text stringlengths 1 1.11k | source dict |
|---|---|
particle-physics, scattering
This is related. explaining how similar plots can be made after the interaction:
In particle physics E_k is the kinetic energy of an outgoing pion from a pi proton scattering.Such plots are usually constructed by scattering a beam of pions with fixed energy and looking at two body interactions with the protons at rest, identifying the outgoing proton and pion by ionization. One would have plotted the invariant mass distribution and then the resonances would have been easily labeled. | {
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# Simplifying Expression with Like Denominators and Factorials
• November 1st 2013, 04:33 PM
SC313
Simplifying Expression with Like Denominators and Factorials
Hi,
I am having difficulties simplifying the expression $\frac{1}{k!}-\frac{1}{(k+1)!}$.
To obtain like denominators, $\frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!}$
• November 1st 2013, 04:43 PM
emakarov
Re: Simplifying Expression with Like Denominators and Factorials
Since (k+1)! = (k+1)k! = (k+1)k(k-1)!,
$\frac{1}{k!}-\frac{1}{(k+1)!} = \frac{1}{k!}\left(1-\frac{1}{k+1}\right)=\frac{1}{k!}\cdot\frac{k}{k+1 } =\frac{1}{(k-1)!(k+1)}$
• November 1st 2013, 04:52 PM
SC313
Re: Simplifying Expression with Like Denominators and Factorials
Hi,
The answer according to Maple is $\frac{k}{(k+1)!}$.
Is it wrong?
• November 1st 2013, 05:08 PM
emakarov
Re: Simplifying Expression with Like Denominators and Factorials
Quote:
Originally Posted by SC313
Is it wrong? | {
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java, matrix, mathematics
@Test
public void testGet() {
double[][] data = new double[ROWS][COLS];
Utilities.populate(data, random);
Matrix matrix1 = new Matrix(data, false);
Matrix matrix2 = new Matrix(matrix1);
assertEquals(matrix1.getRows(), ROWS);
assertEquals(matrix1.getCols(), COLS);
assertEquals(matrix2.getRows(), ROWS);
assertEquals(matrix2.getCols(), COLS);
assertEquals(matrix1.getRows(), matrix2.getRows());
assertEquals(matrix1.getCols(), matrix2.getCols());
for (int i = 0; i < ROWS; i++) {
for (int j = 0; j < COLS; j++) {
assertEquals(matrix1.get(i, j), matrix2.get(i, j));
}
}
assertThrows(() -> matrix1.get(-1, 0));
assertThrows(() -> matrix2.get(0, COLS));
} | {
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ros2, c++, opencv, cv-bridge
publish(std::unique_ptr<MessageT, MessageDeleter> msg)
^~~~~~~
/opt/ros/galactic/install/include/rclcpp/publisher.hpp:187:3: note: no known conversion for argument 1 from ‘sensor_msgs::msg::Image_<std::allocator<void> >::SharedPtr {aka std::shared_ptr<sensor_msgs::msg::Image_<std::allocator<void> > >}’ to ‘std::unique_ptr<sensor_msgs::msg::Image_<std::allocator<void> >, std::default_delete<sensor_msgs::msg::Image_<std::allocator<void> > > >’
/opt/ros/galactic/install/include/rclcpp/publisher.hpp:211:3: note: candidate: void rclcpp::Publisher<MessageT, AllocatorT>::publish(const MessageT&) [with MessageT = sensor_msgs::msg::Image_<std::allocator<void> >; AllocatorT = std::allocator<void>]
publish(const MessageT & msg)
^~~~~~~ | {
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dft, finite-impulse-response, filtering, group-delay, fast-convolution
Title: position of useful filtered data inside DFT based filtered output I'm trying to filter a 400 samples signal with various bandstop FIR filters (constant group delay), one at a time to see which one gives me the desired result. Every filter was build using the Kaiser Window method in Matlab. Each filter has a different number of coefficients. The smaller one has 782 and the biggest one 2460. I'm performing the filtering via fast convolution using the following Matlab code:
temp = ifft( fft(filterCoeffs,ffsSize)' .* fft( signalSamples, ffsSize) );
FilteredSignal = temp(offset+1:offset+length(signalSamples)); | {
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java, programming-challenge, time-limit-exceeded
Let's take the given example with \$n = 10\$ and the queries
a b k
1 5 3
4 8 7
6 9 1
Instead of maintaining an array of the \$ n \$ numbers
index: 1 2 3 4 5 6 7 8 9 10
initial array: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
after (1, 5, 3): [3, 3, 3, 3, 3, 0, 0, 0, 0, 0]
after (4, 8, 7): [3, 3, 3,10,10, 7, 7, 7, 0, 0]
after (6, 9, 1): [3, 3, 3,10,10, 8, 8, 8, 1, 0]
we maintain an array of \$ n+1 \$ differences, where each entry \$ d[i] \$ holds the difference \$a[i+1] - a[i] \$:
index: 0, 1 2 3 4 5 6 7 8 9 10
initial diffs: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
after (1, 5, 3): [3, 0, 0, 0, 0,-3, 0, 0, 0, 0, 0]
after (4, 8, 7): [3, 0, 0, 7, 0,-3, 0, 0,-7, 0, 0]
after (6, 9, 1): [3, 0, 0, 7, 0,-2, 0, 0,-7,-1, 0]
final array: [3, 3, 3,10,10, 8, 8, 8, 1, 0] | {
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homework-and-exercises, oscillators
2. The train is going up a constant slope with constant speed.
If the train travels along a straight line (and a constant slope is a straight line) at constant speed, then the pendulum is in an inertial frame. Therefore, there are no fictitious or other forces adding to gravity. Hence, $T_2 = T_{rest}$.
3. The train moves over a hill of radius R with constant speed.
This one is tougher. The centrifugal force is still pointing radially outwards - but now it is not perpendicular to the gravitational force any more. For instance, at the top of the hill, it is pointing upward, and thus cancelling gravity partly. At other positions it will point diagonally upward. This case needs its own somewhat more involved treatment, because the total $F$ is not constant. Therefore, $T_3$ will also change gradually. In particular, it will be maximal at the top of the hill, and will decrease symmetrically down both sides of the hill. | {
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astrophysics, astronomy, redshift
The Dark Ages
Regarding the second interpretation, the seeds for the structure were there from the beginning (i.e. since the Big Bang), so if that's what you mean by "protogalaxy", they were present at $z=100$. In this case, however, there is virtually no light emitted — we're in the Dark Ages where the Universe still consists of its primordial hydrogen and helium. The temperature of the gas at this time is 275 K, around the freezing point of water, and the density is too low for frequent collisions of atoms. The only light emitted is the occasional hydrogen spin flip transition, but the wavelength of this light is 21 cm, i.e. in the radio regime which James Webb cannot detect. In fact, with current technology not even a radio telescope would be able to detect such faint light. | {
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exoplanet, star-systems, metallicity, elemental-abundances
The problem with this scheme is that individual kilonovae or supernovae probably don't make much of a mark in star formation that has taken place over the last many billions of years. That is because the interstellar medium is already enriched by hundreds of millions of such events in the first billion years or so of our Galaxy's existence. Thus when you do identify stars with big r-process element enhancements (e.g. Gull et al. 2021), they are old, generally metal-poor stars and thus have absolute r-process abundances that are still rather low. Nevertheless, if solid material and planets etc. can form around these stars then it could be that those solids are also r-process-rich and these planetary systems could be literal gold-mines. | {
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general-relativity, special-relativity
But is it possible to show this without general relativity, by considering a uniformly accelerated rocket and looking at how much light bends? Under the equivalence principle this situation should be the same as light in a uniform gravitational field. | {
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electrical-engineering, battery
448*74*2.2 = 73kWh
uboat.net
The US Navy "Balao" type submarine (1944/45) was fitted with 4 four
Elliot Main (Electric Motors) two on each shaft, with a total
horsepower of 2,740. While submerged, these motors were powered by two
massive (each cell weighing 1650#) 126-cell batteries (in series)
capable of delivering 5,320 Amp/Hrs each.
Assuming they meant Amp-hrs, and guessing 2.2 V per cell, 2.2 * 2*5320 = 23.4 kWh | {
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navigation, move-base, ros-kinetic
Original comments
Comment by jayess on 2018-06-27:
Have you taken a look at the navigation stack?
Comment by jayess on 2018-06-27:
Please don't crosspost. It wastes time and effort of everyone. Please see our support page.
Comment by aarontan on 2018-06-27:
Oh okay, sorry I did not know about that but I understand now.
And yes I have looked at the navigation stack however I am not sure how to integrate move_base with a published path message. I am wondering if anyone have experience with this before and could provide me with some pointers.
Thanks
Comment by Ahmed_Desoky on 2020-07-20:
I have the same problem.
Are you finished and solve this problem?
Thanks
Comment by gregbowers on 2023-06-22:
Hello
To make your robot follow a published path:
-Publish the desired path using the package you have.
-Subscribe to the published path in your robot's control system.
-Extract the necessary path information from the received message. | {
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halides, dipole
Title: Why does chloromethane have a larger dipole moment than chloroform?
Why does $\ce{CH3Cl}$, methyl chloride, have a larger dipole moment than $\ce{CHCl3}$, chloroform?
Let us consider $\ce{C-Cl}$ bond moment to be $x$ and that of $\ce{C-H}$ bond to be $y$.
In $\ce{CCl4}$, the dipole moment of any three $\ce{C-Cl}$ atoms is balanced by the dipole moment of the fourth $\ce{C-Cl}$ bond dipole moment. This means the resultant of dipole moment of three $\ce{C-Cl}$ bonds in tetrahedral structure is equal to $x$, which is equal and opposite to the dipole moment $x$ of the fourth $\ce{C-Cl}$ bond.
Now, when we replace one of the chlorines with a hydrogen, the net dipole moment equals $x+y$, (as they are in the same directions). Similarly if we carry out the same procedure for $\ce{CH3Cl}$ it can be proved that its dipole moment is also $x+y$. | {
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navigation, mapping, gmapping
Title: how can I build probability maps with gmapping
I want build probability maps, but I find that gmapping can only generate maps with 3 values (unknown, free and obstacle). How can gmapping work with probablility maps. Thank you.
Further, I change source code, but it doesn't work too.
the code I changed is below:
if(occ < 0)
map_.map.data[MAP_IDX(map_.map.info.width, x, y)] = -1;
else
map_.map.data[MAP_IDX(map_.map.info.width, x, y)] = (int)round(occ*100.0); | {
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} |
homework-and-exercises, newtonian-mechanics, harmonic-oscillator, differential-equations
$$ \ddot x + \omega^2 x = C $$
and rewrite it as:
$$ \ddot x + \omega^2 \left(x - \frac{C}{\omega^2}\right) = 0 $$
Then define a new variable $y$ by:
$$ y = x - \frac{C}{\omega^2} $$
and differentiate twice to get:
$$ \ddot{y} = \ddot{x} $$
Finally substitute into your original equation to get:
$$ \ddot y + \omega^2 y = 0 $$
And this is just the usual SHO equation with the solution:
$$ y = A\cos(\omega t) + B\sin(\omega t) $$
The last step is simply to substitute for $y$ to get:
$$ x - \frac{C}{\omega^2} = A\cos(\omega t) + B\sin(\omega t) $$
or:
$$ x = A\cos(\omega t) + B\sin(\omega t) + \frac{C}{\omega^2} $$
which is the equation that Andrei finished with. | {
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ros, slam, navigation, environment, teleoperation
Title: slam and virtual environment?
I wanna know if this feasible. In the client side, I have a virtual environment where the slave is considered as a point that populates the virtual environment. In this environment, there are some fixed obstacles, so the slave will send the current position of the point as well as the positions of the obstacles. Based on the aforementioned data, I will visualize a map in the master side. Where should I start if this possible?
Originally posted by CroCo on ROS Answers with karma: 155 on 2013-07-23
Post score: 0
The short answer is "yes, this is possible."
It's hard to give a better answer without knowing more details of your system. What do you already have working? Are you starting from scratch? Have you worked through the basic ROS tutorials?
Originally posted by lindzey with karma: 1780 on 2013-07-23
This answer was ACCEPTED on the original site
Post score: 0 | {
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ocean, ocean-currents, tides
A number of modelling studies have been made of more realistic scenarios for early deployment (e.g. Admadien et al 2012). These typically predict local changes to current speeds of up to 30%, which fade after a few km. Typically there is a decrease in the flow speed in line with the TECs, and an increase to either side of the farm/array, as some of the flow diverts around the added impedence. Effects on residual velocity (i.e. that which is left in the long term when the tidal cycles are averaged out over a period of time), which is relevant to sediment transport processes, are predicted up to at least 15km away.
Some baroclinic modelling (Yang & Wang 2013) has suggested increased mixing, and thus decreased stratification, as a result of the turbulence introduced by TECs. It is also conceivable that in other scenarios, stratification might be increased as a result of reduced flow speeds. | {
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electric-circuits, electrical-resistance, batteries
Title: What if I keep holding a household battery? I have a question in my mind. What if I keep holding a Duracell(1.5V Battery) touching both the terminals with my fingertips? Will the circuit be completed and the battery drain? Or will it have no effect? Why? You have to consider the electrical resistance of your fingers. This Aricle shows that the human body has a resistance of up to 100,000 ohms. Air has a resistance of about 2 x 10^16 ohms. While a very long thick wire has a resistance of 4 ohms. So, it looks like your fingers would indeed conduct electricity faster than air, but you'd have to hold the battery for a REALLY long time before you drain it. | {
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quantum-mechanics, quantum-entanglement
Nope. When an alien scientist measures particle A, it does not have an effect on particle B. Specifically, it does not induce any sort of change in particle B that can be detected.
Of course, we would say that the quantum state of particle B (technically: the combined quantum state of the pair AB) has changed, but changes of this nature in the quantum state can't be detected. For example, suppose the entangled state, written in a basis of energy levels, is
$$|\psi_1\rangle = \frac{1}{\sqrt{2}}\left(|1\rangle|2\rangle + |2\rangle|1\rangle\right)$$
so that if particle A has an energy of 1 (in unspecified units), particle B will have an energy of 2, and vice versa. Suppose the alien scientist measures the energy of particle A to be 1. The state of the pair AB becomes
$$|\psi_2\rangle = |1\rangle|2\rangle$$ | {
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c#, performance, wpf
this.Ctx.ChangeState(new PointMovingState(this.Ctx, point));
}
}
public partial class MainWindow : Window, IChartContext
{
private State state;
public MainWindow()
{
InitializeComponent();
// initialization logic
this.state = new IdleState(this);
}
public ChartPoint AddPoint(Point position) {
// add logic for adding point
}
void Select(ChartPoint chartPoint) {
// add logic for select single point
}
void ToggleSelection(ChartPoint chartPoint) {
// add logic for toggle selection
}
void ExtendSelection(ChartPoint chartPoint) {
// add logic for extend selection
}
void ChangeState(State state) {
this.state = state;
} | {
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useful in the case where it hard to compute the probability of an event, but it is relatively easy to compute the probability of "not" the event. The formula for calculating independent events: P(A and B) = P(A) x P(B|A). Determine the following probabilities if each of the following are independent. Choosing a marble from a jar. The Switch Delay Time property is related to repairable systems and is considered in BlockSim only when using simulation. How many different ways can one select the Champion AND the Runner-Up?. Dependent 11. Formula for Joint Probability. Two events A and B in a probability space S are known to be dependent events if the occurrence of one affects the occurrence of the other. Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other (equivalently, does not affect the odds). Example: the probability that a card is a four and red =p (four and red) = | {
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differential-expression
What is the best way to compare treated/control samples?
And here:
What is the best way to compare differences in DEG expression between two populations?
If this is 'ANOVAerable' (its not a word, but if ANOVA is appropriate) than a two-way ANOVA would provide all the answers (however see below regarding the trend in the data). This analysis will perform a 'A' versus 'B' and each 'gen' compared to the other, as well as the interaction between A/B and 'gen'. You would need the raw data, you can't do anything with means (unless you are assuming the distribution). The distribution within each 'gen' is important (if its normal - conforms to the normal distribution, you're in business).
There is nothing further to add and I do feel we need to draw this series of questions to a close. | {
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sorting, balanced-search-trees
As for your more general question about when changing a black node to a red one is allowed: a set of nodes can be repainted if afterwards, the black-height criterion is still satisfied. For example, if a row of the tree is saturated (if it is at depth $k$, there are $2^{k}$ nodes), then you can colour that row black. You may not colour only part of a (saturated or not) red row black. Another example: in the tree $1:red \rightarrow\{ 2:black, 3:black\rightarrow\{4:red, 5:red\}\}$, you can flip the colours of $\{3,4,5\}$ and still have a legal RB tree. This is true generally for subtrees of even height and alternating colours. Maybe you can think of more examples. | {
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comment box. If we de ne s 1(t) = s(t t 0); then S 1(f) = Z 1 1 s(t t 0)e j2ˇftdt; Z 1 1 s(u)e j2ˇf(u+t 0)du; = e j2ˇft 0 Z 1 1 s(u)e j2ˇfudu; = e j2ˇft 0S(f): There is a similar dual relationshp if a signal is scaled by an exponential in the time domain. 下载 常用傅里叶变换对. The Fourier transform. The detection process is mainly based on the wavelet transform. Digital signal processing (DSP) vs. Fourier transform that f max is f 0 plus the bandwidth of rect(t - ½). If any argument is an array, then fourier acts element-wise on all elements of the array. We describe this FFT in the current section. The trigonometric Fourier series of the even signal cos(t) + cos(2. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. Fourier analysis is a method of defining periodic waveform s in terms of trigonometric function s. Inverted frequency spectrum | {
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time
This means that at the equinoxes, a "day" is at least six minutes longer than the twelve hours we are naively taught (and incorrectly reported twice a year by naive weathermen). As an aside, days are always longer than nights at the equator. | {
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javascript, typescript, angular-2+
What I does not like is that I have too many IF and ELSE, anybody knows some better solution, thanks? Defaults & Ternaries
There is nothing wrong with statements.
However Javascript and most C like languages allow a wide variety of branching styles.
For example you can default the navigation, and use a ternary ? (Conditional operator) to select the redirect option.
This has only 1 statement if to allow the local storage key to be set. The default navigation removes the last else and the ternary removes the inner if else.
ngOnInit(): void {
this.activatedRoute.url.subscribe(() => {
var navTo = "/error/404";
const token = this.activatedRoute.snapshot.queryParamMap.get('token');
if (token) {
localStorage[AuthenticationStorageKey] = token;
const redirectPage = this.activatedRoute.snapshot.queryParamMap.get('redirectPage');
navTo = redirectPage ? redirectPage : "";
}
this.router.navigate(navTo);
});
} | {
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Note: I assumed that we know the circumference is $2πr$
-
"..but we need to have infinite divided parts to get an exact rectangle shape". Can you prove that? :) – Sawarnik May 13 at 17:36
Just one remark. The so-called onion proof is a special case of the co-area formula. This formula is a rigorous justification of all those computations that we learned in the first course of general physics. It is a "curvilinear" generalization of Fubini's theorem: instead of slices, you integrate over hypersurfaces like a sphere. And also the fact that "differentiating the volume gives the area" is a consequence of the same theorem.
-
What you have here:
$$\int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2$$ | {
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quantum-mechanics, homework-and-exercises, quantum-information, density-operator, quantum-states
The question:
Is the reduced density matrix itself a density matrix? E.g. can it be treated as a density operator of a system just like density operators from the definition (1)? In my opinion this doesn't follow from (2) necessarily, because it's not clear, that the reduced density matrix is (mathematically) an object in the shape of (1).
Since no book explicitly answers this question, I am tempted to believe that the reduced density matrix is a density matrix because of (quote) - so because the calculation of an expected value is analog to the case of density matrices, which implies they are objects of the same "shape". Density matrices are, by definition, positive semi-definite trace-class operators of unit trace, see for example this. In the following we will only discuss the finite-dimensional case, so we have to show that the reduced density matrix is positive semi-definite and of unit trace.$^\dagger$ | {
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c++, linked-list
This allows us to check every time that the node is not nullptr. It's also easier to visually parse advance(tmp) or advance(node) than tmp = tmp->next. Note that uses a reference to a pointer (i.e. alters a pointer outside the function's own scope), so be careful not to pass head into it directly.
Most of the code in the insert function is actually getting the last node in the list. This could be factored out into a separate function.
printList should be const.
printList could be a little neater: The node variable is only needed if head isn't null. (It also prints an extra line for empty lists). Generally, it's best to treat special cases at the start of the function, and return early:
void printToStdOut() const
{
if (head == nullptr)
{
std::cout << "Empty List \n";
return;
}
Node* node = head;
while (node != nullptr)
{
std::cout << node->data << " ";
advance(node);
}
std::cout << "\n";
} | {
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c++, object-oriented, tree, iterator, c++17
template <typename T>
typename Tree<T>::iterator& Tree<T>::iterator::operator++() {
if (cur == nullptr) {
throw std::out_of_range("No more nodes for traversal");
}
for (auto& child : cur->children_) {
s_.push(child);
}
if (s_.empty()) {
cur = nullptr;
} else {
cur = s_.top(); s_.pop();
}
return *this;
}
template <typename T>
typename Tree<T>::iterator const Tree<T>::iterator::operator++(int) {
Tree<T>::iterator tmp(*this);
++*this;
return tmp;
}
template <typename T>
bool Tree<T>::iterator::operator==(Tree<T>::iterator const& other) const {
return *cur == *other.cur;
}
template <typename T>
bool Tree<T>::iterator::operator!=(Tree<T>::iterator const& other) const {
return !(cur == other.cur);
}
template <typename T>
Tree<T>::Tree(T const& val, Tree<T>* const parent)
: val(val), parent_(parent) {
if (parent) {
parent->children_.push_back(this);
}
} | {
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the Mixed Nash Equilibria in Larger Games • Dominated strategies are never used in mixed Nash equilibria, even if they are dominated by another mixed strategy. 11:00-11:20, Paper WeA01. •Purification theoreme (Harsanyi 1973) - Almost all Nash equilibria in mixed strategies can be reconceptualized as close approximations of Bayes Nash equilibrium in pure strategies •Evolutionary games (Maynard Smith, Price 1973) - A suitable gaming model defines the system of ODE - replicator dynamics that simulates evolutionary. If a player is supposed to randomize over two strategies, then both. We call this concept \perfect" Next solution concept: subgame-perfect Nash equilibrium (SPNE). Mixed strategy means a situation where a saddle point does not exist, the maximin (minimax) principle for solving a game problem breaks down. The concept that determines the optimal solution in a non-cooperative game in which each player lacks any incentive to change his/her initial strategy. 2 Strategies in | {
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"url": "http://pank.vitadaprof.it/mixed-strategy-nash-equilibrium-2x2-calculator.html"
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quantum-mechanics, hilbert-space, quantum-entanglement, quantum-interpretations, superposition
What we don't get is interference at macroscopic scales. One reason for this is that macroscopic objects simply have very short wavelengths. So, e.g., you can't observe double-slit diffraction with a baseball, because the diffraction angles would be too small. Also, the baseball's phase gets randomized too rapidly by interactions with its environment. This is called decoherence. | {
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graphs
My question is: How to solve this problem? Is there any specific name of this problem?
Edit: The graph I am dealing with is very sparse. Coloring based approximations may not work as the complement graph would be huge (not able to store it in memory). This is called the Minimal Clique Cover Problem, and is NP-hard: as a matter of fact, the decision version ("can I do it with only $k$ subgraphs?") is one of Karp's original 21 Problems, the ones that first defined NP-completeness.
Since it's linked to graph coloring, you can't even get a good approximation in polynomial time, unfortunately: everything about this problem is hard.
For further reading, your "super-strongly connected components" are generally called cliques, and a clique cover is a way to "cover" the entire graph with non-overlapping cliques. A minimal clique cover uses the smallest possible number of cliques to do it. | {
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c#, game, linq, sudoku
Utility methods used on the code, I do not require a review on those. But feel free to comment.
public static class CollectionUtils
{
public static IEnumerable<T[][]> ViewAsNByM<T>(this T[][] values, int m, int n)
{
var count = values.GetLength(0)*values[0].GetLength(0);
for (int i = 0; i < count/(n*m); ++i)
{
var matrix = new T[m][];
for (int j = 0; j < m; ++j)
{
matrix[j] = new T[n];
for (int k = 0; k < n; ++k)
{
int idxRow = j + (i/(values[0].GetLength(0)/n))*m;
int idxCol = (k + i*n)%values[0].GetLength(0);
matrix[j][k] = values[idxRow][idxCol];
}
}
yield return matrix;
}
} | {
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A perfect linear relationship yields a correlation coefficient of +1 (or -1 for a negative relationship) and no linear relationship yields a correlation coefficient of 0.
Partial correlation is the correlation between two variables after removing the effect of one or more additional variables. This command is specifcally for the the case of one additional variable. In this case, the partial correlation can be computed based on standard correlations between the three variables as follows:
$$r_{12.3} = \frac{r_{12} - r_{13}r_{23}}{\sqrt{(1 - r_{13}^2)(1 - r_{23}^2)}}$$
with rxy denoting the correlation between x and y.
As with the standard correlation coefficient, a value of +1 indicates a perfect positive linear relationship, a value of -1 indicates a perfect negative linear relationship, and a value of 0 indicates no linear relationship.
It may be of interest to determine if the partial correlation is significantly different than 0. The CDF value for this test is | {
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data-analysis
These are my guesses. But in reality - do studies differ much for open astronomical data compared to proprietary data? I can only tell from my experience, which might be not the ideal type of answer, but I will leave my 2 cents anyway:
For such cases I have seen essentially three different approaches: | {
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We still need to account for the fact that $T(0)=1, T(1)=1$ and $T(2)=1.$ A simple calculation shows that $$T(n) = S(n) - \frac{197}{7} 7^{\lfloor \log_2 n\rfloor} d_{\lfloor \log_2 n\rfloor} + 78 \times 7^{\lfloor \log_2 n\rfloor-1} d_{\lfloor \log_2 n\rfloor-1}.$$ This formula is exact and holds for all $n\ge 3$.
Finally to get the asymptotics look at the two leading terms from the lower and upper bounds. The first is $$\Theta\left(7^{\lfloor \log_2 n\rfloor}\right) = \Theta(2^{\log_2 7 \log_2 n}) = \Theta(n^{\log_2 7}).$$
$T(n)$ fluctuates around this value with the coefficient at most for strings of ones $$\frac{1}{7} \left(\frac{441}{5} - \frac{197}{7} + \frac{78}{7} \right)= \frac{356}{35}$$ and at least $$42 - \frac{197}{7} = \frac{97}{7}.$$ The next term in the asymptotic expansion is $$\Theta\left(4^{\lfloor \log_2 n\rfloor}\right) = \Theta(n^2),$$ with the coefficient between $24$ and $96.$ | {
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newtonian-mechanics, forces, rotational-dynamics, friction, equilibrium
Deriving Eq. 1
When researching, I wasn't able to find an equation for the tipping vs. slipping threshold height $h$. So here's my derivation for Eq. 1.
The block is barely remaining in equilibrium. It's right on the verge of tipping/slipping. Hence, the normal force and static friction force both act at the corner. First, we balance the forces:
$$N=mg\text{cos}\theta$$
$$F_\text{push}=F_\text{s max}-mg\text{sin}\theta$$
$$F_\text{push}=\mu(mg\text{cos}\theta)-mg\text{sin}\theta$$
Then we balance the torques:
$$\tau_\text{push}=\tau_\text{gravity}$$
$$F_\text{push}h=mgr_\perp$$
$$(\mu mg\text{cos}\theta-mg\text{sin}\theta)h=mgr_\perp$$
$$(\mu\text{cos}\theta-\text{sin}\theta)h=r_\perp \text{ }(\text{Eq. 2})$$
We can find the lever arm $r_\perp$ from a little geometry. The distance $V$ forms a right triangle where:
$$\text{tan}\theta=\frac{d_1}{V}$$
$$d_1=V\text{tan}\theta \text{ }(\text{Eq. 3})$$
Next we draw the lever arm $r_\perp$ for the gravity force: | {
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frequency-response
I've tried to find whole system's frequency response by shifting $H(e^{j\omega})$ with value of $\pi$ and summed it up with itself which then got $H_{total}(e^{j\omega}) = 1$ for all $\omega$ and if I'm correct with that I think the system is a wire which makes $y[n] = x[n]$ The system is not just a wire, it is in fact time-varying, so it has no frequency response in the conventional sense. The sequence $w[n]$ is of course just a low-pass filtered version of the input signal. The filtered signal is band-limited to half the Nyquist frequency. Now just write the output $y[n]$ in the time domain as the sum of $w[n]$ and the modulated version of $w[n]$. You should be able to see that effectively every other (odd) sample of $w[n]$ is replaced by a zero. But there occurs no aliasing due to the low pass filter at the input. | {
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catkin
Original comments
Comment by William on 2014-01-15:
You could use the CMAKE_CURRENT_BINARY_DIR variable to get ~/catkin_ws/build/gperftools/. See: http://www.cmake.org/Wiki/CMake_Useful_Variables Also, this is not going to work for the install case. This is one of the reasons we don't do this normally.
For what you are describing I would recommend CMake's ExternalProject system:
http://www.kitware.com/media/html/BuildingExternalProjectsWithCMake2.8.html
It is functionally equivalent to the download_unpack_build like macros in rosbuild. The best thing to do would be to setup a REP-0136 compliant non-catkin package. That way it can be released and it can be built as part of a catkin workspace without downloading during the build:
http://www.ros.org/reps/rep-0136.html
However, if you are not going to release this or release things which depend on it, then it probably is fine to use CMake's ExternalProject system.
EDIT: | {
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java, object-oriented, design-patterns, immutability, fluent-interface
I'd rename PersonBuilder to Builder. It's an inner class of Person so there really is no need for it. Person.Builder is perfectly fine.
I'd also rename the individual building methods for id, name, surname and isOccupied to setId, setName, setSurname and setOccupied, since they are all effectively setters.
Reworked Code
When incorporating these points into your code, it should look a little something like this.
public final class Person {
private int id;
private String name;
private String surname;
private boolean isOccupied;
private Person() {}
public static class Builder {
private Person personToBuild;
Builder() {
personToBuild = new Person();
}
Person build() {
Person builtPerson = personToBuild;
personToBuild = new Person();
return builtPerson;
}
public Builder setId(int id) {
this.personToBuild.id = id;
return this;
} | {
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perturbation-theory
$$
and then say that this last term is also zero? Or would this then be the perturbation of $0$ and therefore not equal to $0$?
I actually think it is the latter, but I don't want to miss a nice opportunity to simplify things and hope someone could validate my thoughts. ;)
Thank you very much in advance. TL;DR: No, it cannot be done as asked.
Perturbing the 0th order, i.e. $\phi\rightarrow\phi+\delta\phi$ gives
$$
a(\phi)+\partial_a\partial^a\phi +\partial_\phi V(\phi) + \partial_\phi a\delta\phi+\partial_a\partial^a\delta\phi+\partial_\phi^2V\delta\phi=0+\partial_\phi a\delta\phi+\partial_a\partial^a\delta\phi+\partial_\phi^2V\delta\phi
=0+\epsilon\neq0
$$
It's like doing
$$
0=x\qquad x\rightarrow x+\delta x=x+1\qquad\nRightarrow0=1.
$$ | {
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fourier-transform
The operation to be applied in order to induce a shift by a number c of samples to $s[n]$ and obtain $s[n+c]$ is :
$$
e^{\frac{-i2\pi kc}{N}}S[k]
$$
Which in matlab code would be done with :
exp(-j*2*pi*(1:(N-1))*c/N).*S_fdt; | {
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algorithms, algorithm-analysis, search-algorithms, lower-bounds
From the top of my head, there is a solution that can support these type of queries in $O(\log n)$ time, while only using linear space and linear preprocessing time to build the data structure. The idea is to build a complete binary search tree on top of the string where the leafs correspond to individual characters of $S$ and in every internal node we store the sum of all the characters that are leafs in the sub tree defined by that node. In this way we can trivially support both of the updates in $O(\log n)$ time.
However, I found a paper proving a lower bound for this problem, stating that you can not do better than $O(\frac{\log n}{\log \log n})$ for the updates, and I also found the following paper http://link.springer.com/chapter/10.1007%2F3-540-51542-9_5, and a direct link to the pdf, giving an algorithm achieving that bound, thus being optimal.
I would like to understand this algorithm however the explanation is like 1 page, and a lot of details are missing. | {
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quantum-optics, intensity, quantum-measurements
We'll assume that the detector has a 100% quantum efficiency. So, if a term has a certain number of photons located at the detectors, we'll assume that all those photons are detected, giving a photon-current proportional to that number of photons. All the other photons located at other points are not detected. Therefore, we can trace over them. The result is a mixed state given by
\begin{equation}
\hat{\rho}_N(\mathbf{x}_0)= \text{tr}_{\mathbf{x}\neq\mathbf{x}_0} \left\{ |N_{\psi}\rangle \langle N_{\psi}| \right\}
= \sum_{n=0}^N |n(\mathbf{x}_0)\rangle \frac{N!}{n!(N-n)!} |\psi(\mathbf{x}_0)|^{2n} \left[1-|\psi(\mathbf{x}_0)|^2\right]^{N-n} \langle n(\mathbf{x}_0)| ,
\end{equation}
where $|n(\mathbf{x}_0)\rangle$ is a Fock state with $n$ photons located at $\mathbf{x}_0$ and we used the normalization of the wave function to say that
\begin{equation}
\sum_{\mathbf{x}\neq\mathbf{x}_0} |\psi(\mathbf{x})|^2 = 1-|\psi(\mathbf{x}_0)|^2 .
\end{equation} | {
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python, python-2.x, unit-testing, reinventing-the-wheel, iterator
Title: Implementing xrange I wanted to learn about Iterable and Iterators and hence thought about getting my hands dirty rather than reading theory and forgetting it.
I am curious how I can improve readability and error handling of my code. I also thought to practice and include test cases.
class foo_range(object):
""" Custom range iterator which mimics native xrange. """
def __init__(self, start, stop, step=1):
try:
self.current = int(start)
self.limit = int(stop)
self.step = int(step)
except ValueError:
raise
if step == 0:
raise ValueError("Step can't be 0")
def __iter__(self):
return self
def next(self):
if self.step > 0 and self.current >= self.limit:
raise StopIteration()
if self.step < 0 and self.current <= self.limit:
raise StopIteration()
oldvalue = self.current
self.current += self.step
return oldvalue | {
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"url": null
} |
black-holes, astrophysics, stars
A star is first made of hydrogen, it undergoes nuclear fussion reaction combining into helium and releasing a large amount of energy. This process continues till star is made up of iron core as iron has largest value of binding energy per nucleon, after this if mass of the star is above the value of Chandrasekhar limit it becomes a black star, what is the reason for this and why is certain mass limit required? You are a little confused in your stellar evolution model. After the ignition of hydrogen fusion in the core of a star, it will next progress to helium fusion, then to carbon/oxygen fusion via the triple-alpha process (I've skipped a lot of steps and details there, if you want the details you can look at either Hansen & Kawaler's Stellar Interiors text or Dina Prialnik's Introduction to Stellar Structure text). What happens next is mass-dependent (using $M_\odot\simeq2\cdot10^{33}$ g and the mass of the star as $M_\star$): | {
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"tags": "black-holes, astrophysics, stars",
"url": null
} |
recursion, quicksort
Title: Number of calls with length 2 array in quick-sort I need to find average number of recursive calls in quick-sort with array of length 2. I established and solved the following recursion:
$$T_N = \frac{1}{N}\sum_{k=1}^N\left(T_{k-1}+N_{N-k}\right) = \frac{N+1}{N}T_{N-1} = \frac{N+1}{3}.$$
Where $T_0 = T_1 = 0, T_2 = 1$. But:
\begin{align*}
T_3 &= 1/3(1 + 1) &= 2/3\\
T_4 &= 1/4(2/3 + 1 + 1 + 2/3) &= 5/6\\
T_5 &= 1/5 (5/4 + 2/3 + 2 + 2/3 + 5/4) &= 7/6\\
T_6 &= 1/6(7/6 + 5/6 + (1+2/3) + (1+2/3) + 5/6 + 7/6) &= 11/9
\end{align*}
The results are different. How to do it right?
EDIT: The code for quick-sort in question is:
void quicksort(int[] a, int lo, int hi)
{
if (hi <= lo) return;
int i = lo-1, j = hi;
int t, v = a[hi];
while (true) {
while (a[++i] < v) ;
while (v < a[--j]) if (j == lo) break;
if (i >= j) break;
t = a[i]; a[i] = a[j]; a[j] = t;
}
t = a[i]; a[i] = a[hi]; a[hi] = t;
quicksort(a, lo, i-1);
quicksort(a, i+1, hi); | {
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"tags": "recursion, quicksort",
"url": null
} |
php, form, symfony2
if (!$friend)
{
throw $this->createNotFoundException('There is no such friend');
}
$edit_fr_form = $this->createForm(new FriendType(), $friend);
if($request->getMethod() != 'POST')
{
return $this->render('EMMyFriendsBundle:Friend:edit.html.twig', array(
'id'=>$id, 'friend'=>$friend, 'fr_form' => $edit_fr_form->createView()));
}
}
/*
* @Edits the friend with the current id
*/
public function editAction($id, Request $request)
{
$em = $this->getDoctrine()->getEntityManager();
$friend = $em->getRepository('EMMyFriendsBundle:Friend')->find($id);
$edit_fr_form = $this->createForm(new FriendType(), $friend); | {
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"tags": "php, form, symfony2",
"url": null
} |
java, minecraft
iterate over the components map. for each item in the map, if it's a Recipe, get its instance from the catalog and traverse it (in recursion fashion).
how to build recipe instances and the catalog
this can be done in the code like you did. but a more convonient way is to have the details in a text file in a standard format. (for example, json or xml) then you can use many of the available libraries to load the text into memry as instances of Recipe and build the catalog. This is also useful for support of mods (you can have different text files for mods)
multiple constructors
This is a technical tip: Item class has multiple constructors with duplicate assignment of instance variables. yuo can call constructor from another constructor to have the assignment written once:
// constructor with no args assigns default values
public Item() {
this.name = "";
this.count = 1;
this.isPopulated = false;
this.isItem = true;
} | {
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"tags": "java, minecraft",
"url": null
} |
rust, math-expression-eval
struct Tokenizer {
number: Regex,
}
impl Tokenizer {
fn new() -> Tokenizer {
Tokenizer {
number: Regex::new(r"^[0-9]+").expect("Unable to create the regex"),
}
}
/// Tokenizes the input string into a Vec of Tokens.
fn tokenize(&self, mut input: &str) -> Vec<Token> {
let mut res = vec![];
loop {
input = input.trim_left();
if input.is_empty() { break } | {
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"id": 21882,
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"openwebmath_score": null,
"tags": "rust, math-expression-eval",
"url": null
} |
gazebo, gazebo-plugin
Originally posted by josephcoombe with karma: 609 on 2018-07-03
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by pk on 2018-07-06:
Thanks a lot for the explanation. | {
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"url": null
} |
regular-languages, regular-expressions
Prove that R.H.S regular expression is same as L.H.S
$$ (a+b)^* a (a+b)^* b (a+b)^* = (a+b)^* ab (a+b)^* $$ Regular expressions stand for languages. To show that two languages are equal, we show that every string in the first also belongs to the second, and vice versa.
As an example, let us prove the identity
$$
a(ba)^*b = ab(ab)^*.
$$
Denote the language of a regular expression $r$ by $L[r]$. A word $w$ belongs to $L[a(ba)^*b]$ if there exists $n \geq 0$ such that $w = a(ba)^nb$. Now $w = a(ba)^nb = (ab)^{n+1} = ab(ab)^n \in L[ab(ab)^*]$. Similarly, a word $w$ belongs to $L[ab(ab)^*]$ if there exists $n \geq 0$ such that $w = ab(ab)^n$. Now $w = ab(ab)^n = (ab)^{n+1} = a(ba)^nb \in L[a(ba)^*b]$.
(You can prove the various identities I used, such as $(ab)^{n+1} = a(ba)^nb$, by induction on $n$.) | {
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"tags": "regular-languages, regular-expressions",
"url": null
} |
quantum-mechanics, operators, hilbert-space, symmetry, unitarity
Now, if your question is whether every operator compatible with a symmetry is (anti-)unitary, then the answer is yes, and it is so by definition. In fact, the whole point of Wigner's theorem is to prove the converse, namely that for any symmetry transformation there is an (anti-)unitary operator that is compatible with it. | {
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} |
re-writing non-linear into linear form, how to transform slope and intercept?
This is a pretty basic question for all of you I'm sure, but I'm doing some intro classes on linear regression, where we are working some data which fits into a linear model as such:
$$W_i = 10.386 - .038*G_i$$
We're now asked to evaluate this same data using a nonlinear model:
$$W_i = \gamma e^{\beta G_i}$$
where it is stated that we can rewrite this into linear form such that: $$log(W_i) = \alpha + \beta G_i + \varepsilon_i$$ with with $$G_i$$ and $$\alpha = log(\gamma)$$
With this said, our previous model becomes:
$$W_i = 2.341 - 0.0038*G_i$$ I've dropped the error terms for simplicity, but all of these are given as correct, and I'm having trouble figuring out why the intercept term was onverted from 10.386 to 2.341 (I know they just took the log, but why?) and why the slope term went from 0.038 to 0.0038 (no idea why).
I would be grateful for any insight/help. Thanks! | {
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"url": "https://math.stackexchange.com/questions/2968897/re-writing-non-linear-into-linear-form-how-to-transform-slope-and-intercept/2969058"
} |
java, linked-list, generics, comparative-review
return true;
}
public Node<T> searchNodeByData(T var) {
if (this.isEmpty()) {
return null;
}
if (var == root.getData()) {
return root;
}
Node<T> currentNode = root;
while (currentNode.getNextNode() != null) {
currentNode = currentNode.getNextNode();
if (var == currentNode.getData()) {
return currentNode;
}
}
return null;
}
public Node<T> searchNodeByIndex(int index) {
if (this.isEmpty() || index > countSize - 1) {
return null;
}
int count = 0;
Node<T> currentNode = root;
while (currentNode.getNextNode() != null) {
if (count == index) {
break;
}
currentNode = currentNode.getNextNode();
count++;
}
return currentNode;
}
class Node<T> {
private Node<T> prevNode;
private Node<T> nextNode;
T data;
public Node() {
}
public Node(T var) {
this.data = var;
} | {
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"openwebmath_score": null,
"tags": "java, linked-list, generics, comparative-review",
"url": null
} |
cc.complexity-theory, spectral-graph-theory, expanders, unique-games-conjecture
The result you mention discussed in my lecture notes is from Section 8 in my paper with Brandao, Harrow, Kelner, Steurer and Zhou ( https://www.cs.cornell.edu/~dsteurer/papers/hypercontract.pdf ). What we show there, roughly speaking, is that a graph is a small set expander if and only if the span of eigenvectors corresponding to low eigenvalues of its Laplacian does not contain an "analytically sparse" vector.
The intuition is the following: consider the following two extremes:
1) A random vector $w$. In this case, the distribution of entries of $w$ is approximately the Gaussian distribution, and so this satisfies that $\mathbb{E}_i w_i^4 = O(\mathbb{E}_i w_i^2)^2$.
2) A vector $w$ that is the characteristic vector of a set of measure $\delta$ (i.e. it has $1$ in the coordinates belonging to the set, and $0$ in the others). In this case, $\mathbb{E}_i w_i^4 = \delta \gg \delta^2 = (\mathbb{E}_i w_i^2)^2$. | {
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with non-zero volume (finite density). Significant changes have been made since then, and the manual used during the current academic year is in NOT available yet on the WEB. The objective of this experiment is to experimentally find the moment of inertia of this system with a sensor that will measure the angular velocity. Solution Pre-Lab13. In particular, we have the parallel axis theorem: the moment of inertia about any axis through some point equals that about the parallel axis through the center of mass plus , where is the perpendicular distance between the axes. The moment of inertia was experimentally determined to by measuring its angular acceleration as a function of applied torque. Tell students that this demonstrates inertia. Experiment 8. If you would like to use a typical density value for a material,. her moment of inertia increases and her angular speed increases. The moment of inertia about the a, b, and c lab reference frame. Theory In this experiment we will determine | {
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"url": "http://angelaleone.it/imhu/moment-of-inertia-lab.html"
} |
proof-techniques, induction, software-verification, lists
Now, if we follow your original blueprint, your inductive hypothesis establishes that
$$
\lfloor\!\!\lfloor f \diamond x ::l \rfloor\!\!\rfloor = (f \diamond) \diamond \lfloor\!\!\lfloor l \rfloor\!\!\rfloor +\!\!\!+ ~(f ~x ::) \diamond ((f \diamond) \diamond \lfloor\!\!\lfloor l \rfloor\!\!\rfloor)
$$
In order for this proof to proceed, you need to tame the term on the righthand side of the concatenation. But what would you want it to look like?
Well, recall that maps distribute into concatenations:
$$
(f \diamond l_1) +\!\!\!+ ~(f \diamond l_2) = f \diamond (l_1 +\!\!\!+~ l_2)
$$
which suggests that we might be able to get our equation into an amenable form if we can rewrite the second term so that it is of the form $(f \diamond) \diamond \cdot$.
Here's where our helper lemma comes in to save the day:
Lemma ("Commutativity"):You have to use finger quotes when you read the name of this lemma out loud.
$$ | {
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EDIT. I forgot to write that, in case (a), there are no more than $$2^n$$ square roots because, necessarily $$AB=BA$$.
You showed that $$A$$ has at least $$2^n$$ square roots. Assume $$B^2 = A$$. Since $$\sigma(B)^2 =\sigma(B^2) = \sigma(A)$$, it follows that $$\sigma(B) = \{\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n}\}$$ for some $$\varepsilon_1, \ldots, \varepsilon_n \in \{-1,1\}$$ where $$\sigma(A) = \{\lambda_1, \ldots \lambda_n\}$$. Set $$D = \operatorname{diag}(\varepsilon_1\sqrt{\lambda_1}, \ldots, \varepsilon_n\sqrt{\lambda_n})$$. $$A$$ has $$n$$ distinct eigenvalues so it is diagonalizable, i.e. there exists an invertible matrix $$P$$ such that $$P^{-1}AP = \operatorname{diag}(\lambda_1, \ldots, \lambda_n) = D^2$$. | {
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Signedness: Signed. Matrix elements: unit stride Vector elements: indirect access for the source vector (the one multiplied by the matrix) This leads us to propose three categories for SpMV problems: Small: everything fits in cache Medium: source vector fits in cache, matrix does not Large: source vector does not fit in cache. inverse_element and override the is_scalar_element function. To evaluate this, we perform scalar multiplication. having the same number of rows and columns respectively. Use the times function to perform element-by-element multiplication of a fi object and a scalar. ^ for powers. col(0), b, c); But it is not working. 3d matrices 1. which replicates the vector (1,2,3) three times. Naive Approach: The idea is to iterate over each query of the array and for each query iterate over the elements of the [l, r] range and find the sum of each element multiplied by x. In order to spatially characterize mosquito abundance, we interpolated the number of mosquitoes sampled | {
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"lm_q2_score": 0.8670357494949105,
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"openwebmath_score": 0.6701095700263977,
"tags": null,
"url": "http://salernoeventicultura.it/cysr/multiply-each-element-of-vector-in-r.html"
} |
each experimental probability. What is the probability of getting a number other than 6?. Notice how for two or more dice the number of combinations equals the sum of combinations one column to the left, starting from one row higher to seven rows higher. When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. California would be the lead Group, as it has two opposed alignments. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). I hope this helps, Harley Go to Math Central. Here we consider two events: A - (finding a sum of 8) & B (getting at least one 4) A : Probability of A is 5/36 because Now, the probability of either of the incidents happening is P(AUB)=P(A)+P(B)-P(AnnB) i. Major changes in Python environment : . There are 6*3 = 18 ways to get two numbers of the same parity (the first can be any of the 6 numbers, and the | {
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"lm_q1q2_score": 0.8399712517041412,
"lm_q2_score": 0.8519527963298946,
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"openwebmath_score": 0.6622301936149597,
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"url": "http://farmaciacoverciano.it/dhti/when-two-dice-are-rolled-find-the-probability-of-getting-a-sum-of-5-or-6.html"
} |
c#, tree, interview-questions, iterator, breadth-first-search
public class TreeNodeIterator<T> : IEnumerator<T>
{
private TreeNode<T> _tree;
private TreeNode<T> _current;
private Queue<TreeNode<T>> Q = new Queue<TreeNode<T>>();
public TreeNodeIterator(TreeNode<T> treeNode)
{
_tree = treeNode;
if (_tree != null)
{
Q.Enqueue(_tree);
}
}
public bool MoveNext()
{
if (_tree == null)
{
Reset();
return true;
}
else
{ | {
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} |
experimental-realization, quantum-state
You can use RGB encoding to map elements of $\mathbb R^3$ to a single color. This can be useful if for example you are dealing with a state in which you have a two-dimensional system (a qubit) interacting with an high-dimensional one. An example of such a system are vector vortex beams, which are states of light in which the polarization is entangled with the orbital angular momentum, and therefore the polarization state depends on the position in the transverse propagation plane (unfortunately, the only example of this kind of RGB encoding actually being used that I know of is as of yet unpublished).
Regarding using this kind of thing to represent the state of a qutrit, I guess you can do it, but I don't know how useful it would be.
The idea is that given a qutrit state of the form
$$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle+\gamma|2\rangle$$ | {
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• (+1) Really nice, and conceptually clear! Plus it gets at the (implied) question about LEM and "noncomputability", which the semantic answers don't address Nov 23, 2021 at 19:43
• Maybe for part of the answer, it might be useful to mention the Goedel translation which is used to show that if intuitionistic first-order logic is consistent, then classical first-order logic is also consistent... In other words, if IFOL could prove the negation of LEM, then IFOL would be inconsistent and thus not useful in the first place. (Though in the case of propositional logic asked for in the original question, the Goedel translation of $P$ is equivalent to $\lnot\lnot P$.) Nov 23, 2021 at 22:41
• @DanielSchepler Indeed. I edited my answer to include a sketch of a direct argument showing that the negation of LEM is not derivable in intuitionistic logic. Nov 24, 2021 at 9:11 | {
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quantum-mechanics, dirac-delta-distributions, fermis-golden-rule
Title: Limit of the $\sin^2$ function in the derivation of Fermi's golden rule In the derivation of Fermi's golden rule one typically arrives at an expression of the form
$$
\frac{\sin^2(\omega t)}{\omega^2}
$$
which is then converted to
$$
\pi t\delta(\omega).
$$
I cannot follow this step. I know the following identity
$$
\delta (\omega) = \lim_{t\rightarrow \infty}\frac{\sin^2(\omega t)}{\pi |t|\omega^2}
$$
from which i would assume that one extends the first expression by $\frac{t}{t}$ and then does the limit. But how can you pull the $t$ out of the limit ? Is this rigorous or is this an approximation ?
I think it should be like this
$$
\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} \not = \pi t\delta(\omega)
$$
and the equation should be
$$
\lim_{t\rightarrow\infty} \frac{t}{t}\frac{\sin^2(\omega t)}{\omega^2} = \pi \delta(\omega) \lim_{t\rightarrow \infty}t.
$$ | {
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newtonian-mechanics, energy, rotational-dynamics, friction, rigid-body-dynamics
Clearly, ω = ω'
Here, I : MOMENT OF INERTIA OF THE BALL ABOUT ITS CENTER OF GRAVITY;
ω = INITIAL ANGULAR VELOCITY;
ω' = FINAL ANGULAR VELOCITY;
v = initial velocity;
m = mass of the ball
In this case, we are assuming that the paper is moving in the direction where the ball was moving initially, that is why velocity increased and angular velocity decreased. If we assume the opposite the velocity will decrease and angular velocity will increase. If we move the paper in any horizontal way such that it has velocity then the ball will have either its velocity or angular velocity increased and the other increased. (while it is on the paper)
We are taking the frame of reference as the table which is inertial. | {
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situations, such as finding the diameter of a broken dish inExample 5. Note: Formulas can only reference the same sheet, using standard notation (='sheetname'!cell). In that case, each trapezoid will include a small amount of area that’s above the curve. · Midpoint Rule (Rectangular Rule) · Euler-Maclaurin Formula · Richardson Extrapolation · Trapezoidal Rule · Simpson’s Rule There are a number of numerical methods that can be used to approximate an integral. after you have selected all the formulas which you would like to include in cheat sheet, click the "generate pdf" button. Non-overlapping rectangles are used to estimate the area. 1: A rectangular quadrature A variation on the rectangular rule is the midpoint rule. centre, center. The problem can be solved with the following worksheet (the formulas are shown below):. Let h= b abe the length of the integration interval. Distance and midpoint calculator This online calculator will compute and plot the distance and midpoint for two | {
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So, plugging that into the equation
g'(2) = 1/f'(1)
g'(2) = 1/4
Good luck, ntretheway. I hope this helps.
degeneratecircle | High School Teacher | (Level 2) Associate Educator
Posted on
This isn't a completely rigorous proof, but it captures the idea of the general proof of the theorem relating the derivatives of inverse functions, and makes it easy to see why the theorem is true.
By definition,
`g'(2)=lim_(x->2)(g(2)-g(x))/(2-x).`
Note that since `f(1)=2,` we have `g(2)=g(f(1))=1,` since `g=f^(-1).` Also, since `f` is one-to-one and onto, any `x` can be written as `f(t)` for exactly one value `t.` Making these replacements, we get
`(g(2)-g(x))/(2-x)=(1-g(f(t)))/(f(1)-f(t))=(1-t)/(f(1)-f(t))=((f(1)-f(t))/(1-t))^(-1).`
Remembering the relationship between `x` and `t,` and relying on continuity, we see that as `x->2,` `t->1.` Using a basic fact about limits and reciprocals, we see that
`g'(2)=lim_(x->2)(g(2)-g(x))/(2-x)=\lim_(t->1)((f(1)-f(t))/(1-t))^(-1)=1/(f'(1)).` | {
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quantum-algorithms, nielsen-and-chuang, hadamard, deutsch-jozsa-algorithm
These formulas are not linked. The first one simply describes that, for a single qubit state $|x\rangle$, we have:
$$H|x\rangle=\frac{|0\rangle + (-1)^x|1\rangle}{\sqrt{2}}.$$
Indeed, we know that:
$$H|0\rangle=\frac{|0\rangle + |1\rangle}{\sqrt{2}}$$
and:
$$H|1\rangle=\frac{|0\rangle - |1\rangle}{\sqrt{2}}$$
Thus we can write:
$$H|x\rangle=\frac{|0\rangle + (-1)^x|1\rangle}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left[(-1)^{x\cdot 0}|0\rangle+(-1)^{x\cdot 1}|1\rangle\right]=\frac{1}{\sqrt{2}}\sum_z(-1)^{x\cdot z}|z\rangle.$$
Note that this formula is always true, it is not specific to the Deutch-Josza algorithm, but is a property/definition of the Hadamard gate.
The second formula describes the state just after we've applied the function $f$ to the state:
$$\left|\psi_2\right\rangle=\frac{1}{\sqrt{2^n}}\sum_x(-1)^{f(x)}|x\rangle\left[\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right]$$ | {
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waves, electromagnetic-radiation, thermal-radiation
Say we take the volume of this one-eighth of a sphere. Since the points corresponding to modes have a density of one, this volume will be equivalent to the total number of modes at or below frequency $\nu$ (approximately, but since for light we usually work with very high frequencies, the error is negligibly small). Thus the number of modes below frequency $\nu$ is
$$N(\nu)=\frac{1}{8}\cdot\frac{4}{3}\pi R^3=\frac{\pi}{6}\cdot\left(\frac{2L\nu}{c}\right)^3=\frac{\pi}{6}\cdot\frac{8L^3\nu^3}{c^3}=\frac{4}{3}\frac{\pi L^3\nu^3}{c^3}$$
Second Question
The equation we previously derived isn't very useful as it is. It shows the number of modes of frequecy equal or below $\nu$. For blackbody analysis, it's be more convenient to work with the number of modes in a small interval of frequencies. This will be
$$dN=\frac{dN}{d\nu}d\nu$$
And $\frac{dN}{d\nu}$ is the mode density per unit frequency. So we differentiate and obtain
$$dN=4\pi\frac{L^3\nu^2}{c^3}d\nu$$ | {
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java, minecraft
# True makes the materials list the only blocks that are modified by this plugin based on previous configuration options. (BLACKLIST)
# False makes the materials list the only blocks that are not modified by this plugin based on previous configuration options. (WHITELIST)
# Options: [true/false]
materials-blacklist: true
# List of materials to be modified based on previous configuration options.
# NOTE: The larger this list is, the larger the impact the plugin will be on performance.
# Options: block names in list format
# Example:
# - DIAMOND_ORE
# - IRON_ORE
# - EMERALD_ORE
materials:
- DIAMOND_ORE public static NoFullBrightMining plugin;
This static field seems unused, in which case it can be removed.
// Global variables
ArrayList<String> block_list = new ArrayList<String>();
boolean perm_nfbm_disabled = false;
// Configuration variables
boolean nfbm_enabled; | {
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python, performance, programming-challenge
Let's fit these results to the equation \$t = am^4\$:
Gnuplot finds \$a = 1.11 × 10^{-5}\$, so we predict that when \$m=100\$, the runtime will be about 1110 seconds (about 19 minutes).
4. Performance improvement
The code here is general-purpose — for example polygon_area computes the area of any polygon. But in this problem we don't need the general case — we only have quadrilaterals whose vertices are at lattice points on the coordinate axes. So the obvious thing to do is to reduce the amount of work by only computing the bare minimum, and reusing as many results as we can.
The key observation is that we can compute the number of interior points in the quadrilateral from the number of interior points in the four right-angled triangles in the four quadrants, like this: | {
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c, game, curses
"squares in the cardinal directions, or 1 square in the diagonal directions."
"\n\n"
"To move a piece, use the [m]ove command. That command takes four parameters: the coordinates (x,y) of both "
"locations. For example, to move the piece at (05,10) to (06,10), the command would be 'm 05 10 06 10'."
"\n\n"
"ABILITIES:\n"
"The Dragon can breathe fire on an enemy piece and destroy it, if there is a maximum of 1 square in between "
"the Dragon and the enemy piece. Breathing fire does not move the attacking piece. "
"\n\n"
"The Archer can shoot arrows at pieces 4 squares away. The Wizard follows the same rules, but it can only "
"shoot 3 squares. Shooting arrows or magic also does not move the attacking piece."
"\n\n"
"To make use of ranged attacks, use the [s]hoot command. That command takes four parameters: the coordinates "
"(x,y) of the attacker, and the coordinates of the victim. For example, to shoot the piece at (05,07) with " | {
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ros, kinect, mit-ros-pkg, skeletal-tracker
and I get the following error:
[ rosmake ] Packages requested are: ['kinect_tools']
[ rosmake ] Logging to directory/home/gsaponaro/.ros/rosmake/rosmake_output-20110525-194202
[ rosmake ] Expanded args ['kinect_tools'] to:
['kinect_tools']
[ rosmake ] Generating Install Script using rosdep then executing. This may take a minute, you will be prompted for permissions. . .
Failed to find stack for package [eigen3] | {
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Department of Mathematical Science University of Alabama in Huntsville Principles of Fourier …. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Convert the real Fourier se-ries of the square wave f(t) to a complex series. Rad225/Bioe225 Ultrasound Fourier Series (review) Fall 2019. This section explains three Fourier series: sines, cosines, and exponentials e ikx. 5 The complex form of the Fourier series. 3 Square Wave–High Frequencies One application of Fourier series, the analysis of a “square” wave (Fig. Fourier series of the elementary waveforms. PDF The Exponential Form Fourier Series. It builds upon the Fourier Series. For the periodic bipolar, 50% duty-cycle square wave, the θ -averaging of this waveform over one θ -cycle is: QED. This subtle property is due to the symmetry of waveforms (except for the sawtooth, which is not symmetric). 01:6; >> fexact=4*(t<=3)-2*(t>=3); >> plot(t,fexact) and results | {
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c, queue
enqueue(queue,15);
enqueue(queue,4);
enqueue(queue,8);
enqueue(queue,13);
enqueue(queue,7);
listQueue(queue);
peekQueue(queue);
struct QueueNode node;
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
node = dequeue(queue);
printf("\"%d\" came out of the queue\n",node.data);
closeQueue(queue);
return 0;
} | {
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thermodynamics, gas
Now let's suppose to fill half of my bottle with liquid Freon at a temperature of 70F, seal it and I cool it off to -15F (note -15F is the boiling point at 14.7 PSI). What happens to its volume, vapor pressure, and boiling point? The vapour pressure of R134a as a function of temperature is readily Googlable. For example there is a table here. Comparison of this data with the picture of the gauge shows that the gauge is showing the relationship between the vapour pressure and the temperature. So it will give the correct readings only as long as some liquid remains in the cylinder. | {
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It is reasonable to ask: why is computing the angle useful? The answer comes in the kind of invariance we expect data to have. Consider an image, and a duplicate image, where every pixel value is the same but $$10\%$$ the brightness. The values of the individual pixels are in general far from the original values. Thus, if one computed the distance between the original image and the darker one, the distance can be large.
However, for most ML applications, the content is the same—it is still an image of a cat as far as a cat/dog classifier is concerned. However, if we consider the angle, it is not hard to see that for any vector $$\mathbf{v}$$, the angle between $$\mathbf{v}$$ and $$0.1\cdot\mathbf{v}$$ is zero. This corresponds to the fact that scaling vectors keeps the same direction and just changes the length. The angle considers the darker image identical. | {
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... I think it is worthwhile to point out that $A$ has a more structural property than just omitting one particular type. If you had in mind something else, I’d be happy to hear more details, it’s perfectly possible I’m missing something. – Emil Jeřábek Nov 27 '13 at 12:31 | {
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python, neural-network, scikit-learn, rmse
# param_list = {"hidden_layer_sizes": [(10,),(50,)], "activation": ["identity", "tanh", "relu"], "solver": ["lbfgs", "sgd", "adam"], "alpha": [0.00005,0.0005]}
# gridsearchcv = GridSearchCV(estimator=model, param_grid=param_list) | {
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(b) As previously, $I_{40}=40I_1$. Therefore,
$I_{40}=10\,{\log \left(\frac{I_{40}}{I_0}\right)\ }=10\,{\log \left(\frac{40\frac{{10}^{8.25}}{20}I_0}{I_0}\right)\ \ }=85.5\ {\rm dB<165\ dB}$
Your are trying to understand important instructions from someone who is soft-spoken. From your distance $r_1$ her voice sounds like an average whisper of $20\, {\rm dB}$. So you move to a position where you are a distance $r_2$ from her and the sound level is at $60\, {\rm dB}$.
(a) If the intensity is $I_1$ at distance $r_1$ and $I_2$ at distance $r_2$, calculate the ratio $I_2/I_1$.
(b) Use your answer to part (a) to calculate the ratio $r_2/r_1$. | {
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"url": "http://physexams.com/exam/Waves/4"
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electromagnetism, optics, electromagnetic-radiation, electric-fields, poynting-vector
Title: Energy of Monochromatic Beam of Light A monochromatic beam of light has energy
$$ E_{\text{beam}} = N \hbar \omega, $$
$ N $ being the number of photons in the beam and $ \omega $ their frequency.
Another way to evaluate this energy is through the Poynting vector. In order to do that, we assume an electric field given by
$$ \textbf{E}(t) = \textbf{E}_0 \text{cos}\left[\omega (t-t_i)\right] $$
in the dipole approximation. As a plane-wave, the magnetic field is related to the electric field by
$$ \textbf{B} = \frac{1}{c} \hat{\textbf{k}} \times \textbf{E}, $$
where $ \hat{\textbf{k}} $ is the propagation direction. Thus, the energy is
$$ E_{\text{beam}} = \int_{t_i}^{t_f} \oint_A \textbf{S} \cdot d\textbf{a} \ dt = \frac{1}{\mu_0 c} \int_{t_i}^{t_f} \oint_A \textbf{E}^2 \ \hat{\textbf{k}} \cdot d\textbf{a} \ dt = \frac{A \textbf{E}_{0}^2}{4 \mu_0 c \omega} \left[2 \omega \tau + \text{sin}(2 \omega \tau) \right], $$ | {
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angular-diameter
If you only have RA and Dec of a single point as @JamesK's answer supposes you won't have a size.
But if you have the celestial coordinates of several points around its perimeter like a nebula or even a constellation, then yes! you can get an angular size!
Let's say there's a "rectangular" patch that extends from 05h 33m to 05h 37m in RA and from -6. to -5 deg in dec.
Since angles in declination are real angles, it will be 1 degree tall.
4 minutes of RA is also 1 degree if it's on the equator, but we have to multiply by cos(dec) because the lines squeeze together at the top and bottom.
In this case cos(5.5 degrees) is almost 1 (0.995) so we can ignore it. But if your patch of sky is farther from the celestial equator, just don't forget to convert RA to degrees then multiply by cos(dec).
If you have a complicated shape defined by several vertices, like "How many square degrees does constellation X cover?" That will be an excellent new question! | {
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python, python-3.x, programming-challenge, strings
Itertools
The pairwise function could be better described using the pairwise recipe.
If you utilize itertools.groupby the challenge is super easy, as it makes slices for you.
def compress_string(s: str) -> str:
if s == "":
return ""
return "".join(
f"{sum(1 for _ in g)}{k}"
for k, g in itertools.groupby(s)
) | {
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} |
Sketching Polar Curves Examples. Graph the point Q, (r; ) = 2; ˇ 6. c Double Integrals in Polar Coordinates (r; ) Let us suppose that the region boundary is now given in the form r = f( ) or = h(r), and/or the function being integrated is much simpler if polar coordinates are used. 2-3-17: We continued to convert between polar and rectangular equations. These will all be positive X,Y rectangular coordinates in Quadrant I of the Cartesian plane (X headed right from 0 and Y headed up from 0). ) The graph of = , where is a constant, is the line of inclination. The angular dependence of the solutions will be described by spherical harmonics. Suppose that X is a random vector with joint density function f X(x). PreCalculus. Key Concept: Constellations can be represented graphically. All the coordinates are made beforehand. i have this function f(r,theta) which i want to graph in polar/cylindrical coordinates. In polar coordinates the position of an object $$R$$ distance from the origin as | {
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"url": "http://gesuitialquirinale.it/rdil/polar-coordinates-pdf.html"
} |
notation, path-integral
\phi_i)$, You would have $e^{-f(\phi_1,...\phi_N)} = \prod _{i=1}^Ne^{-f_i(\phi_i)}$, leading to your last equation:
$$\int \mathcal D \phi \ e^{-F[\phi]} = \lim_{N \rightarrow + \infty} \prod_{i=1}^N \int_{\mathbb R} d\phi_i \ e^{-f_i(\phi_i)}$$
However, for a general functional $F$, and its corresponding discretized multivariable function $f$, the $e^{-f(\phi_1,...,\phi_N)}$ cannot be necessarily written as a product of exponentials of single variables. So in general:
$$e^{-f(\phi_1,...\phi_N)} \neq \prod _{i=1}^Ne^{-f_i(\phi_i)}$$
Meaning that your final equation wouldn't work anymore. | {
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quantum-chemistry, spin, epr-spectroscopy
$$| \langle i | H' | j \rangle |^2 \tag{2}$$
(this is part of Fermi's golden rule). For radiative transitions, the perturbation $H'$ is proportional to the position operator $\vec{r} = (x, y, z)$ (this integral is then called the transition dipole moment). It's not important to actually bother with what this exactly is. The important bit is that this operator depends only on spatial components (it's independent of spin). So, the matrix element in eq. $(2)$ can be rewritten as
$$\begin{align}
\langle i | H' | j \rangle &= \left< \chi_\text{spin}^{(i)}\phi_\text{orbital}^{(i)} \middle| H' \middle| \chi_\text{spin}^{(j)}\phi_\text{orbital}^{(j)} \right> \tag{3} \\[5pt]
&= \left< \chi_\text{spin}^{(i)} \middle| \chi_\text{spin}^{(j)} \right> \left< \phi_\text{orbital}^{(i)} \middle| H' \middle| \phi_\text{orbital}^{(j)} \right> \tag{4} \\
\end{align}$$ | {
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"url": null
} |
javascript, jquery, asynchronous
Title: Processing a collection of objects, one at a time, with an asynchronous method. I currently use setInterval and a wait flag to process this collection. Is there a cleaner way?
var wait = false;
var processInterval = setInterval(function(){
if(!wait){
var currentVideo = videos.shift();
if(currentVideo){
wait = true; | {
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java, performance
Pulling the step response calculation requiring a call to exp() out of the inner loop provided the greatest time savings so far, but I'm running out of ideas to further tune. Are there any optimizations I can make to improve execution speed of update()?
import static java.lang.Math.exp;
import static java.lang.Math.sin; | {
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c#, classes, serial-port, device-driver
{
[Description("D{0}0000{1}")]
StartAutoRead,
[Description("YCC0000{1}")]
ReadMyStructure,
[Description("U")]
SuspendAutoRead,
[Description("Z")]
ResumeAutoRead,
[Description("B{0}0000{1}")]
ReadBlock,
[Description("E{0}0000{1}H{2}")]
WriteBlock
} | {
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"id": 9558,
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"tags": "c#, classes, serial-port, device-driver",
"url": null
} |
Imagine an extreme case of shopping where you have 3 Louis Vuitton bags costing £1,000,000 each. Which is cheaper, getting a discount of £1 per item or putting one bag back? This tells you that 3 x 999,999 is larger than 2 x 1,000,000. And that tells you whether increasing the little term or the big term has a greater effect.
For the layman
Think of the question as big vs small; 35000 (big) vs 25000 (small), respectively.
When comparing column A to B, 35K is being multiplied by small in both situations However, in column A, it's being multipled ~400 times vs ~40 times in B. That's an entire order of magnitude larger (a.k.a extra digit).
Thus, you can intuit that A is bigger.
Note: Once they are both in the same order of magnitude, e.g. if B was >100, you'll probably want to use calculator. | {
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"url": "https://math.stackexchange.com/questions/408759/which-one-is-bigger-35-043-%C3%97-25-430-or-35-430-%C3%97-25-043"
} |
quantum-mechanics, hilbert-space, interactions, quantum-states
This means that if you start with a non-entangled state $\psi = \psi_A\otimes \psi_B\in \mathcal{H}_A\otimes \mathcal{H}_B$, the time evolution of this state will forever remain non-entangled as
$$\mathrm{e}^{\mathrm{i}Ht}\psi = \left(\mathrm{e}^{\mathrm{i}H_At}\psi_A\right)\otimes\left(\mathrm{e}^{\mathrm{i}H_Bt}\psi_B\right),$$
so effectively you can describe the state of the system by pairs $(\psi_A,\psi_B)$ from the direct sum $\mathcal{H}_A\oplus\mathcal{H}_B$ since they can never become entangled.
However, there's rarely a reason to do this: If the two systems don't interact, why would you want to model them as subsystem of a combined system and not just as two separate system? | {
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algorithms, optimization
All these files must be in the same folder.
FILES STRUCTURE
A single text file must have as many rows as daily slots (usually 10) and as many columns as working days (usually 5).
Each cell must contain a preconfigured char: | {
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"tags": "algorithms, optimization",
"url": null
} |
temperature, luminosity, wavelength
$$B_f\ df = B_\lambda\ d\lambda\ , $$
where $B_f$ and $B_\lambda$ are the expressions for the flux per unit frequency and flux per unit wavelength of a blackbody.
The rest is just maths I'm afraid. The frequency where $B_f$ is maximised is found by differentiating with respect to frequency and equating to zero. We can then replace $B_f$ with $B_\lambda\ d\lambda/df$. i.e. At $f_{\rm max}$ it must be that
$$\frac{dB_f}{df}= \frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = 0 $$
But since $f = c/\lambda$, we can replace $df$ with $df = -c\ d\lambda/\lambda^2$.
$$\frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = \frac{\lambda^2}{c}\frac{d} {d\lambda}\left(B_\lambda \frac{\lambda^2}{c}\right) = 0. $$
Eliminating $\lambda = 0$ as a useless solution, multiplying both sides by $c^2/\lambda^2$ and differentiating the product in the bracket, we get
$$ \lambda^2 \frac{dB_\lambda}{d\lambda} + 2\lambda B_\lambda = 0$$
and so
$$\frac{dB_\lambda}{d\lambda} = -\frac{2 | {
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} |
forces, classical-mechanics, pressure, fluid-statics, estimation
Title: Help me understand extreme pressure So we have the tragedy of the submersible being lost near the Titanic site.
I am trying to understand the pressure the vessel was under. I have heard the "$1$ atmosphere for every $10$ meters" approximation. So if the implosion happened $1$ $km$ down, $100$ atmospheres of pressure were acting on the sub.
Is it also valid to think of the column of water on top of the sub? If the sub is $20'\times7'$, and $0.6$ miles down, this comes out to $28.3$ $million$ pounds ($12.9$ $million$ $kg$) of water bearing down on the hull.
My numbers may not be right, but they aren't crazy - there's a lot of water on that sub. How would a physicist explain the force of the implosion to a non-physicist in a way that conveys the effect? TLDR: To understand it, don't think about pressure. Think about energy. | {
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"url": null
} |
java, swing
numberButton = new JCheckBox("Numbers");
getContentPane().add(numberButton);
numberButton.setBounds(121, 131, 97, 23);
}
ActionListener method:
@Override
public void actionPerformed(ActionEvent event) {
Random random = new Random();
if (event.getSource() == genButton) {
boolean boxesArentChecked = !numberButton.isSelected() && !symbolButton.isSelected() && !uppercaseButton.isSelected()
&& !lowercaseButton.isSelected();
boolean numIsZero = lengthChooser.getValue() == 0;
if (boxesArentChecked && numIsZero) {
passField.setForeground(Color.RED);
passField.setText("Select a box and a size!"); | {
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"tags": "java, swing",
"url": null
} |
### Show Tags
29 Oct 2012, 04:41
1
rajathpanta wrote:
Is x^2+y^2>100??
(1) 2xy<100
(2) (x+y)^2>200
To me its only B. because statement 2 boils down to x+y>$$\sqrt{200}$$
Can someone explain the OA
Merging similar topics. You are right, OA should be B, not C. See here: is-x-2-y-108343.html#p859197
_________________
Intern
Joined: 31 Oct 2012
Posts: 2
Re: Is x^2 + y^2 > 100? [#permalink]
### Show Tags
31 Oct 2012, 15:49
It's B for me.
$$(x + y)^2 = x^2 + 2xy + y^2$$
The largest possible value $$2xy$$ can reach is $$(x + y)^2/2$$, that only occurs when $$x = y$$.
When $$x = y$$, it turns out that $$x^2 + 2xy + y^2 > 200$$ is $$x^2 + 2xx + x^2 > 200$$, which can be rearranged in $$2x^2 + 2x^2 > 200$$, meaning that $$x^2 + y^2$$ is at least half of the stated value, while $$2xy$$ can be at most the other half.
Any value above 200 will require that $$x^2 + y^2 > 100$$, making 2) sufficient.
Intern
Joined: 06 Feb 2013
Posts: 21
Re: Is x^2 + y^2 > 100? [#permalink]
### Show Tags | {
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"url": "https://gmatclub.com/forum/is-x-2-y-108343-20.html"
} |
zoology
A more straightforward and well known example that directly contradicts the ears claim is the dolphin. Dolphin ears are small holes on the sides of their heads but they give birth to live young as do all sea mammals. In fact, I am pretty sure that all sea mammals lack external ears (and cetaceans at least also lack external testicles) but give birth to live young. So seals, sea lions, walruses etc also contradict the ears rule.
In conclusion, no it is not true that all animals with internal ears lay eggs. | {
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proof-techniques, lambda-calculus
$$
\dfrac{
x\# N, y\# M \vdash M=[y:=x]N
}{
\vdash \lambda x.M = \lambda y.N
}
$$
We indeed have $x\# N$, since the $x$ variable was renamed as $y$ in $N \equiv [x:=y]N$. We also have $y\# M$ by our choice of $y$. Induction hypothesis states $M=M$, which is $M=[y:=x]N$. Hence, the conclusion of the rule holds, which is $\lambda x. M = \lambda y. [x:=y]M$. If terms are identified up-to $\alpha$, then this equality is the same as $\lambda x.M = \lambda x.M$.
If $A=MN$, the induction hypotheses and rule 3 suffice. | {
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"url": null
} |
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