text stringlengths 1 1.11k | source dict |
|---|---|
newtonian-mechanics, forces, friction
Image with all forces labelled I suppose that when the system was assembled, some small tension $T$ was applied on the string to attach it to the wall, so that it can be straight. That is before $F$ comes to play.
When some $F$ is applied, while $F - T < \mu mg$ nothing moves. And the tension in the string doesn't change because the string is not affected by $F$.
As soon as $F - T = \mu mg$ block A can move. Let's say it moves a small displacement $\Delta L$.
Now, the tension in the string is $T' = T + E\frac{\Delta L} {L}$, where E is the modulus of elasticity and $L$ the total length.
But the meaning of a string is being rigid, so we can translate this to an infinite $E$.
So, even for a very small displacement, $T'$ in the string becames big enough to move block B, what means: $2T' = (2\mu)3mg + \mu mg => T' = \frac{7}{2}\mu mg$.
Back to block A, when it starts to move $T -> T'$, so: $F - T' = \mu mg => F = \frac{9}{2}\mu mg$ | {
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energy, angular-momentum, mass, torque, moment-of-inertia
Title: Are translational KE and rotational KE exactly analogous? My textbook states that translational KE and rotational KE are completely analogous. The author states "They both are the energy of motion involved with the coordinated(non random) movement of mass relative to some reference frame. I can understand why this applies when comparing rotational velocity and translational velocity, but how can we justify that mass is equal to moment of inertia? It appears to me that moment of inertia isn't exactly analogous to mass in that it involves both direction and a physical quantity. Keep in mind what an analogy is. It's a comparison of forms, not of identical quantities. So if we look at the two forms, $$\frac{1}{2}mv^2 \text{ and } \frac{1}{2}\mathcal{I}\omega^2, $$ | {
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c#, wpf
public void InstantiateSelectionsAsProperties(Dictionary<string, string> selectionsFromFixSelector)
{
foreach (var key in selectionsFromFixSelector.Keys)
{
switch (key)
{
case "Fix Class 1 Full Name":
this.FixClass1 = new Fixes.FixClass1();
break;
case "Fix Class 2 Full Name":
this.FixClass2 = new Fixes.FixClass2();
break;
// etc.
}
}
} A couple of small pointers. I'm skipping over larger architecture issues because it seems that this code is not the complete example of working code. | {
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optics, electronics, radar, chirp
As it turns out, however, the radar technique is not quite the same, apparently because the peak-power restrictions hold all the way to the antenna. What they do is something rather clever: they stretch the pulse using a dispersive delay line and then they amplify it, and they send it off uncompressed to interact with whatever planes or ships or aliens are out there. The echo will return at a much lower intensity, but crucially it will retain the original chirp, which means that you can still compress it, without running the risk of having unreasonable powers in your wires or your antennas. | {
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matched-filter, equalizer
How does a practical receiver structure looks like with adaptive filters after the sampler on the digital side? What am I missing here?
Thanks
strong text You are right that in scenarios where the channel is unknown and/or time-varying, we cannot implement a fixed matched filter. The first filter in the signal chain is usually an analog band pass filter eliminating out-of-band noise. Then after A/D conversion one would use an adaptive equalizer that performs part of the function of the matched filter, but that will usually also be optimized for further noise suppression. This optimality criterion for equalizers that combines minimization of intersymbol-interference and noise power is commonly referred to as minimum mean-square error (MSE) criterion. | {
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javascript, node.js, web-scraping, raspberry-pi
});
Also I wrote up this gist on how I deployed it to a raspberry pi. That works, so suggestions on how to improve it would be greatly appreciated
In broad strokes this program:
Listens for a message
Validates the message
Gets the page content at the message url
Parses the page content
Re-formats the page content
Converts the page content into a message
Sends the message Here's how I would do it. Move the logic about which pages a parser consumes into the code for said parser. Also organizing your helper files into objects could reduce complexity in terms of the number of variables you pass around but that's more a matter of personal preference.
class SpecificPageParser {
match(url) {
return (
url.startswith('https://www.d20pfsrd.com/magic/') &&
!url.endsWith('/magic/')
);
}
format(response) {
// do your parsing stuff
const formatted = {...response,'extra':'info'};
return formatted;
}
} | {
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electrostatics, potential, vectors, conventions
On moving from the outer shell $A$ to the inner shell $B$, the vector $d\vec r$ points towards the centre whereas the electric field is radially outward. This implies $\vec E$ and $d\vec r$ are $\pi$ radian apart (antiparallel) and so the dot product is negative. But the author doesn't seem to include this negative sign in step $(2)$. In order to check whether the final result given by $(3)$ is correct or not, I determined the potential difference between the shells by an alternate method*. I found it to be same as that of $(3)$. But here, if I considered the negative sign in $(2)$ then I'll get the same magnitude of the potential difference but it'll be of opposite sign to that of $(3)$. So, how did the author obtain the correct result even after neglecting the negative sign due to the dot product of the antiparallel vectors? | {
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python, playing-cards
def best_completion(self, cards_to_complete: Sequence[Card], available_cards: Sequence[Card]) -> list[Card]:
completions = []
for available_cards_to_test in self._prepare_available_cards_to_test(available_cards):
completion = self._helper_step_size.best_completion(cards_to_complete, available_cards_to_test)
if self.is_match(completion):
completions.append(completion)
return max(completions, default=[])
@staticmethod
def _prepare_available_cards_to_test(available_cards: Sequence[Card]) -> Iterator[list[Card]]:
"""
Prepares available cards for testing by grouping them by color.
"""
cards_by_color = defaultdict(list)
for card in available_cards:
cards_by_color[card.color].append(card)
for available_cards_to_test in cards_by_color.values():
yield available_cards_to_test | {
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dna, genetics, definitions
Title: Do apes and humans share 99% of DNA or 99% of genes? What is the difference? I made an answer on the Scifi.SE that can be read here. It is about how the characters in the story Jurassic Park might have gotten DNA for all the species shown.
In my answer, I said this:
Apes and Humans, for example, share over 99% of their genes. That means the difference between our species is less than 1% of our genes. In fact, all life on Earth shares about 50% of it's genes.
but in the original posting (before someone edited it) I chose to use the word DNA instead of genes.
He left this comment in the section to explain the edit:
Sorry, I'm a biologist, I can't help it. Humans and apes share 99% similarity in the coding sequences of their DNA, the ~5% that codes for genes, not on all the DNA. I simplified this to genes for the answer. | {
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ros-kinetic
If possible, my suggestion is to make an independent process only to control the wheels. So in this process, you will implement the PID for each wheel. And the communication with this process could be through a UDP socket for example. Another solution is to make the control process in a microcontroller and communicate with it through a serial interface like UART. | {
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formal-languages
The alphabet ("words of the language") is finite by definition. The language may or may not be finite. A trivial example of a grammar producing an infinite language is $G' = (\{ S \}, \{ a \}, P, S)$ with $P = \left\{ S \to a|aS \right\}$. $G'$ produces the infinite language $L(G') = \{ a^n : n > 0 \}$.
Both finite and infinite languages exist and can be generated by grammars. | {
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python, python-3.x, programming-challenge
def score_direct(score: int) -> Iterator[tuple[int, ...]]:
scores = [1] * score
while True:
yield tuple(scores)
new_ones = 0
while True:
new_ones += scores.pop()
if len(scores) == 0:
return
if scores[-1] < N:
break
scores[-1] += 1
new_ones -= 1
scores.extend((1,) * new_ones)
METHODS = (score_orig, score_permute, score_direct) | {
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radius of and centered at the left corner of the semi-circle (O) with radius. In order to describe the shape of an object, we give the object appropriate dimensions. If it passes through the center it is called a Diameter. Options: A. #Program to draw tangent circles in Python Turtle import turtle t = turtle. Each of the three circles in the adjoining fiture is externally tangent to the other two and each side of the triangle is t. 6 13 11 A B Not tangent 3) 12 20 16 B A Tangent 4) 15. Solution; Write the equation of the circle with radius $$\sqrt 7$$ and center $$\left( { - 1, - 9} \right)$$. if the area of the curvilinear formed by the point of tangency of the three circles is 142 cm2, compute the radius of each circle. The tangent in between can be thought of as the transverse tangents coinciding together. Calculus: Taylor Expansion of sin(x) example. Key Words • chord • diameter p. A circumscribed circle is a circle that encompasses a polygon such that the circle touches all the | {
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algorithms, heuristics, traveling-salesman
Title: Nearest Insertion Traveling Salesman Heuristic: is it faster to insert nearest nodes first? I am trying to implement the nearest insertion TSP heuristic. However, I am wondering if it matters which node I insert into the subgraph first.
For example, should I start with one node; calculate the distance between this node and the other nodes; and sort the other nodes by this distance? Then, I will insert the closest node (to the first node) first. Will this be faster than randomly choosing nodes to insert? Inserting a random point instead of a nearest point can be faster, but then you are not following the nearest insertion heuristic anymore. The key is to realize that the point is not the runtime efficiency of the algorithm, but rather the quality of the solution obtained. Indeed, remember that we are dealing with an NP-hard problem. | {
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c#, strings
StringBuilder sb = new StringBuilder();
int mo = 0;
int tu = 0;
int we = 0;
int th = 0;
int fr = 0;
int sa = 0;
int su = 0;
sb.Append("<table class=\"openinghours\" cellpadding=\"0\" cellspacing=\"0\">");
sb.Append("<tr class=\"days\">");
sb.Append("<th class=\"void\"><a href=\"\" class=\"saveEdit\" id=\"save-openingHours\" data=\"hastwo\">Save</a></th>");
sb.Append("<th><span>M</span></th>");
sb.Append("<th>T</th>");
sb.Append("<th><span>W</span></th>");
sb.Append("<th>T</th>");
sb.Append("<th>F</th>");
sb.Append("<th>S</th>");
sb.Append("<th>S</th>");
sb.Append("</tr>"); | {
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javascript, performance, animation, raphael.js
Title: Animating Coloured Cogs The script is for rotating 5 different coloured gears (tandwiel) on my site.
I think the code can be improved and improved performance. 5x var paper isn't that smart for speed?
Getting the current code working faster in the normal state is the primary concern, but there are two smaller issues I know of that I would appreciate help with:
The code works most of the time, but when changing the browser size I have a problem with the rotating left and right: 1 gear will rotate the other direction, but starts to rotate very fast and finally slows down to normal speed. The longer you let it rotate at normal speed and then change browser, the longer the gear rotates too fast. It seems to me it rotates backwards all the missing angles til it will be normal again??
How to stop the JavaScript when the browser size is mobile. So it won't overload for mobile. The animation should start again when changing to bigger browser size. | {
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arduino, slam, kinect
Title: How computationally powerful is an Arduino Uno board? What can an Arduino board such as the Uno really do? Of course simple things like controlling a couple servos is very easy for it. However, I don't think an Uno board would be able to preform real-time 3D SLAM from point cloud data gathered from a Kinect sensor on a mobile robot, right? If the robot had any speed at all the Arduino wouldn't be able to keep up, correct? Could it do 2D SLAM while moving and be able to keep up? What about taking 1/10 of the points from the Kinect sensor and processing only those?
Basically, what are some examples of the resource limitations of such an Arduino board? It depends - on the number of landmarks in the feature map, and how much time you're willing to invest tuning the algorithm for speed, and a number other parameters which you may or may not be able to control for a given application. | {
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organic-chemistry, isotope, hydrocarbons
Several methods and results are summarized in O’Keefe, F. R.; Fet, E. V.; Harris, J. M.; Compilation, Calibration, and Synthesis of Faunal and Floral Radiocarbon Dates, Rancho La Brea, California Contrib Sci 2009, 518, 1–16. All techniques involve a combination of solvent washes (e.g. Soxhlet treatment) followed by the isolation of the bone collagen (since bone collagen does not exchange carbon with the environment) under different hydrolysis conditions (time, temperature, acid strength, etc.) | {
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downsampling
This answer at this other post IQ Mismatch and Image shows exactly how we can have image problems with the Zero-IF receiver if there is quadrature or amplitude imbalance, and for complex modulated signals such as QAM, this would be an issue whether there is a carrier offset or not. Please refer to that post which explains how quadrature and amplitude error leads to images. Below I will further explain the differences between a Zero-IF Down-converter (or Zero-IF receiver, ZIF) and Image Reject Down-Converter (or Image-Reject Mixer, IRM) which helps to further understand how images are handled and also demonstrates the benefit of using complex signal representation to evaluate signal processing waveforms ($e^{j\omega t}$ rather $\cos(\omega t)+j\sin\omega t)$).
A side by side comparison of the two architectures is below | {
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In other words, we can fill all values at row $$F[k+1]$$ if we know all values at row $$F[k]$$.
If $$F[n][j]$$ becomes True in the end for any $$j$$, then we return yes. Otherwise, return no.
### Explanation on 5, 1, 7, 2, 8
0 1 2 3 4 5 6 7 8 9
t 0 T T T T T T T T T T
5 1 T T T T T T T T T T
1 2 T T T T T T T T T T
7 3 T T T T T T
2 4 T T T T T T T T
8 5 T T
The fourth row $$F[3]$$ corresponds to chains of the form $$t$$, $$\pm5$$, $$\pm1$$, $$\pm7$$. Their lengths can only be 0, 1, 2, 7, 8, 9. Now if you add $$\pm2$$ to them, you can get lengths
2, 3, 4, 9, 10, 11 (for +2)
-2, -1, 0, 5, 6, 7 (for -2)
So the possible lengths of chains of the form $$t$$, $$\pm5$$, $$\pm1$$, $$\pm7$$, $$\pm2$$ are 2, 3, 4, 9, 0, 5, 6, 7, which correspond to row $$F[4]$$.
### How to find the smallest realizable $$L$$? | {
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cosmology, space-expansion, redshift, bao
$$\text{d}_{a}=\dfrac{\text{d}_{h}(z)}{1+z}\quad\quad(4)$$
with $\text{d}_{h}(z)=c\int_{0}^{z}\dfrac{\text{d}z}{H(z)}\quad\quad(5)$
So from $eq(5)$, what I can only write is (by considering a little $\Delta$ and a curvature parameter $\Omega_{k}=0$) :
$$c\Delta z=\text{d}_{h}(z)H(z)\quad\quad(6)$$
Now, taking the expression of $\text{d}_{h}(z)$ into $eq(6)$ :
$$c\Delta z=\text{d}_{a}(z)(1+z)H(z)\quad\quad(7)$$
Then :
$$c\Delta z=\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)\quad\quad(8)$$
As you can see in $eq(8)$, this is not the same form as in $$eq(1)$$.
How can I make disappear the factor $(1+z) /\Delta\theta$ in order to have simply for the right member : $$H(z)\Delta \chi$$ instead of $\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)$ into $eq(8)$ ?
$\Delta\chi$ represents for me the variation $\Delta$ of radial coordinate, doesn't it ?
CORRECTION 1 : | {
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101111 4. Convert the following decimal values into signed binary numbers using the sign-magnitude format:. To use this decimal to binary converter tool, you should type a decimal value like 308 into the left field below, and then hit the Convert button. Sign-extending means copying the sign bit of the unextended value to all bits on the left side of the larger-size value. Sign-Magnitude Representation. Categories: Coordinate Geometry | Physics. Because of these limitations, sign-magnitude representation is rarely used to represent integers in a machine. One's Complement binary representation: Convert to/from decimal. Excess to N. The converter allows you to; convert binary to decimal numbers; convert hexadecimal to decimal numbers; convert hexadecimal to binary numbers; convert binary to hexadecimal numbers; convert decimal to hexadecimal numbers; convert decimal to binary numbers. Binary to Signed decimal in MATLAB. This section presents a brief review of positional or placement | {
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kinematics, acceleration, mathematics, displacement
% RmaalaaabaGaaGymaaqaaiaaigdacaaIYaGaaGimaaaacqGHRaWkca
% GGUaGaaiOlaiaac6caaaa!6F1D!
$
But this situation looks very unreal.
However, there should be a real-life example, because you can easily get displacement = e (if I am not mistaken) when the acceleration affects only direction, not modulus. Take for example a satellite that is 1 unit away from a gravitation source and that moves tangentially by inertia at v = 1 of whatever units. If gravity acts perpendicularly to the tangential motion and the acceleration caused by gravity is $\frac{v^2}{r}$ (that is to say, with these numbers, 1), then the displacement after 1 time unit will be circular and of modulus 1 radian = $e^i$, wouldn’t it? This situation looks to me very much like compound interest, because you start with a capital (radius) and an interest rate (velocity) and then you make the interest compound and continuous, although only in terms of direction (centripetal acceleration). | {
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organic-chemistry, aromatic-compounds
I don't quite understand where you are going with that, but the lone pair of nitrogen in aniline certainly has overlap with the aromatic system. In general amines and phenols are quite alright nucleophiles. There are other factors determining why they react the way they do.
Your statement suffers from the most-stable-resonance-structure-syndrome, i.e. you take resonance structures much too seriously. Please see: What is resonance, and are resonance structures real? If there is resonance, then this effect is always stabilising, because otherwise it would adopt a molecular structure where resonance would not be possible (compare cycloocta-1,3,5,7-tetraene).
Additionally formal charges are exactly what they are called: formal. There is little to no truth to them other than keeping track of all the electrons. | {
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electromagnetism, forces, electric-current
Title: Force on a current carrying conductor and Hall effect If we consider a thin wire on which flows current, inside a magnetic field, we observe a force
$\mathbf{dF}=i\mathbf{ds} \land \mathbf{B}$ on each $\mathbf{ds}$ of the wire. This force is caused by electrons, on which is acting Lorentz force, which bump into the Crystal structure of the metal. However after a while (assuming $\mathbf{B}$ constant and uniform and the wire firm) Lorentz force generates a separation of charges between two opposite side of the wire (Hall effect), and the force on each electron becomes zero, so they should stop bumping (all togheter and all in the same direction) into the Crystal structure of the metal. Then how is possible that we still observe a force on the conductor? Where I am wrong?
Then how is possible that we still observe a force on the conductor? Where I am wrong? | {
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special-relativity, spacetime, metric-tensor, notation, covariance
\end{align}
Since $\delta$ is the identity, the dual vector will have the same components as the original vector $\vec{v}$, so we can think about them as the same vector.
In spacetime, our metric tensor $\eta$ is no longer the identity. Instead, we have $\eta_{00} = -1$ and $\eta_{11} = \eta_{22} = \eta_{33} = 1$. (The off-diagonal components are zero just like in the Euclidean case.) We can still get the components of a dual vector by contracting with the metric tensor, but now the components of the dual vector are not the same as the components of the original vector!
\begin{align}
x_{\mu} = \sum_{\nu=0}^{3}\eta_{\mu \nu}x^{\nu} \rightarrow x_0 = -x^0, x_1 = x^1, x_2 = x^2, x_3 = x^3
\end{align}
Note the extra minus sign in front of the $0$-component. So in spacetime, vectors and dual vectors are different. | {
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c#, multithreading
if (!cancel2)
{
Task.Factory.StartNew(() =>
{
int count = 0;
while (!cancel2)
{
Log.Append("GUI", "log to GUI " + count++);
Thread.Sleep(200);
}
cancel2 = false;
});
}
if (!cancel3)
{
Task.Factory.StartNew(() =>
{
int count = 0;
while (!cancel3)
{
Log.Append("Error", "log to Error " + count++);
Thread.Sleep(300);
}
cancel3 = false;
});
} | {
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c#, beginner
double side = 0;
// ReSharper disable once SwitchStatementMissingSomeCases
switch (dimensions)
{
case PerfectSquareReturns.Perimeter:
side = this.FromPerfectArea(firstDimension);
this.Perimeter = this.CalculatePerimeter(side, side);
this.Area = firstDimension;
break;
case PerfectSquareReturns.Area:
side = this.FromPerfectPerimeter(firstDimension);
this.Perimeter = firstDimension;
this.Area = this.CalculateArea(side, side);
break;
}
this.Length = side;
this.Width = side;
} | {
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of the corresponding sides. It is stated that it should only take six steps. Similar to a triangle’s perpendicular bisectors, there is one common point where a triangle’s angle bisectors cross. See the derivation of formula for radius of This point is another point of concurrency. s. Expert ... To compensate for the problems of heat expansion, a piston is ... 1/14/2021 7:34:34 PM| 5 Answers. Read and complete the proof . 180 degree the point where a triangle are concurrent c. orthocenter d. circumcenter.... ( E ) 1400 1000 28 the law of sines you want to a... And orthocenter lie at the same point to find the value of.! You calculate the three angle bisectors its circumcenter coincides with the circumcenter 4ABC. And right ) a polyhedron ( when they exist ) derivation of formula radius., there is one common point where the angle bisectors intersect incircle for a triangle calculate the vertical. 'S 3 angle bisectors cross r r, of other math skills incenter and it is call. Second | {
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Hope that helps.
Andre | {
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# Could someone verify my answers?
1. Jan 31, 2005
### Matt.D
Hey guys, I've been given a maths assignment thats due tomorrow and I'd just like to verify the answers before submitting. I've put the question in bold and have just put the answer where I feel confident. The rest I have shown my workings.
Q.1
The length, d m(meters), of a retangular field is 40m greater than the width.
The perimeter of the field is 400m
i) Write this info in the form of an equation for d
ii) Solve the equation and so find the area of the field
Ans i) 4d - 80 = 400
Ans ii) 4d = 480
d = 480 / 4
d = 120m
120 * 80 = 9600m2
Q.2
A) In the Formula S = ut+ 1/2 at2 make 'u' the subject
B) In the Formula T = 2Pi SQRT l/g make 'l' the subject
Ans A) u = s - 1/2 at
Ans B) T = 2Pi SQRT l/g
T / 2Pi = SQRT l /g
T2/4Pi2 = l/g
l = T2 g / 4Pi2
(sorry this looks so complicted, but I don't know how to add the right symbols) | {
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ros, communication, nodes, ros-indigo
Is this the right way to do things? Should I change the structure? Also, I'm not quite sure I understand how the publishing/subscribing will work considering the nodes will be sending messages back and forth. I have already implemented the Python code (without the subscribing/publishing) - However, I don't really understand how to add it to a node.
I'm really eager to learn so please do not hesitate to point me in the right direction, or to point me towards a good read or similar example. In particular, any example where two nodes communicate back and forth and waiting for a signal before executing part of the code would be appreciated! | {
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c++, c++11, computational-geometry
hides the fact that cin, cout, vector and stack are coming from the namespace std where cin, cout,
vector and stack are used in the code. This can cause confusion of where the code is actually
coming from.
As the software becomes more complex and uses more libraries this becomes a bigger problem.
For a more detailed discussion of why it is a bad idea to use using namespace std; see this stackoverflow question and stackoverflow question.
Reduce Complexity, Follow SRP
The Single Responsibility Principle states that every module or class should have responsibility over a single part of the functionality provided by the software, and that responsibility should be entirely encapsulated by the class. All its services should be narrowly aligned with that responsibility.
Robert C. Martin expresses the principle as follows:
A class should have only one reason to change. | {
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equilibrium, aqueous-solution, solubility
You must be wondering that:
Why does the new equilibrium constant $K'=K^n$?
The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction: $\ce{A +B —> C + D}$, $K_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$
Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.
So, if we multiply our reaction with $n$, it actually can be rewritten as:
$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K_\mathrm{c}'$ will be
$$K_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K_\mathrm{c}^n$$
Hence, proved. | {
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quantum-mechanics, quantum-information, bells-inequality, locality
$$\lvert S_\lambda\rvert\le \lvert E_\lambda(B_0)+E_\lambda(B_1)\rvert
+ \lvert E_\lambda(B_0) - E_\lambda(B_1)\rvert = 2 \max(E_\lambda(B_1),E_\lambda(B_2)).$$
Using again that by definition $0\le E_\lambda(B_i)\le 1$, we conclude that $\lvert S_\lambda\rvert \le 2$.
The full $S$ is now defined by averaging $S_\lambda$ over the hidden variable $\lambda$, and because a convex mixture of numbers in $[-2,2]$ remains in $[-2,2]$, we reach the conclusion:
$$\lvert S\rvert\le 2.$$ | {
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microbiology, biotechnology
Title: Ethanol production by fermentation? As bacteria are involved in the production of ethanol through fermentation and ethanol is also used as antiseptic that kills bacteria, so how and why bacteria are involved in the synthesis of such alcohol which is lethal for itself(bacteria)? Yeast can produce up to 16-17 v/v% ethanol without dying according to this article. Yeast is a primary industrial ethanol producer, it produces ethanol even under aerobic conditions, in contrast to bacteria that usually ferment sugar to ethanol under anaerobic conditions. Bacteria can also withstand up to several percent v/v concentration ethanol in their media. Check this paper for details. They produce ethanol because it is a natural byproduct of their cellular metabolism. Under anaerobic conditions sugar as the primary energy source cannot be oxidized and utilized as well as in aerobic conditions. | {
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mathematical-physics, quantum-optics
The second one is similar, but here there can be many photons in one state. So the total energy inside the cavity is a counting of how many photons are inside, and the creation and annihilation operators can be seen here for the harmonic oscillator (probably relevant here) and other examples. | {
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ros, slam, navigation, rplidar, hector-slam
I'll get this error after running the launch file.
{{{[ WARN] [1457932080.622102713]: No transform between frames /map and scanmatcher_frame available after 20.003506 seconds of waiting. This warning only prints once.}}}
I tried viewing the tf tree, the output I get is "No TF data received".
Appreciate any help I can get on this issue as I am new to both linux and ROS. I am running ros jade and linux unbuntu. I haven't edit any other launch file from hector slam. Thanks in advance!
Originally posted by bongkw on ROS Answers with karma: 1 on 2016-03-14
Post score: 0 | {
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matlab, signal-analysis, stft, time-frequency, real-time
On the other hand, the STFT is a widely used method in signal processing, and is useful for analyzing signals that are stationary or have slow time-varying properties. It provides a high degree of frequency resolution, but its time resolution is limited. Unlike WVD, it is not able to distinguish between closely spaced frequency components, and it does not show the instantaneous frequency of a signal to the degree in which WVD can.
The major disadvantage of the WVD is the computational cost. It has a computational complexity of O(N^2) which makes it impractical for large data sizes.
Is the above response true or false?
I had originally added this as a possible answer, removed it due to objection from a member (Jdip) due to inaccuracy and adding it back as part of question as asked by a member (OverLordGoldDragon) in the comments. Hope posting ChatGPT responses is not against the rules or anything. | {
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newtonian-mechanics, forces, classical-mechanics, potential-energy, conservative-field
Yes, that's correct, both conservative and non-conservative forces will contribute to the work done. But the mathematical difference is still as clear as day. One of them depends on the path, the one doesn't. Of course if you start with a general force and do some work with it around a closed path and the work isn't zero, then you can't define a potential energy (there are generalizations of it that you can define in some situations, like generalized potentials, but that's another story) and you can't talk about some decomposition of your force in conservative + non-conservative components. What you can do is start with that supposition and see where it leads you, but from a general non-conservative force you can't say anything like that. | {
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java, design-patterns, reinventing-the-wheel
Does the set of table names really deserve to be decided at compile time? Why not column names, too? I didn't see any place where you're switching on the table [oh, see next item], and if you are that should be moved into a proper class with polymorphic behavior instead. An enumeration should model a truly fixed set of items. QueryType is a great example.
OracleSpecifics seems like it will become a monster as it grows to deal with a) Oracle and b) every table. Create a proper Table class with either a subclass for each table or read the configuration from a file so that OracleSpecifics can be 100% Oracle-specific and you don't have to repeat it in MySqlSpecifics and SqlServerSpecifics. | {
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newtonian-mechanics, mass
My question then is about the force $\mathbf{F}$. In the derivation, we remarked that this has to be the total external force on all the particles under consideration, not just the particles inside the control volume itself. However, this seems quite unreasonable from another perspective. If I fire a machine gun in space, I will experience some recoil and accelerate as a result of it. Neglecting gravitational influence, we might be tempted to set $\mathbf{F}=\mathbf{0}$. But now suppose that someone puts a plate in the way of my bullets, so that the bullets eventually hit this plate and experience some recoil as a result. This would amount to a nonzero average force on the bullets and thereby on the entire system under consideration, affecting my acceleration according to the equation of motion. However this seems quite absurd. The plate might be pretty far away, and my acceleration is evidently not going to change when the bullets hit the plate.
What am I missing here?
EDIT | {
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### A generalization
Here is a natural generalization.
Exercise. Generalize to the case when each variable appears in at most $$r$$ clauses and each clause contains at least $$r$$ distinct variables, where $$r\ge1$$. | {
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equilibrium, spectrophotometry
$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$
$$\mathrm{K} = \frac{\pu{6.39e-5}}{\pu{} \pu{0.94e-3}\cdot \pu{0.336e-3}}$$
$$\mathrm{K} = 202$$ | {
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modulation, qpsk, timing
Thank you, Your Gardner Loop would properly "Flywheel" in the absence of symbol transitions. So it would slowly drift off of synchronization, but would not spiral out of control as you describe. This is ideal for a control system with "missing updates"; to have the contributions for those updates with no additional information to contribute 0 to the accumulated error, resulting in a "flywheel" behavior. There are other detectors used to discriminate the control term in loops (such as a digital phase-frequency detector used in PLL implementations (such as this device: Maxim Phase Frequency Detector) that produce maximum error in the absence of transitions, and these systems would move off of lock at the maximum speed possible - this would be bad for timing recovery where we expect to have missing bits! | {
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general-relativity, frequency, gravitational-waves, polarization
The amplitude of the waves increases with the total mass of the system and depends on the relative masses of the components (equal masses give stronger waves). Amplitude also increases with the frequency of the orbit and decreases with distance to the source. The amplitude of each polarisation component then also varies with inclination angle, as discussed above.
Edit: For a circular orbit, the observed strain amplitude can be written:
\begin{equation}
h = -\frac{4G \omega_{\phi}^2 \mu a^2}{rc^4} \left[\frac{(1 + \cos^2 i)}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\0 & 0 & -1& 0\\0 & 0 & 0 & 0 \end{pmatrix} \cos(2\omega_{\phi} t)
+ \cos i \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0\\ 0 & 1 & 0& 0\\0 & 0 & 0 & 0 \end{pmatrix} \sin(2 \omega_{\phi} t) \right],
\end{equation} | {
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cosmology, big-bang
To be honest the real answer is that nobody knows for sure how the universe started, however strict adherents to quantum mechanics will tell you that classical approaches to the problem are inadequate. | {
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Thank you in advance for any help provided.
• $|f(x)-0|=|f(x)|$, but this isn't true if you replace $0$ by $l$. – vadim123 Aug 2 '13 at 23:29
• There are some things you want to change: You want to start with an $\epsilon$, and then find out what the corresponding value of $\delta$ would be; and your $\delta$ should not involve x at all. – user84413 Aug 2 '13 at 23:33
• @user84413, why? That is only the case if he wants to show that $f(x)$ is uniformly continuous, if I'm not mistaken. – Alex Wertheim Aug 2 '13 at 23:37
• Yes, I don't want to show that f(x) is uniformly continuous. – mauna Aug 2 '13 at 23:53
• The reason $\delta$ can't involve $x$ is because in the rigorous definition of limits, it wouldn't logically make sense. The $\delta$ is chosen before the $x$. The definition is: $L$ is the limit of $f$ at $a$ if, for all $\epsilon > 0$, there is a $\delta > 0$, so that for all $x$, $|x - a| < \delta \implies |f(x) - L| < \epsilon$. – Pratyush Sarkar Aug 3 '13 at 2:49 | {
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"lm_q1q2_score": 0.8301724518938393,
"lm_q2_score": 0.8519528057272543,
"openwebmath_perplexity": 93.61100108274752,
"openwebmath_score": 0.9508238434791565,
"tags": null,
"url": "https://math.stackexchange.com/questions/458375/how-to-show-that-a-limit-cannot-be-another-number"
} |
algorithms, matching, assignment-problem
I want to find an $A$-optimal matching (e.g, for any $(a_1 \rightarrow b_1), (a_2 \rightarrow b_2)$, then either $b_2$ is lower than $b_1$ or does not appear in $a_1$'s preferences, and likewise for $a_2$) between $A$ and $B$, respecting $A$'s preference lists, and where ties between members of $A$ are broken randomly. I've been trying to apply the Gale Shapely algorithm to this problem, with random preferences for B, but it appears to have degenerate cases where the above criterion is violated. | {
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c++
shapes[5]->print();
cout << shapes[2]->getID() << " --";
shapes[2]->print();
}
init2 = true;
break;
case 4:
if (selection == 1) {
compareArea(&shapes[5], &shapes[3]);
}
else if (selection == 2) {
compareVolume(&shapes[5], &shapes[3]);
}
else {
cout << shapes[5]->getID() << " --";
shapes[5]->print();
cout << shapes[3]->getID() << " --";
shapes[3]->print();
}
init2 = true;
break;
case 5:
if (selection == 1) {
compareArea(&shapes[5], &shapes[4]);
} | {
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python, beginner
Title: Implementing a history of user action for a calculator I decided to make a calculator as a project. Implementing basic addition, subtraction, division, and multiplication was fairly easy.
I wanted to add more functionality so I decided to implement a list of results the user view. However, I had a difficult time keeping track of the results numerically. I wrote a maze of if statements that are functional but seem to be overwrought with code. I am sure there is a better way to handle this. Any advice?
def add(x, y):
return x + y
def sub(x, y):
return x - y
def mul(x, y):
return x * y
def div(x, y):
value = None
while True:
try:
value = x / y
break
except ZeroDivisionError:
print('Value is not dividable by 0, try again')
break
return value | {
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Regarding PCA, it says: "Linear dimensionality reduction using Singular Value Decomposition of the data to project. SVD computation example Example: Find the SVD of A, UΣVT, where A = 3 2 2 2 3 −2. Hence, if you hold a reference to it for faster access, you'll need to get a new reference to it using GetArray. 0 | ii TABLE OF CONTENTS and deals with dense matrix factorization and solve routines such as LU, QR, SVD and LDLT, as well as. Out of memory SVD solver for big data @article{Haidar2017OutOM, title={Out of memory SVD solver for big data}, author={Azzam Haidar and Khairul Kabir and Diana Fayad and Stanimire Tomov and Jack J. Finding the optimal/best rotation and translation between two sets of corresponding 3D point data, so that they are aligned/registered, is a common problem I come across. tridiag, to solve non-square, banded and tridiagonal linear systems of equations. SVD definition: (Simultaneous Voice and Data) The concurrent transmission of voice and data by modem over a | {
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"url": "http://yjcg.cesenaticoopen.it/svd-solver.html"
} |
human-biology, biochemistry, endocrinology
In spite of this, I can think of quite a few examples of organs which do not act as endocrine glands. That is because there is a difference between the organs secreting hormones and endocrine glands. The definition of endocrine glands is as follows1, 5, 6, 7, 8: | {
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ros, vslam, vslam-system
Was there a change in the code which is not reflected in the bag-file?
Nikolas
Originally posted by NikolasEngelhard on ROS Answers with karma: 106 on 2011-06-26
Post score: 1
Hi Nikolas,
I think I had the same problem: the bag file was probably recorded using cturtle and you are probably running diamondback. In any case, you may try to fix the bag file:
rosbag fix vslam_tutorial.bag vslam_tutorial_fixed.bag
and play it as follows:
rosbag play --clock vslam_tutorial_fixed.bag
Regards
Jordi
Originally posted by Jordi Pages with karma: 245 on 2011-06-26
This answer was ACCEPTED on the original site
Post score: 2
Original comments
Comment by NikolasEngelhard on 2011-06-27:
Thanks. no highlighting in comment, so see next answer... | {
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c#, com
private void AddTestWindowsMenu(CommandBarPopup menu)
{
_windowsTestExplorerButton = AddMenuButton(menu);
_windowsTestExplorerButton.Caption = "&Test Explorer";
_windowsTestExplorerButton.BeginGroup = true;
_windowsTestExplorerButton.FaceId = 305; // a "document" icon, with a green checkmark and a red cross
_windowsTestExplorerButton.Click += OnTestExplorerButtonClick;
}
private int FindMenuInsertionIndex(CommandBarControls controls)
{
for (int i = 1; i <= controls.Count; i++)
{
// insert menu before "Window" built-in menu:
if (controls[i].BuiltIn && controls[i].Caption == "&Window")
{
return i;
}
}
return controls.Count;
} | {
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topology, quantum-hall-effect, topological-phase
unitaries, because if there exists a local unitary $U$ such that
$|\psi_1\rangle = U |\psi_0\rangle$ and $|\psi_1^{\prime}\rangle = U
|\psi_0^{\prime}\rangle$, then
$$\langle \psi_1 | \hat{o} | \psi_1 \rangle - \langle \psi^{\prime}_1 |
\hat{o} | \psi^{\prime}_1 \rangle = \langle \psi_0 | U^{\dagger} \hat{o}
U | \psi_0 \rangle - \langle \psi_0^{\prime} | U^{\dagger} \hat{o} U |
\psi_0^{\prime} \rangle.$$ One can show that if $U$ is a local unitary
then $U^{\dagger} \hat{o} U$ is a local observable if and only if
$\hat{o}$ is, so it followsj that $|\psi^{\prime}_1\rangle$ and
$|\psi_1\rangle$ are locally indistinguishable if and only if
$|\psi_0\rangle$ and $|\psi^{\prime}_0\rangle$ are locally
indistinguishable. Therefore, the four-fold "topological degeneracy" of
the toric code cannot be removed as the parameters of the Hamiltonian
are varied without closing the gap. | {
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python, python-3.x, sorting, reinventing-the-wheel
Also, a source where I can find standardized code for most algorithms in Python would be nice?
By standardized; I mean code that is deemed efficient and the most up-to-date by the Python community. PEP-8
The Style Guide for Python Code enumerates a set of rules that all Python programs should follow. Of particular note is the Naming Conventions, which indicates that function names and variable names should all be snake_case, not CapitalizedWords. So MergeSort should be renamed merge_sort, and MyList should be renamed my_list.
Private functions
merge is an internal helper function for MergeSort. PEP-8 also recommends naming internal helpers with a leading underscore, to indicate it is private, and should not be used externally, so merge should be named _merge.
Repeated computations and lookups
while len(left) > i and len(right) > j:
if left[i] > right[j]:
sorted_list.append(right[j])
j += 1
else:
sorted_list.append(left[i]) | {
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ros, c++, rviz, moveit, ros-melodic
Title: Cartesian path won't achieve full trajectory
Hi,
Using this code:
geometry_msgs::PoseStamped current_pose;
current_pose = move_group_interface_arm.getCurrentPose("ee_link");
geometry_msgs::Pose target_pose1;
target_pose1.orientation = current_pose.pose.orientation;
//target_pose1.orientation.w = 1.0;
target_pose1.position.x = 0.3;
target_pose1.position.y = 0.5;
target_pose1.position.z = 0.8;
move_group_interface_arm.setPoseTarget(target_pose1);
success = (move_group_interface_arm.plan(my_plan) == moveit::planning_interface::MoveItErrorCode::SUCCESS);
ROS_INFO_NAMED("Spraying", "Visualizing plan 2 (pose goal) %s", success ? "" : "FAILED");
move_group_interface_arm.move();
ros::Duration(0.5).sleep();
moveit::planning_interface::MoveGroupInterface::Plan my_plan_gripper;
// 3. close the gripper
move_group_interface_gripper.setJointValueTarget(move_group_interface_gripper.getNamedTargetValues("closed")); | {
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• @Subramani: Welcome to the site. Have a nice time over here.
– anonymous
Feb 19, 2011 at 4:28
Suppose $$Z \subset X \times Y$$ is closed, and suppose $$x_0 \in X \setminus \pi[Z]$$. For any $$y \in Y, (x_0, y) \notin Z$$, and as $$Z$$ is closed we find a basic open subset $$U(y) \times V(y)$$ of $$X \times Y$$ that contains $$(x_0, y)$$ and misses $$Z$$. The $$V(y)$$ cover $$Y$$, so finitely many of them cover $$Y$$ by compactness, say $$V(y_1),\ldots,V(y_n)$$ do. Now define $$U = \cap_{i=1}^{n} U(y_i)$$, and note that $$U$$ is an open neighbourhood of $$x_0$$ that misses $$\pi[Z]$$: suppose that there is some $$(x,y)\in Z$$ with $$\pi(x,y) = x \in U$$. Then $$y \in V(y_i)$$ for some $$i$$, and as $$x \in U \subseteq U(y_i)$$ (as $$U$$ is the intersection of all $$U(y_i)$$) we get that $$(x,y) \in (U(y_i) \times V(y_i)) \cap Z$$ which contradicts how these sets were chosen to be disjoint from $$Z$$. So $$U \cap \pi[Z]=\emptyset$$ and $$\pi[Z]$$ is closed. | {
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RANDOM-SAMPLE(m, n)
if m == 0
return Ø
else S = RANDOM-SAMPLE(m - 1, n - 1)
i = RANDOM(1, n)
if i ∈ S
S = S ∪ {n}
else S = S ∪ {i}
return S
We prove that it produces a random $m$ subset by induction on $m$. It is obviously true if $m = 0$ as there is only one size $m$ subset of $[n]$. Suppose $S$ is a uniform $m − 1$ subset of $n − 1$, that is, $\forall j \in [n - 1]$, $\Pr[j \in S] = \frac{m - 1}{n - 1}$.
If we let $S'$ denote the returned set, suppose first $j \in [n − 1]$,
\begin{aligned} \Pr[j \in S'] & = \Pr[j \in S] + \Pr[j \notin S \wedge i = j] \\ & = \frac{m - 1}{n - 1} + \Pr[j \notin S]\Pr[i = j] \\ & = \frac{m - 1}{n - 1} + \left(1 - \frac{m - 1}{n - 1}\right) \frac{1}{n} \\ & = \frac{n(m - 1) + n - m}{(n - 1)n} \\ & = \frac{nm - m}{(n - 1)n} = \frac{m}{n}. \end{aligned}
Since the constructed subset contains each of $[n − 1]$ with the correct probability, it must also contain $n$ with the correct probability because the probabilities sum to $1$. | {
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"url": "https://walkccc.me/CLRS/Chap05/5.3/"
} |
thermodynamics, equilibrium
Physical intuition suggests that if the boundary $\partial U$ of the film has a given temperature distribution $f\colon\partial U\to\Bbb R$, then the temperatures at interior points are uniquely determined.
Not physical intuition, but real-world experiments suggest that if the boundary $\partial U$ of the film has a given temperature distribution $f\colon\partial U\to\Bbb R$, then the temperatures at interior points are uniquely determined. Physical intuition is no more than accumulated experience from such real-world experiments.
Historically, physicists have found this intuition strongly compelling, although it is surely not mathematically convincing.
I doubt a physicist would say that. | {
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$\geq$ on $\mathbf { R }$ an equivalence class have... They are equivalent ( under that relation ) For specifying whether or not two quantities the. Purposes, it is true that, so reflexivity never holds ned in example 5, will. It provides a formal way For specifying whether or not two quantities are same. The same with respect to a given setting or an attribute: is the relation? For real. A key mathematical concept that generalizes the notion of equality a key mathematical concept that generalizes the notion of.! Equivalent ( equivalence relation examples that relation ), but is false a6= b be to! Relations as subsets of a a. di erent names purposes, equivalence relation examples may be to..For example, is transitive the equivalence relation: is the relation $\geq$ on $\mathbf { }. Have many di erent names equal to modulo if is a multiple of, i.e on the set Z aRbif. Numbers is not an equivalence relation? For no real number x is it that! Will say that they are equivalent ( under that | {
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"openwebmath_score": 0.8596251010894775,
"tags": null,
"url": "http://news.shirone.co.jp/h2p2a942/fc493d-equivalence-relation-examples"
} |
orbit, binary-star
What makes a binary stable? We'll what would make it unstable? I'm not sure why you think they should spiral in towards each other. This can't happen unless there is some dissipative mechanism, like tidal interactions. Gravitational waves are ineffective in all but close binaries involving compact stellar remnants. They can't "break away" because they are gravitationally bound.
EDIT: Note that it isn't true that most systems are binaries. The majority of "systems" are in fact single stars. The binary frequency for solar-type stars is about 50% - i.e. as many singles as binaries; but the binary frequency for the much more numerous M-dwarfs is probably around 30% and so single stars outnumber binary systems, though it is not clear whether that is true at birth Duchene & Kraus (2013). | {
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java, homework, swing, sqlite, jdbc
Title: Searchable database with Java and SQL What does the code do?
Albeit unfinished (it does work, it's just not complete yet), the code creates a database (members.db) where telephone numbers, IDs, and names are stored. After which, the user can input what they'd like to do, e.g. show every telephone number in the db, every surname etc. The plan further down the line is to implement a backup function as well.
The code works as it is now, but I think it could be refactored to shorten it, which is what I'd like.
import static javax.swing.JOptionPane.*;
import static java.lang.Integer.*;
import java.util.*;
import static java.lang.System.*;
import java.sql.*;
import java.io.*;
public class memberList {
private static File dBfile = new File("members.db"); | {
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ros
in a terminal before staring LGSVL in the same terminal.
Originally posted by Josh Whitley with karma: 1766 on 2020-10-25
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by Mackou on 2020-10-30:
Thanks ! It works indeed !
What is the reason to use ROS2 Native instead of the classical ROS2 bridge ?
Did you run into any difficulties with the ROS2 bridge ?
No ROS2 bridge was available when you started ?
Thanks !
Comment by Josh Whitley on 2020-11-02:
@Mackou The primary reason was that we were seeing very high CPU usage and dropped messages through the websockets-based bridge. | {
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For $n\ge 1$ let $a_n=r_{n-1}=p_{n-1}$. Then $a_1=1$, and it follows from $(3)$ that
$$a_{2n}=r_{2n-1}=r_{n-1}=a_n$$
and
$$a_{2n+1}=r_{2n}=r_{n-1}+r_n=a_n+a_{n+1}$$
and hence that $\langle a_n:n\in\Bbb Z^+\rangle$ is the sequence defined in this question. In my answer to that question I showed that if $q_n=\frac{a_{n+1}}{a_n}$ for $n\in\Bbb Z^+$, the sequence $\langle q_n:n\in\Bbb Z^+\rangle$ is a bijection from $\Bbb Z^+$ to $\Bbb Q^+$, and the rational numbers $q_n$ are just the reciprocals of the numbers $g(n)$.
• Good work, +1! Unfortunately, even though this proof demonstrates that it works, I still find myself wondering how it works, intuitively. But one can't have everything ... – hmakholm left over Monica Mar 8 '15 at 22:48 | {
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cosmic-microwave-background
http://www.rssd.esa.int/index.php?project=Planck
http://www.sciops.esa.int/index.php?project=PLANCK&page=Planck_Published_Papers
The second link contains 25 papers on the arXiv. Again, the maps and timelines will be released in January 2013 according to the current plan - see the conclusions of the 1101.2022 paper above. | {
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quantum-state, density-matrix, linear-algebra
Title: What is the conjugate transpose of $|0\rangle_{A}|1\rangle_{B}$? Suppose a composite state is in $|0\rangle_{A}|1\rangle_{B}$. Then their conjugate transpose would be
$\langle 0|_{A}\langle 1|_{B}$?
Note: Why this question? Because I was checking MIT's "Quantum Information Science's 2010 class's homework solution of 3" (link) Page 2
Here is the corresponding problem set.(In case you feel curious) (link) Yes, the conjugate is $\langle 0|_A\langle 1|_B$. This is also other times written as $\langle 0_A|\langle 1_B|$, or $\langle 0_A|\otimes \langle 1_B|$, or just $\langle 01|$, or similar ways. These are all just notational differences. They are all equivalent as long as one knows what is being discussed. | {
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ros, roswtf
Title: Which roswtf plugins can be installed and how?
Which roswtf plugins can I install? How can I install roswtf plugins? (I just found this tutorial for roswtf which did not address the installation of plugins.)
Originally posted by thinwybk on ROS Answers with karma: 468 on 2017-10-16
Post score: 0
I can't say for certain, but I doubt there are any plugins for roswtf other than the one in tf.
Six reasons for that: | {
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spring, oscillators, phase-space, coupled-oscillators, normal-modes
$\left[s^2 \underline{\underline{M}} + \underline{\underline{K}} \right]\underline{\hat{x}} = \underline{0} $.
In your example,
$\underline{\underline{M}} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}$
$\underline{\underline{K}} = \begin{bmatrix} 2 k & -k \\ -k & 2 k \end{bmatrix}$
and the determinant of the matrix $s^2\underline{\underline{M}} +\underline{\underline{K}} $ reads
$\det (s^2\underline{\underline{M}} +\underline{\underline{K}} ) = (s^2 m + 2k)^2 - k^2 = m^2 s^4 + 4 km s^2 + 3 k^2$
the eigenvalues reads
$s_{1,2}^2 = - \left[ 2 \pm
1 \right]\dfrac{k}{m} $
and thus
$s_1^2 = - \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{1\pm} = \pm j \sqrt{\dfrac{k}{m}}$
$s_2^2 = - 3 \dfrac{k}{m} $$\quad \rightarrow \quad$$s_{2\pm} = \pm j \sqrt{3\dfrac{k}{m}}$,
while the corresponding eigenvectors are:
$\underline{\hat{x}}_{1} =\begin{bmatrix} 1 \\ 1 \end{bmatrix}$,$\quad$$\underline{\hat{x}}_{2} =\begin{bmatrix} 1 \\ -1 \end{bmatrix}$. | {
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} |
image-processing
Title: Region growing algorithm I have been trying to come up with a region growing algorithm but I'm not sure that I fully understood the region growing segmentation method (for gray-scale images). Please correct me if I'm wrong:
I start from a seed point chosen by me (brightest value that fits the wanted region ,because the segmentation target is a girl's face).
From the location of that seed point I compare the adjacent points and if their value is between a given threshold and the seed point value then I add it to the seed point map(matrix).
I must try all the valid seed points for adjacent neighbors. The result is just plotting the seed point map (matrix). Almost there: | {
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ros, kinect, openni.launch
NODES
/camera/depth/
rectify_depth (nodelet/nodelet)
metric_rect (nodelet/nodelet)
metric (nodelet/nodelet)
points (nodelet/nodelet)
/camera/rgb/
debayer (nodelet/nodelet)
rectify_mono (nodelet/nodelet)
rectify_color (nodelet/nodelet)
/
camera_nodelet_manager (nodelet/nodelet)
camera_base_link (tf/static_transform_publisher)
camera_base_link1 (tf/static_transform_publisher)
camera_base_link2 (tf/static_transform_publisher)
camera_base_link3 (tf/static_transform_publisher)
/camera/
driver (nodelet/nodelet)
register_depth_rgb (nodelet/nodelet)
points_xyzrgb_depth_rgb (nodelet/nodelet)
disparity_depth (nodelet/nodelet)
disparity_depth_registered (nodelet/nodelet)
/camera/ir/
rectify_ir (nodelet/nodelet)
/camera/depth_registered/
rectify_depth (nodelet/nodelet)
metric_rect (nodelet/nodelet)
metric (nodelet/nodelet) | {
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ros, boost, camera-drivers, rosconsole
Any ideas of what might be wrong?
/Ola
Originally posted by Ola Ringdahl on ROS Answers with karma: 328 on 2011-08-29
Post score: 2
I found out that the problem is with different versions of Boost. The driver from Basler works with Boost 1.38, but not 1.42 which is the version used by ROS. The manufacturer also seems reluctant to include support for newer versions of Boost in their diver.
/Ola
Originally posted by Ola Ringdahl with karma: 328 on 2011-09-02
This answer was ACCEPTED on the original site
Post score: 1 | {
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c++, beginner, object-oriented, constructor
void Wine::getBottles()
{
using std::cin;
using std::cout;
cout << "Enter holdings for " << yearsHeld << " years.\n";
for (int i = 0; i < yearsHeld; i++)
{
cout << "Year: ";
cin >> holdings.first()[i];
cout << "Bottles: ";
cin >> holdings.second()[i];
}
}
int Wine::sum() const
{
int sum = 0;
for (int i = 0; i < yearsHeld; i++)
{
sum += holdings.second()[i];
}
return sum;
}
… along with the author's Pair<> template:
/**
* @file pair.h
* @brief Defining & using a Pair template
*/
#ifndef PAIR_H_
#define PAIR_H_
#include <iostream>
#include <string>
template <class T1, class T2>
class Pair
{
private:
T1 a;
T2 b;
public:
T1 & first();
T2 & second();
T1 first() const { return a; }
T2 second() const { return b; }
Pair(const T1 &aval, const T2 &bval) : a(aval), b(bval) {}
Pair() {}
};
template <class T1, class T2>
T1 & Pair<T1, T2>::first()
{
return a;
} | {
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python, django
This avoids a couple of problems of the indexing approach:
It is easy to make the mistake of using a wrong number
If you need to add or remove a variable in the middle, you would need to update many numbers, possibly making a mistake.
Unpacking may also be slightly faster, but the difference is rather negligible. | {
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particle-physics, standard-model, neutrinos, beyond-the-standard-model
My question: How many parameters are required to take into account the oscillation? Does it work with only 3 mass-parameters for the 3 different known neutrinos, or are there additional parameters necessary (if so, what are they standing for)? Or is it possible to describe the effects with only 1 or 2 parameters (if so, what idea reduces the effect of the three masses to less free parameters)? The mass and weak eigenstates are related by the Pontecorvo-Maki–Nakagawa–Sakata matrix, which is a $3\times 3$ unitary matrix. Therefore, it can be parametrized by four parameters. The usual choice is the three mixing angles $\theta_{12}$, $\theta_{13}$ and $\theta_{23}$, and a CP-violating phase $\delta_{CP}$ (or, if neutrinos are Majorana fermions, 3 different CP-violating phases). Oscillation experiments provide information about the mixing angles, but not the phase. | {
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} |
java, fizzbuzz, rags-to-riches
Now compare that diff with what it looks like if the bracing is right to start with: | {
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our feedback page is to use the circle... Parallel to the radius completing the CAPTCHA proves you are a human and gives you temporary to... Called the point where it intersects is called the point of tangency is the of. A straight line that intersects a circle Pythagorean Theorem touches ( intersects ) the circle at only point! Compass and straightedge or ruler prove tangent and O P ¯ is line! ) state all the tangents to circles form the subject of several and. Line is perpendicular to the radius welcome to this video explains how to an! It works by using the Two-Tangent Theorem the step-by-step explanations can be an number... | {
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"url": "http://travel.danimgroup.com/1vg9y34/b38cbb-tangent-of-a-circle"
} |
Any number that's fully 1's of composite length is composite. If the length is $$c\cdot b$$, then that means, it contains the lower number with c 1's exactly b times, in non overlapping substrings. This makes it divisible by the number with c 1's . In this case, $$91=7\cdot 13$$. This means, the 91 ones, can be grouped as follows: $$1111111111111,1111111111111,1111111111111,111111111111,1111111111111,1111111111111,1111111111111$$ or $$1111111,1111111,1111111,1111111,111111,1111111,1111111,1111111,111111,1111111,1111111,1111111,1111111$$
But these substrings have values (in base 10) of $$\frac{10^{13}-1}{9}$$ and $$\frac{10^{7}-1}{9}$$ respectively.
91 ones represents S=$$\sum_{k=0}^{90}10^k$$
$$S=\frac{1*10^{91}-1}{10-1}=\frac{10^{91}-1}{9}$$
One less than any power of 10 is divisible by 9 but that factor is cancelled by the denominator.
$$91=7 \times 13$$
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+b^{n-1})$$
$$10^{91}=(10^7)^{13}$$
Let $$a=10^7$$ and $$b=1$$. | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/3164613/how-to-find-divisibility-of-a-very-large-number"
} |
polyfit (linear regression) works by minimizing ∑ i (Δ Y) 2 = ∑ i (Yi − Ŷi) 2. The built-in fitting functions are li ne, polynomial, sine, exponential, do uble-exponential, Gaussian, Lorent- zian, Hill equation, sigmoid, lognormal, Gauss2D (two-dimensional Gaussian peak) and Poly2D (two-. The fourth and final argument is for variable bounds; for example, for the exponential curve fit, the optimized qi value will be between 0 and the calculated qi value, found using the get_max_initial_production() function. , approaches an asymptote), you can fit this type of curve in linear regression by including the reciprocal (1/X) of one more predictor variables in the model. We start by selecting the data, producing a scatterplot, and adding a best fitting curve using an exponential model. If this coefficient equals 1, then the explanation is perfect. An exponential curve, on the contrary, is typical of a phenomenon whose growth gets rapidly faster and faster - a common case is a process that | {
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9820137874059599,
"lm_q1q2_score": 0.8204563465522612,
"lm_q2_score": 0.8354835309589074,
"openwebmath_perplexity": 996.5791857850418,
"openwebmath_score": 0.46903663873672485,
"tags": null,
"url": "http://lzxc.umood.it/exponential-curve-fitting.html"
} |
thermodynamics, entropy
Title: Trying to grasp why internal energy decreases in an isochoric, isentropic process So there's this equation:
$dU_{V,S}≤0$
which comes from
$dS≥\frac{dQ}{T}$
$dQ = dU$ since isochoric
$dS≥\frac{dU}{T}$
$TdS≥dU$
$dU≤0$ if $dS=0$
Firstly, are these set of equations valid? If so, how can the internal energy of a system decrease if there is no work done and no change in entropy? I'm having a hard time wrapping my head around this concept so does anyone have a simple viusalization or example of this process? Thermodynamic inequalities
The inequalities involving $U$, or $H$, or $A$, or $G$ require careful explanation to avoid confusion. First, we have
$$dU = T dS - P d V \tag{1}$$
which is true (we are assuming closed system, $dN=0$ throughout). But then sometimes we also see
$$dU \leq T dS - P d V \tag{2}$$
which, as written, contradicts (1). The proper way to write Eq (2) is
$$
\Big(\Delta U\Big)_{S,V} \leq 0
$$
and the proper way to read it is: | {
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performance, object-oriented, json, go
// ObjectGenerator struct that implements Generator. Used to
// generate random object
type ObjectGenerator struct {
EmptyGenerator
generators []Generator
}
// Value returns a random object
func (g ObjectGenerator) Value(r rand.Rand) interface{} {
m := bson.M{}
for _, gen := range g.generators {
if gen.Exists(r) {
m[gen.Key()] = gen.Value(r)
}
}
return m
} | {
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"url": null
} |
c++, linked-list, reinventing-the-wheel
Title: C++ Generic Linked List My goal is to write a reasonably robust c++ linked list class. The current code represents the latest phase in this attempt. My future goal is to implement a doubly linked list along with more member functions. Before I continue further, I would like some input on my overall design thus far. Also, any input on improving the code is very much welcomed and is of great concern to me. I am hoping to learn other people's approach so that I can carry what I learn into this and future projects.
template <typename T>
struct Node
{
Node(): link(nullptr), value(0) {}
Node(const T val) : link(nullptr), value(val) {}
Node(const Node<T>* node) : link(node->link), value(node->value) {}
Node<T> *link;
T value;
};
#include "node.h"
#include <cstddef>
#include <iostream>
template <typename T>
class List
{
public:
// constructors
List() : head(nullptr), sz(0) {}
List(const size_t s);
List(const List& rhs);
~List(); | {
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python, django
But you won't be able to use it for bulk deletions. | {
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quantum-field-theory, hamiltonian-formalism, quantum-anomalies
$$\partial_\mu \hat{j}^{\mu 5} \sim - \frac{g^2}{16 \pi^2} \epsilon^{\mu\nu\rho\sigma} \hat{F}_{\mu\nu} \hat{F}_{\rho \sigma}$$
where the $\sim$ indicates that only some matrix elements of the operators on both sides are shown to match. Is this in fact an operator equality? If so, where does the extra term slip in above? Short answer: The operator $\hat j$ is meaningless as written, because it contains divergences. In order to manipulate this operator (and take its divergence) one must introduce a regulator which, when removed, happens to leave a finite piece: the anomaly.
The usual narrative is that $\hat j$, defined as
$$
\hat j^\mu(x)\overset?=\bar\psi(x)\gamma^\mu\psi(x)\tag1
$$ | {
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} |
# derivative of arctan | {
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"url": "http://chodel.gmina.pl/c0jfp8/ee20ae-derivative-of-arctan"
} |
Short of making the choices that give the right answer, it's not obvious how we should decide how $\varepsilon$ approaches $0$. This is the point where the fact that $$\arctan(A)+\arctan(B)=\arctan\left(\frac{A+B}{1-AB}\right)$$ is false starts being relevant. You can see this simply by setting $A=1$ and $B=2$. Because $\arctan$ is an odd, increasing function, the left hand side is necessarily positive, but the right hand side is $\arctan(-3)$ which is necessarily negative. The formula is only valid when $0 < 1 - AB$. For the above formulas to be valid, then, $\frac{-\varepsilon}{x}$ must be positive. For the given ranges, this means $\varepsilon$ must have the opposite sign as both $x$ and $x+1$ giving the desired positive $\frac{\pi}{2}$. We also see that for $x\in(-1,0)$, the constraint requires $\varepsilon$ to be positive but $x+1$ will positive as well which will lead to $-\frac{\pi}{2}$ which is indeed what you see in a plot. | {
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"url": "https://math.codidact.com/posts/286829"
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general-relativity, differential-geometry, metric-tensor, symmetry, vector-fields
$$
and
$$
\frac{\partial x^\alpha}{\partial x'^\mu} = \delta^\alpha_\mu - \varepsilon \partial_\mu\xi^\alpha
$$
so, only keeping terms to the first order in $\varepsilon$,
\begin{align*}
g'(x + \varepsilon \xi)_{\mu \nu} &= (\delta^\alpha_\mu - \varepsilon \partial_\mu\xi^\alpha)(\delta^\beta_\nu - \varepsilon \partial_\nu\xi^\beta)g(x)_{\alpha \beta} \\
g'(x)_{\mu\nu} + \varepsilon \xi^\rho \partial_\rho g_{\mu \nu}(x)&= g(x)_{\mu \nu} - \varepsilon \partial_\mu\xi^\alpha g(x)_{\alpha \nu} - \varepsilon \partial_\nu\xi^\beta g(x)_{\mu \beta}
\end{align*}
Therefore, under this infinitesimal diffeomorphism of translation along the vector field, we can see that the change in the metric at point $x$ is
\begin{align*}
\delta g_{\mu \nu} &= \tfrac{1}{\varepsilon}(g'(x)_{\mu \nu} - g(x)_{\mu \nu}) \\
&= -\xi^\rho \partial_\rho g_{\mu \nu} - \partial_\mu\xi^\alpha g_{\alpha \nu} - \partial_\nu\xi^\beta g_{\mu \beta}.
\end{align*}
Next, note that the Lie derivative of the metric along $\xi$ is | {
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c#, object-oriented, snake-game
Name the rules controlling game play. Move isBodyCollision declaration out of the loop.
bool isBodyCollision = false
while (! isBodyCollision) {
// ...
isBodyCollision = Snake.Move(direction, Board.Width, Board.Height);
if (isBodyCollision) continue;
// ...
}
Let the loop conditional control execution. It tends to help prevent temperal coupling as code changes and complexity increases.
Leaky Abstraction
I like the variable name "Position"
public class Snake { ... Position = new List<Point>(); ... }
Playing the game we talk about the Snake's position, not a collection of positions. But then the abstraction leaks its implementation somewhat with a method name:
UpdateSnakePositionList
Change that to
UpdateSnakePosition | {
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terminology
Title: What makes a computer program modular? Consider:
Constants and variables
If-then-else-elseif statements
Procedures and loops
The very fact that a program is comprised of two or more files (instead just from one large file)
Each one of these features indicates "modularity" but what really makes a program modular by any rigorous standard of computer science (if there is one)? Barbara Liskov has done research in the area of modularity:
A program is a collection of modules
Each module has an interface, described by a specification
A module’s implementation is correct if it meets the specification
A using module depends only on the specification
Modules provide the advantage of:
Local reasoning
Modifiability
Independent development | {
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physical-chemistry, electrons, atoms
at zero velocity relative to the observer). This is a bit more complicated than it seems: consider a gas. The center-of-mass velocity may be zero (relative) but obviously each molecule has its own velocity not dependent on center-of-mass. Fortunately, in the range of temperatures chemists are concerned with, relativistic velocities aren't seen, so no correction is necessary between a molecule's rest mass and its apparent mass. Apparent mass, observed mass, relativistic mass, dynamic mass, are (usually) all different ways of expressing the fact that the energy of an object affects its mass; E= mc² and that energy is the sum of potential and kinetic terms, where the kinetic term is ½mv². Hence the mass you will observe depends on how fast the object is traveling relative to you (or your observational equipment). The Special Relativity equation for mass is m = m0 ÷ √( 1 - (v/c)²) where v is the particle's speed (magnitude of its velocity), c is the speed of light, and m0 is its mass when | {
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error-correction, stabilizer-code
Title: CSS codes are the only stabilizer codes with transversal CNOT? Given a stabilizer code $\mathcal{C}$ then
$$
\mathcal{C} \text{ is CSS} \iff \text{CNOT} \text{ is transversal}.
$$
The forward implication is well known, see for example Transversal logical gate for Stabilizer (or at least Steane code). On the other hand, Adam Zalcman claims in the comments in Small codes with transversal Hadamard that the reverse implication can be shown.
Can someone share a proof of this claim? TL;DR: There are a few inequivalent ways to define the transversal construction for a logical gate. The precise statement of the relationship between transversal CNOT and the CSS codes depends on the choice of definition.
Conventions | {
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machine-learning, classification, statistics, probability
Title: Using classifier's probabilities as independent variables to predict Y Suppose you have a classification task y~X with (n_samples,m_features). A colleague told me that it is correct to run r different classifiers to predict y based on X and then use the probabilities given for each classifier as a new matrix Xnew (n_samples_probabilities,r_columns) to train a new classifier y~Xnew
My questions are:
1) Is this something reasonable to apply?
2) If so, is there any mathematical support on this method? This is called "stacked ensembling," or just "stacking."
Wikipedia
I'm not aware of any theoretical mathematical support, but intuitively it may smooth out the results of the $r$ initial models, possibly (depending on the final model) promoting models on data segments where they perform best. Stacking consistently wins Kaggle competitions, though the added complexity might make it less attractive for certain applications. | {
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r, visualization, rstudio
Title: Which graphical tools can be used to display uni- or bivariate continuous data? There are 4 options to this multiple question,
Scatterplots
Conditional density plots
Histograms
Boxplots
I chose Scatterplots and Histograms but the answer is either wrong or not enough, I haven't seen anything about conditional density plots in my textbook and boxplots I'm not so sure. Tried googling it but there is no mention of boxplots. I think I might have misunderstood the question. It’s asking you about these and these, both of which can visualise continuous data. | {
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pressure, air
My guess is that the seal between the sheet of paper and the ground is not great, allowing air to flow underneath. However, consider the following demonstration: a ruler lies on a table with 5 inches hanging off the side. If I spread a sheet of newspaper over the part of the ruler on the table, I increase the surface area for air pressure to press down on the ruler (without the sheet of newspaper, the air pressure is just acting on the ruler which doesn't have a lot of surface area). Now I can karate chop the ruler and break it in half due to atmospheric pressure giving a counter torque. However, why is it that I can easily lift the ruler with the newspaper on it (meaning, instead of karate chopping, I grab the ruler and just lift)? My guess is that, even if there is 3000 lbs of force acting downwards on the newspaper, I can still lift the ruler newspaper ensemble with say 1 lb of force because air instantly rushes into the cracks. But why don't I feel 2999 lbs of resistive force | {
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matrix is matrix the contains the euclidean distance between each point across both matrices. It works for (easier to reason through) 1, 2, or 3 dimensions, plus 4, 5, and 6 dimensions as well. The raw euclidean distance is 109780.23, the Primer 5 normalized coefficient remains at 4.4721. Learn more about Euclidean distance analysis. Distance Between Two Points Calculator This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces. The formula is derived from the hypotenuse of a right angle triangle – if you drew two line segments from the points that met at a 90 degree angle, the opposite side length (our distance) called the hypotenuse, is easier to find. Since the distance … Euclidean Distance Calculator The distance between two points in a Euclidean plane is termed as euclidean distance. "1 Dimension" distance is just a straight line distance on a single axis. Euclidean space was originally | {
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"lm_q2_score": 0.8397339736884711,
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"openwebmath_score": 0.6504970788955688,
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"url": "https://www.abbaselama.org/all-john-kado/13a95a-euclidean-distance-calculator"
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from state,! Chain transitions in a simulation to move from one state to another reaching. 'M trying to figure out this problem geeksforgeeks.org to report any issue with the DSA Paced! Reaching a vertex, with equal probability for each the best answers voted! Our website to the top ccccc } & 4 & 7 & 9 10! It take to reach either 3 or 7 like Counterspell asking for,. There are four states in this Markov chain from the output of two other Markov chains how stop... A registered trademark of Wolfram Research, Inc Totalus ) without using the counter-curse let p the... Generate link and share the link here probability, an absorbing state probabilities from state 1, we make! How long on average does it take to reach either 3 or 7 0.5 & 0.5 & 0 ej... Called the transition markov chain probability of reaching a state, and I was interested in only one transient state Gun when in! Clarification, or responding to other answers when it is bothering me mathematica Exchange! Least one such state is | {
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"lm_q2_score": 0.8438950966654774,
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"openwebmath_score": 0.6366798877716064,
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"url": "https://absfinishers.com/economic-development-zqt/markov-chain-probability-of-reaching-a-state-1d371d"
} |
# 11.4: The Integral Test
It is generally quite difficult, often impossible, to determine the value of a series exactly. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem.
If all of the terms $$a_n$$ in a series are non-negative, then clearly the sequence of partial sums $$s_n$$ is non-decreasing. This means that if we can show that the sequence of partial sums is bounded, the series must converge. We know that if the series converges, the terms $$a_n$$ approach zero, but this does not mean that $$a_n\ge a_{n+1}$$ for every $$n$$. Many useful and interesting series do have this property, however, and they are among the easiest to understand. Let's look at an example.
Example 11.3.1
Show that $\sum_{n=1}^\infty {1\over n^2}$ converges.
Solution | {
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javascript
.main {
width: 100vw;
height: 100vh;
display: flex;
flex-direction: column;
justify-content: space-between;
align-items: flex-start;
color: white;
background-color: rgba(0, 0, 0, 0.65);
}
.title {
margin-top: 7%;
height: 80px;
width: 100%;
display: flex;
flex-direction: column;
align-items: center;
justify-content: space-evenly;
/* animation: display_opacity_zoom 1s ease-out; */
}
.year_title {
margin-left: 5px;
font-size: 40px;
letter-spacing: 5px;
color: lightsalmon;
text-align: center;
}
.month_title {
margin-left: 15px;
font-size: 25px;
letter-spacing: 15px;
text-align: center;
}
.calendar {
height: 75%;
width: 100vw;
display: flex;
flex-direction: column;
justify-content: space-between;
align-items: center;
}
.month_days {
margin-top: 10px;
width: 100%;
height: 50%;
/* animation: opacity 1s ease-in-out; */
}
table {
margin-top: 20px;
width: 100%;
font-size: 22px;
} | {
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mathematics, celestial-mechanics, ephemeris, equinox, archaeoastronomy
Meeus Astronomical Algorithms table 27.B gives a similar polynomial optimized for years 1000 to 3000.
The periodic terms in table 27.C, modeling perturbations by the Moon and other planets, add up to 29 minutes at most. | {
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"url": null
} |
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