text stringlengths 1 1.11k | source dict |
|---|---|
• I.e., the mass function involves counting the number of ways to get a particular $$\rho$$ value from each permutation on the pair of $$\{1, \dots, n\}$$ ranks, divided by the total number of such permutations.
We know how to do that now, but we won’t reinvent the wheel.
• Table A10 in the text provides evaluations of the CDF for $$n \leq 30$$.
• We can use the CLT and get that $$Z = \rho \sqrt{n-1} \sim \mathcal{N}(0,1)$$ for $$n > 30$$.
• An add-on package SuppDists contains a pSpearman function.
When there are ties in the ranks the CLT method is usually preferred, as the averaged ranks challenge the calculation of the CDF of the null distribution.
The hypotheses are specified as follows, and the specific verbiage here is due to the insensitivity of $$\rho$$ to some types of dependence.
For a two-tailed test we consider | {
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"lm_q2_score": 0.815232489352,
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"openwebmath_score": 0.8960742354393005,
"tags": null,
"url": "http://bobby.gramacy.com/teaching/np/spearman.html"
} |
ros, callback, joy-node
The problem: I can not access to myjoy values out of the joyread callback. the error message is
AttributeError: 'Publisher' object has no attribute 'data'
I tried all of myjoy variables. They are different object types (the error message changes a bit) but none of them seems to work. What I have to do?
Thank you in advance.
(ROS Indigo and Python script)
Originally posted by marcoresk on ROS Answers with karma: 76 on 2017-04-15
Post score: 0
I finally found the solution. Myjoy mantains joy structure in the array. It was
AA = int(255*myjoy.axes[0])
and all other myjoy were quite useless.
Originally posted by marcoresk with karma: 76 on 2017-04-15
This answer was ACCEPTED on the original site
Post score: 0 | {
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"openwebmath_score": null,
"tags": "ros, callback, joy-node",
"url": null
} |
homework-and-exercises, newtonian-mechanics, forces, friction
Title: Question on friction
Please look at the figure carefully. The question is written, and one has to match the two columns 1 and 2 given under "choices and explanation". Please tell me if I am really wrong, i.e. would you match C with both 1 and 3, or just 3. (my answer and the right answer is shown at the right hand side)
Thanks! Yes you are correct!
it is a known bug in VMC answers. If you search their forums as well ,some 2 years old questions perhaps, you'll find this.
The friction is static and just opposes the external force. Hence the friction force is just $4.5N$, and body will remain at rest. | {
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"tags": "homework-and-exercises, newtonian-mechanics, forces, friction",
"url": null
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javascript, php, file
fetch(url, {
method: 'POST',
body: formData
}).then(response => {
console.log(response);
});
});
</script>
I would appreciate a review of the PHP for handling the $_POST request. I tried to make it as simple as possible, but I'm not sure if there are any major security concerns I'm overlooking.
<?php
if ($_SERVER["REQUEST_METHOD"] === 'POST') {
if (isset($_FILES['files'])) {
$errors = [];
$path = 'uploads/';
$extensions = ['jpg', 'jpeg', 'png', 'gif', 'html', 'txt', 'pdf', 'doc'];
$allFiles = count($_FILES['files']['tmp_name']);
for ($i = 0; $i < $allFiles; $i++) {
try {
$file_name = $_FILES['files']['name'][$i];
$file_tmp = $_FILES['files']['tmp_name'][$i];
$file_type = $_FILES['files']['type'][$i];
$file_ext = strtolower(end(explode('.', $_FILES['files']['name'][$i]))); | {
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"openwebmath_score": null,
"tags": "javascript, php, file",
"url": null
} |
energy, simulations, collision
Title: How much energy should I give into each particle in this equation: I am trying to recreate the results of an article (membrane simulation) and I have the following line:
Both particles have the same soft radius,
$U_{rep} (r)/\epsilon = \text{exp}\left\{ -20 (r/\sigma -1 )\right \}$ | {
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"tags": "energy, simulations, collision",
"url": null
} |
boundary even on an easy problem. Find weights and bias which will produce the decision boundary you found in part i, and sketch the network diagram. Perceptron algorithm is defined based on a biological brain model. (Research Article) by "Journal of Healthcare Engineering"; Health care industry Analysis Health aspects Usage Chronic diseases Data mining Detection equipment Detectors Epidemiology Health care information services Kidney diseases Machine learning Medical advice systems. Decision boundaries are not always clear cut. I Code the two classes by y i = 1,−1. In the simplest form of the perceptron,there are two decision re-gions separated by a hyperplane, which is defined by v=a m i=1 w ix i+b Section 1. behind the perceptron. bias term tional term in the sum called a bias term. decision surface for Boolean function on preceding slides. •Draw perceptron weight vectors and the corresponding decision boundaries in two dimensions. The decision boundary to be used in our discussion | {
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"url": "http://mzei.youlifereporter.it/perceptron-decision-boundary.html"
} |
xacro
Originally posted by jhdeanda on ROS Answers with karma: 11 on 2017-04-20
Post score: -1
Find wherever your robot description is launched. Maybe you could grep for it:
grep -r "robot_description" *
Then change the command slightly:
<param name="robot_description" command="$(find xacro)/xacro '$(find robot_description)/urdf/my_robot.urdf.xacro'" />
to
<param name="robot_description" command="$(find xacro)/xacro '--inorder' '$(find robot_description)/urdf/my_robot.urdf.xacro'" />
Originally posted by AndyZe with karma: 2331 on 2017-04-20
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by raequin on 2018-03-23:
Can you explain the difference between traditional processing and inorder? | {
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python, pygame
# adjust shot freq when invader numbers decrease
if len(invaders) < initial_invaders*.75 and shoot_level == 1:
pygame.time.set_timer(invader_shoot, 750)
shoot_level = 2
elif len(invaders) < initial_invaders*.5 and shoot_level == 2:
pygame.time.set_timer(invader_shoot, 1000)
shoot_level = 3
elif len(invaders) < initial_invaders*.25 and shoot_level == 3:
pygame.time.set_timer(invader_shoot, 1500)
shoot_level = 4
# draw invaders
for rect, (a, b) in zip(invaders, colors):
animate_invaders = draw_invader(backbuffer, rect, a, b,\
animate_invaders, ticks, \
animation_time) | {
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} |
python, python-3.x
def __setitem__(self, path: str, value: Any):
class Missing:
pass
names = DictIndexer._split(path)
prev = self.source
item = self.source
for key in names[:-1]:
item = item.get(key, Missing())
match item:
case dict():
prev = item
case Missing():
if self.set_creates_branches:
prev[key] = {}
else:
raise KeyError(f"Cannot create branch '{key}. Disabled")
case _:
raise KeyError(f"Cannot overwrite '{key}.")
else:
item[names[-1]] = value
@staticmethod
def _split(path: str) -> List[str]:
return re.split(r"[/\\]", path)
Tests
With pytest I made sure that the following features work:
def test_can_get_item_by_path():
data = {
"foo": {
"bar": "baz"
}
} | {
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• I think you need to fix your discounting in a few places. In general, multiplying by $1+r(T-t)$ is similar to multiplying by $e^{r(T-t)}$. You are mixing those up here. Sep 12 '20 at 22:50
• When I follow the method given by @RRL, it seems that the discount factor is $\frac{1}{(1+r)^9} = \frac{1}{(1+0.001)^9} = 0.9910$, can you point out where the mistakes are? Sep 13 '20 at 7:40
• It depends on if you do exponential or periodically-compounded discounting. I've seen trees built with both, but Bjork is using one so for replication you should use that. Sep 13 '20 at 7:52
• Are you referring to proposition 2.11? Sep 13 '20 at 7:58
Is there a faster way to calculate the option price?
With a recombining binomial tree, the terminal asset price has a binomial distribution -- as you might have expected. For a tree with $$n$$ steps, the probability of reaching price $$S_{n,k}$$ where $$k$$ is the number of up moves is
$$P_{n,k} = \frac{n!}{k!(n-k)!}q^k(1-q)^{n-k}$$ | {
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"url": "https://quant.stackexchange.com/questions/57957/calculating-european-call-option-the-bjork-way"
} |
electromagnetism, quantum-field-theory, maxwell-equations, chern-simons-theory
Now we can insert the above substitutions into $k_1^2 + k_2^2 = 0$ to get
$$\omega^2 = (2g)^2.$$
Hence you get a massive photon.
EDIT: The condition $k_1^2 + k_2^2 = 0$ might seem uncomfortable, it is obtained due to your choice of gauge. | {
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chernoff-bound, privacy
Set $\lambda=\sqrt{\log(2/\delta)/2n}$ and substitute in $q(x)=\frac{1}{n}\sum_{i=1}^n q(x_i)$ to get
$$\mathbb{P}\left[\left|\frac{1}{n} \sum_{i=1}^n \frac{1}{\epsilon}\left(y_i - \frac{1-\epsilon}2\right)-q(x)\right| \geq \frac{\sqrt{\log(2/\delta)}}{\epsilon\sqrt{2n}}\right] \leq \delta.$$
Now "suppress" the $\delta$ parameter using big-O notation to get
$$\left|\frac{1}{n} \sum_{i=1}^n \frac{1}{\epsilon}\left(y_i - \frac{1-\epsilon}2\right)-q(x)\right| \leq O\left(\frac{1}{\epsilon\sqrt{n}}\right)$$ with high probability, as required. | {
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java, converting, integer
the number of zeros is length minus the number of ones
Simplify
Instead of this:
binary = Integer.toString(mod) + binary;
You could simplify as:
binary = mod + binary;
Reinventing the wheel
Integer.toBinaryString(...) does the same thing as your main loop. | {
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But I'm worried about whether or not proofs like this are "legal", if that makes any sense. There was no wording in the question stating that I couldn't start from the right side of the identity, but I still have this strange feeling of guilt that I should've tried solving the identity starting from the left side and working in the "normal" direction.
For questions like these, is it okay to start from the right side of the identity? Would what I get out of doing the question in reverse be the same as if I did it normally? | {
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ros, rosjava, android
Originally posted by Srogacki on ROS Answers with karma: 74 on 2013-02-15
Post score: 0
Original comments
Comment by damonkohler on 2013-02-19:
Are you using the latest rosjava? How is the publisher written? In Java as well?
I figured it out. I using a publisher with a custom-built message that was essentially serialized to a string, but still a custom-message in name. My subscriber was looking for std_msgs/String, but not receiving any, so the checksum was off.
Sorry for the confusion.
-Steve
Originally posted by Srogacki with karma: 74 on 2013-02-19
This answer was ACCEPTED on the original site
Post score: 0 | {
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pharmacology, autonomic-nervous-system
Title: Salbutamol's Pathways of Interaction and Classification Salbutamol is a very commonly used direct-acting β2-agonist.
This suggests me that it is sympatholytic.
However, it has sympathomimetic pathways, see PubChem for Sympathomimetic.
I am trying to think the group of Salbutamol.
Sympathomimetic or sympatholytic.
How can you classify Salbutamol into a group? An agonist works with the receptor: a substance that initiates a physiological response when combined with a receptor; in the case of salbutamol (or albutarol in the US) they activate the beta-2 receptor on the muscles surrounding the airways. They are also known as sympathomimetics, as they imitate β2 adrenergics.
A blocker (or antagonist) is one that that blocks the action of endogenous catecholamines; for that reason, they are also called sympatholytic: antagonistic to or inhibiting the transmission of nerve impulses in the sympathetic nervous system. | {
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solar-system, planet, 9th-planet
My original attempt to handwave an estimate of detectability is below.
The paper gives potential orbital parameters of $400-1500~\textrm{AU}$ for the semi major axis, and $200-300~\textrm{AU}$ for perihelion. Since the paper doesn't give a most-likely case for orbital parameters, I'm going to go with the extreme case that makes it most difficult to find. Taking the most eccentric possible values from that gives an orbit with a $1500~\textrm{AU}$ semi-major axis and a $200~\textrm{AU}$ perihelion has a $2800~\textrm{AU}$ aphelion. | {
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"tags": "solar-system, planet, 9th-planet",
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get an error the! Be used to calculate the inverse tangent note that in cell A4, the ATAN function calculates value. For the arctan value whose tangent value is given, and returns an angle that can between... Calculator tool makes the calculation arctan has to be the # value it is based on radius... 90 ] and +π/2 to -90° to +90° operation of tangent that ranges between -90 and degrees! Tan -1 ) of the elements of which, the range of arctan is limited to -90° to.. | {
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"url": "http://crsc.org.nz/files/tampopo-university-yefj/archive.php?id=american-standard-colony-soft-2-handle-kitchen-faucet-db07eb"
} |
The alternatives according to my book are as follows:
$\begin{array}{ll} 1.&\textrm{15 min}\\ 2.&\textrm{0 min}\\ 3.&\textrm{25 min}\\ 4.&\textrm{60 min}\\ \end{array}$
This part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\textrm{6 p.m}$ the time which would be seen by Vincent will be:
$18\times 50= 900\,s$
which would be $60$ minutes
But Vincent seen in his watch was:
$\textrm{5:45 pm}$
hence until 5 pm would be:
$17\times 50=850\,s$
$56\frac{2}{3}$ minutes
and Vincent's watch would have seen:
$\textrm{5:56:40 pm}$
At this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?
skeeter
Math Team
50 seconds an hour = 5 minutes in 6 hours | {
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"openwebmath_score": 0.41406530141830444,
"tags": null,
"url": "https://mathforums.com/threads/how-can-i-find-the-time-when-two-people-agree-to-meet-if-both-of-their-watches-have-an-offset.348121/"
} |
ros, rosdep, package
Title: satisfying package dependencies
I have package that depends on the willow garage pr2_cockpit package. Is there a way to inform a user that this dependency exists if the package is not already installed (besides looking through the output of rosmake or the package manifest file and checking manually)? I tried adding "ros-diamondback-pr2-cockpit" as an external dependency in the manifest file. However, rosdep install for my package failed with the following error:
rosdep install <package_name>
Failed to find rosdep ros-diamondback-pr2-cockpit for package <package_name> on OS:ubuntu version:10.04
ERROR: ABORTING: Rosdeps [u'ros-diamondback-pr2-cockpit'] could not be resolved
Any suggestions of a structured way to install external ros packages would be nice. Is rosinstall the only other option?
Thanks for any help.
marc
Originally posted by mkillpack on ROS Answers with karma: 340 on 2011-05-03
Post score: 0
At the moment installation instructions usually consist of | {
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If $a$ is a web page, let $V(a)$ be the set of people who have visited $a$. Then $\langle a,b\rangle\in R$ if and only if $V(a)\subseteq V(b)$. Now it’s always true that $V(a)\subseteq V(a)$, even if $V(a)=\varnothing$, so it’s always true that $\langle a,a\rangle\in R$. Thus, $R$ really is reflexive.
Here’s a slightly different way to look at it. If $a$ and $b$ are particular web pages, how could you prove that $\langle a,b\rangle\notin R$? You would have to find someone who has visited $a$ but has not visited $b$. If no one has visited $a$, you can’t do this: you can’t find anyone who has visited $a$, let alone someone who has visited $a$ but not $b$! This shows that if $a$ has never been visited, then $\langle a,b\rangle\in R$ for all $b$ (and hence certainly for the particular case $b=a$).
$R$={$(x,y) | \forall p (F(p,x) \to F(p,y))$}, where $F(p,x)$ means "p has visited x". Is it right? – Freewind Jan 7 '13 at 4:39 | {
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} |
ros, tutorials, beginner
'nodelet_tutorial_math': ['/opt/ros/indigo/share/nodelet_tutorial_math/srv'], 'qt_gui': ['/opt/ros/indigo/share/qt_gui/srv'], 'filters': ['/opt/ros/indigo/share/filters/srv'], 'hector_quadrotor_controller_gazebo': ['/opt/ros/indigo/share/hector_quadrotor_controller_gazebo/srv'], 'hector_map_tools': ['/opt/ros/indigo/share/hector_map_tools/srv'], 'controller_manager_tests': ['/opt/ros/indigo/share/controller_manager_tests/srv'], 'pointcloud_to_laserscan': ['/opt/ros/indigo/share/pointcloud_to_laserscan/srv'], 'smclib': ['/opt/ros/indigo/share/smclib/srv'], 'roslib': ['/opt/ros/indigo/share/roslib/srv'], 'roscpp_serialization': ['/opt/ros/indigo/share/roscpp_serialization/srv'], 'diagnostic_msgs': ['/opt/ros/indigo/share/diagnostic_msgs/srv'], 'rosbuild': ['/opt/ros/indigo/share/rosbuild/srv'], 'rosclean': ['/opt/ros/indigo/share/rosclean/srv'], 'ps3joy': ['/opt/ros/indigo/share/ps3joy/srv'], 'tf': ['/opt/ros/indigo/share/tf/srv'], 'rqt_publisher': | {
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"tags": "ros, tutorials, beginner",
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homework-and-exercises, newtonian-mechanics, orbital-motion, velocity, celestial-mechanics
$$w \mathrel{{+}{=}} v_2 - v$$
$$E = \cos^{-1}{ \frac{e + \cos{v}}{1 + e \cos{v}} }$$
$$M = E - e \sin{E}$$
$$n = \sqrt{ \frac{GM}{a^3} }$$
$$t = \frac{M}{n}$$
$${\rm calculateOrbit}()$$
I noticed if I were to $v = v + \pi$ if ever $v_2 < 0$ and $v > 0$, then the weird orbit/position flip won't occur until the $v = \pi$ Otherwise I do not know what the problem is. I would really appreciate it if someone can point me the way, as I have been scratching my head over this for a long time now. A quick note, your equation for the radius $r$ as a function of the true anomaly is missing the semi-major axis $a$. I prefer to use the symbol $\theta$ for the true anomaly instead of $v$, since I will use that for the total velocity. So:
$$
r(\theta)=\frac{a(1-e^2)}{1+e\cos\theta}
$$
And the way I would calculate my new trajectory parameters after a velocity change is by evaluating the semi-major axis $a$ and eccentricity $e$:
$$
a=\frac{\mu r}{2\mu-rv^2}
$$
$$ | {
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python, python-3.x
places: required number of places after the decimal point
curr: optional currency symbol before the sign (may be blank)
sep: optional grouping separator (comma, period, space, or blank)
dp: decimal point indicator (comma or period)
only specify as blank when places is zero
pos: optional sign for positive numbers: '+', space or blank
neg: optional sign for negative numbers: '-', '(', space or blank
trailneg:optional trailing minus indicator: '-', ')', space or blank
https://docs.python.org/3/library/decimal.html#recipes
"""
q = Decimal(10) ** -places # 2 places --> '0.01'
sign, digits, exp = value.quantize(q).as_tuple()
result = []
digits = list(map(str, digits))
build, next = result.append, digits.pop
if sign:
build(trailneg)
for i in range(places):
build(next() if digits else '0')
if places:
build(dp)
if not digits:
build('0')
i = 0 | {
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of a triangle is the point Time and work word problems. Incenter- Imagine that there are three busy roads that form a triangle. Log in for more information. Similar to a triangle’s perpendicular bisectors, there is one common point where a triangle’s angle bisectors cross. 1. Triangle has , , , and .Let , , and be the orthocenter, incenter, and circumcenter of , respectively.Assume that the area of pentagon is the maximum possible. Let , , for convenience.. Circumcenter of a right triangle is the only center point that lies on the edge of a triangle. The internal bisectors of the three vertical angle of a triangle are concurrent. Recall that the incenter of a triangle is the point where the triangle's three angle bisectors intersect. Only in the equilateral triangle, the incenter, centroid and orthocenter lie at the same point. The point of intersection of angle bisectors of a triangle is called the incenter of the triangle. Remember that each side of the triangle is tangent to the | {
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Recall that the period of sin(kt) is 2π/k
1. 0
2. 0
posted by Steve
2. What I ended up getting was
v(t)=3 sin( pi/3 (t- ((4-pi/2)))) -1
so B= pi/3 and C=4–pi/2
1. 0
2. 0
posted by Stephen
3. I'll have to see where you went wrong. Too bad you didn't show your work, like me. wolframalpha says you are off:
http://www.wolframalpha.com/input/?i=3+sin(+pi%2F3+(t-+((4-pi%2F2))))+-1
1. 0
2. 0
posted by Steve
## Similar Questions
1. ### Calculus
The position of a particle moving on a horizontal line is given by s(t)=2t^3-15t^2+24t-5, where s is measured in feet and t in seconds. a: What is the initial position of the particle? b: What is the average velocity of the
2. ### Math: Calculus
The velocity function is v(t)= t^2-5t+ 4 for a particle moving along a line. The position function s(t) is an antiderivative of the velocity function. Find the displacement, s(6) - s(-3), by the particle during the time interval
3. ### Calc I | {
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opposite angles. It is classified into two types : concave and convex. © Academo.org 2021. • Area of a quadrilateral • Area of a Rectangle by the length of one side and the diagonal • Total square • Math section ( 240 calculators ) local_offer area convex coordinates kite Math parallelogram quadrilateral rectangle rhombus square trapezoid. There is no way that the side of length 100 can fit into the available space. We can apply Brahmagupta formula to find area of inscribed quadrilateral when we know the lengths of all sides of a quadrilateral a, b, c and d. Maths Geometry Area. Quadrilateral Area. Perimeter defines the outer boundary of a geometric shape. There are many cases in which it is useful to calculate the area of a quadrilateral. Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. Side - A side is a line segment which joins any two adjacent vertices. Calculates the area and perimeter of a quadrilateral given four sides and two | {
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"openwebmath_score": 0.7066324949264526,
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"url": "https://dariuszmakowski.com/j2uwnu/e776f9-quadrilateral-area-calculator"
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We also define and give a geometric interpretation for scalar multiplication. Vector Calculus is explained in these GATE 2019 notes. 1 An Introduction to Vectors. Accordingly, you are urged to read (or reread) Chapter 1 of “Proofs and Problems in Calculus” UPSC Mains Mathematics optional Syllabus is available for Download as PDF. There are three types of products used with vectors. Vectors & Scalars in Math Chapter Exam Instructions. Scalar d)The magnitude of the force required to push a pumpkin o an abandoned overpass. CLASSIFICATION OF VECTORS : There is a very important classification of vector in which vectors are divided into two categories. but in mathematics as well. We will see that many questions about vector spaces can be reformulated as questions about arrays of numbers. 1 Matrix — A Mathematical Definition Chapter Three considers algebraic operations on tensors. set V with two operations, like the addition and scalar multiplication from above, and call it a real vector | {
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physiology, brain, metabolism, endocrinology, glucose
When the body starts using the fatty acids for energy production, one problem arises in the brain and that is the blood-brain barrier( barrier that separates the circulating blood from the brain extracellular fluid in the central nervous system )Brain cannot use long chain fatty acids because they cannot cross the blood-brain barrier due to their bondage with albumin. As ketones are medium chain fatty acids which can effectively cross the barrier. So brain uses the ketones for energy production by a process called as ketosis. The ketone bodies are then incorporated into acetyl-CoA and used in the citric acid cycle.
The citric acid cycle – also known as the tricarboxylic acid (TCA)
cycle or the Krebs cycle – is a series of chemical reactions used by
all aerobic organisms to generate energy through the oxidation of
acetate derived from carbohydrates, fats and proteins into carbon
dioxide and chemical energy in the form of adenosine triphosphate
(ATP).
Reference | {
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# If $(a,b,c)=1$, is there $n\in \mathbb Z$ such that $(a,b+nc)=1$?
In the book Lectures on modular forms, one finds the statement at page 8 that
If $(a,b,c)=1$ then there is $n\in \mathbb Z$ such that $(a,b+nc)=1$.
I know that, if $(a,b)=1$, then we can take $n=0$. But, if $(a,b)\not=1$, then what could we do? Further, I tried to look at the linear combinations of $a, b, c$ which are $=1$, but to no avail have I discovered anything. The main difficulty I encountered is that the coefficient of $b$ might not be $1$.
Any hint is well-appreciated. | {
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} |
algorithms, distributed-systems
Title: Algorithm of Communication with Failures I am interested in Distributed Algorithms especially in communication in network with failures.
I look for the proof of the following randomized algorithm of communication in network with failures. For me it seems like very general result in the communication, nevertheless I haven’t found the proof yet.
Algorithm: Initially only vertex $v_0$ has the message, at the end of the algorithm every vertex of the network should have the message.
On every round every vertex that has the message choice the neighbour randomly and sends it the message.
Assumptions: only $f$ failures might happen on the edges between the vertices.
$T = O(\log n)$ - time complexity and the entire network will know the message with high probability, when $f<n/3$, where n - number of vertices. | {
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of the minimum surface potential. The electric field is related to the charge density by the divergence relationship. LaPlace's and Poisson's Equations. Our analysis will be in 2D. Homogenous neumann boundary conditions have been used. Solving the Poisson equation almost always uses the majority of the computational cost in the solution calculation. Two-Dimensional Laplace and Poisson Equations In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. This has known solution. Poisson equation. Furthermore a constant right hand source term is given which equals unity. The homotopy decomposition method, a relatively new analytical method, is used to solve the 2D and 3D Poisson equations and biharmonic equations. Finite Element Solution fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Either approach | {
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"openwebmath_score": 0.814243495464325,
"tags": null,
"url": "http://arcivasto.it/lpmo/2d-poisson-equation.html"
} |
$$=5\sqrt{2}\big[\sin(2x)\cos(-\pi/4)+\sin(-\pi/4)\cos(2x)\big]$$
$$=5\sqrt{2}\sin(2x-\pi/4).$$
(Of course, $-\pi/4$ works the same as $2\pi-\pi/4=7\pi/4$.)
Note that $\cos(7\pi/4) = 1/\sqrt{2}$ and $\sin(7\pi/4) = -1\sqrt{2}$ so your expression $$=5\sqrt{2}(\cos(7\pi/4)\sin 2x +\sin(7\pi/4)\cos(2x) = 5\sqrt{2}\sin(2x+7\pi/4).$$
• Thank you for helping. I have been able to work this problem out to 5sqrt2(1/sqrt2sin2x - 1/sqrt2cos2x), but not sure how to derive result to be in quadrant 4 with sine positive and cosine negative (this would be quadrant 2 and 3pi/4, not 7pi/4). What am I missing? – Michael Servilla Oct 19 '16 at 12:34
• I think I see this now, using the example below from artic tern. Thanks! – Michael Servilla Oct 19 '16 at 12:40
Write
$5(\sin2x-\cos2x) = a \sin(bx+c)$
Then
$5\sin2x-5\cos2x = a \sin(bx)\cos(c)+a \sin(c)\cos(bx)$
Choose $b=2$, $a\cos(c)=5$ and $a\sin(c)=-5$. So | {
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} |
rotational-kinematics, rigid-body-dynamics
Torque is the manifestation of force at a distance
A force $\mathbf{F}$ applied at a distant location $\mathbf{r}$ has an equipollent torque of $$ {\boldsymbol \tau} = \mathbf{r} \times \mathbf{F} $$
Here $\mathbf{r}$ is the position of the force relative to the measuring point.
A force couple is a special case of the above
When two equal and opposite forces act on a body, it is equivalent to a single zero force at infinity with $\mathbf{F} \rightarrow 0$ and $\mathbf{r} \rightarrow \infty$. | {
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group-theory, representation-theory, rotation, lie-algebra
I believe there is a fundamental misunderstanding here, so I'll try to quickly review the story.
I: Representations of SO(3)
The group $SO(3)$ is the rotation group in 3D space. It is defined by its fundamental representation as the $3\times 3$ real, orthogonal matrices with determinant $1$. This representation allows us to act on elements of $\mathbb R^3$ with elements of $SO(3)$ via standard matrix multiplication.
From here, it is natural to ask what effect a rotation has on something which isn't an element of $\mathbb R^3$. This leads us to the representation theory of Lie groups. Given some vector space $V$, we seek a map $\rho: SO(3) \rightarrow GL(V)$ (where $GL(V)$ is the set of invertible linear maps from $V\rightarrow V$) which has the following property:
$$\forall R_1,R_2 \in SO(3), \qquad \rho(R_1\circ R_2)= \rho(R_1)\circ \rho(R_2)$$ | {
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oceanography, plate-tectonics, sea-level, paleontology
When this happens to a rock a glacier picks up, it is called a "dropstone".
During glacial maxima, icebergs can survive much farther south than now. The presence of dropstones in what were supposedly equatorial sediments led to the theory of a Snowball Earth a billion or so years ago.
Based upon CO2 data, 100,000 years ago, the Earth was working its way into the most recent glacial maximum, called the Wisconsin glaciation in North America ( the likely source of this iceberg).
Source: Wikimedia Commons, Author Tom Ruen, CC BY-SA 3.0
Notice that the article did not call the tusk a "fossil". You can only call it that after a specific process of mineralization after being buried in sediment. They might very well have found original ivory from the mammoth.
But this is not necessary. There are two ways the tusk got into the glacier: | {
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r, python, bioconductor, structural-variation, simulated-data
Again. all the examples above have random chrA and chrB. Setting these specifically, only seems possible with the GRanges and IRanges, for which I tried tons of versions already.
It should be a very simple thing according to the manual, but actually doing it doesn't seem to get the desired results, while my version of Insertions is just the normal one in my opinion? (why would you want to create extra DELS or DUPS while making insertions?).
On their own support page, it is only possible to ask a question on the forums, but they never answer it (only look at it) Answer eventually found on the BioConductor post:
https://support.bioconductor.org/p/114354/#114585
It basically comes down to not using RSVSim, but other R packages.
Final code snippit:
genome = DNAStringSet(
c(chr1 = "AAAAAAAAAAAAAAAAAAAATTTTTTTTTTTTTTTTTTTT",
chr2 = "GGGGGGGGGGGGGGGGGGGGCCCCCCCCCCCCCCCCCCCC")) | {
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ros, rviz, ros-melodic, translation
Rotate the "stereo_camera" frame by +90 around world's X axis (so that the pattern would overlay with the "calibration_pattern" frame).
Now, having spotted these, I tried on Python to apply these rotations and then using static tf publisher to visualize on RViz whether or not it was correct. But no matter what I tried, I never got the correct rotations applied. Somehow what I see as "X" is not really "X" for Rviz apparently, so when I rotate around a particular axis, some strange stuff happens. One wastes his time keep on rotating.
Is there a tool on ROS, like a feature of RViz maybe I don't know, or a way to make this easy transformation hassle-free? | {
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After Year 2 17393.75 17393.75
GREAT JOB, XxmathguyxX !!!!!
P.S - you can take a breath now.....
Feb 18, 2022
edited by CPhill Feb 18, 2022
#7
+360
+1
THANK YOU CPHILL for adding proof to my solution. Also thanks to @Builderboi for providing an answer. Thanks ya'll!
Feb 18, 2022
#8
+360
+1
finally i can breath....
i've been holding my breath for 10 minutes now
Feb 18, 2022
#9
0
wie bite kanst du das auf deutsch schreiben
Feb 25, 2022
#10
0
arschloch
Feb 25, 2022 | {
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computability, closure-properties, semi-decidability
Title: A and B are Turing recognizable, is A - B Turing recognizable? If A and B are Turing recognizable, is A - B Turing recognizable?
I think that A - B would be Turing recognizable because they're both in the space of Turing recognizability. For example, if A is context free and B is a regular language A - B would result in a language that is sill Turing recognizable.
However, does this become a question about emptiness? Two Turing recognizable languages equal to each other leave an empty set. Is the empty set Turing recognizable? I would still say yes.
Not sure if I'm thinking about this correctly... Suppose that $L$ is a Turing-recognizable language and $L^C$ is its complement. Let's assume, for the purposes proof by contradiction, that $L^C$ is also Turing-recognizable. This means that a recognizer exists for each of these two languages: we will call them $M_L$ and $M_{L^C}$. How might we use these two recognizers to do something absurd? | {
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python, python-2.x
main() – I've run some test cases to document how the different call scenarios could look like, which I hope is not too far from how you are going to call your code. The output is:
force_buyant = 1800
Error: Missing argument in force_buyant - volume
Error: Too many missing arguments - density, volume
force_friction = 200
force_friction_v2 = 200
force_friction_v3 = 200
Error: Invalid equation id - 10
force_friction_v3 = 200
force_friction_v3 = 1800
TypeError - force_friction_v3() takes at least 1 argument (2 given)
force_centripetal = 133.333333333
TypeError - force_centripetal() takes exactly 3 arguments (2 given) | {
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c#, design-patterns
Title: Design issues, tendency to write custom methods I seem to have this tendency to write custom methods, i.e. code that isn't reusable. The following seems to be one example of trying to do too much in a single method.
But the point I'm not sure about is, in order to split it up, I'd have to redo the for loop each time. Is that acceptable?
Any general pointers on how this should be structured would really be appreciated.
public static void ShowInputs(DataTable dtSProcList, List<Label> inputLabels,
ComboBox cbSProcList, List<TextBox> inputTxtboxes,
List<TextBox> selectedSProcInputs,
List<String> selectedSprocDataTypes, List<string> selectedSprocParamterNames)
{
int counter = 0; | {
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# In how many ways can we split 6 boys and 6 girls to 6 tables such that in every table there's a boy and girl (only 2 seats per table)
In how many ways can we split 6 boys and 6 girls to 6 tables such that in every table there's a boy and girl (only 2 seats per table)?
I saw this post and I think it's kind of the same of my question
In how many ways if no two people of the same sex are allowed to sit together?
And the answer is $$2*(6!)^2$$
But then I saw this one (e) -
Discrete Mechatronics - sequences with repeats and no repeats
Which says the answer is $$2^6*(6!)^2$$
I would like to know which one is correct in my case and why? (like what are the differences) | {
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Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
`'red'``'r'``[1 0 0]``'#FF0000'`
`'green'``'g'``[0 1 0]``'#00FF00'`
`'blue'``'b'``[0 0 1]``'#0000FF'`
`'cyan'` `'c'``[0 1 1]``'#00FFFF'`
`'magenta'``'m'``[1 0 1]``'#FF00FF'`
`'yellow'``'y'``[1 1 0]``'#FFFF00'`
`'black'``'k'``[0 0 0]``'#000000'`
`'white'``'w'``[1 1 1]``'#FFFFFF'`
`'none'`Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
`[0 0.4470 0.7410]``'#0072BD'`
`[0.8500 0.3250 0.0980]``'#D95319'`
`[0.9290 0.6940 0.1250]``'#EDB120'`
`[0.4940 0.1840 0.5560]``'#7E2F8E'`
`[0.4660 0.6740 0.1880]``'#77AC30'`
`[0.3010 0.7450 0.9330]``'#4DBEEE'`
`[0.6350 0.0780 0.1840]``'#A2142F'` | {
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quantum-mechanics, electromagnetic-radiation, visible-light, temperature, blackbody
You can calculate spectral radiance. Spectral radiance is like an (spectral radiance isn't fully intensity, it's just like an intensity) intensity of emitted light with frequency $f$ from object with temperature $T$ (in kelvins) of emitted light from an object with temperature $T$ (in kelvins) using this equation:$$\beta(T) = \frac{2hf^3}{c^2}\frac{1}{e^{\frac{hf}{kT}}-1}$$ Earth is blue because it reflects light, not emits it. But the Sun doesn't reflect light (it's a blackbody, but not perfect), it just emits it, that's why it's hotter. | {
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opposite sides must be congruent and it must satisfy all other properties of parallelograms ≅ ΔCDA // Side-Angle-Side.... About it so that we will rely on the properties of parallelograms, there. From scratch, repeating the proofs we did for parallelograms, but there ’ no! Service and Privacy Policy 같고 서로 다른 것을 이등분한다 upvote ( 17 ) how are. Star below, SA =5, what is the value of x in rectangle STAR below 'd like... The dimensio … n of the other two sides dot product of vectors, that. Square are perpendicular are the two diagonals of a rectangle equal why each other in a parallelogram are equal. 1 square. Since a rectangle Share it on Facebook Twitter Email can do the same rectangle alwayssometimesnever Write are the two diagonals of a rectangle equal why column-proof... In Computer Engineering and an MBA degree 대각선은 길이가 같고 서로 다른 것을 이등분한다 explain to. ) ASK a question of parallelograms from scratch, repeating the proofs we did parallelograms! The rectangle along diagonal NL, we will show | {
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This is autonomous...
For $$y > 1$$ or $$y < -1$$ we get $$y = \frac{-1}{\tanh \frac{x}{2}}$$ which blow up in finite time, one in each direction. Also translates.
For $$-1 < y < 1$$ $$y = - \tanh \frac{x}{2}$$
• @thanks a lot for your answer and the picture...that makes things more clear... – Isham Sep 23 '18 at 1:01 | {
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rotational-kinematics, relative-motion
I've gone around in circles trying to decide if one would actually move. This top answer seems to think you would: Earth moves how much under my feet when I jump?
Any insight would be appreciated. Thank you! Assume that you jump straight up, standing on the equator.
As soon as your feet leave the ground, you are in a highly elliptical orbit around the center of the earth. At that point you have the same angular velocity as the point you jump from. As you rise toward your one and only apogee, conservation of angular momentum requires that your angular velocity decrease very slightly, then increase again as you drop. Of course, the orbital motion will stop when you hit the ground again. Overall, the ground will have rotated slightly further. I doubt that the effect could ever be measured.
At the poles, you land where you jumped from. Mid-latitude jumps are left as an exercise for the reader... | {
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radio-astronomy, photography, radio-telescope
In the case of single dish instruments, are the elements - roughly speaking - used as pixels? Despite the longer wavelength, it's still optics and it is a telescope. If there are N individual, uncoupled feeds, does one build up an image roughly N times faster? Is the relative phase between the feeds ever used (for single dish instruments)? | {
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When we study logic, is it not true that we impose certain axioms and rules that we follow, and from those conclude other things? We don't just say "Once there was nothing, and then.... LOGIC!"
However, in regards to the question you're asking... I understand this from a historical point of view moreso than I know about the mathematical details, but Hilbert tried to prove the consistency of mathematics using nothing but the axioms of mathematics in the early 30s (as part of the whole constructivist vs intuitionist math war that was all the rage). It was a big thing at the time, but broke down after Godel's incompleteness theorems showed that it is impossible to prove the consistency of any formal system from the interior of that system. And I think this is where your example "This sentence is false" falls under. You can read more at http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems | {
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Because v (the velocity of S' is 0.6c, gamma = 1/SQRT[1 - 0.6^2] = 1/0.8 = 1.25
Using Lorentz for distance, x'_2 = gamma[(x_2 - vt_2)) = 0.8[7.2 - 9*0.6) = 2.25
Thus x'_2 - x'_1 = 2.25 - 0 = 2.25. This is less than the 7.2 lt-yr difference cited above for x_2 - x_1. I've set t_1 = t'_1 = 0 and t_1 = t'_1 (because we used the same instantaneous measurement of the x_2 - x_1 as you cited in one of your explanations.) I made no such assumption for t'_2.
Clearly, t'_2 = gamma[t_2 - vx_2/c^2] = 1.25 (9 - 0.6*7.2) = 5.85 years. This not = to t'_1 of 0.
5.85 - 0 = 5.85 or t'_2 - t'_1 which is not 9 or t_2 - t_1.
But, there is another question associated with this:
Assume a rod is of sufficient length that at 0.8c moving to the right, it measures 7.2 lt-yr. in the S F.O.R. This rod would have to be 12 lt-yr in length in the S' F.O.R. for it to do this.
It appears that this rod is longer in S' (12) or x'_2 - x'_1 than in S (7.2) or x_2 - x_1.
What gives?
H-E-L-P-! | {
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c++, linked-list, homework
When passing values here:
void create_contact(std::string first, std::string last, std::string cellNumber)
You are passing the parameters by value. This means a copy is passed to the function. It is more normal to pass values by const reference to avoid the copy.
void create_contact(std::string const& first, std::string const& last, std::string const& cellNumber)
You should simplify your code with a constructor (or use the new list initializer).
node *temp = new node;
temp->firstName = first;
temp->lastName = last;
temp->phoneNumber = cellNumber;
temp->next = NULL;
Easier to write as:
Node* node = new Node(first, last, cellNumber, nullptr);
Or you can use the list initialization if you don't want to write a constructor:
Node* node = new Node{first, last, cellNumber, nullptr}; | {
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organic-chemistry, reaction-mechanism, amines
But how exactly does the reaction proceed?
The first step (what I think) could be that $\ce{NOCl}$ dissociates into $\ce{NO+}$ and $\ce{Cl-}$ ions and then the $\ce{Cl-}$ ion attacks the amine group and replaces it by $\mathrm{S_N2}$ mechanism.
But according to paperplane.io (via the Internet Archive), a diazonium product is produced. Even though a diazo product is formed, how does it produce an alkyl halide according to my textbook? Why would the chloride ion attack the electron rich amine? The amine is the better nucleophile in the system, and will attack the $\ce{NO+}$. This gives a $\ce{RNH-NO}$ nitroso species which is then protonated on oxygen, loses water, and gives the diazonium cation. The resultant diazonium cation loses nitrogen rapidly and get attacked by chloride to give the primary alkyl chloride.
More can be read about this process here (which is where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine" | {
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c, library, bloom-filter
H_a(data, len, hash);
for(i = 0; i < bf->nhashes; ++i)
if(!TEST_BIT(bf, G(hash, i, data, len, bf->mbits)))
return false;
return true;
}
hashes.h
/* Murmur hash implementation: modified from - https://github.com/PeterScott */
/* FNV1A hash implementation: modified from - http://encode.ru/threads/612-Fastest-decompressor!?p=22184&viewfull=1#post22184 */
#ifndef HASHES_H
#define HASHES_H
#ifdef __GNUC__
#define FORCE_INLINE __attribute__((always_inline)) inline
#else
#define FORCE_INLINE inline
#endif
#include <stdint.h>
static FORCE_INLINE uint32_t rotl32(uint32_t x, int8_t r)
{
return (x << r) | (x >> (32 - r));
}
#define ROTL32(x,y) rotl32(x,y)
#define getblock(p, i) (p[i])
static FORCE_INLINE uint32_t fmix32(uint32_t h)
{
h ^= h >> 16;
h *= 0x85ebca6b;
h ^= h >> 13;
h *= 0xc2b2ae35;
h ^= h >> 16;
return h;
} | {
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c++, catkin, roscore, rosrun, ros-groovy
Comment by lindzey on 2013-12-04:
please edit your original question to include this information, formatted nicely.
Comment by Eman on 2013-12-05:
Thanks for your help . God bless you amen
Ok , I will edit my original question to include this information, formatted nicely. | {
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quantum-field-theory, condensed-matter, mass, definition
What I am confused is that there are some quadratic dispersion but gapless, e.g. magnon of FM, but some books argue it is massive excitation, which is different from massless excitation of AFM. This kind of dispersion is corresponding to another form of Green function:
$G\sim\frac{1}{-\omega +k^2}$.
So I want to know what the definition of "massive excitation" in the CMP?
By the way, I am also confused the hidden reason of such distinction between magnon dispersion of AFM and FM, i.e. linear v.s. quadratic at small $k$. The term "massless" originates in relativistic quantum field theory, where mass-energy equivalence means that an energy gap in the Hamiltonian's spectrum is equivalent to a mass for the corresponding particle. | {
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php, wordpress, twitter-bootstrap
<?php endforeach; ?>
</ol>
<div class="carousel-inner">
<?php
// Get posts (tweak args as needed)
$i=0;
$incNum = 0;
$args = array(
'post_type' => 'post',
'posts_per_page' => -1,
'orderby' => "date",
'order' => "desc"
);
$posts = get_posts( $args );
?>
<?php foreach (array_chunk($posts, 4, true) as $posts) : ?>
<div class="item <?php if ($i==0){echo 'active';}?>">
<div class="row">
<?php foreach( $posts as $post ) : setup_postdata($post); ?> | {
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"tags": "php, wordpress, twitter-bootstrap",
"url": null
} |
machine-learning, deep-learning, cnn, word2vec, overfitting
Epoch 00005: val_loss did not improve
Epoch 00005: ReduceLROnPlateau reducing learning rate to 1.0000000656873453e-06.
Epoch 6/10
254/253 [==============================] - 443s 2s/step - loss: 0.7099 - acc: 0.9560 - val_loss: 0.4127 - val_acc: 0.9744
Epoch 00006: val_loss did not improve
Epoch 00006: ReduceLROnPlateau reducing learning rate to 1.0000001111620805e-07.
Epoch 7/10
254/253 [==============================] - 443s 2s/step - loss: 0.7099 - acc: 0.9560 - val_loss: 0.4127 - val_acc: 0.9744
Epoch 00007: val_loss did not improve
Epoch 00007: ReduceLROnPlateau reducing learning rate to 1e-07.
Epoch 8/10
254/253 [==============================] - 443s 2s/step - loss: 0.7099 - acc: 0.9560 - val_loss: 0.4127 - val_acc: 0.9744
Epoch 00008: val_loss did not improve
Epoch 00008: ReduceLROnPlateau reducing learning rate to 1e-07.
Epoch 9/10
254/253 [==============================] - 444s 2s/step - loss: 0.7099 - acc: 0.9560 - val_loss: 0.4127 - val_acc: 0.9744 | {
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• I see what you are saying, but I would characterize the root cause as $a_n$ is not less than 1 as $n \rightarrow \infty$, rather it converges to 1. – Shree Mar 29 '18 at 6:10
• @Shree I don't understand your comment. Explain why you think $1 - 1/n^2 \not\lt 1$ for $n \ge 2$? As I have defined it, $a_n = 1 - 1/n^2$ is always strictly between $0$ and $1$ for $n \ge 2$, and thus furnishes an example of a sequence of numbers that are all less than $1$, yet their product is not zero--i.e., a counterexample to your flawed reasoning that an infinite product of numbers less than $1$ would somehow "eventually" tend to $0$. – heropup Mar 29 '18 at 6:13 | {
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mvc, objective-c, calculator, finance
Title: Model class representing a financial calculator Here is my Model class's new header file:
@interface MMCalculator : NSObject
@property (readonly) CGFloat calculatedPay;
@property (readonly) CGFloat calculatedSavingsForStuff;
@property (readonly) CGFloat calculatedSavingsForProfitFormula;
@property (readonly) CGFloat calculatedSavingsForTaxes;
- (void)calculateValuesWithMonthlyRevenue:(CGFloat)monthlyRevenue;
@end
Here is my Model class's new implementation file:
#import "MMCalculator.h"
#pragma mark - Const Variables
static CGFloat const kPercentageToPayYourself = 0.50;
static CGFloat const kPercentageToSaveForStuff = 0.20;
static CGFloat const kPercentageToSaveForProfitFormula = 0.20;
static CGFloat const kPercentageToSaveForTaxes = 0.10;
@interface MMCalculator () | {
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quantum-mechanics, quantum-electrodynamics, superconductivity, quantum-computer
In this sense, the transmon is very unlike the charge qubit, which has a wide wave function in the flux basis.
So in the end, while the transmon and charge qbits share a certain theoretical similarity in the form of their Hamiltonians, for all practical purposes the transmon is actually more like a flux qubit with a large $C$ and biased so that the flux qubit only has one potential well.
Note that the width of the wave function in the flux basis decreases as we increase $C$.
The whole reasons the transmon was invented was that narrowing the wave function by increasing $C$ leads to less sensitivity to charge noise. | {
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• you should use the complex analysis notation : $\int_0^\infty e^{i x^2}dx = \lim_{a \to 1,Im(a) > 0} \lim_{R \to \infty}\int_0^{a R}e^{i z^2}dz$ and $\int_0^{a R}+\int_{a R}^{e^{i \pi/4}R } +\int_{e^{i \pi/4}R}^0 e^{i z^2}dz = 0$ but $\lim_{R \to \infty}\int_{a R}^{e^{i \pi/4}R } e^{i z^2}dz = 0$ while $\lim_{R \to \infty}\int_{e^{i \pi/4}R}^0 e^{i z^2}dz = -e^{i \pi/4} \int_0^\infty e^{-x^2}dx$ – reuns Oct 9 '16 at 5:41
• Great contribution, thanks :) @Felix Marin – Tom Oct 9 '16 at 5:41
• @Tom It's nice it was useful for you. It's a standard evaluation you can check in Wikipedia ( I gave the link in the text ). Thanks. $\left(\bullet\,\,\,\,\, \bullet\atop {\mid \atop \smile}\right)$. – Felix Marin Oct 9 '16 at 5:45 | {
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c++, performance, beginner
The usual advice is to pick a style and use it consistently. I'd add one item to that though: unless you have a really good reason to do otherwise, pick one of the styles that's already reasonably well know and widely used (e.g., K&R, OTBS, etc.)
Standard containers
Right now, your a is basically acting as a set. That it, it simply records what numbers are present/absent in a range. The standard library already provides std::set (and std::unordered_set, though it's probably not the best choice here) specifically for that purpose. It might be worth at least considering using one of these.
Simplification
Since we care about the numbers that are missing (rather than the ones that are present), we probably want to invert the sense of our set. Rather than inserting numbers as we encounter them, then searching for what's missing, we can start by inserting all the numbers from 100 to 999, then remove the numbers as we read them from the file. | {
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machine-learning, data-mining, nlp, text-mining, similarity
LDA - Is also a technique used for topic modeling, but it's different from LSA in that it actually learns internal representations that tend to be more smooth and intuitive. In general, the results you get from LDA are better for modeling document similarity than LSA, but not quite as good for learning how to discriminate strongly between topics.
Pachinko Allocation - Is a really neat extension on top of LDA. In general, this is just a significantly improved version of LDA, with the only downside being that it takes a bit longer to train and open-source implementations are a little harder to come by | {
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php, security, database, authentication
}
?> Ok, just a few quick thoughts, will be adding to this answer later today (ie after work):
One class, One task
Chances are, you've already heard of the SOLID principles. Well, here they are again: your code is in violation of the SRP (Single Responsability principle). A class has one task. No more, no less. Your class does 4 things: it stores data (all those public properties), messes with sessions, connects to the DB and performs queries.
The Database connection and queries are all things that should be dealt with in another class, that simply connects to the DB and executes queries.
This other class could then have a method like this:
public function login(User $user)
{
$query = $this->db->prepare("SELECT username, password FROM blog_users WHERE username = ? AND password = ? LIMIT 1");
$query->bind_param('ss', $user->username, $user->password);
$query->execute();
$query->bind_result($user->username, $user->password);
$query->store_result(); | {
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in the array is, by definition, sorted. n 6, Along with the binary, the science of computers employs bases 8 and 16 for it's very easy to convert between the three while using bases 8 and 16 shortens considerably number representations. {\displaystyle X} , we have: Here, {\displaystyle n} 5 The CooleyâTukey algorithm of FFT is a A Split and combine algorithm. x {\displaystyle O(n)} : But from this division into smaller parts, it's not clear how we can multiply these parts such that we can combine the results for the solution to the main problem. ( Here, the base case is when the array only has one element. However, in our above algorithm we've been using four multiplications total. <= Thus sooner or later the first argument will be less than the second and the process will start emerging from the recursion, still a step at a time. Here, we loop n times, and in the worst-case, we make n calls to add. I would like to have a function say called Conversion that takes two arguments M and | {
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javascript, jquery, html, css, game
// Create new row
tr = $("<tr/>").addClass("nonogram_row");
// Create definition for the current row
$("<td/>")
.attr("id", this._idOfRowDefinition(y))
.addClass("nonogram_definition nonogram_row_definition")
.html(this._rowDefinitionToHTML(this._model.getRowDefinition(y)))
.appendTo(tr);
for (x = 0; x < width; x++) {
// Separate groups of five columns
if (x && x % 5 === 0) {
$("<td/>")
.addClass("nonogram_separation_column")
.appendTo(tr);
} | {
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mechanical-engineering, control-engineering, pneumatic
Figure 1. A very nice but expensive SMC electric pressure regulator.
If you add this in before your solenoid valve you will have the same pressure on advance and retract and this will ease your speed / cushioning control. The alternative is to place it on the actuator "advance" line that requires regulation but that means that actuator retraction will be at higher pressure and may make speed control difficult with a noticeable initial jerk on moving home.
If you place it before the valve then bear in mind that the valve may require 1 to 2 bar minimum pressure to move the spool in the valve. If you need to pass low pressures through the valve then you'll probably need a pilot operated valve. | {
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momentum, mass-energy, explosions, asteroids
Super-heated compressed air will be pushed at hypersonic (or orbital) speeds - fast enough and hot enough to cut through rock- and contact with the ground will squirt material out at those speeds as well. Molten white hot material mixed with cooler material pushed by it and it will travel at hypersonic speeds for great distances with some going into space and coming down anywhere on Earth. The atmospheric shockwave will be supersonic for a long ways then become the speed of sound. Note the speed of sound is determined by temperature, not density. Compression by the shock wave will have some strange heating effects.
All in all, a very bad day but quite something to see from a safe distance. By the way, it would looke rather slow. Hollywood explosions of this sort move faster than light if you do thumbnail calculations. | {
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# If I know the order of every element in a group, do I know the group?
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one. | {
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organic-chemistry, hydrolysis
Title: Relative rate of hydrolysis of the carbon chlorine bond in thioethers and ethers I am solving problems based on rates of hydrolysis of compounds but I don't have a general approach to such questions.
Which of the following compounds undergoes hydrolysis faster than the second under comparable conditions? It may be acid catalysed or base catalysed hydrolysis but the conditions must remain the same for each pair.
In my view since oxygen is more electronegative than sulfur, hence hydrolysis should be easier for second. But I am not sure about it.
see part 2 here
see part 3 here This is a classic case of neighboring group participation. The hydrolysis of the thioether is significantly faster because the sulfur can displace the chlorine to give a three member ring with a sulfur in it. The ring is quite reactive and will easily be opened by water. | {
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simulation
$$
48\times 2^n\leq 4\times 10^9
$$
Rearrange for $n$ and you have $n\leq26$ qubits.
Note that applying gates in a quantum circuit is relatively inexpensive memory-wise. See the "Efficiency Improvements" section in this answer. From that strategy, one should be able to estimate the time it takes to apply a single one- or two-qubit gate to an $n$-qubit system, and hence how many gates you might expect to fit within some times limit (an hour is very modest, but would certainly serve for illustrative purposes). | {
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the-moon, earth, near-earth-object
Title: Does the Earth have another moon? I was just wondering what are the chances that there is a small object (say less than 1 km but more than few meters) that orbits the Earth but has remained undetected by us? Are we actually scanning the space around the Earth continuously for orbiting bodies? Not strictly satellites/moons, but certainly companions are 2010 TK7 with a diameter of ~300 m, an Earth trojan at the L4 point, and the ~5 km 3753 Cruithne in a peculiar orbit locked to the Earth's. | {
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$f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).$
Note by Calvin Lin
7 years, 5 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. | {
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displayed will be "2" because 23/3 = 7 with a remainder (modulo) of 2. h” header files support all the arithmetic. print "107 % 4 = ", print 107 % 4 Related examples in the same category. 11 mod 4 = 3, because 11 divides by 4 (twice), with 3 remaining. Explicitly, if we write x ∈ ℝ in polar form, x = r e i ϕ , r > 0 , ϕ ∈ [ 0 , 2 π ) , then ϕ = 0 or ϕ = π , so e i ϕ = ± 1. The quotient is then, 6 + 3 i 10 + 8 i = ( 6 + 3 i) ( 10 − 8 i) 164 = 60 − 48 i + 30 i − 24 i 2 164 = 21 41 − 9 82 i. Divisor is the numeric expression to divide the dividend. Seach Every Job, Everywhere with Adzuna. But when the modulus is 5, the inverse of 2 is 3. $\endgroup$ – Ritam Bhaumik Nov 23 '16 at 14:04 $\begingroup$ @fgrieu: Yes, well, that distinction is a matter of convention. Of course, one can do this with any number as modulus, not just 12. With some bits of maths, the remainder is the important bit, and if you take a MODULUS, that means you're getting the remainder. Choosing a math program is | {
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quantum-mechanics, hamiltonian, quantum-measurements
$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & 0 \\ 0 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$
$\psi_1$ and $\psi_2$ are the observer and observed system states respectively, and may have many dimensions; even infinitely many. The matrix $H$ should be thought of as a block diagonal matrix. The zero blocks top-right and bottom-left in $H$ express the fact that the two systems do not interact. Each can vary completely independently of the other.
Now we modify the equation to introduce a weak interaction between the systems. The off-diagonal terms are set to non-zero values, expressing how each system affects the other. This represents the measurement process, and the details of the constant $\epsilon$ values define what sort of measurement it is.
$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & \epsilon_1 \\ \epsilon_2 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$ | {
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mobile-robot, kinematics, motion, two-wheeled, forward-kinematics
Title: Calculate position of differential drive robot How do you calculate or update the position of a differential drive robot with incremental sensors?
There is one incremental sensor attatched to each of the two differential wheels. Both sensors determine the distance $\Delta left$ resp. $\Delta right$ their wheel has rolled during a known time $\Delta t$.
First, let's assume the center between both wheels marks the position of the robot. In this case, one could calculate the position as:
$$
x = \frac{x_{left}+x_{right}}{2} \\
y = \frac{y_{left}+y_{right}}{2}
$$
"Deriving" those equations under the assumption that both wheels rolled in a straight line (which should be approximately correct for small distances) I get:
$$
\frac{\Delta x}{\Delta t} = \frac{1}{2}\left( \frac{\Delta left}{\Delta t} + \frac{\Delta right}{\Delta t}\right)cos(\theta) \\
\frac{\Delta y}{\Delta t} = \frac{1}{2}\left( \frac{\Delta left}{\Delta t} + \frac{\Delta right}{\Delta t}\right)sin(\theta)
$$ | {
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I think that you're missing boundary conditions. Anytime you want to solve partial derivative equations, you have to provide boundary conditions, or the solution is undefined.
To get a standing wave, the most commun situation is to use "strict boundary conditions", where the wave's amplitude has to be zero on both extremities. If you take $$y(x,t)=2A\cos(\omega t)\cos(kx)$$ with $$y(0,t)=y(l,t)=0$$, you naturally get the quantization condition similar to the one you mentioned.
Another possibility is "periodic boundary conditions": the wave's amplitude must be equals on both ends, but not necessarily zero.
So far it's mostly math, so choosing the proper boundary conditions is where physics starts. | {
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c, embedded
You need some manner of semaphores from protecting the caller from reading part of the data, then get interrupted in the middle of it. I like to provide these as a feature in the ring buffer ADT itself, making that one intrinsically interrupt safe.
Alternatively, you could temporary disable interrupts in the caller while you grab the data, but that only works if the caller can do this in less time than it takes for the serial bus to send another byte.
Usually this is done by providing double-buffering (no matter if you have a ring buffer or a linear one). You have one software buffer where the incoming data is getting written, and another buffer which contains the latest completely received data. When the ISR is done receiving, it only swaps the pointers between these two buffers. | {
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ros, moveit, kuka
Originally posted by gvdhoorn with karma: 86574 on 2021-02-17
This answer was ACCEPTED on the original site
Post score: 3
Original comments
Comment by gvdhoorn on 2021-02-17:
I would also recommend not to convert the files to URDF, but keep working with the .xacros directly. Xacro is much more flexible, as it allows to combine different xacro fragments in an easy way. URDF cannot be combined with anything else.
Comment by Ranjit Kathiriya on 2021-02-18:
Thanks for the help! and once again sorry was not knowing about the ticking. Thanks for helping me
Comment by gvdhoorn on 2021-02-18:
No need to apologise. | {
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programming, qiskit, cirq
I'm going to change $\sin$ and $\cos$ from your state. You can fix this by changing the angles with appropriate angle redefinitions. $\theta \to \frac{\pi}{2}-\theta$.
$$
| \psi_1 \rangle = (\mathrm{R}(\theta_1) \otimes I_4) | 0 0 0 \rangle = \cos \theta_1 | 0 0 0 \rangle + \sin \theta_1 | 1 0 0 \rangle\\
| \psi_2 \rangle = (\mathrm{CR}(0,\theta_2) \otimes I_2) | \psi_1 \rangle = \cos \theta_1 \cos \theta_2 | 0 0 0 \rangle + \cos \theta_1 \sin \theta_2 | 0 1 0 \rangle + \sin \theta_1 | 1 0 0 \rangle\\
| \psi \rangle = \mathrm{CCR}(00,\theta_3) | \psi_2 \rangle
$$
where $\mathrm{R}(\theta)$ is to indicate a 2 by 2 rotation matrix.
$\mathrm{CR}(0,\theta)$ is to indicate controlled $\mathrm{R}(\theta)$ on the second index but controlled on 0 instead of 1 on the first.
$\mathrm{CCR}(00,\theta)$ is to indicate controlled $\mathrm{R}(\theta)$ on the third index but controlled on 00 instead of 11 on the first two. | {
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is found by adding up the two Fibonacci numbers before it. Now, this expression is fairly easy to understand and quite sufficient to produce any Fibonacci number by plugging the required value of $n$. By taking out a factor of $x$ from the second expansion, we get, $$\left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right ) + x \left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right).$$, Using the definition of $F(x)$, this can finally be written as. This pattern turned out to have an interest and … Our job is to find an explicit form of the function, $F(x)$, such that the coefficients, $F_n$ are the Fibonacci numbers. I know that the relationship is that the "sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term", but I don't think that is worded right? If you got 4 correct answers: You made it! ( Using power of the matrix {{1,1},{1,0}} ) This another O(n) which relies on the fact that if we n times … If you got between 0 and 1 correct answer: You can do it next | {
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electric-fields
Title: Electric Field Around a Biased Material I feel like this is a fairly straight forward question, but I cant seem to come to a good understanding.
Imagine I have a very large metal box. Inside that box I have a suspended metal plate. I have some voltage generator in which I attach the hot lead to the suspended plate and the ground lead to the large metal box. If I set the voltage generator to 100V, what is the electric field around the plate? Also how can you set the surface of a material to a set voltage like that? My understanding was as you got closer and closer to point charges your potential diverges? | {
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rviz, moveit
Originally posted by danielsnider with karma: 203 on 2017-05-10
This answer was ACCEPTED on the original site
Post score: 1 | {
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newtonian-mechanics, forces, reference-frames, flow
Title: How can the center of mass accelerate in this case? I have asked this question earlier and got some conflicting answers in the beginning but then got the answer that the friction force is what accelerating the center of mass.
Here's the question
Take the example of a man (say of mass 50kg) and a tank (of negligible mass) filled with water kept at some distance (whose total weight is also about 50 kg). Now suppose there is a pipe of uniform cross section (whose mass can be ignored) connecting the mouth of the man to an opening at the bottom of the tank. Also the surface on which they are standing is sufficiently rough (so that they don't move).
At $t=0$ the system is released from rest. So the center of mass of the system is exactly in the midway at $t=0$ (represented by the dashed line). But as water comes out, the center of mass of the system shifts towards the man . This means that the center of mass accelerated from its initial position both horizontally and vertically. | {
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algorithms, string-matching
if $P'[i]==0$,
if $i== 1 $ or $P'[i-1]==0$, let $P''[i] = (555555, -1)$
otherwise, let $P''[i] = (55555, -2)$
Otherwise,
if there is no nonzero element before index $i$ in $P'$, let $P''[i] = (55555, -3)$.
else, suppose $j$ is the index of last such element. Let $P''[i] = (P'[i]/P'[j], i-j)$
When $P''[i]$ is one of $(55555,-1)$, $(55555,-2)$ and $(55555, -3)$, the second parts, $-1, -2, -3$ carries all the information in $P''[i]$. The first parts, $55555$, are padded so that $P''[i]$ is always a pair of numbers for all $i$.
Create $T''$ from $T'$ in the same way.
What does "match" mean?
How can we apply/tweak KMP algorithm to process $P''$ and $T''$ so that we will find all affine transformation of $T'$ in $P'$? The key point is to customize the meaning of "$P''[i]$ matches $T''[j]$" to suit our purpose.
They match if | {
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"tags": "algorithms, string-matching",
"url": null
} |
newtonian-mechanics, momentum, conservation-laws, collision
Title: Can the velocity of the center of mass of two spheres change after a collision? I'm curious as to whether or not the velocity of the center of mass of a system comprised of two spheres can change after the two spheres collide. Looking at the equation for the velocity of the center of mass for a system of particles:
$$V_\text{CM} = \frac{m_1v_1 + m_2v_2 + \cdots + m_nv_n}{m_1 + m_2 + \cdots + m_n}$$
It looks like, if after the collision, one of the particles changes direction, and the negative terms outweigh the positive terms in the numerator, the velocity of the center of mass would change direction (and possibly magnitude). However, I can't think of any examples where this would happen in elastic or inelastic collisions. I'm not even sure if its speed can change. It would make sense that it would if kinetic energy (velocity) was lost in an inelastic collision, however, I can't come up with any conditions to make this happen. | {
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quantum-mechanics, operators, hilbert-space, group-representations, unitarity
in fixing the map $g \mapsto U_g$ concerns only the arbitrary phase and not the charcter of the operator $U_g$.
A map satisfying (a) and (b) (with a precise choice of the operators $U_g$) is called unitary projective representation of $G$.
In some cases, especially when $G$ is equipped with further structures as the Lie-group one, it is possible to redefine the operators $U_g$ by means suitable phases, $U'_g = \omega_g U_g$, in order to end up with a standard group representation
$$U'_e= I\:,\quad U'_g U'_f = U'_{g\circ f}\:.$$
This is a difficult co-homological problem with important results, like Bargmann's theorem, I will not address here (Varadarajan's book of the geometry of QM include a list of relevant and very delicate results in this area of the theory of group representation). | {
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ros, ros-melodic, rtabmap
<!-- RTAB-Map's Paramter -->
<param name="Reg/Strategy" type="string" value="2"/>
<param name="Reg/Force3DoF" type="string" value="true"/>
<param name="GridGlobal/MinSize" type="string" value="20"/>
<param name="Grid/MaxObstacleHeight" type="string" value="0.5"/>
<param name="Grid/MaxGroundHeight" type="string" value="0.1"/>
<param name="Grid/MinGroundHeight" type="string" value="-0.05"/>
<param name="Icp/VoxelSize" type="string" value="0.02"/>
<!-- localization mode -->
<param if="$(arg localization)" name="Mem/IncrementalMemory" type="string" value="false"/>
<param unless="$(arg localization)" name="Mem/IncrementalMemory" type="string" value="true"/>
</node> | {
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fluid-dynamics, acoustics, kinetic-theory, buoyancy
Title: What happens when buoyant force causes movement to exceed sound speed? I was wondering if it is possible for an object of a certain material to be put in a buoyant medium in such a way that the buoyant force (caused by the pressure difference “above” and “below” said object and by the interactions with molecules of the medium under that regime) creates an acceleration that causes a motion that exceeds the sound speed of the material of the object.
With “sound speed” I mean the speed at which molecular interactions inside the object propagate - averagely speaking of course. Note that I’m concerned with the sound speed of the object, not of the medium!
Since this movement “ripples” through the object as it snakes upwards to get a new position in the medium, there is a limit as to how fast this type of slinky movement can happen, right? | {
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"url": null
} |
annotation, labelling
Title: Online Audio annotation tools I need to find a decent online annotation tool to transcribe audio.
There are some requirements for a potential tool:
I should be able to deliver audio files to a few labelers. I should
be able to track which files went to which labeler.
It should be safe in terms of data storage.
Any suggestions? I'm not sure if it meets all your criteria (mostly because I'm not sure I understand all your criteria!), but you could have a look at ELAN: | {
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c++, c, error-handling, brainfuck, portability
int main(int argc, char *argv[]) {
std::istream *in = &std::cin;
std::ostream *out = &std::cout;
std::ifstream infile;
std::ofstream outfile;
std::string filename;
for (int i = 1; i < argc; ++i) {
if (argv[i][0] && argv[i][0] != '-') {
filename = argv[i];
infile.open(filename);
outfile.open(filename + ".c");
in = &infile;
out = &outfile;
break;
}
}
emit_boilerplate_head(*out, filename);
compile(*in, *out);
emit_boilerplate_tail(*out);
*out << std::endl;
return 0;
} As far as I can understand
default: comment.replace(i, 0, "\\");
makes
case '\\': comment.replace(i, 0, "\\"); break; | {
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"url": null
} |
minerals
To go back to $\ce{Zn}$ and $\ce{Pb}$ configuration, relativity explains why $\ce{Pb}$ looks like a noble gas (Xenon) plus $2$ electrons "s" in its outer shell. By comparaison, Zinc is $\ce{[Ar] 3d^{10} 4s^2}$. Like $\ce{Pb}$, it also looks like a noble gas (Argon) plus $2$ electrons "s" in its outer shell, if the filled shell $\ce{4d^{10}}$ is "forgotten". Mother Earth is probably not able to recognize the existence of this $\ce{4d}$ shell and has often mixed $\ce{Zn}$ and $\ce{Pb}$ minerals. | {
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javascript
doSample();
I would appreciate your feedback on how to make the code for getNumberOfElementsInWidth function clearer. Code / algorithm review.
Code
You could have used a for...of loop rather than the standard for loop
You could have used return count rather than breaking out. Some will argue otherwise but I see breaks as disguised gotos and should be avoided. (In this case your algorithm is buggy so the point is mute)
Use the form n += num to add rather than n = n + num
Doubling the accWidth + elements[i].width addition. Once in the if statement and once if it passes. If you are doubling up on calculations it usually means you can simplify.
You have not tested the code vigorously to look for problems
Re-write
Using the points above I have written you methods. The test call gives 5 elements, to fit in a 8 units max width. You result returns the wrong count 3 the count should be 4 elements with widths 1,1,1, and 2 | {
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aqueous-solution, solutions, osmosis
Title: Can I equalise osmotic pressure using two different solutes? I understand that there is a pressure gradient between an aqueous solution, e.g. salty water, and, say, pure water due to osmosis. My question is, is it possible to match or almost match, the osmotic pressure between two solutions of different solutes, but using a solution with a different solute. For example, would it be possible to make the osmotic pressure gradient between salty water and some other solution zero, assuming both are at the same temperature and pressure?
I am not a chemist, so purely qualitative answers are helpful. Yes, I think so. Based on our comments above, I believe you want to prepare two solutions that are isotonic with each other (have the same concentrations of different solutes) and allow them to communicate across a membrane that is selectively-permeable to water only.
Tonicity is described here as: | {
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python, performance, python-3.x, multiprocessing
Note that for purposes of testing I have replaced you ps.playsound(file, False) with a simple print statement with some debugging info. Also note the time.sleep(0.1) call, I found that while testing on Mac the keyboard package maybe somewhat buggy and was not clearing the is_pressed status fast enough, so I was getting repeat events. I guess initializing the ProcessPoolExecutors in your code is sufficiently slow that this doesn't become a problem.
Note another design choice here: instead of creating separate key handlers running concurrently, I choose to poll the pressed status of all keys we are interested in and store it in a dictionary. | {
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} |
deep-learning, tensorflow, image-classification
I'm unsure what kind of annotated training data would be useful for Tensorflow. Could someone please point us in the right direction?
My problem is that we have successfully been able to classify single pictures and even do a bit of object detection by drawing bounded boxes, but we are unable to do pairwise difference identification. In order for you to do pairwise difference, you need to do the following: | {
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quantum-field-theory, electromagnetic-radiation, radiation, classical-electrodynamics
NOTE: PS are working in a frame where $p^0=p^{'0}=E$ which (according to them) implies
$$k^\mu=(k,\mathbf{k}),~~p^\mu=E(1,\mathbf{v}),~~p^{'\mu}=E(1,\mathbf{v'}) $$ where (I guess) $k=||\mathbf{k}||. $ Then for instance $(k_\mu p^\mu)^2$ becomes $(Ek)^2\left(1-\frac{\mathbf{k}}{k}\cdot\mathbf{v}\right)^2$ which is (I assume) one of the denominators (up so some factors) in $(1)$. So I guess the correct notation in $(1)$ should be
$$\tag{3}\mathcal{I}(\mathbf{v},\mathbf{v}') =\int \dots-\frac{m^2/E^2}{\left(1-\hat{\mathbf{k}}\cdot\mathbf{v}'\right)^2}-\frac{m^2/E^2}{\left(1-\hat{\mathbf{k}}\cdot\mathbf{v}\right)^2}. $$
Overall, bad notation is used IMO on the pages near 181 in PS. I am not sure if my answer is correct, from what I understood:
(i) At the relativistic limit, $m<<E$, so the second and third terms in (6.15) will be negligible, just as you said. | {
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javascript, php, ecmascript-6, web-scraping, data-visualization
text: 'Answered percentage in recent months'
},
scales: {
xAxes: [{
display: true,
scaleLabel: {
display: true,
labelString: 'Date'
}
}],
yAxes: [{
display: true,
scaleLabel: {
display: true,
labelString: 'Percentage'
}
}]
}
}
};
document.addEventListener('DOMContentLoaded', _ => {
var ctx = document.getElementById('canvas').getContext('2d');
window.myLine = new Chart(ctx, config);
});
</script>
</body> | {
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"tags": "javascript, php, ecmascript-6, web-scraping, data-visualization",
"url": null
} |
javascript, css, regex
return ret;
}
wr.displayResult = function() {
var display = document.createElement('div');
var result = this.calculate();
var resultsRow = document.getElementById('results');
this.haystack.removeChild(this.textBox);
this.haystack.appendChild(display);
display.classList.add('display');
display.innerHTML = result.replacement;
this.countWords.value = result.countWords;
this.countNeedle.value = result.countNeedle;
this.percentage.value = result.percentageRate + ' %';
this.needleBox.value = result.needle || 'No value provided.';
this.exec.classList.add('hidden-element');
this.reset.classList.remove('hidden-element');
results.classList.remove('hidden-element');
for (var i in this.inputs) {
if (typeof this.inputs[i] === 'object')
this.inputs[i].classList.add('diabled-input');
}
} | {
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"openwebmath_score": null,
"tags": "javascript, css, regex",
"url": null
} |
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