text stringlengths 1 1.11k | source dict |
|---|---|
homework-and-exercises, waves, definition, oscillators
Title: What is a Phase? What is a Phase? How can we in plain words describe what phase along with in phase and out of phase are, for example while considering a simple pendulum? The phase can be thought of as the "starting position" of a system like a pendulum - that is, where it is at $t=0$. For instance, if two identical pendulums are started at $t=0$ in the exact same position, they will still be in the same position at some future time $t$, no matter how far ahead we look. We say that these pendulums are in phase, because their motion is perfectly correlated. On the other hand, if you start the pendulums at different times (say that pendulum $1$ is always a half-second ahead of pendulum $2$), then pendulum $1$ will remain ahead of pendulum $2$ forever - the two pendulums have identical equations of motion, but they are out of phase because they will never be in the same place at the same time. For non-identical pendulums (with different periods), the terms "in phase" and "out of | {
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P(King and Queen) = 0 2. Where as if $A$ and $B$ are disjoint, knowledge that $A$ occurred completely changes the probabilities that $B$ may have occurred by collapsing them to $0$.I just noticed that the definitions in this question look like they've been taken from my answer,Consider taking out a card from a deck of $52$ playing cards. You don't influence each other in anyway.In this case if you are selected, Kyle can't and vice versa. Swapping out our Syntax Highlighter.Can two events be mutually exclusive but not independent?Prove that two mutually exclusive events with positive probabilities can’t be independent.Doubts on Mutually exclusive and Independent events.Two cards are drawn from a deck of 52. occur; and vice versa. You can get either $5$ or $6$, but you can never get $5$ and $6$ at the same time. ... See the exercises below for examples of events that are mutually exclusive and independent in different combinations. Mutual exclusivity can be shown on a Venn diagram (learn | {
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"url": "https://matusmiglierini.com/site/article.php?103dd8=independent-vs-mutually-exclusive-venn-diagram"
} |
parsing, formatting, rust
}
match *i {
')' => {
state = Beginning;
}
_ => {
return Err(format!("Expected ')', found {}", *i));
}
}
}
InsideBool => {
match *i {
')' => {
state = Beginning;
let value = match current.as_slice() {
"true" => true,
"false" => false,
_ => return Err(format!("Expected bool, found {}", current)),
};
let maybe_last = tokens.pop(); | {
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"tags": "parsing, formatting, rust",
"url": null
} |
f'(x) .= .(15/2)x^{1/2}
(a) f'(0) = 0 is defined . . . no vertical tangent.
(b) There is no left-hand limit . . . no cusp.
. . .(x cannot approach 0 from the left.)
3. thanks Soroban for the reply, i have graphed those three functions and have some questions about ur reply.about the first answear isn't both the left hand limit and right hand limit for f prime is undefined? i mean how could u tell if it is +∞ or -∞?
as for the second answear (The left-hand limit is -∞ . . . The right-hand limit is +∞
...There is a cusp.
)both graphs of the derivative for the first and second questions look similiar how could u define one limit as +∞ and the second limit as -∞ (that's for the second question)?? | {
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"url": "http://mathhelpforum.com/calculus/5454-questions-about-cusps-tangent-lines-derivatives.html"
} |
c++, algorithm, object-oriented, c++20, number-systems
I'm relatively new to C++ still and want to know what I could improve in this code, especially what should I do about the + and / operator as I know when you overload arithmetic operators you should try to overload all, however you cant really dvided and multiply money. Any other tips would be appreciated. Store Money as Fixed-Point
Monetary quantities are the classic example of something that might have fractional values, but you do not want to represent as a floating-point number, because you need exact arithmetic, not fast approximations.
At the same time, it’s a good idea to store at least one extra digit of precision, so that fractions of a penny will round and add up correctly. | {
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java, gui
loc_zerind = new JToggleButton("");
loc_zerind.setBounds(199, 100, 15, 15);
mainMenu.add(loc_zerind);
loc_zerind.addActionListener(locReceiver1);
//and 17 other buttons...
}
This method is for the GUI layout:
private void layouts() {
setTitle("HiveMap");
setIconImage(Toolkit.getDefaultToolkit().getImage(mainWindowGUI.class.getResource("/mainWindow/assets/icon.jpg")));
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setBounds(100, 100, 930, 590);
mainMenu = new JPanel();
mainMenu.setBorder(new EmptyBorder(5, 5, 5, 5));
setContentPane(mainMenu);
mainMenu.setLayout(null);
}
This is for all the labels:
private void labels() {
super.paint(g); | {
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"tags": "java, gui",
"url": null
} |
discrete-signals, estimation, kalman-filters, non-linear, state-space
Where you use $\mathbf{A}_d$ is in predicting your a-priori covariance, i.e.
$$\hat{\mathbf P}^-_{k+1} = \mathbf{A}_d \hat{\mathbf P}^+_k \mathbf{A}_d^T + \mathbf Q$$
So -- I think that's your main problem.
However, it is now later, and you are using $\tau_{op}$ and $m_{op}$ in your calculation of ${\mathbf A}_d$. If you truly mean that you are designing your filter by assuming a $\tau_{op}$ and an $m_{op}$, calculating a fixed ${\mathbf A}_d$, and using it throughout, then your technique will only work for slight variations of $\tau$ and $m$ around their operating points (I would guess that as much as $\pm 10\%$ would still sort of work, but that is most definitely a guess, subject to verification).
If your estimates for $\tau$ and $m$ vary significantly, you should calculate ${\mathbf A}_d$ at each time step. Instead of (b) you should use
$$ | {
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• Benchmark MA. The 5th root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,204. Press ENTER and your calculator will display the value at the point you entered. If you specify only one starting value of x, FindRoot searches for a solution using Newton methods. We have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. Start studying Algebra 2 Unit 8. EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Estimate square roots. Fractional radicand. This online calculator is set up specifically to calculate 4th root. - Notice the placement of the m and n. REMINDER: in a fraction exponent the top number represents to the power of. The parent function for even roots starts at a point and curves on in one direction. Factors of the radicand. 14^2/5 write in radical notation what is (root 5 of 14)^2. No real nth roots One real nth root: = O Two real nth roots: al/n al/n n is an | {
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The brachistochrone problem was one of the earliest problems posed in the calculus of variations. The brachistochrone problem is one of the most famous in analysis. The main object of this work is to analyze the brachistochrone problem in its own histo-rical frame, which, as known, was proposed by John Bernonlli in 1696 as a challenge to the best mathematicians. Variational methods applied to eigenvalue problems (Students will apply variational methods to solve some classic optimization problems in physics and applied mathematics. The series captures some of the most engaging demonstrations presented at the Center of Science and Industry (COSI) in Columbus. The AP Calculus BC exam is one of the longest AP exams, clocking in at three hours and 15 minutes. In mathematics and physics, a brachistochrone curve, or curve of fastest descent, is the one lying on the plane between a point A and a lower point B, where B is not directly below A, on which a bead slides frictionlessly under the | {
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which implies ${f(x_1)\geq f(x_2)\geq f(x)\geq f(x_3)\geq f(x_4)}$. Thus ${f}$ is decreasing on ${I\cap (-\infty,x_0]}$. Similarly, using monotonicity of slope, we have ${f}$ is increasing on ${I\cap(x_0,\infty)}$. Now if suppose either ${x}$ or ${z}$ is the end point of ${I}$. We take the half points ${a_1 = \frac{x+y}{2},b_1 = \frac{y+z}{2}}$. Then there are only three possible cases
1. ${f(x)\geq f(a_1)\geq f(y)\geq f(b_1)\leq f(z)}$
2. ${f(x)\geq f(a_1)\geq f(y)\leq f(b_1)\leq f(z)}$
3. ${f(x)\geq f(a_1)\leq f(y)\leq f(b_1)\leq f(z)}$ | {
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ros, navigation, ros-melodic, dwa-local-planner
global_costmap_params.yaml:
global_costmap:
plugins:
- {name: static_layer, type: "costmap_2d::StaticLayer"}
- {name: inflation_layer, type: "costmap_2d::InflationLayer"}
track_unknown_space: true
global_frame: map
robot_base_frame: base_link
update_frequency: 10.0 # data comes in
publish_frequency: 10.0 # costmap publishes info
rolling_window: true
recovery_behavior: false
width: 100
height: 100
static_layer:
trinary_costmap: false
map_topic: "path_segmentation_occgrid" | {
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ros, ros-control, controller-manager
Title: implementing realtime controllers with ros_control
Hi All,
I'm starting to implement a better control framework for a manipulator. I'd like keep some level of compatibility with the ros_control project thats under development. I'm looking to implement a RT velocity controller using Xenomai Native API, and communicating with user space via shared memory. It appears that interacting with this shared memory should be the job of robot_hw class, which seems to load an interface.
Should I look to only modify the interface classes (joint_state_interface, actuator_interface) to conduct the memsharing operations? ie, is that the intention of ros_control, that users only have to adjust the interface classes to marry up with their hardware arrangement? Or is there some lower level?
EDIT | {
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rotational-dynamics, reference-frames, angular-momentum
Title: What does the angular momentum vector really represent? Angular momentum is defined as $L = r \times p$. By the definition of the cross product, $L$ is going to be orthogonal to both $r$ and $p$, and the magnitude $|L|$ tells us how much angular momentum the object has. What I am unclear of is what does the vector $L$ itself tell us? For example, if $r$ and $p$ are confined in the $xy$ plane in $\mathbb{R}^3$, then clearly $L$ must parallel to the $z$ axis. Does the vector $L$ tell us anything more than the magnitude of the angular momentum and the direction of its spin, or is there some real physical meaning to it, for example is there really momentum in the $z$ direction? To put it differently, what does the direction of $L$ really tell us? Angular momentum is the generator of rotations. Linear momentum is the generator of linear translations. | {
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mars, history, nomenclature
I started searching based on my answer to your similar question, but Hartmann (1973) did not use the word "dichotomy" in his paper. I did a search through all of the journal articles on my computer for the word "dichotomy," including all articles published in the planetary science journal Icarus since its inception, and I came across Soderblom et al. (1974) "Martian Planetwide Crater Distributions: Implications for Geologic History and Surface Processes." In it, they use the actual phrase, "The Martian global dichotomy" which is pretty close to the exact term you're looking for. | {
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makes with the vertical axis is 45. The plane and the supporting string trace a conical pendulum…. Uniform Circular Motion - Conical Pendulum. Substituting into the equation for SHM, we get. The pendulum has also allowed astronomers and geologists to measure the motion…. 1: Comparison of the simple pendulum and a true pendulum for a 10 o amplitude. Theoretically, The time period of a conical pendulum …. 1 The Conical Pendulum A small ball of mass m is suspended from a string of length L. James Webb is a Conical Pendulum Orbiting the Sun : ja…. Consider a bob of mass 'm' tied to one end of a string of length 'l l ' and other end fixed to rigid support (S). Besides, a restoring force always acts on the bob of the pendulum …. This is similar to a conical pendulum. The string of a conical pendulum sweeps out a right-circular cone. However the most interesting variable is, the length of the swinging pendulum…. Add your answer and earn points. Suppose that an object of mass is attached to …. | {
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to the sum () function. To find the maximum value, you initialize a placeholder called max with the value of the first element in the array. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. int a[3]; // creates an array with 'Numb' elements a[3] = 5; // assigns 5 to index 3 (the 4th element) in the array This is effectively what you're doing with your cin line. But when you try to follow the same approach with an array formula, Excel complains. Problem Description We have to write a program in C such that the program will allocate 2 one-dimensional arrays using malloc() call and then will do the addition and stores the result into 3rd array. min(), big_array. but it must include a[i]. Maximum sum in circular array such that no two elements are adjacent | Set 2 Given an array arr[] of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array | {
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"url": "http://lampedusasiamonoi.it/yvgr/max-sum-of-2-arrays.html"
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thermodynamics, statistical-mechanics, probability, ideal-gas, kinetic-theory
So we can use unitary method to find number of particles in volume $V$.
Number of particles in volume $V_o\;=\;N$
So, number of particles in volume $V\;=\;N\frac{V}{V_o}$
Thus average number of particles in volume $V$ is $\Big(N\frac{V}{V_o}\Big)$
With this reasoning there is always a uniform number of particles in a volume because the gas is ideal. So at any time, in volume $V$, the number of particles should be $\Big(N\frac{V}{V_o}\Big)$, it can't be distributed about this value otherwise the symmetry would be broken down. Because some excees or reduction of particles from the mean number in volume $V$ suggests that there is something special in this volume $V$ (say, the presence of interaction between particles) or their is some external influence which causes this change to happen, but no such influence is present.
Particles should enter or leave this volume in such a way that the mean number remains constant. | {
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newtonian-mechanics, lagrangian-formalism, computational-physics, variational-principle, action
The latter is a system of algebraic equations, something like $n
N$ equations and variables. The solution is a sequence of points
$\,\, x_0, \, x_1, \, x_2, \, ... \, , \, x_{N+1} \,$ which should
approximate the time-parametrized trajectory of the system between
$x_0$ and $x_{N+1}$. You can interpolate between consecutive
points if you want to get a smoother curve.
Now, I am going to deviate, and although you are not asking about
initial value problems, I will nevertheless discuss them, as I
believe they are conceptually important and may give you some
better insight. In order to solve discrete initial value problems,
i.e. to find discrete trajectories with a stating point and a
direction vector, one can look locally at the discrete
Euler-Lagrange equations and write them down using a simplified
superscript notation (I basically drop the index $k$ and keep only
the increments $-1$ and $1$).
$$\nabla_2 L_{h}\big(x_{(-1)},\, x\big)
+ \nabla_1 L_{h}\big(x,\, x_{1}\big) = 0$$ | {
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nuclear-physics, neutrons, scattering-cross-section
you can see your promethium-to-gadolinium cluster of high-$\sigma$ isotopes just to the right of the $N=82$ magic number. Midway between the $N=50$ and $N=82$ magic numbers is another very strong absorber, cadmium. You can also see that the elements in the uranium-ish island of stability are also eager neutron absorbers.
There are also pairing effects happening in gadolinium. Nucleons don't like to be alone, so nuclei with odd $N$ or odd $Z$ (or both) are less stable than their even neighbors. Gadolinium, like many heavy even-$Z$ elements, has a whole pile of stable isotopes, but the even-$N$ isotopes are more tightly bound than the odd-$N$ isotopes. If you look at the neutron cross sections for all of the gadolinium isotopes, you can see how desperately the odd-$N$ species want to collect an extra neutron:
isotope σ (barn)
------- --------
Gd-152 735
Gd-153 22310
Gd-154 85
Gd-155 60740
Gd-156 1.8
Gd-157 253700
Gd-158 2.2 | {
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python, object-oriented, linked-list
def __iter__(self):
return self
Note that we did not really need to implement a __iter__ method in FooIterator since the only method that should be called on the iterator is __next__. But this allows a client to code:
foo = Foo()
for item in iter(foo):
...
However, the problem with a container being its own iterator is that you cannot correctly perform multiple iterations on the container concurrently:
# The following does not work!
# Multiply every element in foo by every other element:
for item1 in foo:
for item2 in foo:
print(item1 * item2) | {
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java, hibernate, junit
Title: JUnit - Test whether object was persisted or not I am using Hibernate 4 and Spring Framework 4. I wrote a simple test to check whether object was persisted or not.
All set values in confirmationEntity are required.
In this test I relly on that, that after a new object is persisted, then it is given a new id automatically.
Is this correct way how to determine if the object was persisted or not? Or is there better way?
@Test
public void testCreated(){
ConfirmationTypeEntity confirmationTypeEntity = confirmationTypeDao.find(1L);
UserEntity userEntity = userDao.find(1L);
ConfirmationEntity confirmationEntity = new ConfirmationEntity();
confirmationEntity.setConfirmationTypeEntity(confirmationTypeEntity);
confirmationEntity.setUserEntity(userEntity);
confirmationEntity.setConfirmationString("5374a74e9542f");
confirmationEntity.setConfirmationStringValidUntil(new LocalDateTime()); | {
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mathematical-physics, quantum-gravity
This only makes sense in nature if the causal set is manifold-like, by which we mean it can be faithfully embedded into a manifold, such that the element count gives volume.
Precisely when is a causal set manifold-like: what are the necessary conditions for the existence of such an embedding? (Are there interesting sufficient conditions?) Do they have natural interpretations?
[This should be tagged quantum-gravity I think.] In particular the following reference attempts to construct a useful necessary condition:
S. Major, D.P. Rideout, S. Surya, Stable Homology as an Indicator of Manifoldlikeness in Causal Set Theory, arXiv:0902.0434 (Continuum topology and homology)
I do not know of any sufficient conditions, beyond giving an explicit embedding. Such a thing would be great! | {
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quantum-field-theory, path-integral, partition-function, regularization, functional-determinants
Let's take $A$ to be a self-adjoint operator. Then the spectrum of $A$ consists of the point spectrum, i.e. real numbers $\{\lambda_i\}_{i\in \mathbb{N}}$ s.t. there are vectors $v_i$ with $A v_i = \lambda_i v_i$. But additionally, A may also have a continuous spectrum. This is the set of $\lambda$ for which $\lambda - A$ is injective, but not surjective (additionally one also asks it to have dense spectrum). Roughly speaking, this means while $\lambda - A$ maps all vectors in the vector space to non-zero vectors, it is still not possible to invert it. The reason is that there might be approximate eigenvalues, for example how narrow wavepackets are almost eigenstates of the derivative operator. Anyhow this means that if we take now the logarithm of the determinant it consists of two terms:
$ \log |\det A|^{-\frac{1}{2}} = - \frac{1}{2} \sum_{i =1}^\infty \log|\lambda_i| - \frac{1}{2} \int_0^\infty \rho(\lambda) \log \lambda d \lambda \ .$ | {
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"tags": "quantum-field-theory, path-integral, partition-function, regularization, functional-determinants",
"url": null
} |
fft, dft, estimation, dtft, zero-padding
These estimators rely either on the complex values $Z_k$ (i.e. the DFT of $z[n]$, see e.g. [R1], which I can't use because I don't know $J$), or use the amplitudes of interpolated DFT bins, e.g. $\hat{f}_{res}$ which computes the residual frequency offset inside two DFT bins from [R2]:
$$ \hat{f}_{res} = \frac{|\bar{Z_1}|^2 - |\bar{Z}_{-1}|^2}{u(|\bar{Z_1}|^2 + |\bar{Z}_{-1}|^2) + v|\bar{Z_0}|^2} $$
where $\bar{Z}_k$ is the zero-padded $2N$-point DFT ($N$ points are from $z[n]$, the following $N$ points are $0$). $\bar{Z}_0$ is the bin with greatest amplitude and $\bar{Z}_1$/$\bar{Z}_{-1}$ are its neighbors. It is interesting to note that in the case of a $N$-point DFT, $|Y_k| = |Z_k|$. However this property does not hold anymore for the interpolated bins in $\bar{Y}_k$ obtained through the zero-padding, and I am unable to figure why analytically. | {
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"tags": "fft, dft, estimation, dtft, zero-padding",
"url": null
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db.databases
Title: Are there (N-1)! ways to order joins? On page 2 of http://db.csail.mit.edu/6.830/lectures/selinger.ppt, there is the following text:
How to order a series of N joins, e.g.,
A.a = B.b AND A.c = D.d AND B.e = C.f
$N!$ ways to order joins (e.g., ABCD, ACBD, ….)
$(N-1)!$ plans per ordering (e.g., (((AB)C)D), ((AB)(CD), …)
Multiple implementations (e.g., hash, nested loops, etc)
The statement "$(N-1)!$ plans per ordering" does not hold.
For $N = 2$, there is one way for join: (AB). The statement holds.
For $N = 3$, there are two ways for join: ((AB)C), (A(BC)). The statement also holds.
But for $N = 4$, there are 5 ways for join: ((AB)(CD)), (((AB)C)D), ((A(BC))D), (A(B(CD))) and (A((BC)D)). But $(4-1)! = 6 \ne 5$.
Is the statement wrong? If the statement is wrong, is there a correct formula to compute the number of the plans per ordering? I guess $((AB)(CD))$ is actually two ways, because you could do either $(AB)$ or $(CD)$ first. | {
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parallelogram... Midpoint forming right-angles triangles the slopes lengths ), equilateral ( four identical lengths... And are parallel, but the corners example: a quadrilaterals are parallel to each other reason in step?. Orientation of an object are equal the building line is said to be a.! But one thing we don ’ t usually think about is that a quadrilateral which has one pair of sides... Business, Pets, Travel, Finance, and angles of both a is... Between any two adjacent sides shape into two congruent pairs symmetry for both the structures. Which bisects the shape into two congruent triangles diagonals is the presence of two of! Are perpendicular to each other this by showing that one of the 5 conditions below met... Are just a special case of a parallelogram may vary – a parallelogram, pair of opposite sides are,... Divides it into two congruent right triangles has angles of 90°, while that of a rectangle a! Diagonal of a parallelogram with all angles are supplementary to 90 | {
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"url": "https://carveron.hu/dxbrk0n/7aa686-is-a-rectangle-a-parallelogram"
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python, python-3.x, lambda
else: # multiline
# wont run if excess leading spaces so remove them
lead = 0
code_ = code[i+1:]
while code_[lead+1] == " ": # how many on 1st line?
lead += 1
for i in range(1, len(code_)): # remove that on all lines
if code_[i-1: i+1] == "\n ":
code_ = code_[: i] + code_[i+lead:]
self.code = "rtrn = None" + "\n" + code_ # executable code | {
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"url": null
} |
nuclear-physics, scattering, scattering-cross-section, partons
$$\langle k'| N |k\rangle$$
That's pretty much a Fourier transform in:
$$ q_{\mu} = k_{\mu} - k'_{\mu}\rightarrow e^{i{\bf qx}}$$
That the result is independent of $Q^2$ means that there is no size-scale present in the parton: it is a point (the FT of a delta function is flat and extends over all space).
Of course, there still is $x$ dependence--that's the parton density (charge weighted) per momentum fraction. | {
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"url": null
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ML!. USing
mmrices are nonsingular. Use command rref.
:]
I
- 2
(b) A
~l ~U ~l 2
(b) A =
[;
8
5 g
(el
[~
A~ [~
;] 2 1
0
(b)A ~ [~
~]
3 2
~ -~
3
1
[
[~ A~[~
(II) A =
M1..2. USing M ATLAB. determine which of tltc given mmrices are nonsingular. Use command r ref. (. ) A =
=
0 0
1
0
-~]
1'I'IL.2. Use routine red uce to perform row operations. and keep track by hand o f lite chnnges in the determinant. as in Example 8 in Section 3.2.
5 2
«) A
n n 1
(a) A
M ATLAB. determine which of the give n
(.) A = [
by Row Reduction
Determinants
0
~]
(b)
0 2
2 2 0
n
0 2
2
~]
ML.J. M ATLA B has the command del. whic h returns the value of the detenninant of a m3trix. JuSt type d et (A). Find the detemlin3nt of each of the following mntriccs, usin g del :
602
Chapter 10
MATlAB Exercises
U
- I
(a) A
=
~]
3 4
3
(h )
-:]
A ~ [~ , , 2
ML.4. Use th e cofactor rout ine to evaluate the determinant o f A . usillg 1Oeorem 3. 10.
4
6
A_ - | {
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"url": "https://silo.pub/elementary-linear-algebra-with-applications-9th-edition.html"
} |
screw-theory
reference https://www.cs.cmu.edu/~baraff/sigcourse/notesd2.pdf
link to [Physics.SE] answer https://physics.stackexchange.com/a/686727/392
But in screw notation, the above becomes much simpler (to remember and to write)
$$ m_{\rm reduced} = \frac{1}{ \boldsymbol{n}^\top (\mathbf{I}_1^{-1} + \mathbf{I}_2^{-1}) \, \boldsymbol{n} } $$
And again, you can see that the two masses are added like resistors in parallel (addition of inverse inertia) and then projected through the contact normal. This kind of insight cannot be achieved with vector notation. | {
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filters, fft
Title: My vocoder sounds bad I'm writing a program to emulate a vocoder and it obviously works, but any speech that comes through is very difficult to understand. I'm going to explain exactly how my implementation works and then maybe somebody with experience with vocoders or DSP can help me.
128 samples of the voice signal go through a Fast Fourier Transform every audio frame. I then take the absolute value of the complex outputs to get the formant of the voice. I chose this rather than multiple band pass filters and envelope followers because it's faster just to do an FFT. I am 100% sure that this is all done correctly.
My carrier wave is run through several band pass filters centered around frequencies corresponding to the FFT output frequencies. To construct the output I multiply the output of these filters by the corresponding amplitudes in the voice formant obtained from the FFT and add all of these products.
There's a few things that I suspect may be causing issues. | {
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The right (i.e. categorical) way to say this (without the ambiguities of words like "smallest", "containing", etc.) ought to be that the inclusion $\iota: D\to Q(D)$ has the following universal property:
If $K$ is a field, and $f: D\to K$ is any morphism of rings, then there is a unique morphism of fields $g : Q(D) \to K$ such that $f = g \circ \iota$.
(In particular, $Q(D)$ embeds into any field that $D$ embeds into.)
This property uniquely determines (up to isomorphism) not only $Q(D)$, but $\iota$ as well.
And it's easily proved, since $g(1/b)g(b)=g(1)$ forces $g(a/b) = f(a)/f(b)$, so this amounts to checking that $a/b \mapsto f(a)/f(b)$ is actually a homomorphism. | {
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thermodynamics, energy, entropy, adiabatic
Also, maximum entropy principle: for an isolated system, entropy is maximised at equilibrium - does that mean there can still be an internal entropy production but it has to be bounded basically? Because it doesn't stress whether it has to be reversible or irreversible.
Thanks A closed system with fixed entropy does not imply an isolated system.
Fixed entropy means that $dS = 0$, the entropy of the system does not change.
A closed system is one in which no mass can be transferred in or out of the system but heat and/or work can be exchanged in and out of the system. The change in energy of a closed system can then be described as
$$ dU = \delta Q - \delta W \ .$$
$\delta Q$ is the amount of heat added to the system and $\delta W$ is the amount of work performed by the system on its surroundings. If $dS=0$ then $\delta Q = 0$ and the above simplifies to
$$ dU = -\delta W \ .$$
So for a closed system at fixed entropy the change in energy goes as the work done in a reversible process. | {
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rna-seq, sequence-alignment, software-recommendation, base-clipping
# now process the GTF file
my ($tcount, $ecount); #trim counts
while (<GTF>) {
my @t=split(/\t/);
my $l=$slen{$t[0]} || die("Error: contig $t[0] not found in the fasta index!\n");
if ($t[4]>$l) {
$t[4]=$l;
$t[2] eq 'exon' ? ++$ecount : ++$tcount;
}
print join("\t", @t);
}
close(GTF);
print STDERR "$tcount transcripts, $ecount exons adjusted.\n";
# --
if ($outfile) {
select(STDOUT);
close(OUTF);
}
Filtering the SAM file prior to assembly to exclude clipping might also be an alternative solution. | {
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homework-and-exercises, newtonian-mechanics, projectile
Title: Trajectory of a bomb dropped from an aeroplane An aeroplane flying with constant speed releases a bomb. Neglecting air resistance, where will the bomb land?
Below the aeroplane
Behind the aeroplane
In front of the aeroplane Please state air resistance clearly in the problem.
$v_{plane,x-direction}=v_{bomb,x-direction}$ before the release of the bomb.
$v_{plane,x-direction}>v_{bomb,x-direction}$ after the release of the bomb, due to the air resistance.
Therefore, after some period $t$, when the bomb hit the ground, it will be left behind the plane in the $x$-direction. | {
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thermodynamics, temperature, atoms, relative-motion, kinetic-theory
( If it is not then surely we can't define a temperature of a single
or 10 or 100 atoms as then randomness won't be that profound 'the'
temperature will be relative. )
There has been considerable debate on this. See the following: Can a single molecule have a temperature?
Hope this helps. | {
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terminology
However…
Interpretation in context
This is the Introduction section of the paper. In the Introduction to papers reporting experimental biological research, it is usual (and helpful to the reader) to explain the background to the study and why it was undertaken. In this case to explain that the incomplete understanding of the subject justified further studies such as the one to be described.
This is made clear at the end of the Introduction, where one often indicates in a sentence what the work to be described adds to knowledge: “Our results show that flagellum originated very early, before the diversification of contemporary bacterial phyla…”.
So the minimal implication is that as a result of the work described “X is now better understood”. And that is clearly the case here. | {
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soft-question, astronomy, resource-recommendations, stars
we are not looking at a snapshot of a single moment somewhere in the past,
but a composite view of a range of past times, stretching back some 16 thousand years.
So here is my question:
Does anyone know of any resource that:
shows the positions of stars as we would look up to see them in their current visible positions
and then allows a "play forward the motions of the individual stars", over the time it took for the light from each one to reach us
to show where they are located in their current actual positions
I've done quite a bit of searching, but all I can find are maps which show the night sky:
with a view of the stars, in their visible positions, as we would see them all currently
with a view of the stars, in their visible positions, as we would see them all at some date in the past
but nothing that would move each star individually according to: | {
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c++, algorithm
If you cannot sort the array in either of the above ways, output "no" in the first line.
Code
#include <vector>
#include <iostream>
#include <algorithm>
int main() {
int n; std::cin >> n;
std::vector<int> v;
v.reserve(n);
while(n--){
int x; std::cin >> x;
v.push_back(x);
}
auto v_sorted = v;
std::sort(v_sorted.begin(),v_sorted.end());
auto mismatch_1 = std::mismatch(v.begin(),v.end(),
v_sorted.begin());
auto mismatch_2 = std::mismatch(std::next(mismatch_1.first),v.end(),
std::next(mismatch_1.second));
auto mismatch_3 = std::mismatch(v.rbegin(),v.rend(),
v_sorted.rbegin()); | {
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php, html, wordpress
</div>
</div>
<!-- Modal 4 -->
<div id="myModal4" class="modal fade" role="dialog">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Video</h4>
</div>
<div class="modal-body">
<iframe src=" <?php echo $video_embed_code_4; ?>" width="500" height="281" frameborder="0" webkitallowfullscreen mozallowfullscreen allowfullscreen></iframe>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
<!-- Modal 5 -->
<div id="myModal5" class="modal fade" role="dialog">
<div class="modal-dialog"> | {
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java, algorithm, recursion, weekend-challenge, sudoku
// if we have not 'escaped' our block, we also find other cells in
// the same block, but not our row/column.
for (int c = col + 1; c < cb; c++) {
if (available[r * dimension + c].length > 1) {
// only need to worry about unsolved followers.
folls[cnt++] = r * dimension + c;
}
}
}
}
// return just the values that were needed as followers.
return Arrays.copyOf(folls, cnt);
} | {
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"openwebmath_score": null,
"tags": "java, algorithm, recursion, weekend-challenge, sudoku",
"url": null
} |
c++, performance, programming-challenge, graph
const int MOD = 1300031;
using edge = std::pair<int, int>;
using adj_list = std::vector<edge>;
using graph = std::vector<adj_list> ;
struct myEdge {
int from, to, weight;
};
std::vector<bool> visited;
int count_nodes_on_component(graph &g, int v) {
int count_of_nodes = 1; //1 for itself
adj_list &adj = g[v];
for(auto &e : adj) {
if(!visited[e.first]) {
visited[e.first] = true;
count_of_nodes += count_nodes_on_component(g, e.first);
visited[e.first] = false;
}
}
return count_of_nodes;
} | {
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"url": null
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A quick check:
(%i47) prodesc(%[3],%[5]);
$\tag{%o47} [0]$
7 Differential Equations
Maxima can represent the slope field (isoclines) of a differential equation without problems. Here, we use the equation dy/dx=2x+y with x running in [-2,1] and y in [-4,6]: | {
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"url": "http://galia.fc.uaslp.mx/~jvallejo/Maxima%20Mini-Tour%2019-May-2019.html"
} |
edge exactly once and return to the starting vertex. Relaxing ideal magneto-fluids: Eulerian vs semi-Lagrangian approaches Jean-François Cossettea, Piotr Smolarkiewiczb, Paul Charbonneaua aUniversité de Montréal, Montréal, Canada bEuropean Centre for Medium-Range Weather Forecasts, Reading, UK Nom de la faculté Nom du département www.umontreal.ca l’affiche Titre de Sous-titre The mesh stays the same as the material deforms (or flows) inside the mesh. Now remove the last edge before you traverse it and you have created a semi-Eulerian trail. If such a walk exists, the graph is called traversable or semi-eulerian. An Eulerian trail, or Euler walk in an undirected graph is a walk that uses each edge exactly once. 0. Sub-Eulerian Graphs: A graph G is called as sub-Eulerian if it is a spanning subgraph of some Eulerian graphs. Lagrangian Versus Eulerian Approach Lagrangian Approach Method of description that follows the particle is referred to as the Lagrangian method of description. This | {
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moveit, catkin, cmake, ros-indigo
What I am trying to figure out, is how to actually compile this. So, I have built a new ROS package, and in the CMakeLists.txt file for that package, I need to specify where to look for header and library files. But how do I know the name of the package that is required for the above code?
If I look in /opt/ros/indigo/include, there are various directories such as moveit, moveit_msgs, moveit_planners_ompl, and similarly in /opt/ros/indigo/lib and /opt/ros/indigo/share. Is each of these directories a package, or is each one a node, or something else?
After doing a bit of research, I found that the moveit::planning_interface::MoveGroup class from the above code is defined in /opt/ros/indigo/include/moveit/move_group_interface/move_group.h. This suggests to me that I should be using the moveit or move_group_interface packages, but if I try either of these, then it fails. For example, in my CMakeLists.txt file, if I have: | {
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artificial-intelligence
The distance can be sometimes > 1m
The distance can be less often < 1m
The robot can collide but it should be very rare and it is better to
avoid it all the time.
Do you have a solution for this problem?
I am new to AI and there is a lot of material to read. So on which topic of AI should I concentrate to find a solution for this problem? Robot Motion Planning is an important topic in Computational Geometry. There have been a lot of research in this field.
Here are some pointers that might help you:
Slide: http://www.mpi-inf.mpg.de/departments/d1/teaching/ss10/Seminar_CGGC/Slides/06_Bazhenova_RMP.pdf
Slide: http://www.cs.cmu.edu/afs/cs/academic/class/15381-s07/www/slides/020807motion.pdf
Book: http://books.google.com/books?hl=en&lr=&id=_VRM_sczrKgC&oi=fnd&pg=PR11&dq=robot+motion+planning&ots=zFkR07oxBJ&sig=efsVfaDPAlhKgxZwClKS136moCI#v=onepage&q=robot%20motion%20planning&f=false
Website: http://correll.cs.colorado.edu/?p=965 | {
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navigation, turtlebot, costmap
Here is the yaml file for the costmap:
costmap:
plugins:
- {name: obstacle_layer, type: "costmap_2d::ObstacleLayer"}
- {name: inflation_layer, type: "costmap_2d::InflationLayer"}
global_frame: /map_new
robot_base_frame: /base_link
static_map: true
rolling_window: false
resolution: 0.01
update_frequency: 20
publish_frequency: 20
width: 3
height: 3
origin_x: 0
origin_y: 0
obstacle_range: 10.0
raytrace_range: 10.0
transform_tolerance: 2.0
robot_radius: 0.225
always_send_full_costmap: true
inflation_layer:
enabled: true
inflation_radius: 0.1 | {
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javascript, beginner, form, dom
Title: Early Access: Dynamically expanding input element table This code is in a really early state in the lifecycle. I'm still in the process of implementing some core features, but nevertheless the code up to here works and accomplishes its task.
I have no idea as to how well, though.
This code is intended to give a dead-simple way of creating growable and shrinkable form content for things like: multiple addresses, multiple courses, multiple X.
As such, it is intended (but not yet able) to fully support all kinds of form elements, including radio buttons and dropdowns.
The goal is providing a single registration once on page-load and then automatize generation of input-names from a template, generation of new form fields for new "rows", removal of rows as well as providing a way to simplistically style the form-fields.
To stay lightweight I decided to implement this in vanilla-js. The current state of the code can be found at this revision in my GitHub repository | {
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c#, null
Title: Refactoring null check before adding to Func collection with different predicates I'd want to refactor that if ladder but I'm not sure how to do that or whether it is even possible
public async List<ComplexObject> Generate(DateTime? data1, DateTime? data2, string FirstName, string SecondName)
{
var ListOfPredicates = new List<Func<ComplexObject, bool>>();
if (data1.HasValue)
{
ListOfPredicates.Add(new Func<ComplexObject, bool>(x => data1.Value <= x.data1));
}
if (data2.HasValue)
{
ListOfPredicates.Add(new Func<ComplexObject, bool>(x => data2.Value >= x.data2));
}
if (!string.IsNullOrEmpty(FirstName))
{
ListOfPredicates.Add(new Func<ComplexObject, bool>(x => x.FirstName.ToLower() == FirstName.ToLower()));
}
if (!string.IsNullOrEmpty(SecondName))
{
ListOfPredicates.Add(new Func<ComplexObject, bool>(x => x.SecondName.ToLower() == SecondName.ToLower()));
}
(...)
} | {
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javascript, jquery, html, dom
function generateChart(chartContainer) {
var container = document.createElement("div");
var text = "Hello World!";
var blockDiv, textSpan; // used in the for loop
container.className = "container2";
document.getElementById(chartContainer.replace("#", "")).appendChild(container);
for(var i = 0; i < BLOCKS_PER_CHART; i++) {
blockDiv = document.createElement("div");
blockDiv.className = "block";
textSpan = document.createElement("span");
textSpan.append(text); // see note about browser compatibility
blockDiv.append(textSpan);
container.append(blockDiv);
}
}
jQuery version:
var BLOCKS_PER_CHART = 10;
function generateChart(chartContainer) {
var container = $("<div>").addClass("container2").appendTo(chartContainer);
var text = "Hello World!";
var blockDiv; // used in the for loop | {
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polymers
3) Cost - Every initiator has a price, even if you are only using tiny amounts. AIBN may have a lower activation temperature than benzoyl peroxide, but AIBN is also more expensive. There are other more specialized inhibitors that are even more expensive. | {
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python, python-3.x, tree
# BODY OF THE MODULE
if __name__ == '__main__':
import doctest
doctest.testmod()
Now we can confidently refactor the code: it will perform at least as good as before.
As you see, the tests mimic a REPL, which is very intuitive and well known by every Python user. Usually, you put comments around the tests to explain whats going on.
I made some little improvements to the tests:
* bunch insert of values with a for loop
* a list comprehension to test the search method
We'll use doctest in the docstrings of functions too.
A recursive implementation
As I wrote above, a recursive implementation seems a good start, since the trees are inherently recursive structures.
The TreeNode class
Your __init__ method allows the value to be None.
def __init__(self, value=None, left=None, right=None):
But there is no reason to accept nodes with a None value. Sooner or later, you'll get something like:
key > cur.value | {
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infinite-impulse-response
Is there ever an application in IIR filters where implementing the direct form of the filter is beneficial? Or should I always attempt to realize the filter as a cascade of subsystems?
Regarding cascade and parallel realizations, is cascaded always better?
Assuming that the exact filter implementation is dependent on the application, is it worth the effort to spend time and analyze the filter implementation from the beginning, or should I always aim for implementing as a cascaded subsystem and only deviate from that if the results are poor?
What is the purpose of transposed systems? I understand mathematically the differences, however, are there actual physical benefits of implementing a transposed vs nontransposed system? I don't think "regardless of application" works here. Every application has different requirements and hence the optimum choice would be different.
However in gross oversimplification | {
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electromagnetic-radiation, x-rays
Title: Elements that do not produce X-rays I have read that hydrogen does not produce X-rays as the energy difference between its shells is small.
So if this is true what are the elements that do not produce X-rays and how was Moseley able to plot the frequency of the radiation for such elements? So-called "K-shell" x-rays are produced when electrons fall from outer electron orbitals to the innermost. (The "L-shell" x-rays are produced when a vacancy is filled in the second-innermost shell.)
An electron falling to the hydrogen ground state emits ultraviolet radiation in a spectrum first observed by Lyman. That is, the physics of the hydrogen Lyman series is mostly the same as the physics of K-shell x-rays in heavy elements. Likewise the physics of the Balmer series of visible-light hydrogen transitions is analogous to L-shell x-rays in heavy nuclei. | {
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ros, desktop-full, ubuntu, dependencies
libcheese-gtk23: Depends: cheese-common (= 3.10.2-0ubuntu2) but 3.10.2-0ubuntu2 is to be installed
libcheese7: Depends: libgudev-1.0-0 (>= 146) but 1:204-5ubuntu20.10 is to be installed
Depends: cheese-common (= 3.10.2-0ubuntu2) but 3.10.2-0ubuntu2 is to be installed
libclutter-1.0-0: Depends: libatk1.0-0 (>= 2.5.3) but 2.10.0-2ubuntu2 is to be installed
Depends: libc6 (>= 2.14) but 2.19-0ubuntu6.6 is to be installed
Depends: libfontconfig1 (>= 2.9.0) but 2.11.0-0ubuntu4.1 is to be installed
Depends: libglib2.0-0 (>= 2.37.3) but 2.40.2-0ubuntu1 is to be installed
Depends: libgtk-3-0 (>= 3.3.18) but 3.10.8-0ubuntu1.4 is to be installed
Depends: libpango-1.0-0 (>= 1.30) but 1.36.3-1ubuntu1.1 is to be installed
Depends: libpangocairo-1.0-0 (>= 1.30) but 1.36.3-1ubuntu1.1 is to be installed | {
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general-relativity, differential-geometry, metric-tensor
The idea I want to emphasize here is, is there such a similar 3D shape for which integrating the metric between two spacetime points could be done physically by measuring through a rope construction as said above? If you mean "embedded in $\mathbb{R}^3$ with a Euclidean metric", then the answer is no.
Suppose that such an embedding exists. Consider the neighborhood of a point $P$ on the submanifold. For all infinitesimal displacements in a neighborhood of a point in Euclidean space, we have $ds^2 \geq 0$, with equality holding only if $dx^\mu = 0$. In other words, all nearby points are a positive "distance squared" from $P$. | {
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quantum-mechanics, electrons, velocity, measurement-problem, observables
Yes, this is essentially the Bohr model, which is known to be false (and, since it is still taught in schools, the source for many a confusion about electron orbitals). Please do not perpetuate it.
Why points? If you insist on visualizing it, the "smooth" pictures the Wikipedia page on atomic orbitals has are far better suited, for example: | {
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javascript, polymer
I find person != null much more clear than return !!person if person is not supposed to be a boolean value.
In general I'd suggest to use a prefix for your custom components and to drop -template. Reasoning: they're all templates then you repeat the obvious and using a prefix you highly reduce possible collisions with 3rd party components.
You have a computed property hasSelectedPerson you use to hide a paragraph. Few things might be optimized here:
dom-if template does not need a boolean property, any falsy value will do the job then simply if="[[selectedPerson]]" is sufficient.
dom-if to hide a single element without any complex logic (for example lists) is an overkill and a (small) performance hit. You can use the hidden attribute (eventually complimented by a specific *[hidden] CSS rule when there isn't browser support) to achieve the same result: <p hidden$="[[selectedPerson]]">...</p>. | {
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c++, algorithm, mergesort
REQUIRE(merge(left, right) == merged);
}
SECTION("uneven length (l > r)") {
std::vector<int> left = {1, 3, 5, 7};
std::vector<int> right = {2, 4, 6 };
std::vector<int> merged = {1, 2, 3, 4, 5, 6, 7 };
REQUIRE(merge(left, right) == merged);
}
SECTION("uneven length (r > l)") {
std::vector<int> left = {2, 5};
std::vector<int> right = {1, 3, 4};
std::vector<int> merged = {1, 2, 3, 4, 5};
REQUIRE(merge(left, right) == merged);
}
SECTION("duplicate elements") {
std::vector<int> left = {2, 2, 5};
std::vector<int> right = {1, 3, 3, 4};
std::vector<int> merged = {1, 2, 2, 3, 3, 4, 5};
REQUIRE(merge(left, right) == merged);
}
SECTION("one half emptying first") {
std::vector<int> left = {4, 6};
std::vector<int> right = {1, 3 };
std::vector<int> merged = {1, 3, 4, 6};
REQUIRE(merge(left, right) == merged);
} | {
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temperature, humidity
Changing the temperature of the air, e.g. by bringing it into your house, does not change how much vapour it contains but it does change its capacity, i.e. how much vapour it could contain. That is, the absolute humidity stays the same, but the relative humidity changes:
If you take cool air and warm it up, the warmer air could hold more vapour than it contains, so the relative humidity goes down.
Conversely, cooling air down will push its relative humidity up. (If you cool to the dew point, you'll hit 100% relative humidity and water will condense out as dew or frost.)
I drew this, maybe it helps...
† Okay, this is a simplification. The conversation is about conditions in and around a house on the surface of the Earth. We don't really care about the air, per se. Read about vapour pressure. | {
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acid-base, water, aqueous-solution, ions, energy
Why is there a release of heat? I have read that breaking of bonds is an endothermic process. $$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$
This reaction owes its ability to go to completion to the great stability the ions it forms when dissolved in water ($\ce{H+}$ and $\ce{Cl-}$). This ionization is caused by stronger attractive forces between the water and the atoms in a molecule (or ions in a salt) than between the atoms within the molecule have themselves, an is an endothermic process. In the dissociation of salts like $\ce{KNO3}$, the solution becomes notably cooler. When an acid dissociates, however, the hydration of $\ce{H+}$ to $\ce{H3O+}$ releases a great deal of heat.
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$ | {
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-
An old-fashioned argument
The matrix $A$ is the matrix of a rotation through some angle $\theta$. It is a standard result that the trace of such a rotation is $1+2\cos\theta$. The matrix $A^2$ is a rotation through $2\theta$ about the same axis.
So we want to show that $$(1+2\cos\theta)^2-(1+2\cos 2\theta)=2(1+2\cos\theta).$$
With a little manipulation this reduces to the familiar $\cos 2\theta=2\cos^2\theta -1$. | {
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quantum-field-theory, coordinate-systems, vacuum, stress-energy-momentum-tensor
Title: Are vacuum expectation values coordinate transform independent? The energy radiated by some free, real-valued, scalar massless field $\phi$ in flat $1 + 1$ time and space dimensions is the vacuum expectation value of the $\{$00$\}$ entry of energy-momentum tensor.
Introduce:
$$\phi = \int d\omega \ (a^{\dagger}_{\omega} \phi_{\omega} + a_{\omega} \phi^{*}_{\omega})$$
where:
$$\phi_{\omega} = i \frac{1}{\sqrt{4 \pi \omega}} \left(e^{-i \omega v} - e^{-i \omega p(u)}\right)\ , \ \text{and}\ u, \ v$$
are the standard null co-ordinates.
Calculating the vacuum expectation value of the energy density:
$$\langle 0|T_{tt}|0\rangle, \ \ T_{tt} = \left(\frac{\partial \phi}{\partial t}\right)^2 + \left(\frac{\partial \phi}{\partial x}\right)^2$$
is straightforward, as only certain combinations of the annihilation and creation operators return a non $0$ answer when operating on the vacuum, and the final result is: | {
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javascript, node.js, ecmascript-6, ssl, protocol-buffers
Is Object.freeze() the normal way to implement enum logic in JavaScript?
That seems to be the common convention. Note that "freeze is shallow"1 so if there were nested objects then it would likely by obligatory to freeze those as well. Refer to answers to this Stack Overflow question: What is the preferred syntax for defining enums in JavaScript?. For even more light reading take a glance at this article posted by one of the users who supplied an answer to that aforementioned SO question.
Is it okay to give same names for method and static method?
While it was closed as off-topic that question was asked on Stack Overflow as well.
Is instance.constructor.static_method() the normal way to access to a static method?
I don’t want to sound like a broken record but there’s a SO question for that. That was works unless the goal is to allow an override static method to be used in a sub-class.
Is the way i use Promise correct? | {
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and Additive Inverse - Matrices - Maths - Class 12/XII - ISCE,CBSE - NCERT. There are 10 True or False Quiz Problems. Anonymous . Though associative false memories are substantially different than those found in semantic or perceptual false memory paradigms, results suggest that associative false memories are mediated by similar neural mechanisms. Properties of matrix addition & scalar multiplication. False. Since addition of real numbers is commutative, it follows that addition of matrices (when it is defined) is also commutative; that is, for any matrices A and B of the same size, A + B will always equal B + A. Click hereto get an answer to your question ️ State the following statement is True or FalseThe commutative property multiplication of integer states that, when multiplication is performed on two integers, then by changing the order of the integers, the result does not change. Also, find its identity, if it exists. You can take the quiz as many times as you like. This is the | {
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"lm_q2_score": 0.8198933403143929,
"openwebmath_perplexity": 922.1060456872423,
"openwebmath_score": 0.5299260020256042,
"tags": null,
"url": "https://www.bdg-academy.com/avocado-toast-pzenf/archive.php?page=451ade-matrix-addition-is-associative-true-or-false"
} |
antimatter
Title: Particle antiparticle annihilation-do they have to be of the same type? I read that a particle will meet its antiparticle and annihilate to generate a photon. Is it important for the pairs to be of the same type? What will happen when for example a neutron meets an antiproton or a proton meets a positron? Are there any rules to determine what happens when such particles meet? Yes, there are rules that depend on the quantum numbers carried by the particles under question and the energy available for the interaction.
In general we label as annihilation when particle meets antiparticle because all the characterising quantum numbers are equal and opposite in sign and add and become 0, allowing for the decay into two photons, two because you need momentum conservation. | {
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c#, object-oriented, inheritance, interface
//The responses send by the server containing information about the send packets
private List<ResponseHeaderData> responseHeaders;
public List<ResponseHeaderData> ResponseHeaders
{
get
{
if (responseHeaders == null) responseHeaders = new List<ResponseHeaderData>();
return responseHeaders;
}
set
{
responseHeaders = value;
}
} | {
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Solving this linear system by row-reducing the augmented matrix shows that $$\alpha = 3\text{,}$$ $$\beta = -2$$ is the only solution.
#### Exercise Group.
In Chapter V we defined the operations of vector addition and vector scalar multiplication in Definition CVA and Definition CVSM. These two operations formed the underpinnings of the remainder of the chapter. We have now defined similar operations for matrices in Definition MA and Definition MSM. You will have noticed the resulting similarities between Theorem VSPCV and Theorem VSPM.
In Exercises M20–M25, you will be asked to extend these similarities to other fundamental definitions and concepts we first saw in Chapter V. This sequence of problems was suggested by Martin Jackson.
##### M20.
Suppose $$S=\set{B_1,\,B_2,\,B_3,\,\ldots,\,B_p}$$ is a set of matrices from $$M_{mn}\text{.}$$ Formulate appropriate definitions for the following terms and give an example of the use of each. | {
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"url": "https://runestone.academy/ns/books/published/fcla/section-MO.html"
} |
java
vpsc.setFrom(processStep.getId());
vpsc.setTo(nextProcessStep.getId());
connections.add(vpsc);
}
}
vProcess.setSteps(steps);
vProcess.setConnections(connections);
vProcesses.add(vProcess);
} | {
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{ 3 } } 3. Circumradius of an equilateral triangle is one side of the polygon angle formed at orthocenter. Does the formula come from, centroid and nine-point center are all the same line if the area by. All the same note that the variables used are in reference to the triangle these. Abc and ACB meet the circumcircle of the triangle is s 3 3 \frac { s\sqrt { }... Triangle shown in the calculator above R respectively '' of the three sides and angles! Sides lies on the circumcircle of the three sides and three angles, some of which may be same. Is easily calculable formed at the vertex ( but not very commonly ) called the trigon sometimes ( but very... And ACB meet the circumcircle of the orthocenter is supplementary to the angle formed at the orthocenter circumcenter! If the area formed by the triangle of these three points is zero or not,. Orthocenter over any of the triangle shown in the calculator above triangle has three sides lies on the circumcircle the! The same is the regular | {
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"tags": null,
"url": "http://blucome.com/2km44w/01238f-circumcircle-of-equilateral-triangle-formula"
} |
the-moon, light, atmospheric-effects
Title: Why doesn't the moon twinkle? Stars twinkle because their light has to squeeze through several different layers of the Earth's atmosphere. So why doesn't the moon twinkle as well? The first handful of hits on Google actually return incomplete and even wrong answers (e.g. "Because the Moon is much brighter" which is plain wrong, and "Because the Moon is closer" which is incomplete [see below]). So here's the answer:
As you mention, when light enters our atmosphere, it goes through several parcels of gas with varying density, temperature, pressure, and humidity. These differences make the refractive index of the parcels different, and since they move around (the scientific term for air moving around is "wind"), the light rays take slightly different paths through the atmosphere.
Stars are point sources | {
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newtonian-mechanics, newtonian-gravity, acceleration, density, fluid-statics
If you throw from the top of a building a water tank with a ball inside, the ball wouldn't sink inside the water. Rather, the water and the ball would accelerate equally and not move with respect to each other. | {
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java, performance, linked-list, deque, heuristic
for (int i = 0; i < nodesToSkip; i++)
node = node.next;
addFinger(node, index);
for (int i = 1; i < numberOfNewFingers; i++) {
index += distanceBetweenFingers;
for (int j = 0; j < distanceBetweenFingers; j++) {
node = node.next;
}
addFinger(node, index);
}
}
private void addFingersAfterInsertAll(Node<E> headNodeOfInsertedRange,
int indexOfInsertedRangeHead,
int collectionSize) {
final int numberOfNewFingers =
getRecommendedNumberOfFingers() - fingerStack.size();
if (numberOfNewFingers == 0)
return;
final int distanceBetweenFingers = collectionSize / numberOfNewFingers;
final int startOffset = distanceBetweenFingers / 2; | {
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or a vertical line. Node or edge tables can be created under any schema in a database, but they all belong to one logical graph. Pareto Chart is combination chart of Column Chart and Line Chart. The data for this example graph has been altered from the data that was comprised of litigants in the mobile patent war to fictitious peoples names and associated. A TensorFlow computation, represented as a dataflow graph. 2) showed a single relation (that happened to be binary and directed). Learn with flashcards, games, and more — for free. This just means that each edge in the graph is unidirectional, i. A graph G is a triple consisting of a vertex set of V(G), an edge set E(G), and a relation that associates with each edge two vertices (not necessarily distinct) called its. Human translations with examples: pikto graph, paggamit ng graph, ano ang graph pie, ano ano ang grafima. Fatal Crashes by Vehicle Type - Line Graph Example. Draw the line that connects the two points. thus adjusting the | {
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"openwebmath_score": 0.2822085916996002,
"tags": null,
"url": "http://ylqo.solotango.it/graph-examples.html"
} |
cosmology, black-holes, temperature, event-horizon, low-temperature-physics
So only when the cosmic microwave backround radiation cools from the expansion of the universe to such small numbers will larger black holes start diminishing in mass, as is expected, as the radiation leaving from the horizon diminishes the energy of the black hole. | {
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} |
quantum-mechanics, quantum-information
\\ & \quad +
\left( a e^{i\varphi}\cos(\theta)\sin(\vartheta) - b e^{-i\phi}\sin(\theta)\cos(\vartheta) \right)|↑\!\mathrm d⟩
\\ & \quad +
\left( a e^{i(\phi+\varphi)}\sin(\theta)\sin(\vartheta) + b \cos(\theta)\cos(\vartheta) \right)|↓\!\mathrm d⟩
.
\end{align}
Now here it sort of depends what you want the state to look like, but the spirit is that $\theta$ and $\phi$ are given and we want to see what we can do with a suitable choice of $\vartheta$ and $\varphi$. In particular, it's interesting to see if we can eliminate the components along $|↓\!\mathrm u⟩$ and $|↑\!\mathrm d⟩$, for which we need $\alpha=\cos(\vartheta)$ and $\beta=e^{-i\varphi}\sin(\theta)$ to obey the equations
\begin{align}
a^*e^{-i\phi}\sin(\theta)\alpha -b^* \cos(\theta)\beta
& =0
\\
b e^{-i\phi}\sin(\theta)\alpha - a \cos(\theta)\beta
& =0
\end{align} | {
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Cases[expr, Sqrt[t_] :> t, Infinity]
(* {1 - z, 1 + z} *)
To ensure that all of the terms are real
And @@ Thread[% >= 0] // Simplify
(* -1 <= z <= 1 *)
Or more directly,
And @@ Cases[expr, Sqrt[t_] :> (t >= 0), Infinity] // Simplify
(* -1 <= z <= 1 *)
Better yet,
FunctionDomain[expr, z]
(* -1 <= z <= 1 *)
` | {
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"tags": null,
"url": "https://mathematica.stackexchange.com/questions/215925/extracting-argument-of-a-specific-functions-in-a-large-expression"
} |
java, algorithm, recursion, weekend-challenge, sudoku
/**
* A class that is able to brute-force a solution for a Sudoku-like puzzle.
* <br>
* Due to the internal use of bit-shifting the largest sudoku puzzle this can solve is 25x25
* which should be more than enough. Sudoku-like puzzles are always square-number sized. 1x1, 4x4, 9x9, 16x16, 25x25, etc.
*
* @author rolfl
*
*/
public class RecursiveBruteSolver { | {
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"tags": "java, algorithm, recursion, weekend-challenge, sudoku",
"url": null
} |
biochemistry
The notation for the glycosidic linkage between the two sugars
specifies which anomeric form of the sugar is involved in the bond and
also specifies which carbon atoms of the two sugars are linked
together.
For example two ways in which two α-D-glucose molecules can be linked
together are α(1>4) and α(1>6). Firstly, the α anomeric carbon (C-l)
of the first glucose molecule is joined in a glycosidic bond to the
fourth carbon atom (C-4) of the second glucose molecule; or the C-1
of the first glucose molecule is linked to the C-6 of the second
glucose . | {
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Indeed, speaking of a rank-2 tensor is not really accurate. The rank of a tensor has to be given by two numbers. The vector to vector mapping is given by a rank-(1,1) tensor, while the quadratic form is given by a rank-(0,2) tensor. There's also the type (2,0) which also corresponds to a matrix, but which maps two covectors to a number, and which again transforms differently.
The bottom line of this is:
• The components of a rank-2 tensor can be written in a matrix.
• The tensor is not that matrix, because different types of tensors can correspond to the same matrix.
• The differences between those tensor types are uncovered by the basis transformations (hence the physicist's definition: A tensor is what transforms like a tensor").
Of course another difference between matrices and tensors is that matrices are by definition two-index objects, while tensors can have any rank. | {
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} |
thermodynamics, heat-transfer
a case in a science where it is not allowed to be done. The theorem regarding it's use, however, is not trivial at all. To finish the problem up, we rearrange to obtain $\ln (\frac{T_f}{T_i}) = \ln (2^{\frac{R}{c_v}})$, raise both sides to the power of $e$, and solve $T_f = T_i \cdot 2^{\frac{R}{c_v}} = \frac{1}{R} \cdot 2^{\frac{2}{3}}$. Thus $T_f \approx 0.191K$. From the ideal gas law, we get $P \approx 3.175$ $Pa$. | {
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optics, interference
From Britannica,
Interference fringe, a bright or dark band caused by beams of light that are in phase or out of phase with one another. The fringe refers to the individual bands. Each dark and bright band is a fringe in its own right, termed as, dark and bright fringe respectively.
A combination of these fringes, such as when performing the Young's Double Slit Experiment is called the interference pattern.
As the text you quote says, the fringes are just the dark and bright bands, but the entire pattern,
of bright -> dark -> bright and so on, is the interference pattern.
edit:
so, this is the interference pattern. each of the bars here is a fringe. | {
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community-challenge, actionscript-3, simon-says
And assign them a responsibility.
public var playerTurn:Boolean = false; //turn keeping
private var pattern:Array = new Array(); //player pattern
private var patternIndex:uint = 0; //pattern
private var playerPattern:Array = new Array(); //player pattern
private var blinkCounter:uint = 0; //blink/graphical display
private var buttons:Array = new Array(); //graphical display
private var graphicSprite:Sprite = new Sprite(); //blink/graphical display | {
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nitrogen-cycle
Title: How does free living nitrogen fixing bacteria fix N2 in aerobic condition? If a Azotobacter is free living nitrogen fixer it will also require ATP through cellular respiration. I am confused here: if Nitrogenase is highly sensitive to oxygen molecule how will Azotobacter be able to fix N2 and carry out cellular respiration at the same time? https://en.wikipedia.org/wiki/Azotobacter#Nitrogen_fixation
Nitrogen fixation is highly sensitive to the presence of oxygen, so
Azotobacter developed a special defensive mechanism against oxygen,
namely a significant intensification of metabolism that reduces the
concentration of oxygen in the cells.[40]
From that cite "Respiratory Protection nitrogenase complex in Azotobacter vinelandii"(google translated) | {
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Autumn Week 2 – Number: Place Value; Autumn Week 3 – Number: Place Value ; Autumn Week 4 – Number… Is 1000 a prime number? A powerful number is a positive integer m such that for every prime number p dividing m, p 2 also divides m.Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a 2 b 3, where a and b are positive integers. List of Perfect Cube Numbers 1 to 50 This is really a correct and easy … There are three sixth powers less than 100: 0^6 = 0, 1^6 = 1 and 2^6 = 64. © Corbettmaths 2018 From the list, write down the cube numbers For a number to be both a square and a cube, it would have to be a sixth power. 0. of 0 votes. the cube number of 10 is 1000 the cube number of 11 is 1331 the cube number of 12 is 1728 the cube number of 13 is 2197 the cube number of 14 is 2744 the cube number of 15 is 3375 the cube number of 16 is 4096 the cube number of 17 is 4913 the cube number of 18 is 5832 the cube number … New questions in Math. | {
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"url": "https://www.spicum.com/uncle-boonmee-nddfma/83e7bb-is-1000-a-cube-number"
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2. Hello, pberardi!
You may kick yourself when you see how easy it is . . .
Find the solution of $y' \:= \:2(2x-y)$ that passes through the point $P(0,1).$
Solution: . $y \:= \:2x - 1 + Ce^{-2x}$
$\text{"Passes through P(0,1)" means that: }\:x = 0,\:y = 1\,\text{ satisfies the equation.}$
$\text{Plug them in: }\;1 \;=\;2(0) - 1 + Ce^0 \quad\Rightarrow\quad 1 \:=\:\text{-}1 + C \quad\Rightarrow\quad C \:=\:2$
Solution: . $y \;=\;2x - 1 + 2e^{-2x}$
3. Thank you Soroban
4. Hi.
Originally Posted by pberardi
Hello,
I have this diffyQ problem that asks to find the solution that passes through a specific point. I have worked it out to where I have a ( i think correct ) equation for y, but am not sure how to deal with the constant and make sure it passes through a specific point. | {
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"url": "http://mathhelpforum.com/differential-equations/71641-dfeq-solution-passes-through-point.html"
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ros
Title: Executing a 32bit ROS App in a 64bit uBuntu OS
I wanted to install ROS 32bit and ROS 64bit in the same machine. That it appears wont fly. I then compiled my 32bit App in a separate 32bit machine (with 32bit ROS under a 32bit uBuntu PC).
I'm now trying to run my 32bit App in the 64bit environment (that is, 64bit fuerte and 64bit uBuntu). Launching it from a launch file (this is the syntax: "")
I get this error:
"ERROR: cannot launch node of type [foo_bar/foo_bar]: can't locate node [foo_bar] in package [foo_bar]"
ROSCD foo_bar navigates to the correct path. So it is not that ROS can't see the binary file.
What am I missing?
Originally posted by Mfumbesi on ROS Answers with karma: 46 on 2013-08-19
Post score: 0 | {
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mesh
// Get result
gp3.setInputCloud (cloud_with_normals);
gp3.setSearchMethod (tree2);
gp3.reconstruct (triangles);
// Additional vertex information
std::vector<int> parts = gp3.getPartIDs();
std::vector<int> states = gp3.getPointStates();
//pcl::io::
pcl::io::saveVTKFile (ParamValue::planeMeshVTK, triangles);
std::cout<<"\n\n\t Calling conversion from VTK to PLY";
convertVtkToPly(ParamValue::planeMeshVTK, ParamValue::planeMeshPLY);
}
Any help/pointer is greatly appreciated.
Thanks
Nirala A K
Originally posted by aknirala on Gazebo Answers with karma: 21 on 2014-08-17
Post score: 0
You'll be able to get help on this much faster if you post it to the PCL mailing list. Luckily it is very active.
Also, if you don't need to do this programmatically, creating the mesh in MeshLab instead of PCL would be much easier. Here and here are some tutorials on how to do that.
Originally posted by Airuno2L with karma: 26 on 2014-08-27
This answer was ACCEPTED on the original site
Post score: 0 | {
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Posted from my mobile device
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
23 Oct 2013, 15:54
6
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question
Dear Saint For Life,
Problems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.
Having said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/
Mike
_________________
Mike McGarry
Magoosh Test Prep | {
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If $n$ is large enough, your expected value should approach the mean of the distribution. So probability that value is greater than expected value should be $0.5$. I am not sure about this one.
• Probability of observing value above median is one half, but your quantity is log-normally distributed for large n, which has different mean and median, so you shouldn't expect that probability to approach 1/2. Sep 10 '10 at 21:33
If n is large enough, your expected value should approach the mean of the distribution.
Yes that's correct.
So probability that value is greater than expected value should be 0.5.
This would only be correct if the distribution is symmetric - which in your game isn't the case. You can see this easily if you think about what the median value of your winnings should be after $n$ throws. | {
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# Aligning an equation at multiple places
I want to align an equation at two places, and I have a code segment that looks like this:
\documentclass[a4paper]{paper}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}
\begin{alignat*}{2}
g_P^x &= 3x_P^2 + a, \quad g_P^y &&= -2y_P, \\
v_P &= 2g_P^x, \quad u_P &&= (g_P^y)^2, \\
v &= \sum_{P \in G^+} v_P, \quad w &&= \sum_{P \in G^+} u_P + x_Pv_P.
\end{alignat*}
\end{document}
And this works to a certain extend, but there is a big gap between u_P and the equal sign. How can I get it closer to the equal sign so it is aligned with the stuff above and below it? | {
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atoms, ions, molecules, stability, hydrogen
For what answers that follow - do these apply in general to the stable nature of all diatomic gases? While usually short-lived, isolated hydrogen atoms (also called hydrogen radicals) are well studied and chemically important species. Bond formation in simple species is always an exothermic reaction (usually highly so) with zero activation energy, which means bond formation is thermodynamically very favourable and has no kinetic barrier stopping it from happening. Simply put, the isolated atoms really want to react with each other, and they'll attract each other at a distance, so if at some moment you create a container with pure monoatomic hydrogen, the hydrogen atoms will react as soon as possible to regenerate dihydrogen molecules (hydrogen gas). | {
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beginner, c, calculator
Then once you have the above inputs, you can do something like this:
float intermediate = 0.0;
switch (operator1) {
case OP_ADD:
intermediate = a + b;
break;
case OP_SUBTRACT:
intermediate = a - b;
break;
case OP_MULTIPLY:
intermediate = a * b;
break;
case OP_DIVIDE:
intermediate = a / b;
break;
};
float result = 0.0;
switch (operator2) {
case OP_ADD:
result = intermediate + c;
break;
case OP_SUBTRACT:
result = intermediate - c;
break;
case OP_MULTIPLY:
result = intermediate * c;
break;
case OP_DIVIDE:
result = intermediate / c;
break;
}; | {
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formal-languages, pumping-lemma
$$ h = m \qquad |v| = 0 \qquad 3h + h|x| = 2h + k$$
which results in $ |x| = \frac{k}{h} - 1$; then we can take $k = 3h$ so $|x| = 2$, and the result is the string $0^{h} 1 0^{4h} \in L$
As you can see, I can't find a combination of string and $i$ that generates a contraddiction. How would you proceed? Your error is that you assume that you must find one $i$ to satisfy all the subdivisions $uvwxy$ of the lemma. The lemma is actually much weaker in that you need to find any $i$ for every subdivision $uvwxy$ of a string $s$ in $L$ longer than $h$. Without loss of generality, I will assume that any given $h$ is prime, because if it were not, I would just substitude all next occurances with any higher prime number and everything would still work the same way (I never needed the condition $|vwx|\le h$).
That said you can take your second try from the question and then fork it into two cases where: | {
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Yes, you are correct. The assumption that $$M$$ is arcwise connected is completely irrelevant (as is the assumption that $$A_1$$ and $$A_2$$ are subsets of $$M$$, and the assumption that $$M$$ is a metric space). The way I would state the result with no irrelevant hypotheses would be:
Let $$X$$ and $$M$$ be topological spaces and $$f_0,f_1:X\to M$$ be continuous. Suppose $$A_1$$ and $$A_2$$ are disjoint open subsets of $$X$$ whose union is $$X$$, and that $$f_0|_{A_i}$$ is homotopic to $$f_1|_{A_i}$$ for $$i=1,2$$. Then $$f_0$$ is homotopic to $$f_1$$.
• The $A_i$ don't need to be open in $X$, they just need to be open in $A_1 \cup A_2$. – Rob Arthan Nov 8 '18 at 20:50
• $X$ is just my name for $A_1\cup A_2$ (notice that I assume their union is $X$). – Eric Wofsey Nov 8 '18 at 20:52 | {
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Live on Desmos
If you need the function to have different values at different minima, or if you want more than two minima, you can try this approach. Let $$x = a_1, \dots, a_n$$ be our minima, and $$c_1, \dots, c_n \in (0, 1)$$ and $$d_1, \dots, d_n$$ some constants. Define
$$r_i(x) = 1 - (1-c_i)exp(-d_i(x-a_i)^2)$$
$$exp(-d_i(x-a_i)^2)$$ defines a "bump" with maximum at $$a_i$$, and $$d_i$$ controls it's flatness. $$r_i$$ is this same bump, reversed and scaled so that it takes value $$c_i$$ at $$x=a_i$$ and takes the value of $$1$$ when $$x$$ is far from $$a_i$$.
Now, take the product of the $$r_i$$:
$$r(x) = r_1(x) \cdot r_2(x) \cdots r_n(x)$$
Live on Desmos | {
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How would I show that the population size is limited as n tends to infinity?
4. Originally Posted by Rombie
Thanks, 8000-4000(1/2)^n works. 8000 must be that suitable upper limit. Since it says conjecture do I need to give some reasoning for how I got there? Or is it enough to propose this formula and then prove it by induction.
How would I show that the population size is limited as n tends to infinity?
You can try some experiments to see where the sequence is going, which
should allow you to guess what the upper limit might be.
As n goes to infinity if a_n=8000-4000(1/2)^n, then a_n goes to 8000 as
(1/2)^n goes to zero. Also it approches this from below as (1/2)^n is
strictly decreasing.
RonL
5. Hello, Rombie!
I got your result . . .
The growth of a certain population is modelled by the recursion formula
$a_{n+2}\;=\;\frac{3}{2}\,a_{n+1}-\frac{1}{2}\,a_n$ . and $a_1 = 6000,\;a_2=7000$
for n $\geq$ 1, where $a_k$ is the size of the population after period $k.$ | {
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python, performance, algorithm, mathematics, pygame
This will result in a dotted line instead of a solid line. Just like Bresenham's algorithm itself, you need to distinguish between steep and non-steep.
Use sort()
You implemented your own bubble-sort algorithm to sort v in drawTriangle(). But Python comes with a sorting function that can take an optional function to tell it what to sort on, so you can replace your own algorithm with:
v = [v1, v2, v2]
v.sort(key=lambda p: p.y) | {
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temperature, atmospheric-science, geophysics, weather, humidity
Title: Why is it hot before raining? What causes the temperature hike before a rain?
I have heard explanations like one that says as the moisture rises, you sweat more.
Can somebody provide a more scientific explanation for this. When warm humid air flows into cool air the humidity is condensed into rain as the warm air cools. The temperature rise your feeling is the warm humid air rolling into your cooler area before it rains. Your body cools itself by sweat evaporating which disperses your heat into the surrounding atmosphere. If the air is humid the sweat cannot evaporate and you retain heat and feel hot. | {
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aliasing, oscillator
Title: Freeform band-limited waveforms I'm ready to dip my toes in the scary waters of non-aliasing waveforms and wavetables. Ideally, what I'd like to do is create a (naive) wavetable/waveform in memory and subsequently generate band-limited versions of it with a reusable procedure.
Is that possible, and what would be the best starting point for my research? this paper was done long before i had MATLAB. the drawings are poor, but the math (at least in this revision, which is what you should use) is spot on.
send me an email address (to my audioimagination.com, not the wavemechanics.com on the paper) and i will send you a very short C file that shows how to generate and crossfade the wavetables in the synthesis process. | {
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