and are mutatic poisons. In most cases, it will not be fatal immediately, but given time, yes.
Generally, if the silica gel is colored, it is highly likely that they might be very dangerous. So don't eat them. :) | {
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Since we must have $0, to find the extrema, we may equate:
$2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)=0$
$(8+x)-128\left(1+\frac{8}{x} \right)\left(\frac{1}{x^2} \right)=0$
$(8+x)-128\left(\frac{1}{x^2}+\frac{8}{x^3} \right)=0$
Multiplying through by $x^3$ and distributing, we find:
$x^4+8x^3-128x-1024=0$
Factor:
$(x+8)(x^3-128)=0$
Discarding the negative root, we have:
$x^3=128=2\cdot2^6$
$x=2^2\cdot\sqrt[3]{2}=4\sqrt[3]{2}$
I suspect you have set it up like Soroban did, so your critical value may be different from what I have.
9. ## Re: Application of maxima and minima
Oh yeahhh.! TY for the replies. I got the correct answer but I didn't realized that it just has a different form from the one given by the book.
Didn't know it til I checked their numerical values. I'm such a f*ol! | {
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notation
https://math.stackexchange.com/questions/159005/notations-for-array-operations
https://math.stackexchange.com/questions/20412/element-wise-or-pointwise-operations-notation Based on my experience, I would say that the standard notation is just to have a regular function, and specify that it applies element wise. For example, a common notation for activation functions is $\sigma$, so e.g. you could represent the activations of a regular dense layer as $\sigma(W x + b)$ where $x, b$ are vectors and $W$ is a matrix. I've never seen a special notation for specifying that the function $\sigma$ is applied element wise.
As you suggest in the question, if the function to be applied element wise is linear, then you can use either hadamard product, e.g. $a \circ x$ or the diag function $\text{diag}(a)x$. | {
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ajax, html5
$('#carousel1b').append(d);
}
else {
console.log("image" +i);
var d = "<div class='item'>";
d += "<a href='" +data.pictures[i]['photo'] +"' rel='prettyPhoto'>";
d += "<img src='" +data.pictures[i]['photo'] +"' alt=''>";
d += "</a>";
d += "</div>";
$('#carousel1b').append(d);
$("a[rel^='prettyPhoto']").prettyPhoto();
console.log('pp init');
}
}
}
})
</script> | {
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"tags": "ajax, html5",
"url": null
} |
ros-kinetic, roscpp, quaternion
In fact, I have a cv::Mat object and I want to convert it to a quaternion to populate a geometry_msgs/Pose message.
Comment by EDU4RDO-SH on 2020-07-28:
I'll try! Thanks.
In that case, personally, I'd lean towards constructing a tf2::Matrix3x3 from the cv::Mat. Then you can construct a tf2::Transform from the Matrix3x3, and then use tf2::toMsg to convert to a geometry_msgs/Pose directly. You could also call getRotation as I mentioned earlier to get a Quaternion directly, and then fill out the components of the pose.orientation with the components of the quaternion.
A quick Google also found this which might be helpful: https://gist.github.com/shubh-agrawal/76754b9bfb0f4143819dbd146d15d4c8
EDIT
Since this became the accepted answer, I figured I'd post a little example of what I described. Something like the following should work:
#include <opencv2/calib3d.hpp>
#include <opencv2/core.hpp>
#include <tf2/LinearMath/Matrix3x3.h>
#include <tf2/LinearMath/Transform.h> | {
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machine-learning, deep-learning, neural-network, cnn, convolutional-neural-network
Title: Why first fully connected layer requires flattening in cnn? One can read everywhere on internet or in books that in convoluted neural networks, between convolution layers and the first fully connected layer, you should flatten your data.
I managed to understand that Dense layer (=first fully connected layer) requires 1d (= flattened = linearized) data.
However, I failed to figure out WHY dense layer specificaly requires 1d data.
Could you share your explanation if you have a didactical one? Requiring a fully connected layer to only accept one dimensional (a vector) makes for a consistent interface between layers. Strict inputs makes the the code more straightforward. Otherwise a fully interconnected layer might have to accept arbitrary inputs (e.g., n-dimensional). | {
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"tags": "machine-learning, deep-learning, neural-network, cnn, convolutional-neural-network",
"url": null
} |
php, codeigniter
The sidebar-single.php partial (that displays the post image):
<div class="card-list-group card bg-light mb-3">
<h6 class="card-header text-dark">Featured Image</h6>
<div class="card-body p-0 bg-white">
<?php if (isset($post->post_image) && $post->post_image !== 'default.jpg'): ?>
<img src="<?php echo base_url('assets/img/posts/') . $post->post_image; ?>" alt="Main Image of <?php echo $post->title; ?>" class="img-fluid">
<?php else: ?>
<img src="<?php echo base_url('assets/img/posts/') . 'default.jpg'; ?>" alt="Default Post Image" class="img-fluid">
<?php endif ?>
</div>
<div class="card-footer p-2 bg-white text-center">
<a href="#<?php echo isset($post->post_image) && $post->post_image !== 'default.jpg' ? '' : 'imageUploader' ?>" <?php echo isset($post->post_image) && $post->post_image !== 'default.jpg' ? 'data-pid="' . $post->id . '"' : '' ?> id="postImage" class="smooth-scroll"> | {
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in an Infinite Circular Well; From Bohm to Classical Trajectories in a Hydrogen Atom; Chaos (Wolfram MathWorld). org/trac/boost/changeset/42140 Log: Added expint docs. The implementation used is that of GSL. com/ for more information. Note that the second term is exactly zero for integer v; to improve accuracy the second term is explicitly omitted for v values such that v = floor(v). Bessel's equation arises when finding separable solutions to Laplace's equation and the Helmholtz equation in cylindrical or spherical coordinates. It shows that R is a viable computing environment for implementing and applying numerical methods, also outside the realm of statistics. Here we consider yet another way to get the solutions { we can use an integral transform (like the Fourier transform, or Laplace transform) to simplify the ODE. 9 and 10 of Ref. All of the definitions used here for Bessel functions, and spher-ical Bessel functions, are contained in Chaps. Bessel Function Calculator: x: n: J 0 | {
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"openwebmath_score": 0.834372341632843,
"tags": null,
"url": "http://scdw.ajte.pw/mathworld-spherical-bessel-function.html"
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congruent. In this section we will consider two more cases where it is possible to conclude that triangles are congruent with only partial information about their sides and angles. Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent. The correct option is the AAS theorem. Proof: You need a game plan. Yes, AAS Congruence Theorem 11. Thus the five theorems of congruent triangles are SSS, SAS, AAS, HL, and ASA. Since AC and EC are the corresponding nonincluded sides, ABC ≅ ____ by ____ Theorem. Show Answer ∆ ≅ ∆ ≅ ∠ Example 2. Congruence and Congruence Transformations; SSS and SAS; ASA and AAS; Triangles on the Coordinate Plane; Math Shack Problems ; Quizzes ; Terms ; Handouts ; Best of the Web ; Table of Contents ; ASA and AAS Exercises. Given I-IF GK, Z F and Z K … You also have the Pythagorean Theorem that you can | {
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# How to show this matrix is positive semidefinite?
Let
$$K=\begin{pmatrix} K_{11} & K_{12}\\ K_{21} & K_{22} \end{pmatrix}$$
be a symmetric positive semidefinite real matrix (PSD) with $K_{12}=K_{21}^T$. Then, for $|r| \le 1$,
$$K^*=\begin{pmatrix} K_{11} & rK_{12}\\ rK_{21} & K_{22} \end{pmatrix}$$
is also a PSD matrix. Matrices $K$ and $K^*$ are $2 \times 2$ and $K_{21}^T$ denotes the transpose matrix. How do I prove this?
• I think this question needs the self-study tag. – Michael Chernick Jan 9 '18 at 19:34
• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung Jan 9 '18 at 21:54
• If K is 2x2, does that mean K_21 is a scalar? If so, why are you talking about its transpose? – Acccumulation Jan 9 '18 at 22:51
## 2 Answers
This is a nice opportunity to apply the definitions: no advanced theorems are needed. | {
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homework-and-exercises, momentum, dirac-matrices, trace
Since $p$ doesn't carry spinor indices.
$p_{2\rho}p_{1\sigma}$Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$
using ${\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu}=4g^{\rho\mu}$
Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$= Tr$(4g^{\rho\mu} \gamma^{\sigma}\gamma_{\mu})$=$2$ Tr$( \{ \gamma^{\rho},\gamma^{\mu}\} \gamma^{\sigma}\gamma_{\mu})$
$\implies2$ Tr$( \gamma^{\rho}\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu})$ $+$ $2$ Tr($\gamma^{\mu}\gamma^{\rho} \gamma^{\sigma}\gamma_{\mu})$=$2$Tr$(\gamma^{\rho}\times-2\gamma^{\sigma})$+$2$Tr$(4g^{\rho\sigma})$ where I have used $\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu}=-2\gamma^{\sigma}$
Since $g^{\rho\sigma}$ doesn't have any spinor indicies we can take it out of second trace.
$\implies-4$Tr$(\gamma^\rho\gamma^\sigma)+8g^{\rho\sigma}$Tr$(\delta_{\alpha\beta})=-16g^{\rho\sigma}+32g^{\rho\sigma}$ | {
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java, sudoku
fullJoiner.add("|-------+-------+-------|");
}
j++;
} else if (i % UNIT == 0) {
lineJoiner.add("|");
}
i++;
}
return fullJoiner.toString();
}
} | {
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"openwebmath_score": null,
"tags": "java, sudoku",
"url": null
} |
python, html, error-handling, form, django
The form template (add.html):
{% extends "library/base_site.html" %}
{% block title %}Add book{% endblock %}
{% block content %}
{% if added %}
<div class="alert alert-success" role="alert">
<button type="button" class="close" data-dismiss="alert">
<span aria-hidden="true">×</span>
<span class="sr-only">Close</span>
</button>
Succesfully added book "{{ added.title }}" by {{ added.author }} - ISBN {{ added.isbn }}
</div>
{% endif %}
<form role="form" method="post" action=".">
{% csrf_token %}
<div class="form-group">
<label for="id_isbn">Isbn</label>
<input id="id_isbn" class="form-control" maxlength="13" name="isbn" type="text" placeholder="Enter ISBN or scan barcode">
</div>
<button type="submit" class="btn btn-default">Submit</button>
</form>
{% endblock %} | {
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"url": null
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to display this calculator. Simplifying Imaginary Numbers. So, $$i = \sqrt{-1}$$, or you can write it this way: $$-1^{.5}$$ or you can simply say: $$i^2 = -1$$. 2. Simplify functions thanks to their properties. + ix55! Learn how to simplify imaginary numbers with large exponents in this video. (2x-5)^{\frac{1}{3}}=3. Imaginary and complex numbers--how to add, subtract, simplify. OR. See if you can solve our imaginary number problems at the top of this page, and use our step-by-step solutions if you need them. $$-\sqrt{-225}$$ Add To Playlist Add to Existing Playlist. Square Roots, odd and even: There are 2 possible roots for any positive real number. Algebra calculator mathpapa. Free imaginary number calculator online | [email protected] Polar to rectangular online calculator. A number is real when the coefficient of i is zero and is imaginary And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square | {
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"openwebmath_score": 0.710343599319458,
"tags": null,
"url": "http://nagelstudio-ramona.nl/hae94y/3a8365-simplify-imaginary-numbers-calculator"
} |
java, swing, animation, awt
return rhs;
})
.orElseThrow(AssertionError::new);
final int deltaX = dx / ANIMATION_FRAMES;
final int deltaY = dy / ANIMATION_FRAMES; | {
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gravitational-waves
Title: What is the shape of a gravitational wave form? What is the shape of a gravitational wave as it hits the Earth, particularly the time portion.
Does time start at normal speed, then slow slightly, and then return to normal speed?
Or does it start at a normal speed, slow down slightly, then speed up slightly, and then return to normal speed?
Those other questions only concerned whether time dilation exists. I'm more concerned with the shape of the wave form. So not the same questions at all. A simple monochromatic gravitational plane wave has two possible transverse polarizations. If the wave is traveling in the $+z$ direction, then the metric for one of the polarizations can be written in the simple form
$$ds^2 = -dt^2 + (1+h_+)\,dx^2 + (1-h_+)\,dy^2 + dz^2$$
where the small metric perturbation $h_+$ is wavelike:
$$h_+ = A_+ \cos{(k z-\omega t)}.$$
The metric for the other polarization is similar, but rotated by 45 degrees in the $x$-$y$ plane. | {
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compilers, program-analysis
Let $Stmt$ be the set of statements occurring in a program and let $(\wp(Stmt),\subseteq)$ be the lattice of subsets of statements ordered by subset inclusion. Let us assume for simplicity that a statement in $Stmt$ is either an assignment $x:= e$ or a Boolean expression $b$ (representing the entry of a conditional or head of a loop).
A control flow graph has the form $G = (Stmt,E)$ where $E$ is an edge relation over statements. The following definitions are extremely standard:
$succ: \wp(Stmt) \to \wp(Stmt)$ maps a set of statements to their successors.
$succ(X) = \{ y \in Stmt | \text{ there exists } x \text{ in } X \text{ such that } (x,y) \text{ is in } E\}$
$pred: \wp(Stmt) \to \wp(Stmt)$ maps a set of statements to their predecessors.
$pred(X) = \{ y \in Stmt | \text{ there exists } x \text{ in } X \text{ such that } (y,x) \text{ is in } E\}$
It is a standard exercise to show that these are monotone functions. | {
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javascript, jquery
JS: in the comments
//protect your code inside an immediate function
//that way, you can get away with all the crazy stuff you do
//while protecting yourself from all the crazy stuff other codes do
;(function (window, document, $, undefined) {
//for constants, I'd declare then up top so they are easily configurable
var animationSpeed = 250 //this one's for animation speed
, fn = {} //this is for our methods, explained later
, offset = 12 //the offset caused by the triangle
, dataName = 'tooltip' //our data name
;
//our tooltip methods
fn.show = function (html) {
//cache frequently used values to avoid refetching
var element = $(this)
, tooltip
, offset
, visible
, same
;
//this condition will return true if there was no data found
//thus no tooltip was created beforehand
if (!(tooltip = element.data(dataName))) { | {
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} |
keras, tensorflow
Title: Why does Keras UpSampling2D layer cast to float32? When using the Keras UpSampling2D layer to resize a float64 array the result is cast to float32. The following code snippet demonstrates this behavior.
import numpy as np
from keras import layers
arr = np.ones((4,4), np.float64).reshape((1,4,4,1))
print(arr.dtype) # float64
upsample = layers.UpSampling2D(interpolation='bilinear')
arr_upsampled = upsample(arr)
print(arr_upsampled.dtype) # float32
Why is this happening, and is it possible to keep the resulting tensor float64? This is what happens: UpSampling2D invokes keras.backend.resize_images, which invokes tensorflow.image.resize. In its documentation, we find that:
The return value has type float32, unless the method is ResizeMethod.NEAREST_NEIGHBOR, then the return dtype is the dtype of images: | {
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angular-momentum, rotational-dynamics, reference-frames
However for some arbitrary point the angular momentum will generally not be constant because you will be working in a non-inertial frame. The advantage of choosing the centre of mass as the reference point is that this selects a frame that is inertial so the angular momentum will be constant, and this generally makes analysing the system much simpler.
Unless you have a very good reason to do otherwise you should choose the center of mass as the origin for the position vector. | {
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} |
finite-automata
Title: ε-NFA to DFA - initial state with only epsilon transitions I am having trouble discovering how to convert a ε-NFA to DFA (image below) when all transitions in the initial state are epsilon transitions. I already know how to convert ε-NFA to DFA (common cases), but this case I never saw before.
Thank you guys! Here is an algorithm to convert an $\epsilon$-NFA to an equivalent DFA. | {
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analytical-chemistry, solutions, units
Title: Problem with determining the mass value of a soil sample for gravimetric calculations I'm having some trouble with how to determine the representative mass value of a prepared soil sample for calculating its $\ce{Cl-}$ concentration in ppm (mg/kg). | {
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f(x,y) and g(x,y) when (x,y) is close to. Linearizing two variable function. A new iterative linearization approach for solving nonlinear equations systems Nonlinear equations arise frequently while modeling chemistry, physics, economy and engineering problems. B = A + B and we can show this using the following table. The limit calculator allows the calculation of the limit of a function with the detail. Stone Aug 3 '19 at 17:54. Theorem 10. When the variables are stacked with $$\{v_k\}$$ following $$\{u_k\}$$, the upper and lower bandwidths are $$N$$. 2 Consider the trigonometric function $\sin x$. cos (1 + 2)x −1/2. " Let's remember how exponents work. If they are equal, the process is somewhat more complex. For any differentiable function, the. Specify ParameterValues only when Type = 'Function' and your block linearization function requires input parameters. To start practicing, just click on any link. Stewart/Clegg/Watson Calculus: Early Transcendentals, 9e, is now published. Auto | {
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Mr. Dylan drew 2 hearts, 16 stars, and 16 circles. Below is a basic ratio word problem as an example of how to solve to find the missing value. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Ratio Worksheet Answer Page. Detailed solutions and full explanations to ratio maths problems for grade 9 are presented. Example of a true proportion: 4 24 3 18 = We know this because 4 x 18 = 3 x 24. It can be written with a colon ( 1 : 5 ) , or using the word "to" ( 1 to 5 ) , or as a fraction: 1 5 Example 1: A backyard pond has 12 sunfish and 30 rainbow shiners. ISBN 1-57685-439-6 1. Using proportions to solve word problems This lesson is the continuation of the lesson Proportions of this module. A proportion sets two ratios equal to each other. The number of rotations of the second gear has then to be worked out. Now, I think we're ready to figure out how many total girls there are. Erik ran kilometer more than | {
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python
Remove decimal; you don't need it here.
Add friendly names for compounds in comments.
Consider using the Unicode point for subscript-2 and 3 in your output.
Consider using milligrams instead, since they can show more accuracy with fewer characters in this case.
Since clarity is of paramount importance in pharmacology, write out your two reactions in full.
Don't bother making a constant for a newline.
SI convention is to have a space between the quantity and the unit. | {
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bash, shell, rags-to-riches
Title: Rename files by editing their names in an editor Inspired by a recent question to rename files by editing the list of names in an editor, I put together a similar script.
Why: if you need to perform some complex renames that are not easy to formulate with patterns (as with the rename.pl utility), it might be handy to be able to edit the list of names in a text editor, where you will see the exact names you will get.
Features:
Edit names in a text editor
Use the names specified as command line arguments, or else the files and directories in the current directory (resolve *)
Use a sensible default text editor
According to man bash, READLINE commands try $VISUAL or else $EDITOR -> looks like a good example to follow.
Abort if cannot determine a suitable editor.
Abort (do not rename anything) if the editor exits with error
Paths containing newlines are explicitly unsupported
Perform basic sanity checks: the edited text should have the same number of lines as the paths to rename | {
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logic, first-order-logic, propositional-logic, natural-deduction
Theoremhood for a first-order theory in (classical or intuitionistic) first-order logic is generally not computable. Some particular first-order theories are decidable, but most aren't. They are at least semi-decidable via the argument in the first paragraph. There are, of course, tons of automated theorem provers that will attempt to find proofs, but they may fail. For theorems, they may fail due to resource limitations1. For non-theorems, they just don't have a way of showing that something is contingent. They could find a proof for the negation of your formula, but that technically doesn't rule out your formula being a theorem since your theory could be inconsistent. Model checking techniques could be used to attempt to find counter-models to show that some formula can't be a theorem (but may still be contingent). Most model checking tools are not aimed at full first-order logic though. They have essentially the same theoretical and practical limitations as automated theorem | {
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javascript, node.js, ecmascript-6, asynchronous, cache
mapping.delete(key); // disable mapping for current key
// re-simulate the access, async
access(key,function(newData){
callback(newData);
},aType,value);
});
}
else
{
mapping.delete(key); // disable mapping for current key
access(key,function(newData){
callback(newData);
},aType,value);
}
}
}
else // slot life span has not ended
{
bufVisited[slot]=1;
bufTime[slot] = Date.now(); | {
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"url": null
} |
javascript, array, json
if (m.sender_name === personName) {
stat.messageCount++;
if (m.photos) stat.photoMessageCount++;
if (m.gifs || m.videos) stat.videoGifMessageCount++;
if (m.audio_files) stat.audioMessageCount++;
}
});
stat.richContentMessageCount = stat.photoMessageCount + stat.videoGifMessageCount
+ stat.audioMessageCount;
stats.push(stat);
}); | {
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and bn. The corresponding analysis equations for the Fourier series are usually written in terms of the period of the waveform, denoted by T, rather than the fundamental frequency, f (where f = 1/T). Show that the Laplace transform of the half rectified sine-wave function of period 2π, is. Fourier Cosine series. Then the Fourier cosine series for f(x) is the same as the Fourier series for fo(x) (in the sense that they look exactly the same). Like Example Problem 11. Example of Rectangular Wave. A Fourier Series will describe the output as a function of the input and the output will scale with the input as above. well as by the subject this qestion is about fourier series. | {
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gscam, image-transport
Title: gscam - "illegal operation"
I'm trying to get a camera working on a "roboard" (i386) which is running Ubuntu 10.04 with a full ROS diamondback install.
gscam is falling down at the following line (in http://brown-ros-pkg.googlecode.com/svn/trunk/experimental/gscam/src/gscam.cpp ), and only returning illegal operation.
image_transport::CameraPublisher pub = it.advertiseCamera("gscam/image_raw", 1);
This setup works on a laptop running the same code, has anyone else had this problem or know a solution?
Originally posted by alexsleat on ROS Answers with karma: 41 on 2011-07-01
Post score: 0 | {
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algorithms, optimization, dynamic-programming, scheduling
There is also a $O(n^2)$ solution based on a matrix of subproblems. Define $OPT(i, j)$ as the maximum from day $i$ to $n$ with last reboot done $j$ days before, more exactly on $i-j$ day. For every day $i$ you have to consider these alternatives:
you have a reboot. So on day $i$ you cannot process anything: $OPT(i,j) = OPT(i+1, 1)$
go on with processing: $OPT(i,j) = \min\left\{s_j, x_i\right\} + OPT(i+1, j+1)$
Because it's of no use reboot on last day, you have $OPT(n,j) = \min\left\{s_j, x_n\right\}$.
This last approach is taken from Mangat Rai's work, a user that shared a collection of solutions to Kleinberg's book problems, although the first one i reported is also very close to what you can find there. I suggest you to do a web search for it (no official site as far as i know). | {
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convolutional-neural-networks
Basically, you would train your network with lots of A, B examples, with the input as B and desired output as A.
The limitation is that where you have transformations that are technically irreversible, then the CNN may learn to produce a best "mean" output. The symptom of this will be fuzzy images lacking high frequency components, and matrices which are not representative of the target distribution, but that do solve the transformation (within limits of training accuracy). If you want to improve on that, and produce a more realistic/precise original, then you will probably want to look at adding a generative component - a GAN, VAE/GAN or RBM etc. Note this would not accurately produce the original matrix, but would generate one that both transformed into your given transformed matrix (within some level of accuracy) and was sampled from your input distribution. That is, it could be more of a feasible original than one generated using a simpler CNN architecture. | {
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quantum-mechanics, condensed-matter, topological-order, second-quantization, anyons
of freedom can be written as a product of anyon creation/annihilation operators. This can be explicitly realized in exactly solvable models such as the Toric Code or the Kitaev Honeycomb model. So the answer to whether anyons have creation and annihilation operators is yes. | {
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the-sun
Title: How long does a sunrise or sunset take? From the time that the sun appears on the horizon, or meets it on its setting, to the time that it is fully visible, or no longer visible on its setting, how much time passes?
Secondly, is there a place in the world where a sunrise/sunset occurs over a period of a few days? Meaning, that from the time it begins to appear over the horizon until it is fully visible, a period of a few days pass without night intervening (and the same for the opposite with sunset)? The time it takes depends on various factors: the angle that the path of the sun makes with the horizon is the main one, though there are also optical effects caused by the atmosphere have an effect too.
Generally the closer to the equator you live, the steeper is the angle, and so the faster is the sunset.
Using Stellarium I did a couple of tests: | {
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Question
Polynomial arithmetic
DISCOVER: Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x +5$$
$$Q(x) = (((3x - 5)x + 1)x 3)x + 5$$
Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial
R(x) =x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x + 4\) in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head.
Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division? | {
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"url": "https://plainmath.net/1515/polynomial-polynomials-polynomial-polynomial-calculating-calculating"
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computer-networks, integrity
P.S: I hope "confidentiality" and "integrity" are the right terms in English. We learned "Vertraulichkeit" & "Integrität" in German. Confidentiality means that only the sender and receiver know the message. We ensure confidentiality by encrypting the message.
Integrity roughly means that the message being sent cannot be modified in transit. In more detail, it means that there is a way for the receiver to check that the message has not been tampered with. We ensure integrity by adding a hash or checksum to the message.
The two properties are completely independent: | {
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ros, openni.launch
What is wrong?
Thank you very much for your help.
Originally posted by Chik on ROS Answers with karma: 229 on 2013-04-07
Post score: 1
Thank you very much, Philip. I discovered the reason. I am using Electric in Turtlebot laptop but Fuerte in workstation. Perhaps the datatype changes when the ROS version changes. I run dynamic_reconfigure from the turtlebot laptop, then it is okay. And in rviz (run from workstation) the registered point cloud looks very nice.
Originally posted by Chik with karma: 229 on 2013-04-07
This answer was ACCEPTED on the original site
Post score: 1 | {
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Letting $$H$$ be the height of the medium-sized right triangle, then on the one hand, we have $$\cos\theta=\frac{H}{l},$$ so that $$H=l\cos\theta.$$ On the other hand, $$\cos\theta=\frac{l}{H+h}=\frac{l}{l\cos\theta+h},$$ so $$h\cos\theta+l\cos^2\theta=l,$$ so $$h\cos\theta=l(1-\cos^2\theta),$$ which is not the same as $$h=l(1-\cos\theta).$$ Instead, we can see that $$h\cos\theta=l\sin^2\theta,$$ so that $$h=l\tan\theta\sin\theta.$$
Since $$x=l\sin\theta,$$ then this is exactly equivalent to $$h=x\tan\theta.$$ | {
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ros-melodic
The setup discussed in ros-industrial/abb_driver#3 indeed runs two driver instances per robot (in your case), which exposes two sets of action servers per robot. This removes the possibility to synchronously move all axes. If that's a problem, another solution must be found. If it's not a problem, then depending on how you interact with those action servers, it can be made to look like it's a single robot (MoveIt fi should be able to deal with that, but I'd have to verify, alternatively, a custom node could be created which splits a JointTrajectory in two parts, but presents a single action server to your application). From a JointStates perspective it's not difficult to pretend there's only a single mechanism. | {
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slam, visual-odometry, probability
Math example
\begin{equation*}
(J^T J) \hat{x} = J^T y
\end{equation*}
The solution of our problem is gotten by solving the following linear equation. This is internally what our factor graph solves. $J$ is the Jacobian, $\hat{x}$ is what we are trying to solve for and $y$ is our observations/error. $J^T*J$ is called the information or Hessian matrix, and is equivalent to the inverse covariance matrix you know from Gaussian math.
This information matrix can be broken up into blocks that corresponds to the various nodes we are trying to solve. In the image below it breaks it up into the individual nodes. The block is only filled if there is a connection between nodes, else it is white or all 0s.
We can group those blocks arbitrarily. For our purpose we are going to group it into 3 parts
$$\begin{equation*}
\begin{bmatrix}
A & B & C \\
B^T & D & E \\
C^T & E^T & F \
\end{bmatrix} \begin{bmatrix}
x_A \\
x_D \\
x_F \
\end{bmatrix} = \begin{bmatrix}
y_A \\
y_D \\
y_F \
\end{bmatrix} | {
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fluid-dynamics, flow, chaos-theory, turbulence
Title: How does object move in bottom of swimming pool? Suppose there is an object $O$ (swimming goggles) that has fallen to the bottom of a swimming pool. I have the swimming pool circulation pump turned on.
Initially, the object is at some position $P_1$. I stared at it for a few minutes; it's roughly in the same position. However, after ~5 hours, when I looked at it again, it has definitely moved to another position $P_2$. I stared at it again for a few minutes; it's still not moving at all.
Which gets me curious, how exactly does the object move across the bottom of the pool from $P_1$ to $P_2$? Does it move at some really slow constant speed, or completely chaotically? That is a chaotic system in any real life version. Meaning if you changed the circ pump's starting flow rate by 0.01 L/min, or the goggles' starting position by 1 cm, you might end up with a completely different movement. Basically impossible to model. | {
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java, performance, programming-challenge
while (newr != 0) {
long quotient = r / newr;
long tmp;
tmp = newt;
newt = t - quotient * newt;
t = tmp;
tmp = newr;
newr = r - quotient * newr;
r = tmp;
}
if (t < 0)
t += n;
return t;
} | {
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formal-languages, reductions, polynomial-time
[FALSE]$\ \ $ If $Y\le_p X$ and $X$ is solvable in constant time, then $Y$ is solvable in constant time as well.
You cannot ignore the time spent in the reduction step. $Y\le_p X$ tells us that it takes no more than polynomial time to map (a.k.a. reduce) an instance (a.k.a a word) of $Y$ to an instance of $X$. That polynomial time, as you suspected, is significant.
In fact, the complexity class $\mathbf P$ can be defined as all $Y$s such that $Y\le_p X$ for some $X$ that is solvable by constant time.
[ALMOST TRUE]$\ \ $ Say you have $SORT$ which checks if the list of ints is sorted, then for all $X$ we have $SORT \leq_p X.$ | {
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reinforcement-learning, reward-functions, reward-design, reward-shaping
You may have opinions about valid solutions to the environment that you want the agent to use. In which case you can extend or modify the system of rewards to reflect that - e.g. provide a reward for achieving some interim sub-goal, even if it is not directly a result that you care about. This is called reward shaping, and can help in practical ways in difficult problems, but you have to take extra care not to break things.
There are also more sophisticated approaches that use multiple value schemes or no externally applied ones, such as hierarchical reinforcement learning or intrinsic rewards. These may be necessary to address more complex "real life" environments, but are still subject of active research. So bear in mind that all the above advice describes the current mainstream of RL, and there are more options the deeper you research the topic. | {
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telescope
If you want to keep costs low, look for "hobby" lenses, but make sure they are achromatic lenses and the focal length is 8 or 10 times longer than the diameter.
For example D = 60 mm, f = 500 mm would do.
Then for 20x magnification for example, use an eyepiece with a focal length 20 times shorter, or 25mm.
My first refractor telescope was with simple glass lenses like this (but with the long focal length and smaller diameter like I'm suggesting) and I had big problems with achromatic aberration.
My second one was made from the objective and eyepiece from my grandmother's bird watching binoculars (she didn't need them anymore) Basically they were ideal lenses but I put them in a cardboard tube with a cardboard and tape sliding focusing holder so I could have the feel of making a telescope.
After that I realized that cheap lenses are only good to learn about aberrations, not to look at the Moon. | {
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we can think of a function as a machine, where the input objects are put into the top, and for each input, the machine spits out one output. So there is a perfect "one-to-one correspondence" between the members of the sets. Definition. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Hence it is bijective. That is, we can use functions to establish the relative size of sets. 2^{3-2} = 12$. Cardinality … The idea is to count the functions which are not surjective, and then subtract that from the But your formula gives$\frac{3!}{1!} Surjective Functions A function f: A → B is called surjective (or onto) if each element of the codomain has at least one element of the domain associated with it. The function is They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known. Surjections as epimorphisms A function f : X → Y is surjective if and only if it is right-cancellative: [2] given any | {
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"openwebmath_score": 0.9288099408149719,
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"url": "http://felizcidade.org.br/2rf5m4/cardinality-of-surjective-functions-b8ae3d"
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gravity, general-relativity
Einstein's theory of gravity
Here the role of gravitational potential $\Phi$ is taken by the metric tensor $g_{\mu\nu}$ which is a collection of 16 real numbers, one for every combination of $\mu$ and $\nu$, that can take values 0,1,2 and 3. Actually not all 16 numbers are independent, because $g_{\mu\nu} = g_{\nu\mu}$, so there are only 10 independent numbers. We go from $\Phi$, that was only 1 real number for every point in spacetime, to $g_{\mu\nu}$ which carries 10 real numbers for every point in spacetime. This is already looking more complicated.
A similar argument can be made for $\rho$. It gets substituted with the stress-energy tensor $T_{\mu\nu}$, which again carries 10 independent real numbers. | {
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python, python-3.x, networking, concurrency
Title: Python Port Scanner 2.0 A few months ago I put up a port scanner for review. Now, I've updated it with some new Python knowledge and integraing the feedback I got.
Some things I specifically think might be wrong with it or feel clunky to me: | {
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electrostatics, potential, work, potential-energy
Title: Work needed to assemble charges vs. potential energy I'm reading through an ENM text called Introduction to Electrodynamics by Griffiths, and in the section on work and energy, after giving the expression for work, he says:
"That's how much work it takes to assemble a configuration of point charges; it's also the amount of work you'd get back if you dismantled the system. In the meantime, it represents energy stored in the configuration ("potential" energy, if you insist, though for obvious reasons I prefer to avoid that word in this context.) | {
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$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$
-
I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. – Michael Hardy Apr 11 '12 at 2:42
Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! – Brandon Amos Apr 11 '12 at 11:34
.....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) – Michael Hardy Apr 11 '12 at 21:00
Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$. | {
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optics, differential-geometry, mathematics, calculus
The question:
If $r=0$, there is no base, and thus, there is no cone. Given that this is a simple fact, why is the elimination of the case of $r=0$ referred to as an analysis, when it appears to be a fact trivially establishable? Is this as simple as I think it is, or is there more going on here? The equation being referenced is
$$ R = 1 - \frac r h , \qquad r>0, h>0$$
An upper bound on the ratio $R$ is $1$ because $1$ is equal to or greater than any value of $R$ which can be achieved from any valid choice of $r,h$. It also happens that $1$ is the smallest upper bound or supremum. | {
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compilers, parsers, ambiguity
type
: basetype
| basetype parameter_list
;
arguments
: arraytype IDENTIFIER
| arraytype IDENTIFIER COMMA arguments
;
argument_list
: OPEN_PAREN CLOSE_PAREN
| OPEN_PAREN arguments CLOSE_PAREN
;
parameters
: arraytype
| arraytype COMMA parameters
;
parameter_list
: OPEN_PAREN CLOSE_PAREN
| OPEN_PAREN parameters CLOSE_PAREN
;
expressions
: expression
| expression COMMA expressions
;
expression_list
: OPEN_PAREN CLOSE_PAREN
| OPEN_PAREN expressions CLOSE_PAREN
;
// just a type that can be an array (this language does not support
// multidimensional arrays)
arraytype:
: type
| type OPEN_BRACKET CLOSE_BRACKET
;
block
: OPEN_BRACE CLOSE_BRACE
| OPEN_BRACE statements CLOSE_BRACE
;
function_expression
: arraytype argument_list block
| arraytype IDENTIFIER argument_list block
;
call_expression
: expression expressions
;
index_expression
: expression OPEN_BRACKET expression CLOSE_BRACKET
; | {
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