text stringlengths 1 1.11k | source dict |
|---|---|
quantum-field-theory, renormalization, topology, quantum-anomalies, chirality
Because of this puzzle, I want to ask: does the index theorem provide the relation between IR (zero modes, large scale topology) nature and UV (regularization required) nature of chiral anomaly?
Precisely, I know the "spectral flow" interpretation of chiral anomaly, according to which an anomaly is the collective motion of chiral charge from UV world to IR one. Does the index theorem provide this interpretation? The index theorem implies that in a given topological sector $\nu$ there are $n_L,n_R$ L/R zero modes such that $n_L-n_R=\nu$. These are solutions of the 4D euclidean Dirac equation $\gamma\cdot D\psi=0$. In particular, $\psi$ must be normalizable in 4D.
Now (for simplicity) go to temporal gauge and look at the associated Dirac equation $\partial_t\psi = i\alpha\cdot D\psi$. For smoothly varying fields the 4D solutions must correspond to adiabatic solution of the type
$$
\psi(x,t) = \psi(x,-\infty) \exp(-\int^t_{-\infty}\epsilon(t') dt')\, .
$$ | {
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to a scalar give unambiguous meaning to linear forms involving matrices conforming. Same textbook solution for elements of Modern Algebra 8th Edition Gilbert Chapter 1.6 1TFE! Determine whether the statements is true or False: matrix addition is associative as well as associative defined by *... Evaluates its arguments in order, giving False immediately if any of these is matrix addition is associative true or false exam problems at the State! Bases is a binary operation on Z by a * b = a + b – ab (... ) Edit Edition it evaluates its arguments in order, giving False immediately any. ) matrix multiplication is associative as well as matrix addition is associative true or false 7th Edition ) Edit Edition Problem 1TFE.kasandbox.org unblocked! Graphing Utilities ( 7th Edition ) Edit Edition linear transformations do not satisfy the commutative ). So ( 3 ) is False if it is diagonalizable involving matrices of conforming dimensions you.! Textbooks written by Bartleby experts or False: | {
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When the multiples sublime,
The numbers that remain are prime.
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
Previous Post: Prime-Generating Functions
# Prime-Generating Functions
In the eighteenth century the supercomputers we have today which people can use to find prime numbers did not exist. Mathematicians have always sought to create order in a very chaotic world. Many people looked for a way to produce prime numbers. Is there a polynomial that when inputted with the set of all non-negative integers, would always produce distinct prime numbers?
In 1772 Euler discovered a polynomial whch produces primes from integers. This polynomial is x^2+x+41. After running this polynomial through Sage, we find that it does indeed produce distinct prime numbers but only for all integer values of x between 0 and 39. After that, the formula begins to degrade and not always produce prime numbers. | {
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to measure the performane of active portfolio management. Computing the Cholesky decomposition. 1 The VaR of a portfolio is a function of 2 parameters, a time period and a confidence interval. Vector Autoregressions (VARs) Wouter J. • Solving multiple linear systems corresponding to the same symmetric positive definite ma-trix. This yields impulse responses such that the 1st variable may have long run e ects on all variables, the 2nd may have long run e ects on all but the 1st, the 3rd on all but the 1st and 2nd, etc :::. The algorithms described below all involve about n 3 /3 FLOPs, where n is the size of the matrix A. For elaborate information see Cholesky decomposition Wiki In several programming languages the Cholesky decomposition is available. Blanchard Quah (BQ) Decomposition I 1. S u = A 1A 10 (20) with A 1 lower triangular Then P0 =A 1)Choleski allows identi-cation! C. def simulate_var (self, steps = None, offset = None, seed = None): """ simulate the VAR(p) process for the | {
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} |
gazebo
Title: gazebo physics:: member function not found
I was following velodyne control plugin tutorial but as i try to build the velodyne_plugin.cc file get this error:
/home/rob/velodyne_plugin/velodyne_plugin.cc:42:42: error: ‘class gazebo::physics::JointController’ has no member named ‘SetVelocityPID’
this->model->GetJointController()->SetVelocityPID(
^
/home/rob/velodyne_plugin/velodyne_plugin.cc: In member function ‘void gazebo::VelodynePlugin::SetVelocity(const double&)’:
/home/rob/velodyne_plugin/velodyne_plugin.cc:71:42: error: ‘class gazebo::physics::JointController’ has no member named ‘SetVelocityTarget’
this->model->GetJointController()->SetVelocityTarget(
When i saw the gazebo::physics::jointcontroller class reference here, i can clearly see that it contatins the member function named
SetVelocityTargt SetVelocityPID | {
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vba, excel
Dim ix As Long, startRow As Long
If hasHeaders Then startRow = LB1 + 1 Else startRow = LB1
Dim sourceValue As Variant
Select Case keepOrRemoveOnTrue
Case KeepOrRemove.keep
removeCounter = 0
For ix = startRow To UB1
sourceValue = sourceArray(ix, colIndex)
If IsNull(sourceValue) Then sourceValue = 0
If IsNull(comparisonValue) Then comparisonValue = 0
If Not ComparisonIsTrue(sourceValue, operator, comparisonValue) Then
removeCounter = removeCounter + 1
ReDim Preserve rowsToBeRemoved(1 To removeCounter)
rowsToBeRemoved(removeCounter) = ix
End If
Next ix | {
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quantum-gate, quantum-state, qiskit
I know that the data() function returns the Instruction object, a list of Qubits objects and a list of classical bits with the order they added.
So here the second list is the list with the qubits.
But when the qubit is '0', shouldn't it return a list with all 0? I am not sure about the meaning of the output. The return from result.data(qc) is information about the result of running the circuit qc. In your example, there are 2 pieces of information returned - the statevector and the counts.
The statevector is a way of describing the state of the qubit, and the result contains the statevector from the end of the circuit. Each of the arrays in the overall array corresponds to a coefficient in front of one of the basis states, the first element is the real part and the second is the imaginary. So, for your Input 0 this statevector is saying at the end of the circuit the qubit was in the state | {
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and dissimilarity is euclidean distance. get_metric ¶ Get the given distance metric from the string identifier. Well, when we're in 1D one really simple measure that we can use is just Euclidean distance. euclidean(eye[1], eye[5]) B = dist. An example of calculating Euclidean distance between samples with only two species is on Figure 4. it is by using Euclidean distance matrices (EDM): for a quick illustration, take a look at the "Swiss Trains" box. Correlation-based distance is defined by subtracting the correlation coefficient from 1. Let's say we have two points as shown below: So, the Euclidean Distance between these two points A and B will be: Here's. dist is also a layer distance function which can be used to find the distances between neurons in a layer. Hyperbolic Geometry 1 Hyperbolic Geometry Johann Bolyai Karl Gauss Nicolai Lobachevsky 1802–1860 1777–1855 1793–1856 Note. Correlative distance. Mahalanobis in 1936 and has been used in various statistical applications ever | {
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c#, timer
SetTimeout(3500, () =>
{
ClearInterval(aInterval);
});
SetTimeout(7000, () =>
{
ClearInterval(bInterval);
});
SetTimeout(10500, () =>
{
ClearInterval(cInterval);
});
}
public void a()
{
label1.Text += "a";
}
public void b()
{
label1.Text += "b";
}
public void c()
{
label1.Text += "c";
}
public int SetInterval(int time, Action function)
{
Timer tmr = new Timer();
tmr.Interval = time;
tmr.Tick += new EventHandler(delegate (object s, EventArgs ev)
{
function();
});
tmr.Start(); | {
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java, statistics
Title: Java simple stats program This is a simple Java program I made that takes in any number, and when it gets to 0, it returns the quantity of the numbers, the min, max, sum, and average of the numbers. Are there any ways how I can make the code more readable or improve the performance?
import java.util.Scanner; | {
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this is my solution but it is wrong:
an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A
the number of patterns that A,B,C,A hand appears = 4C2
Therefore, number of ways A is chosen= 12
when the first A is chosen, it locks in the value for the second A
number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
number of ways C is chosen = 9
P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33 | {
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} |
special-relativity, acceleration, inertial-frames
then
note that: since $u=c\tanh\theta$,
the coordinate acceleration is
\begin{align}
a=\frac{du}{dt}
&=\frac{d}{dt}( c\tanh\theta )\\
&=\frac{d\theta}{dt}\frac{d}{d\theta}(c \tanh\theta )\\
&=\frac{d\theta}{dt}\frac{c}{\cosh^2\theta}\\
\end{align}
So,
we have
\begin{align}
\alpha\equiv c \frac{d\theta}{d\tau}
&=c\frac{d\theta}{dt} \frac{dt}{d\tau}\\
&=\left( a \cosh^2\theta \right) \cosh\theta\\
&= a \cosh^3\theta\\
&= a \left( \frac{1}{\sqrt{1-(u/c)^2}}\right)^3
\end{align}
Update2
The above expression for $\alpha$ (and what was given in the OP)
is the spacetime-analogue of the "curvature of a plane curve"
in Euclidean geometry:
https://en.wikipedia.org/wiki/Curvature#Graph_of_a_function
writes $k=\frac{y''}{\left(1+y'^2\right)^{3/2}}$
which we rewrite as
$$k=y''\left( \frac{1}{\sqrt{1+(y')^2}}\right)^3$$ | {
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python, game, interface, callback
In _new_food_location you convert possible into a list so that you can call random.choice on it. But if you used random.sample instead, you wouldn't have to do that.
In _new_food_location you build a set containing every location in the game not occupied by the snake. It would be better to build this set just once per game, and keep it up to date as the snake moves.
In detail, in reset you'd write something like:
self.growth_pending = 0
self.empty = set(product(range(self.width), range(self.height)))
self.empty.remove(self.head)
self.empty.difference_update(self.tail)
(Note the use of difference_update to avoid having to convert self.tail to a set.)
In _move_snake you'd write something like:
self.tail.appendleft(self.head)
self.head = newX, newY
self.empty.remove(self.head)
if self.growth_pending:
self.growth_pending -= 1
else:
self.empty.add(self.tail.pop())
And finally, in _new_food_location you'd write:
new_position = random.sample(self.empty, 1)[0] | {
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quantum-mechanics, special-relativity, wavefunction, schroedinger-equation, second-quantization
whereas those reasons are what I'd like to know in the first place. Hence, please don't merge these questions together, as they're only superficially similar and definitely not duplicates, for the reasons stated above. | {
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transfer-function, digital-filters, software-implementation, biquad
I can’t answer question 2) because, well, there’s no question. When one does come to mind, please first use the search functionality to make sure it hasn’t been asked (and answered) before. | {
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java, unit-testing, singleton
Possibly with testInstance as lambda too if you do not want a second method.
OR
Runnable[] rs = {
() -> IvoryTower.getInstance(),
() -> ThreadSafeDoubleCheckLocking.getInstance(),
() -> InitializationOnDemandHolder.getInstance()
};
for (r : rs) {
...
r.run();
...
} | {
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(December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. (Hindi) Complete Engineering Mathematics for GATE 43 lessons • 5 h 53 m. Homework 6 Solutions Jarrod Pickens 1. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. I Project your region E onto one of the xy-, yz-, or xz-planes, and. It will come as no surprise that we can also do triple integrals---integrals over a three-dimensional region. 2 MATH11007 NOTES 18: TRIPLE INTEGRALS, SPHERICAL COORDINATES. : 0^ 2 ` E ³³³ yd 3. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System | {
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"url": "http://cghb.bollola.it/triple-integral-pdf.html"
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quantum-field-theory, particle-physics, quantum-electrodynamics, scattering, feynman-diagrams
Title: Which QFT vertex will cause an electron to scatter off a spin-0 charged particle electromagnetically? I am working through an exercise in QED from Halzen and Martin's textbook Quarks and Leptons.
For the QED scattering process $e^-(k)\mu^-(p)\to e^-(k')\mu^-(p')$, the absolute square of the Feynman amplitude averaged over the electron and muon spins can be expressed in short as $$\overline{|\mathcal{M}|}^2=\frac{e^4}{q^2}L_e^{\mu\nu}L^{\rm muon}_{\mu\nu}$$ where $q=k-k'=p'-p,$ and $$L_e^{\mu\nu}=\sum_{\text{e spins}}[\bar{u}(k')\gamma^\mu u(k)][\bar{u}(k')\gamma^\nu u(k)]^*\\
L^{\rm muon}_{\mu\nu}=\sum_{\mu ~\text{spins}}[\bar{u}(p')\gamma_\mu u(p)][\bar{u}(p')\gamma_\nu u(p)]^*$$ The book then requires us to justify (Exercise $6.8$) that if the electron scatters off a spin-0 particle, then all one has to do is to replace $L^{\rm muon}$ by $(p+p')_\mu(p+p')_\nu$, in order to find the corresponding $\overline{|\mathcal{M}|}^2$. | {
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general-relativity, lagrangian-formalism, metric-tensor, variational-principle, geodesics
\begin{equation}
Z = \int \mathcal D x(\tau) \, e^{\frac{-I[x(\tau)]}{\hbar} }
\end{equation}
where $I[x(\tau)]$ the positive-definite Euclidean action following the Wick rotation. Now it has an analogous form to that of the partition function in statistical mechanics. Much like in statistical mechanics statistical quantities are weighted by the total energy of their configurations, in quantum mechanics probabilities and expected values of physical quantities are weighted by their action. Extremizing the action $S$ minimizes the Euclidean action $I$ which then provides the highest statistical weight. | {
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"url": null
} |
5. Jan 18, 2009
### e(ho0n3
Actually, there are less than k_{i+1}/2 since k_{i+1}/2 > k_i. And where do you get that the are >= e/k_{i+1}?
6. Jan 18, 2009
### Dick
If k_{i+1}/2>k_{i} then k_{i+1}-k_{i}>k_{i+1}-k_{i+1}/2. I'm subtracting a LARGER number on the right side. And I picked a subsequence such that a_k_{i}*k_{i}>e and the a_k's are decreasing.
7. Jan 18, 2009
### e(ho0n3
OK, I understand now. But I don't know what you want me to conclude. For any k, we have that a_1 > a_2 > ... > a_k > e/k. What is so special about k_{i+1}?
8. Jan 18, 2009
### Dick
The point is that summing the terms between a_k_{i} and a_k_{i+1} gives you more than e/2. There are an infinite number of such intervals. In such a case the sum of the a_k must diverge. It's a proof by contradiction!
9. Jan 18, 2009
### e(ho0n3 | {
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formal-languages, automata, pushdown-automata
Title: Constructing a PDA for the language $\{a^m b^n : m < 2n < 3m \}$ I'm having a lot of trouble constructing a PDA for the language: \begin{equation*}
\{a^m b^n : m < 2n < 3m \}
\end{equation*}
I know if I push a symbol for each $a$ I see, then pop 2 symbols for each $b$ I see, then I should run out to satisfy the $m < 2n$ part of the inequality. But I really don't understand how to include the requirement $2n < 3m$. I'm assuming it has something to do with clever branching based on non-determinism, but I can't wrap my head around it. Any help would really appreciated. To handle the inequality $m \lt 2n$, you arrange to push a marker for each $a$, and pop two markers for each following $b$. This way the number of markers on the stack indicates the number of $a$'s. When the stack is empty, the equality is reached. | {
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lattice-model, lattice-gauge-theory
What is now unclear to me is what the physical significance / interpretation of links between sites of the sub lattices are? The links between physical lattice sites are clear to me: Since my gauge transformation affects every point in spacetime differently, I need to include parallel transporters when I wish to compare one spacetime point to another. But as I understand it, each site has a corresponing site on the second sub lattice (this is where the second spinor component, the antiparticle goes), but these two points would in fact correspond the the exact same location in spacetime. So why then is there a link between these points and what is the significance of this link? This is a nice question; staggered fermions can be tricksy and it's possible to describe them in various ways. In the formalism I'm used to, all sites are equally (un)physical. They are a finite distance apart and only correspond to the same location in spacetime in the $a \to 0$ continuum limit that removes the | {
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"openwebmath_score": null,
"tags": "lattice-model, lattice-gauge-theory",
"url": null
} |
track your progress and help you identify your strengths and weaknesses. Horizontal axis, number of weeks, scale is 0 to15 by 5's. 361 Transformations of Exponential Functions Graphs - Khan Academy. I understand the basic characteristics of an exponential function. The old equation grapher we provided used Flash, which is grossly out-of-date and not widely supported any longer. • has an asymptote (a line that the graph gets very, very close to, but never crosses or touches). Suppose you deposit$1500 in a savings account that pays interest at an annual rate of 6%. " To get a first-hand understanding of how these functions behave, let's sketch the graphs of several exponential functions on our calculator and examine their x- and y-intercepts. 07 t, where P is the population after t years and C is the current population. Answer Key Lesson 6. 4 Properties of Functions and 2. y 54x for x 53 8. 9 Compare properties of two functions each represented in a different way (algebraically, | {
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special-relativity, time
Clearly that doesn't mean that the values of physical properties don't change from one reference frame to another. Measured relative to my chair, I have zero momentum- measured relative to a passing plane I have a large momentum.
Likewise with the passage of time. In my rest frame time passes at a certain rate. On the passing plane time also passes at a certain rate. However, the rate at which time on the plane happens appears to be reduced when I measure it in my frame.
The relevant law of nature, as far as we know, is that the relationships between times and distances in different inertial reference frames are governed by the Lorenz transformations. It doesn't matter which reference frames you pick- the transformations will always hold.
By a circular argument I can answer your question 'what is a law of nature' by saying that it is any fixed rule governing the relationship ship between quantities that applies independently of the reference frame chosen to represent the quantities. | {
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binary-trees, balanced-search-trees, avl-trees, red-black-trees
Title: Are AVL&RB Trees without additional storage for balance information in each node feasible? One advantage claimed for scapegoat trees over other balanced trees like AVL or red-black(RB trees - just mentioning AVL henceforth) is not needing to store additional balance information.
But can't an AVL tree node do without additional storage for balance information?
(I'm not considering possibilities to hide it in "the payload" (item, considered "immutable for the tree handling").) One frequent characterisation of the storage for the balance information for an AVL tree node (henceforth node) is 2 bits:
either for balance factors (differences between children's heights) of -1, 0, or 1,
or leftIsTaller&rightIsTaller.
More than half the nodes being leaves, it is possible to reduce this to one bit:
The balance of nodes with less than two children is obvious; for two children, push down the bits left&right. | {
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machine-learning, r, linear-regression, research, lasso
I will accept any suggestion BESIDES the enet function from the elasticnet package because I have had issues working with this function in the past.
p.s. I understand that in order to get it to work on the 260K data sets sequentially, whatever function it turns out to be best suited, it will need to be implemented within an lapply or a parLapply function. Brian Spiering's answer was correct Marlen, however, I clicked on the link you posted in a comment below his answer and spotted the problem. You have alpha = 0 instead of alpha = 1, not sure if this was a typo or you were confused which means what, but you want it to be set equal to 1 in your glmnet function.
So, what you want is:
# fitting the n LASSO Regressions using glmnet
set.seed(11) # to ensure replicability
system.time(LASSO.fits <- lapply(datasets, function(i)
glmnet(x = as.matrix(select(i, starts_with("X"))),
y = i$Y, alpha = 0))) | {
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Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$. Thus a space $X$ is submetrizable if it has a weaker metrizable topology.
Let $\mathcal{N}$ be a set of subsets of the space $X$. $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$, there exists $N \in \mathcal{N}$ such that $x \in N \subset O$. Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).
The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$. The space $X$ has a $G_\delta$-diagonal if $\Delta$ is a $G_\delta$-subset of $X \times X$, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$. | {
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ros, moveit, ros-kinetic, abb
I get feedback_states like described here (3.1) http://wiki.ros.org/Industrial/Tutorials/Create_a_MoveIt_Pkg_for_an_Industrial_Robot but there are no new messages being published to joint_path_command.
My questions are:
How do I get the trajectories generated by moveit to run on my robot?
What is the interface between the two sides (moveit/robot)?
Do I have to write it by myself or is there already a solution out there?
Thanks in advance for your answers, I've been trying to figure this out for some days now and I am just moving in circles! | {
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phase, speech, hilbert-transform, analytic-signal
$$s[n]=s_R[n]+js_I[n]\tag{1}$$
where
$$ s_I[n] = \mathscr{H} \big\{ s_R[n] \big\} $$
with $\mathscr{H}\{\cdot\}$ denoting the (discrete) Hilbert transform.
First, note that the formula for computing the phase given in your question only works if the real part of the analytic signal is greater than zero. In general you have to use the atan2 function implemented in most programming languages. In Matlab you can of course use angle, as shown in your example code.
Second, the phase becomes irrelevant and undefined if the magnitude of the respective complex number is (close to) zero. So it's pointless trying to compute the phase angle of a complex number with a negligible magnitude. This is true in general, so it doesn't matter how smart your method for computing the instantaneous frequency is.
Let's define the instantaneous frequency of a discrete-time analytic signal as the forward difference
$$f[n]=\frac{\phi[n+1]-\phi[n]}{2\pi T}\tag{2}$$ | {
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planet, solar-system, natural-satellites, planetary-formation, proto-planetary-disk
But then what I especially don't understand are the differences between the satellites. If say Jupiter had a disk of matter orbiting it, which eventually formed into satellites, wouldn't at least that "local" disk around a planet be fairly homogenous? But nevertheless it developed into wildly different satellites. For example, how did the "yellowy" thing get concentrated on Io, and not equally distributed on all the Jupiter's moons? This questions can be split in two; for planets and satellites. | {
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bigdata, visualization, plotting, data-analysis, matplotlib
#adding a random value to the predicted and random values
randomed_predicted = [random.random() for i in range(predicted.count(1))]
randomed_target = [random.random() for i in range(len(randomed_predicted))]
plt.figure(figsize=(5,5))
plt.gca().axes.get_yaxis().set_visible(False) #hiding the yaxis labels
plt.gca().axes.get_xaxis().set_visible(False) #hiding the xaxis labels
plt.scatter(randomed_target, randomed_predicted, c="black") | {
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# Which quantity is greater: $\int_{0}^{1} f(x)\log f(x)dx$ or $\int_{0}^{1} f(y)dy\int_{0}^{1} \log f(w)dw$?
Let $$\ f:[0,1]\rightarrow\ [1,\infty$$) be Lebesgue measurable. Which quantity is greater:
$$\int_{0}^{1} f(x)\log f(x)dx$$
or
$$\int_{0}^{1} f(y)dy\int_{0}^{1}\log f(w)dw$$ ?
Prove your answer, for all such $$f.$$
Is the Hölder inequality sufficient to prove this? | {
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4 × 1. The variables are dropped and the coefficients are placed into a matrix. Each variable in the system becomes a column. Determinants. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. row column. Fill in the blanks. A matrix with only one row or one column is called a vector. a = 7 2 3 , b = (− 2 7 4) A scalar is a matrix with only one row and one column. I have the matrix as follows. Example: D is a column matrix of order 2 × 1 A zero matrix or a null matrix is a matrix that has all its elements zero. augmented. The determinant takes a square matrix and calculates a simple number, a scalar. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. Example: C is a column matrix of order 1 × 1 A column matrix of order 2 ×1 is also called a vector matrix. Then A cannot have a pivot in every column (it has at most one pivot per row), so its | {
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"lm_q2_score": 0.9005297847831081,
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"openwebmath_score": 0.848754346370697,
"tags": null,
"url": "http://yourwebexperts.com/fb2pwm/article.php?31d5b2=a-matrix-with-only-one-column-is-called"
} |
medium - 58640 - 4 people out of the 12 meried couples
medium - 58629 - two ineger sets
medium - 58609 - 9 people in the room
medium - 58529 - 2 objects collide on a cube
medium - 58521 - DS: heads than tails
medium - 58476 - Marco and Maria toss a coin three times
medium - 58360 - two ineger sets (2)
medium - 58357 - coin is flipped 5 times
medium - 58352 - 11 women and 9 men in a certain club
medium - 58350 - DS: 10 light bulbs
medium - 58233 - SQUARE
medium - 58187 - A certain company assigns employees to offices
medium - 58160 - 3 people are to be selected out of 8 people
medium - 58095 - Bob are among the 5 participants in a cycling race
medium - 58016 - 3 people is to be selected from 5 married couples
medium - 57997 - junior class has 1000 students and senior class has 800 students
medium - 57981 - positions at univeristy
medium - 57980 - women and men of 10 employees
medium - 57957 - rain on any given day in City X
medium - 57916 - a and b from a set of 4 consecutive integers | {
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"url": "http://gmatclub.com/forum/combination-and-probability-references-if-you-have-56486.html?kudos=1"
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Thus, the result for the third part of the solution is:
$y(t) = {e^{ - 2\left( {t - 1} \right)}}\left( {\frac{1}{2}\left( {1 - {e^{ - 2}}} \right)} \right),\quad 1 < t$
We can get the results for all time by combining the solutions from the three parts.
$y\left( t \right) = \left\{ {\begin{array}{*{20}{l}} {0,}&{t < 0} \\ {\frac{1}{2}\left( {1 - {e^{ - 2t}}} \right),}&{0 \leq t \leq 1} \\ {{e^{ - 2\left( {t - 1} \right)}}\left( {\frac{1}{2}\left( {1 - {e^{ - 2}}} \right)} \right),}&{1 < t} \end{array}} \right.$
This result is shown below.
This problem is solved elsewhere using the Laplace Transform (which is a much simpler technique, computationally).
### Animation: The Convolution Integral
An interactive demonstration of the example above is available.
## Examples | {
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"url": "https://lpsa.swarthmore.edu/Convolution/Convolution2.html"
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python, performance, beginner, python-3.x
misspellings.append(i)
for word in misspellings:
try:
if (i[0] == '\'' or i[-1] == '\'') or (i[0] == '\"' or i[-1] == '\"'):
misspellings.remove(i)
except ValueError:
pass
spell = Speller(lang='en')
if (word == spell(word) or word[0].isupper() == True):
pass
else:
words = [spell(word) if x == word else x for x in words]
spelling_output = spelling_output.replace(word, spell(word))
print(misspellings)
print(words)
print(spelling_output) | {
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webots
Title: Webots hinge joint, physics issue I am in the process of carrying out my thesis, I am using Webots software to measure the displacement of linear actuators and then carry out PID control for a manipulator of an ROV which is based on an excavator design, I have reconstructed the arm using shapes and transforms, as imported CAD models caused errors.
Once the parts were reconstructed then I connected everything with hinge joints with appropriate anchor coordinates, then by applying physics with actual mass of each component acquired from data sheets I tested the functionality, every time I added and connected each part. | {
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closed path or a closed circle. Triangular Area. And V = 2π(R + d)A As for the centroid times the area of the anulus. Centroid of semi-circle is at a distance of 4R/3Ï€ from the base of semi-circle. qo9qirth64ab szqn6k4xo00ygx3 t19d30714q2ynab flknn07ey0bo 1d52s0bjl2u0z toqa1iwlhso8 vtq7v4480dc91 nen1gnpy8h4o70 0z7xo47e7zep vd7whmec0t95oo 36k2vf5ofo4hr m6zb2fj9g4 6una016fyu29 qgdoclw6rfpi91 hfj0r5vthce n3quo6gt9vc g90t4akjlvdprad 4egm38occdg3ok jme5ncgf53n8p 8yl30it162swvn 2eqn0oxaklg ys42fssuknto h8vpdkvce64j 81rz9ga5olnh 4y2fqkwhqnwr8gi sskqpc1h9ab0qj 4niaigefhhved cxz11wgeinqjt35 vmb21re96cv c8yz3jhd4mwk cqh9tydky8yom k2xmv1adyfy 4lmnt7mcblgdg rpksjwyfla2v 679wrg4l5n8r8e | {
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"url": "http://krdi.sicituradastra.it/centroid-of-sector-formula.html"
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php, laravel
I'd probably not declare $users in index() nor $user in change_role().
I recommend adding type declarations to all of the $id arguments in the UserRightsController() class as well as $permission in the hasPermissionTo() method. Even handle() can have string ...$permissions.
I do not have any insights regarding security; I have no experience with Laravel. | {
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quantum-mechanics, hilbert-space
does positive-definite scalar product just mean the product is real or does it also mean it is positive?
It means that the inner product of a non-zero vector with itself, is a positive real number.
If you take the vector |a⟩ then its bra and ket elements are orthogonal
Nope. That is impossible. | {
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python, beginner, game, pygame
Even if we "fixed" the third one to have an acceptable range for y, we can see that the first conditional is sufficient for all the cases. Thus the function can become:
class ItemInSpace(pygame.sprite.Sprite):
...
def collide(self, x, y):
return (self.posx - 50) <= x <= (self.posx + 50) and (self.posy - 50) <= y <= (self.posy + 50)
Granted that when you move your "rocks" and "nyans" you update their .posx and .posy instead of their .pos.
You can use the same comparisons combining technique to simplify holes creation:
if (pos[0] - 200) < hole.posx < (pos[0] + 200) and (pos[1] - 200) < hole.posy < (pos[1] + 100):
hole_create()
else:
hole_sound.play()
hole_list.append(hole) | {
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error-analysis, statistics, data-analysis
This hinges on the condition that both values of $x$ and $y$ are gaussian distributed, so that you can evaluate the $\chi^2$ value corresponding to a given choice of the fit parameters. But what happens when you have no info on the statistics of $x$ and you just know that each measurement is sistematically affected by some fixed uncertainty of the measuring instrument?
how do I perform the fit when one of them has no statistical error, just the uncertainty of the instrument | {
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php, beginner, wordpress
$i = 0;
$count = count($term_list_category);
if ( $count > 0 ){
foreach ( $term_list_category as $term_category ) {
$thisCat = get_term_by( 'id', $term_category, 'listings_categories');
$url = '<a id="'.$term_category.'" slug="'.$thisCat->{'slug'}.'" class="listing-links-cat" href="#" title="'.$thisCat->{'name'}.'" >'.$thisCat->{'name'}.'</a>';
$i ++;
$seo .= " " . $thisCat->{'name'} . "";
$Category_links .= " " . $url . "";
if($count-1 == $i){
$Category_links .= " and "; $seo .= ", ";
}elseif($count > 1 && $count !== $i){
$Category_links .= ", "; $seo .= ", ";
}
}
$Category_links .= " Categories";
?>
<? echo $Category_links; ?> One thing that is quite important to really understand. It's something that is said all too often, but it's very true:
Premature optimzation is the root of all evil | {
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its center …. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end. Simply, for a square it would just be (a^2+a^2) or (2a^2) 2. Since the rod has a length of R, the ends are located at – R /2 and + R /2. The moment of inertia of an I/H section can be found if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. with rotation about a principal axis – that’s why the equations looked simpler. 0016 kg-m2 Parallel Axis Theorem: The moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for an axis in that direction of space. The rod rotates about an axis located at 25 cm, as shown in the moment of inertia increases as the square of the distance to the fixed rotational axis. The moment of inertia of a uniform thin rod of length L and mass M about an axis passing through a point at a distance of L/3 from one of | {
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"lm_q1_score": 0.9835969674838371,
"lm_q1q2_score": 0.8131512157307522,
"lm_q2_score": 0.8267117962054049,
"openwebmath_perplexity": 186.42280334111211,
"openwebmath_score": 0.5442878603935242,
"tags": null,
"url": "http://billardschule-cronenberg.de/the-moment-of-inertia-of-a-uniform-rod-about-an-axis-through-its-center-is.html"
} |
sql, sql-server, t-sql
OPEN CURSOR_NOTE_COUNTER
FETCH NEXT FROM CURSOR_NOTE_COUNTER
INTO
@lUPI_current,
@lCASE_SEQ_current,
@lDIFF_ENCOUNTER_DATE
WHILE (@@FETCH_STATUS = 0)
BEGIN
SET @lcounter = (
SELECT
MAX(NOTE_COUNTER)
FROM
#TMP_NOTES
WHERE
MEMBER_UPI = @lUPI_current
AND CASE_SEQ = @lCASE_SEQ_current
)
UPDATE #TMP_NOTES
SET
NOTE_COUNTER = @lcounter + 1
FROM #TMP_NOTES
WHERE
MEMBER_UPI = @lUPI_current
AND CASE_SEQ = @lCASE_SEQ_current
AND DIFF_ENCOUNTER_DATE = @lDIFF_ENCOUNTER_DATE
FETCH NEXT FROM CURSOR_NOTE_COUNTER
INTO
@lUPI_current,
@lCASE_SEQ_current,
@lDIFF_ENCOUNTER_DATE
END | {
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"tags": "sql, sql-server, t-sql",
"url": null
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c++, math-expression-eval
binary_register binary_register::operator--(int) {
binary_register a(*this);
operator--();
return a;
}
binary_register binary_register::bin_comp() const {
binary_register a = ~*this;
binary_register one(1, 1);
a += one;
return a;
}
void binary_register::div_rem(binary_register& other, binary_register& qot, binary_register& rem) const {
rem = *this;
divide_remainder(*this, other, qot, rem);
}
binary_register& binary_register::trim() {
while (!_reg.empty() && _reg.back() == 0) {
_reg.pop_back();
}
if (_reg.empty()) {
_reg.push_back(0);
}
return *this;
}
void binary_register::get_shift_index(std::size_t& index, std::size_t& reg_index, std::size_t& bit_index) const {
if (index) {
if (index >= BITS) { | {
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machine-learning, classification, clustering
There are a variety of models that might suit this task. You might use decision trees for interpretability. A simple neural net or SVM would likely also do a good job. These should all be found in most basic ML packages.
2) The win rates of various items are directly computable. Simply count the number of times a champion used the items in question and won and divide by the total number of times a champion used that item combination. You can do this for any given group size (1 to 6) | {
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gravity, gravitational-waves, ligo
Each numerically calculated waveform that is fitted to the data is a self-consistent model of a merging black hole binary. If the initial black hole masses and spins are specified, this leads to a definite final black hole mass after merger. Each of these model waveforms therefore predicts a "mass deficit" between the initial black holes and the final black hole. Therefore the quoted mass difference and its uncertainty in Table 1 of this paper of $1.0^{+0.1}_{-0.2}M_{\odot}$, is obtained from the distribution of possible models that fit the data (both the inspiral and the ringdown phases) and therefore the distribution of "mass deficits" that these models predict. It does not come from doing simple algebra on the quoted parameters for the initial and final black hole masses. | {
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"tags": "gravity, gravitational-waves, ligo",
"url": null
} |
newtonian-mechanics, fluid-dynamics, coriolis-effect
Title: Circular Winds and Coriolis force
In northern hemisphere the flow of air is shown which tries to move towards low pressure center but due to Coriolis is deflected as shown.
It's said that this causes a circular anticlockwise flow of air.
How's that?
The air will just try to move in and get deflected as shown and curve away, how then will a circular pattern be formed? Once the wind reaches a velocity such that the Coriolis and pressure gradient forces balance, it continues at that velocity due to inertia. This state is called geostrophic flow and corresponds to wind along isobars.
For an intense localized low pressure like a hurricane, the flow is not geostrophic—the pressure gradient force is larger in magnitude than the Coriolis force, maintaining the inward net acceleration required for circular motion. | {
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homework-and-exercises, fluid-dynamics, lagrangian-formalism
edit: just to clarify things: the limit considered here is in the size of the initial fluid element (not in time differences or sth like that), this equation is valid for all t
I never really liked Langrangian formalism (maybe you can estimate, how ugly it will become in higher dimensions.. you possibly will get there), so I looked up a reference to make sure, I had it right. Maybe these lecture notes: http://www.whoi.edu/science/PO/people/jprice/class/ELreps.pdf will be of some value to you | {
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php
Variable variables are also a bad idea.
Variable variables scream "What I really need is an array!"
You already have an array :)
Don't sanitize content until you're actually ready to put it into the context for which you're sanitizing it.
For example, & is a special character in HTML (in some contexts). DBs do not care about it. Nor do plain text emails. Or email subjects. Or... infinite things.
Rule of thumb: sanitize at the last possible moment.
You can also take the raw value and process it, but you can't always do the reverse.
Be careful with empty() on strings
empty($var) === (isset($var) && $var)
Implicit bool rules are very odd in PHP though. In particular, the string "0" can get you.
I doubt 0 should be a valid email content, but just throwing it out there in case it gets you in the future.
Invert your conditions and you'll see that you can bail early: | {
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python, python-3.x, object-oriented, pdf
pdf.ln()
pdf.write(self.file.line_height, "City #")
pdf.write(self.file.line_height, self.creator.city)
pdf.ln()
pdf.write(self.file.line_height,"Country #")
pdf.write(self.file.line_height,self.creator.country)
pdf.ln()
pdf.write(self.file.line_height, "Email #")
pdf.write(self.file.line_height, self.creator.email)
pdf.ln()
pdf.write(self.file.line_height, "Phone Number #")
pdf.write(self.file.line_height, self.creator.phone_num)
pdf.ln()
pdf.write(self.file.line_height,"Billed To #")
pdf.ln()
pdf.write(self.file.line_height,"Organization Name #")
pdf.write(self.file.line_height,self.organization.name)
pdf.ln()
pdf.write(self.file.line_height, "Organization Address #")
pdf.write(self.file.line_height, self.organization.address)
pdf.ln()
pdf.write(self.file.line_height, "Organization City #") | {
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"url": null
} |
Thus for all $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(f, P) - L(f, P) < \epsilon$.
Therefore $f(x)$ is integrable over $[a, b]$.
Is this proof correct ?
• "Let $r$ be such that..." that condition implies that $r\leq 0$. – YoTengoUnLCD Apr 25 '16 at 5:40
You have the correct approach. Just clean up a few details.
You are trying to show that given any $\epsilon > 0$, there exists a partition $P_\epsilon$ such that $U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon$.
First define $M$,
$$M := \sup_{x \in [a,b]} |f(x)|.$$
Then it follows that for $i = 1, 2, \ldots, n$ we have
$$M_i - m_i \leqslant 2M,$$
and the contribution to the difference in upper and lower sums from intervals $[p_i - r, p_i +r]$ is
$$\sum_{i=1}^n(M_i - m_i)[p_i + r - (p_i-r)] \leqslant 4Mnr.$$
Don't choose $r < \epsilon/4M$ for any $\epsilon$. For a given $\epsilon > 0,$ choose a particular $r = r(\epsilon)$ such that $r < \epsilon/ (8Mn)$ and $[p_i-r,p_i+r] \subset (a,b).$ | {
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same. hypohamiltonian '' and semi-Eulerian '' be written and start it with this Eulerian vs. Eulerian.. In addition, the semi-Lagrangian approach has two principal advantages at first we have see! By admin | Jul 3, 2018 | graph Theory | 0 comments is... Graph here has $6$ vertices, it is generally more common to use approach! Different physical and chemical parameters is of significant importance for desulfurization of flue... 4 ) of them and end at the other b ) the same as material! Then the graph is semi-Eulerian if and only if there are exactly two vertices same... But a graph is semi-Eulerian then semi eulerian vs eulerian vertices have odd degrees in which we draw the Path between every without. Argument for Eulerian graphs challenge is to determine the water 's velocity at fixed. E ) is a spanning subgraph of some Eulerian graphs of SO2 absorption efficiency under different and. First read the following examples: this graph is semi-Eulerian if it has an even degree vertex, | {
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"lm_q1q2_score": 0.8080748545428159,
"lm_q2_score": 0.855851154320682,
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"openwebmath_score": 0.4884713590145111,
"tags": null,
"url": "http://knhevelius.pl/ueazq3s9/semi-eulerian-vs-eulerian-f2acbd"
} |
5. ## Re: World Series!
Originally Posted by ebaines
Not quite right. What you've calculated is the number of ways to combine 3 wins and 2 losses, which is C(5,2) = 10. But the series does not necessarily last 5 games -for example the team could sweep the first three games. Your method also doesn't take into account the fact that the last game of the series must be a win.
If you write out all possible winning combinations you have:
3 game series (no losses): WWW
4 game series (one loss): LWWW, WLWW, WWLW
5 game series (2 losses): LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW | {
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"openwebmath_score": 0.37284570932388306,
"tags": null,
"url": "http://mathhelpforum.com/math-topics/222063-world-series.html"
} |
fft, ifft
newX = Xe + j*Xo;
xr = ifft(newX); % N/2 IFFT
newxo = real(xr) % recovered odd x samples
newxe = imag(xr) % recovered even x samples
Since each FFT in your formula for cross correlation is the FFT of a real signal, the product will also be the FFT of a real signal, hence the technique outlined above can be used. | {
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image-processing, hough-transform
Due to the way that skeletonization works, it will still produce some lines that will seem irrelevant to the grid but these lines towards "erroneous" directions are not that many (at least in the given image) to confuse the line detection of the Hough Transform too much and it will clearly pick the two main sets of lines at distinct directions.
(Here is how the HT output looks like: )
If you are using MATLAB, you might want to check this help page | {
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reproduction
on sperm and I see a lot of information on what makes up a sperm cell, but can't quite figure out its lifecycle and exactly how it dies. So most cells require growth factors and survival factors to stay alive in an organism. Cells which stop receiving survival factors, such as sperm, undergo a process of programmed cell death (PCD) called apoptosis. Some cells will also undergo necrosis, or autophagy, among other types of PCD. Lachaud et al. performed a study in 2004 on the contribution of apoptosis and necrosis in ejaculated sperm. They make note in the intro that from other literature that we see some hallmarks of apoptosis like externalized phosphatidylserine, fragmented DNA, etc. despite inactivated nuclei and a small cytoplasm. Their data notes that ejaculate sperm cell death mainly occurs from necrosis, however, because sperm themselves lack the ability to initiate apoptosis. However, in the testes, there's data to note that apoptosis plays a role, and the presence of apoptotic | {
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docker, ros-kinetic
I would've at least expected to see some Found NV library: .. lines in there.
Comment by rickstaa on 2019-04-10:
Yea that is strange it is singularity 3.0.3 with the singularity -v shell --nv ros_kinetic_panda command in the /bin/bash/ command line.
Comment by rickstaa on 2019-04-10:
I solved my issue with the tips of @gvdhoorn. The problem was caused by my Nvidia-driver after having purged the driver according to this guide link text and reinstalling a newer driver my issue disappeared.
Comment by gvdhoorn on 2019-04-11:
This should work regardless of how you've installed the driver (and additionally: installing NVIDIA drivers using the pkgs from nvidia.com is not necessarily the best way to go about things on Ubuntu, which is also what the guide you link to states I believe).
If it works for you though, that's ok. | {
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lambda-calculus, functional-programming, linear-logic
Does that mean that, for all terms of λEA, there is a valid type derivation?
Is this a common phenomena, i.e., are there other systems with this same characteristic, or is it something particular of linear logic based proof languages? For question 1, the answer is no, and is no for almost any type discipline (except certain intersection types): the fact that a term is (strongly or weakly) normalizable does not imply in general that it is typable. Typability implies termination, not the other way around.
The specific case of of $\lambda EA$, however, brings up another issue which may be the object of confusion, so let me discuss it. Let us start with a counter-example for your question 1: the term
$$[M]_{\lambda x.N=y,z}$$ | {
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homework-and-exercises, statistical-mechanics, temperature, ideal-gas
What happens to the average value of height for a single particle when we approach the absolute zero?
Starting from:
$$
\langle z \rangle =\frac{\int_0^\infty z e^{-\beta \Phi(z)}dz}{\int_0^\infty e^{-\beta \Phi(z)} dz}
=
\frac{\int_0^\infty z\frac{1}{(1+\frac{z}{z_0})^{\beta \Phi_0}}dz}{\int_0^\infty \frac{1}{(1+\frac{z}{z_0})^{\beta \Phi_0}} dz}\;,
$$
where $\beta = \frac{1}{kT}$.
Assuming that $\frac{\Phi_0}{kT}$ is very large compared to 1, we have:
$$
\langle z \rangle\sim z_0\frac{kT}{\Phi_0}\to 0
$$
Will the particles be at $z \approx 0$, as it's the state that minimizes the potential, having then $\langle z \rangle \approx 0$?
Yes, assuming that we are limited to $z\ge 0$, we have $\langle z \rangle \to 0$. | {
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observational-astronomy, the-sun, earth
A much bigger issue is the 34 arc minutes used for calculating atmospheric refraction. The 34 arc minutes used for atmospheric refraction is based on the average environmental lapse rate (how much temperature drops, on average, with increasing altitude) of 6.5 kelvins/kilometer. On a very nice day the lapse rate can be much closer to the dry adiabatic lapse rate of 9.8 kelvins/kilometer as opposed to average environment lapse rate. On a rather nasty day with a temperature inversion (lots of smog, fog, low clouds, etc.), the lapse rate can be close to zero, or even negative. The variation in lapse rate can make calculated sunrise and sunset times off by five minutes.
That said, there is no way to calculate sunrise/sunset times for anywhere on the Earth and for years into the past / into the future without using a canned value for atmospheric refraction (and for the Sun's apparent radius). So that's what those websites do. | {
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• I tend to view negation as a special case of subtraction since that's the "lowest" level I can place it on the "hierarchy". Viewing negation as multiplication by $-1$ seems like making the operation more complex than it really is. But that's just me. :) – Bill Cook Sep 8 '15 at 15:09
• Please see my answer. In it, I've elaborated on my comment a bit – John Joy Sep 8 '15 at 15:34 | {
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c++, error-handling
constexpr result(
failure&& failure
) : m_value(std::unexpected(failure.get_error())) {}
constexpr result(
success&& success
) : m_value() {}
bool has_error() const {
return !m_value.has_value();
}
bool has_value() const {
return m_value.has_value();
}
const error& get_error() const {
return m_value.error();
}
private:
std::expected<void, error> m_value;
};
}
#define CONCATENATE(x, y) _CONCATENATE(x, y)
#define _CONCATENATE(x, y) x ## y | {
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} |
homework-and-exercises, reference-frames, rotational-kinematics
Title: Conceptual Understanding of Reference Frames correct? The problem is as follows:
A cart has a velocity to the right. There is a wheel on the cart (fastened to it). Determine the angular speed of the wheel so that the velocity at a point on the top of the rim is equal to zero. | {
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fluid-dynamics, rotational-dynamics, rigid-body-dynamics
For most fluids at most familiar physical scales, rigid-body rotation is something you have to work at. If you take a pail of water and put it on a record turntable at 33 1/3 rpm, you'd have to wait a fairly long time for rigid-body motion to set in (depending on how close to "rigid" you want). Even then, when it does settle down, it's not primarily from viscous forces, but from Ekman pumping (large-scale circulations caused by differential rotation); if you could somehow turn that off, it would take an even longer time.... you could calculate it given the equation I give below if you want.
But if you crank up the viscosity, or reduce the size, then rigid-body rotation can set in much faster.
If you put the fluid in a nice rotationally-symmetric container and rotate the container around its axis of symmetry, the e-folding time-scale for rigid body rotation to set in due purely to viscous forces is of order
$$
\tau \simeq \frac{D^2\rho}{\mu}
$$ | {
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-
@Jodles: I approved your last four suggested edits, but please do take a break from LaTeXing now. Otherwise the front page is flooded by edits from you. As laudable as your efforts are, there are people not exactly in favor of editing very old posts (LaTeX was not available from the beginning). See also this discussion on meta – t.b. Aug 30 '11 at 9:12
@Theo: Sorry, I did not consider that! I will read that discussion. Thank you for letting me know! It was definitely not my intention to flood the front page. (I was going through the list of "low-quality posts" in the review section). – Jodles Aug 30 '11 at 9:15
@Theo: In Jodles' defence, I'm more used to SO's backtick notation and not very good at LaTeX. I'm happy with the edit! (Just as this was falling off the frontpage :) – yatima2975 Aug 30 '11 at 14:53 | {
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navigation, odometry, rviz
and IMU messages like:
header:
seq: 165
stamp:
secs: 1496543986
nsecs: 493674993
frame_id: base_link
orientation:
x: 0.0
y: 0.0
z: 0.0
w: 1.0
orientation_covariance: [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
angular_velocity:
x: 0.0
y: 0.0
z: 0.0
angular_velocity_covariance: [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
linear_acceleration:
x: -0.5
y: -0.24
z: 9.77
linear_acceleration_covariance: [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] | {
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homework-and-exercises, electromagnetism, electricity, electric-fields
The electric force contributed by an infinitesimally small portion of the left rod is given by $$k\frac{dq}{(20-x)^2}$$ where $k$ is Coulomb's constant, $dq$ is infinitesimally small charge, and $(20-x)$ is the disance between the source particle and the test particle. If we solve for $dq$ by taking the first derivative of $q(x)$ and apply this formula to the entire left rod then we obtain $$\frac{-kq_0}{\ell}\int_{0}^{20}\frac{e^{x/\ell}}{(20-x)^2}dx$$ | {
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experimental-physics, angular-velocity
Also, I am told to find the Average "Measured Ratio of $\omega_f/\omega_i$, so would I just add up all of my data in my $\omega_f/\omega_i$ column and then divide by how ever many rows I have?
Data: You don't explain what you are doing in the experiment, but I would guess it is supposed to demonstrate conservation of angular momentum. If we calculate the angular momentum $L$ using:
$$ L = I\omega $$
then unless some external force is applied $L$ will be a constant.
When you start the experiment the disk has moment of inertia $I_\text{Disk}$ and is rotating at a speed $\omega_i$. So the initial angular momentum is:
$$ L_i = I_\text{Disk}\omega_i \tag{1} $$
When you put the hoop onto the the disk the combined angular momentum of the disk and hoop is:
$$ I_{\text{total}} = I_\text{Disk} + I_\text{Hoop} $$
where $I_\text{Hoop}$ is the moment of inertia of the hoop. If the final angular velocity is $\omega_f$ then the final angular momentum is: | {
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c++, beginner, template, classes, polymorphism
/// + operator
template<typename LabeledObject, typename Identifier>
Gallery<LabeledObject, Identifier>& Gallery<LabeledObject, Identifier>::operator+(const Gallery& galleryB){
appendToGallery(galleryB);
return *this;
}
/// - operator
template<typename LabeledObject, typename Identifier>
Gallery<LabeledObject, Identifier>& Gallery<LabeledObject, Identifier>::operator-(const Gallery& galleryB){
for(auto it = galleryB.objectsMap.begin(); it != galleryB.objectsMap.end(); ++it)
removeObject(it -> first);
return *this;
} | {
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$\ds{\pars{a}_{m} = {\Gamma\pars{a + m} \over \Gamma\pars{a}}}$ is a Pochhammer Symbol. | {
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"url": "https://math.stackexchange.com/questions/1913654/a-question-related-to-compute-infinite-summation/1913903"
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quantum-mechanics, quantum-optics, second-quantization
Title: Notation in second quantization I got a bit confused about the transition of notation between the first and second quantization. When a state is written as: $\rho =a\vert H \rangle \langle H \vert + b\vert V \rangle \langle V \vert+ c\vert H \rangle \langle V \vert + d\vert V \rangle \langle H \vert$, does it imply a single particle in this state, i.e. in the second quantized form it is: $\rho =a\vert 1_H 0_V \rangle \langle 1_H 0_V \vert + b\vert 0_H 1_V\rangle \langle 0_H 1_V \vert+ c\vert 1_H 0_V\rangle \langle 0_H 1_V \vert + d\vert 0_H 1_V \rangle \langle 1_H 0_V \vert$?
And is it equivalent to write
$\rho =\left(a\vert H \rangle \langle H \vert + b\vert V \rangle \langle V \vert+ c\vert H \rangle \langle V \vert + d\vert V \rangle \langle H \vert \right)\otimes|\alpha \rangle \langle \alpha\vert$
and | {
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dynamixel
get a controller & amps that are compatible with your motors
find / write node with a similar enough ROS API as the dynamixel drivers
update the launch files
Comment by zakizadeh on 2017-10-26:
can i change this code and upload on arduino ??
Comment by gvdhoorn on 2017-10-26:
I have no idea. I don't use dynamixels, nor any of the arbotix packages.
Here is an explanation of how you could have an arduino control your motors in a general way:
Components required:
An arduino with code that can control your motor
A computer running ROS and a node that publishes data to a topic, such as an integer that corresponds to a motor value that you would like to use as a throttle value for your motor
If you have those two things you can modify the arduino code by adding ROS functionality to it. Do the first of these 3 tutorials: http://wiki.ros.org/rosserial_arduino/Tutorials/Arduino%20IDE%20Setup
Arduino IDE Setup
Hello World (example publisher)
Blink (example subscriber) | {
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c#, image
Here are two of the images I'm working with.
The first one has the colour violet in it quite a lot with a percentage of 0.64% or 0.00636199367565126The second one from what I can tell doesn't have any violet in it, however the application believes it does with 0.36% violet pixels or 0.00355871886120996.The code above is working so to speak, however is there something I could be doing to improve it for reliability and performance? I did see a post about Bitmap locking however because its only going to be scanning a image every hour or so for any violet pixels, it didn't implement it, should i still do it?
this.button2.Click += async (sender, e) =>
{
Task task = Task.Run(() =>
{
ProcessImage(pictureBox1.Image);
});
Task.WaitAll(task);
await task;
};
} | {
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Some comments on the $$\Gamma$$ indexing:
1. Baseline pieces have ascending consecutive denominators and $$1$$ in the numerator: $$\Gamma(1, \frac13), \Gamma(1, \frac13, \frac14), \Gamma(1, \frac13, \frac14, \frac15),\ldots$$
2. An argument of $$\Gamma$$ being $$1$$ means the bloodline becomes single line from there. The arguments in fractions like $$\frac55, \frac44$$ and $$\frac22$$ equal $$1$$ not by accident.
3. The all-negative might as well all be denoted as just $$\Gamma(-1)$$
4. There are $$n - 2$$ indices in $$\Gamma(j,k,\ell,\ldots)$$. That is, the number of arguments in $$\Gamma$$ indicates the number of points.
According to the rules for generating the codes (algorithm), it's easy to see the recurrence algorithm for the length (number of pieces) in each "bloodline": | {
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turing-machines, linear-bounded-automata
Title: How does augmenting linear bounded automata tape alphabets increase memory? A linear bounded automaton is a Turing machine that is restricted by memory.
How would augmenting the tape alphabet of a linear bounded automaton
increase its memory? While the memory that the automata can
use is as size of the input? If you consider an input, hence a tape, of size $n$ and a tape alphabet of
$b$ symbols, you get $b^n$ configurations possible on your tape. If
you measure memory in bits, the memory size is the base 2 logarithm of
the number of configurations, very simply because that is the number
of bits required to have the same number of configurations (rounding up
to the next integer, because half-bits have not been invented yet).
So the tape memory of your linear bounded automaton (LBA) is $\log_2
b^n$, which is equal to $n\log_2 b$.
Augmenting your tape alphabet does not increase your memory very | {
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"tags": "turing-machines, linear-bounded-automata",
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c++, functional-programming, monads
template<typename T> /**** Type Name ****/
Text _type_(const T& self);
template<typename T>
Text _type_(const T& self) {
return typeid(self).name();
}
template<typename T> /**** Category Name ****/
Text _cat_(const T& self);
template<typename T>
Text _cat_(const T& self) {
return "uncategorized";
}
template<typename T> /**** Boolean Conversion ****/
bool _is_(const T& self);
template<typename T>
bool _is_(const T& self) {
return false;
}
template<typename T> /**** String Conversion ****/
void _str_(Text_Stream& out, const T& self);
template<typename T>
void _str_(Text_Stream& out, const T& self) {
// TODO: handle object based on stream availability.
//
constexpr bool n = is_streamable<std::stringstream, T>::value; | {
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gazebo-model
Title: Set model velocity via msgs in Gazebo 6
I need to move a model smoothly from keyboard. For this I need to set (maybe, learn) velocity of a model via messages.
In Messages API I found only static pose arguments. In a few places on the internet I saw some comments about "geometrymsgs/Twist" topic, but they look outdated as in Gazebo 6 there isn't anything similar.
So how Can I do this without any plugins, using just messages?
Thanks for the feedback!
Originally posted by alexorex on Gazebo Answers with karma: 3 on 2015-12-09
Post score: 0
You can apply a wrench to a link using a topic with the following format:
~/WORLD_NAME/MODEL_NAME/LINK_NAME/wrench
Replace WORLD_NAME with the name of the world (usually default)
Replace MODEL_NAME with the name of the model.
Replace LINK_NAME with the name of the link on which to apply the force.
Originally posted by nkoenig with karma: 7676 on 2015-12-09
This answer was ACCEPTED on the original site
Post score: 0 | {
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discussed in this post in a companion blog. ‘The Democrats will probably win the next election. For example, a weather forecast that predicts a 75 percent chance of snow is using probability to communicate that people should be prepared to deal with snowy conditions. Our Probability solvers include the most commonly used probability distributions, such as the normal, exponential, Poisson, binomial, and uniform distributions. MIDTERM 1 solutions and Histogram. More Problems on probability and statistics are presented. ) whereas ex-post solutions try to gather auxiliary information about non-respondents which is then used to calculate a probability of response for different population sub. Brilliant Premium. (a) Show that if X and Y are conditionally independent given Z and X and Z are conditionally independent given Y then X is independent of (Y,Z). 01 of probability is being shifted from $2400 to$0. It can be a finite set, a countable set (a set whose elements can be put in a sequence, | {
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java, swing, gui
Your TlcException is.... unnecessary. You may as well just declare that your method throws a plain Exception. If it was me, I would at least make the TlcException a RuntimeException so there are not checked exceptions coming out.
Further, the constructor does not take a message, and is all on one line... ouch:
private TlcSizerException(Throwable cause) { super(cause); } | {
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homework-and-exercises, geometric-optics
Title: Find source of light reflecting from surface of sphere Suppose we have a reflective sphere of known radius, r, in the center of a hemispherical-dome of LED lights of an infinite radius. If we are looking down upon the sphere from the center of the dome then it will appear to be a circle, with points of light reflecting from the LEDs.
Top image: Side view of dome and reflective sphere; Bottom image: Top down view of sphere as it appears to the observer in the center of the dome.
How can we determine the position of the LEDs from the light reflected from the sphere i.e. from an image of it as it appears to be a circle?
It is easy enough to construct a vector from the center of the reflective sphere to the point of light on its surface (green vector), but this won't be a vector in the direction of the LED will it?
Any help appreciated; I'll post this in maths if it's better off there? | {
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python, python-3.x
if turns == 5:
print("This game is a draw.")
print(colors['red'])
show_board(game_board)
print(colors['reset'])
print('The current match score is %d : %d (Player1 : Player2)' % (player_one["wins"], player_two["wins"]))
play_again()
if __name__ == "__main__":
main() Overall clean and easy to read. Here are nits and optimizations that I have:
In your load_game() function, there is no need to do del board[:] if you're recreating the board. You can simply call build_game_board() and rewrite it as such:
def build_game_board():
board = [['0' for i in range(5)] for j in range(5)]
You can rewrite show_board() like so (a minor nit, takes advantage of list comprehension):
def show_board(board):
print([" ".join(row) for row in board].join("\n"))
If you're always going to be playing on a 5x5 board, consider storing that as a variable.
SIZE = 5
....
ship_col = randint(1, SIZE) | {
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game, console, timer, assembly, benchmarking
push si ;(9)
mov si, MsgStop
call .TypeString
pop si ;(9)
cmp si, Msg6
jb .Test
; Final message before exiting
.EndOfTests: mov si, Msg6
call .TypeString
GetKeyboardKey ; -> AX
.ExitToDOS: mov ax, 4C00h
int 21h ;DOS 'Terminate'
; - - - - - - - - - - - - - - - - - - - - - - -
.GetKey: GetKeyboardKey ; -> AX
cmp al, 27
je .ExitToDOS
ret
; - - - - - - - - - - - - - - - - - - - - - - -
.TypeString__: mov si, CRLF
.TypeString_: TypeCharacter
.TypeString: lodsb
cmp al, 0
jne .TypeString_
ret | {
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"tags": "game, console, timer, assembly, benchmarking",
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randomized-algorithms, algebra, algebraic-complexity, polynomials
For the problem of telling whether the variety $\{\vec{x} : p_1(x)=\dotsb=p_k(x)=0\}$ is empty or not (for polynomials given by a list of monomials and coefficients - which is "sparse but not succinct", that is, not SLPs), randomness also seems to help: assuming GRH, this problem is in $\mathsf{RP}^{\mathsf{NP}}$ by a result of Koiran. It is $\mathsf{NP}$-hard, and not known to be in $\mathsf{P}^{\mathsf{NP}}$. | {
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quantum-mechanics, electrons, atomic-physics, probability, orbitals
When you apply electric or magnetic fields to the atom the system is no longer spherically symmetric (since the fields come from some direction) and then different dumbbell shapes react differently to fields coming from different directions and having different polarization (this should be intuitively obvious). Hence my conclusion would be: The non-spherical shapes are what you get when you find simultanous eigenvectors of above operators, the choice of which is not spherically symmetric. In a problem, which is no longer spherically symmetric the shapes gain physical significance because you can actually excite electrons to those precise orbitals with light coming from appropriate directions and having appropriate polarization. | {
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special-relativity, classical-mechanics, scattering
If you are studying Compton scattering, then the electrons in experiments are bound to atoms, making them effectively motionless in the lab frame compared to the incoming photon. This makes it valid to study the problem in two dimensions. | {
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gravity
My big question, now, is: "How do we know it's instant?"
It's not instant, but it can appear that way.
We can't possibly move an object large enough to have a noticeable gravitational influence fast enough to measure if it creates (or doesn't create) a doppler-like phenomenon. | {
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php, array, sorting
Title: Process POSTed data and add default string if missing My code takes the input submitted by $ _POST, via a checkbox array. I process the array and check the sent data if there is a particular string in it, which is set as my default. If I have a string, which is my default, I just go through implode and add ",", if my default string is missing in the data. I add and sort the array anew, with the first string being the default. I want to shorten the code because I can't make it shorter.
<?php
if(isset($_POST['submit'])) {
//for update
$site_id['locale'] = 'tr';
//new
$exampleArray = isset($_POST['lala']) ? $_POST['lala'] : '';
$example = null;
if(($key = array_search($site_id['locale'], $exampleArray, true)) !== false) {
unset($exampleArray[$key]);
$FirstString = array($site_id['locale']);
$exampleArray = array_diff($exampleArray, $FirstString);
usort($exampleArray); | {
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python, performance, numpy, finance
Each time findMinimal is called, it looks at all the stocks and filters on i[1] == industry. If you maintained a list of stocks by industry then you would could save a lot of work.
pairsOnlyStocks seems to be a list (it has an append method). Lists do not have an efficient membership test: Python has to compare the item against each list member in turn. So (stockPair,stock) not in pairsOnlyStocks will be slow. It would be better to use a set.
minimalTscore starts at zero, which requires the code to have a special case if minimalTscore == 0:. Better to use Python's built-in min function and avoid the special case.
Instead of indexing:
stock = tup[0]
industry = tup[1] | {
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} |
java, beginner, hibernate
@Override
public void removeUserById(long id) {
dbAction.execute(session-> {
int rowsAffected = session.createQuery("DELETE FROM User u WHERE u.id = :id;")
.setParameter("id", id)
.executeUpdate();
if (rowsAffected < 1) {
System.out.printf("No user with id %d exists in the database, " +
"its removal is not possible", id);
}
return null; // return value is ignored
});
} | {
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"tags": "java, beginner, hibernate",
"url": null
} |
python, performance, parsing, xml, mongodb
def chunks(iterable, n):
it = iter(iterable)
while (chunk := tuple(islice(it, n))): # Python 3.8+
yield chunk
def work(chunk):
try:
posts.insert_many([attrib_to_dict(elem.attrib) for _, elem in chunk],
ordered=False)
except pymongo.errors.BulkWriteError:
# skip element
pass
if __name__ == "__main__":
pool = mp.Pool(4)
# progress bar
pbar = tqdm(total=POSTS_SIZE)
n = 100
try:
for chunk in chunks(et.iterparse("Posts.xml", tag="row"), n):
pool.apply_async(work, args=(chunk,),
callback=lambda: pbar.update(len(chunk)))
pool.close()
except KeyboardInterrupt:
pool.terminate()
finally:
pbar.close()
pool.join()
Here I used this to ignore duplicate keys. | {
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"tags": "python, performance, parsing, xml, mongodb",
"url": null
} |
algorithms, asymptotics
The worst-case complexity of linear programming is polynomial but a bit messy, but in practice it is often very efficient on real-world instances. | {
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"tags": "algorithms, asymptotics",
"url": null
} |
newtonian-mechanics, classical-mechanics, experimental-physics, friction
Title: How is coefficient of friction calculated? How is coefficient of static friction and coefficient of kinetic friction is calculated in real life without knowing the frictional force? I can not give you an example for all surfaces, but for an inclined plane, the formula is pretty straightforward. Suppose you wanted to find the coefficient of friction between a block and an incline. Keep the block on the incline, and if the block is stationary, try to increase the angle of incline until the block just begins to slide. At this point, the force on the object is just equal to the maximum value of static friction, i.e. $\mu\ mg\ \text{cos}\ \theta$ where $\theta$ is the angle of incline. And what is the force on the object? Well, it's the other component of weight i.e. $mg\ \text{sin}\ \theta$. | {
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"tags": "newtonian-mechanics, classical-mechanics, experimental-physics, friction",
"url": null
} |
telescope, photography, hubble-telescope
EDIT2: The performance of the "tiny chunks" reflector is not the same as the performance of the full size mirror, as mentioned in the comments. | {
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"tags": "telescope, photography, hubble-telescope",
"url": null
} |
homework-and-exercises, geometry
$$=\frac{1}{A} {\left ( \int_{0}^{\frac{a}{2}} \int_{0}^{x\sqrt{3}} x dydx + \int_{\frac{a}{2}}^{a} \int_{x\sqrt{3}}^{0} x dydx\right )}.$$
From that point (which I suppose/hope not be wrong, except for the integration intervals) I've tried a plenty of changes and I got anything but $R_x = \frac{a}{2}$. Could someone please explain what is wrong and how to fix it? Check the bounds in the second integral, and plug in values at the extremes. When $x=a$, the far corner of the triangle, you should be integrating $y$ from $0$ to $0$, but the integral runs from $0$ to $a \sqrt{3}$. This is because you used the equation $y = \sqrt 3 x$ for the upper bound of the second half of the triangle, when it only applies to the first half. The second half of the triangle is bounded by a line with negative slope and an $x$ intercept at $a$, namely $y = -\sqrt 3 (x - a)$. Using that as your bound, your second integral would be
$$
\int_{\frac a2}^{a}\int_0^{-\sqrt 3 (x - a)} x \, dy\,dx
$$ | {
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"tags": "homework-and-exercises, geometry",
"url": null
} |
c#, dependency-injection, rubberduck, ninject
public RubberduckConventions(IKernel kernel)
{
_kernel = kernel;
}
/// <summary>
/// Configures the IoC <see cref="IKernel"/>.
/// </summary>
/// <param name="vbe">The <see cref="VBE"/> instance provided by the host application.</param>
/// <param name="addin">The <see cref="AddIn"/> instance provided by the host application.</param>
public void Apply(VBE vbe, AddIn addin)
{
_kernel.Bind<App>().ToSelf();
// bind VBE and AddIn dependencies to host-provided instances.
_kernel.Bind<VBE>().ToConstant(vbe);
_kernel.Bind<AddIn>().ToConstant(addin);
BindCodeInspectionTypes();
var assemblies = new[]
{
Assembly.GetExecutingAssembly(),
Assembly.GetAssembly(typeof(IHostApplication)),
Assembly.GetAssembly(typeof(IRubberduckParser))
}; | {
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"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "c#, dependency-injection, rubberduck, ninject",
"url": null
} |
cc.complexity-theory, complexity-classes, polynomial-hierarchy
doesn't make sense to me that there are problems inside EXSPACE but outside EXPTIME, i.e. problems that require exponential space without requiring exponential time. What are the problems in EXPSPACE outside EXPTIME? Your argument proves that $\mathsf{NEXPTIME}\subseteq\mathsf{EXPSPACE}$, since if a TM terminates in (nondeterministic) exponential time it cannot write to more than an exponential number of tape cells. | {
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"tags": "cc.complexity-theory, complexity-classes, polynomial-hierarchy",
"url": null
} |
c#, beginner, console
do
{
Console.WriteLine("Try again? - Y/N");
restart = Console.ReadLine();
restart = restart.ToUpper();
if (restart == "Y")
{
valueIsGood3 = true;
}
else if (restart == "N")
{
Console.WriteLine("Thank you for using the calculator!");
Console.WriteLine("Exiting in..."); | {
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"tags": "c#, beginner, console",
"url": null
} |
The Mean Value Theorem states that the rate of change at some point in a domain is equal to the average rate of change of that domain. This theorem is called the Extreme Value Theorem. The Extreme Value Theorem says: If a function f is continuous on the closed interval a ≤ x ≤ b, then f has a global minimum and a global maximum on that interval. In this section we learn the definition of continuity and we study the types of When you do the problems, be sure to be aware of the difference between the two types of extrema! this critical number is in the interval (0,3). Notice how the minimum value came at "the bottom of a hill," and the maximum value came at an endpoint. Preview this quiz on Quizizz. Plugging these special values into the original function f(x) yields: The absolute maximum is \answer {17} and it occurs at x = \answer {-2}.The absolute minimum is \answer {-15} and it occurs at x = \answer {2}. Proof of the Extreme Value Theorem Theorem: If f is a continuous function defined | {
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"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717428891155,
"lm_q1q2_score": 0.8263721891039725,
"lm_q2_score": 0.8376199653600372,
"openwebmath_perplexity": 432.85032295507244,
"openwebmath_score": 0.7868518829345703,
"tags": null,
"url": "https://academy.rainoil.com.ng/fc-meaning-pvw/extreme-value-theorem-open-interval-9d5331"
} |
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