text stringlengths 1 1.11k | source dict |
|---|---|
thermodynamics, entropy, reversibility
I know that if I consider an irreversible adiabatic process from A to B, the change in entropy $\Delta S_{AB}$ is more than $0$ and that if I consider a reversible adiabatic process starting from A, its path will not pass through B (and its difference of entropy will be $\Delta S_{AP} = 0$, where $P$ is its ending point) | {
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"tags": "thermodynamics, entropy, reversibility",
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} |
c++, performance, xml
str_set b;
doc.load_file("new.xml");
// fill set b with just the ids from file b
for(auto&& node: doc.child("site_entries").children("entry"))
b.emplace(node.child("id").text().as_string());
// now use the <algorithms> library
str_set b_from_a;
std::set_difference(a.begin(), a.end(), b.begin(), b.end()
, std::inserter(b_from_a, b_from_a.begin()));
str_set a_from_b;
std::set_difference(b.begin(), b.end(), a.begin(), a.end()
, std::inserter(a_from_b, a_from_b.begin()));
str_set a_and_b;
std::set_intersection(a.begin(), a.end(), b.begin(), b.end()
, std::inserter(a_and_b, a_and_b.begin()));
for(auto&& v: a)
con("a : " << v);
con("");
for(auto&& v: b)
con("b : " << v);
con("");
for(auto&& v: b_from_a)
con("b_from_a: " << v);
con("");
for(auto&& v: a_from_b)
con("a_from_b: " << v);
con(""); | {
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} |
ros, jointtrajectory, pr2
with the time in seconds in the brackets. And your stamp values in your header you just use:
header.stamp = ros::Time::now();
before you publish it.
hope it helps
Originally posted by PeterMilani with karma: 1493 on 2013-03-03
This answer was ACCEPTED on the original site
Post score: 1 | {
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"tags": "ros, jointtrajectory, pr2",
"url": null
} |
filters, filter-design, dft
Assume fundamental sample time is 1 and Step time is 1, the result of
output at time 5 is?
Assuming that you have knowledge of DSP you claim, it is a very simple task.
First of all convert the difference block into proper filter representation (negative powers of $z$ reflect delay of samples), simply divide it by $z$:
$$\dfrac{z-1}{z}=\dfrac{\frac{z}{z}-\frac{1}{z}}{\frac{z}{z}}=\boxed{\dfrac{1-z^{-1}}{1}} $$
Now we have two filter blocks connected one after another, thus the transfer functions have to be multiplied to obtain general transfer function $H(z)$:
$$H(z)=\dfrac{Y(z)}{X(z)}=\dfrac{1-z^{-1}}{1} \cdot \dfrac{1}{1+z^{-2}}=\boxed{\dfrac{1-z^{-1}}{1+z^{-2}}}$$
Having the transfer function of our IIR filter, we can convert back to difference equation, let's multiply both sides of the above equation and use the time-shifting property of the Z-transform:
$$Y(z)\left(1+z^{-2}\right)=X(z)\left( 1-z^{-1}\right)$$
$$Y(z)+Y(z)z^{-2}=X(z)-X(z)z^{-1}$$
$$y[n]+y[n-2]=x[n]-x[n-1] $$ | {
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} |
electrostatics, acoustics, multipole-expansion
Title: Physical intuition for quadrupole source In his Theory of Vortex Sound M. S. Howe defines sources "mathematically" (i.e. dipole is a source that could be described as a vector and than there is proved that it's equivalent to a two point sources with usual features). The problem arises when it comes to quadrupoles:
A source distribution involving two space derivatives is equivalent to a combination of four monopole sources (whose net volume source strength is zero), and is called a
quadrupole. A general quadrupole is a source of the form: $$ F = \frac{\partial \ T_{i,j}}{\partial x_i \partial x_j}$$ in the equation: $$\Box p = F$$ | {
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"tags": "electrostatics, acoustics, multipole-expansion",
"url": null
} |
black-hole, universe, cosmology
Title: Observable universe equals its Schwarzschild radius (event horizon)? The estimated age of the universe is 14 billion years.
The estimated Schwarzschild radius (event horizon) of the observable universe is 14 billion light-years.
What are the ramifications? There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is not equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of $\gtrsim3$.
The Schwarzschild radius of the Universe
Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is $R \simeq 46.3\,\mathrm{Glyr}$.
The mean density of the Universe is very close to the critical$^\dagger$ density $\rho_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$. | {
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"tags": "black-hole, universe, cosmology",
"url": null
} |
Dual; Minimization with constraints 5. feasible region I This feasible region is a colorred convex polyhedron (àıœ/) spanned by points x 1 Toy LP example: brewer’s problem. Let's say a FedEx . Where, , ≥ 0 . 3 Date 2015-09-18 Title Interface to 'Lp_solve' v. To achieve this requirement, convert any unrestricted variable X to two non-negative variables by substituting T - X for X. Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. A company makes two products (X and Y) using two machines (A and B). zx y xy xy xy. Sometimes for integer variables the value is not integer. t. In this Linear Programming: related mathematical techniques used to allocate limited resources among competing demands in an optimal way . ment of linear programming and proceeds to convex analysis, network flows, integer programming, quadratic programming, and convex optimization. 3 THE SIMPLEX METHOD: | {
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"lm_q2_score": 0.8705972684083609,
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"openwebmath_score": 0.5247300267219543,
"tags": null,
"url": "http://makariharlem.com/qt6hqml/linear-programming-2-variables-examples.html"
} |
matrix, matrix-factorisation, notation
Convolutional Neural Networks
CNNs add an additional filtering step before passing through the weights. It passes a filter through the matrices to get a value that represents a neighborhood around a value. The filter takes the Hammond product and then sums all the elements of the resulting matrix.
For example, if we have the matrix in green
with the convolution filter
Then the resulting operation is a element-wise multiplication and addition of the terms as shown below. This kernel (orange matrix) $g$ is shifted across the entire function (green matrix) $f$. Performing a Hammond product at each step, and then summing the elements.
Note: The Hammond product is often defined for matrix of exactly the same dimensions, this is relaxed in CNNs due to the filter moving through the image successively. It is also possible to perform the Hammond product on the edges of the image where different padding techniques can be used. | {
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"url": null
} |
php, design-patterns, dependency-injection, static
/**
* Dependancy Injector Container / Inversion of Control
*
* Learning project for DIC/IoC design pattern.
*
*/
class IoC {
private static $definitions = array();
private static $registry = array();
/**
* Register a Closure for use in creating an object later.
*
* @param string $identifier String to identify the registered definition.
* @param Closure $closure A Closure to be called that will create an object.
*/
public static function register($identifier, Closure $closure) {
if(array_key_exists($identifier, self::$definitions)) {
throw new Exception("Identifier '$identifier' already defined in registry.");
} else {
self::$definitions[$identifier] = $closure;
}
} | {
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"tags": "php, design-patterns, dependency-injection, static",
"url": null
} |
java
public boolean isSupposedToSucceedInDefense() {
return isSupposedToSucceedInDefense;
}
public void setSupposedToSucceedInDefense(boolean supposedToSucceedInDefense) {
isSupposedToSucceedInDefense = supposedToSucceedInDefense;
}
}
TestFightRules:
public class TestFightRules implements FightRules<Chance, TestEntity> {
@Override
public Chance getAttackChance(TestEntity attacker, TestEntity defender) {
if (attacker.isSupposedToSucceedInAttack()) {
return new WinningChance();
}
return new FailingChance();
}
@Override
public Chance getDefendChance(TestEntity attacker, TestEntity defender) {
if (defender.isSupposedToSucceedInDefense()) {
return new WinningChance();
}
return new FailingChance();
} | {
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"tags": "java",
"url": null
} |
sensors
Is ultrasonic even the best choice? I was hoping that the beacon would be kept in a pocket, so I figured optical sensors were out.
Edit: The beacon and robot will both be mobile so fixed base stations are not an option. If you were to place 3 ultrasonic microphones/sensors in a triangle layout you should be able to determine the direction of the beacon based on the time difference the ping arrives at each sensor with some straightforward trigonometry. They'd all have to be facing in the same direction (probably upwards) and you may have to find some way to make them omnidirectional. The further apart you can space the sensors, the more accurate the calculation will be. | {
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} |
php, json, web-scraping, api, client
$allStocksJson=json_encode($allStocks);
// Write the raw file
$symbolsDir= $i->symbolsDir;
if (!is_dir($symbolsDir)) {mkdir($symbolsDir, 0755,true);}
$rawStockFile=$symbolsDir . "/" . $currentTime . ".json";
$fp=fopen($rawStockFile, "x+");
fwrite($fp, $allStocksJson);
fclose($fp);
echo "YAAAY! stock large json file updated successfully! \n";
}
}
?> | {
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"tags": "php, json, web-scraping, api, client",
"url": null
} |
friction
EDIT
On a second thought i realised the frame is not the local one (see this issue), but the fdir does not work in the global frame.
EDIT2 Indeed I was right. The fidir still does not work for all the cases. See here.
Originally posted by SorinV with karma: 41 on 2017-06-22
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by josephcoombe on 2018-09-12:
At least according to the Bitbucket issue (https://bitbucket.org/osrf/gazebo/issues/2068/directional-friction-still-broken), the issue is resolved as of 2017/09/11.
Comment by yuril on 2020-03-05:
do you manage to get the wheels right? if so could you show me how? | {
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organic-chemistry, hybridization
With that out of the way, here is how benzyne works (PBE0/cc-pVTZ gas phase optimisation + NBO analysis). We're going to look at the orbitals that the triply bonded carbon (C1) uses for bonding to its neighbours (C2 and C6, see structure below):†
One sp orbital (50.02% s, 49.75% p) for forming a σ-bond to C2;
One p orbital (0.39% s, 99.38% p) for forming the π-bond to C2 in the plane of the molecule;
One p orbital (0.00% s, 99.84% p) for forming the π-bond to C2 perpendicular to the plane of the molecule;
One sp orbital (49.64% s, 50.17% p) for forming a σ-bond to C6.
On their own, the numbers don't tell you which direction the p-orbitals are pointing in, but you can use visualisation software to figure this out. From left to right, these are the (σ- or π-, not the individual hybrid) orbitals, in the same order as listed above.
Based on the above, the orbitals can be crudely illustrated as such: | {
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rna-seq, differential-expression, quality-control
Title: Removing PCR duplicates in RNA-seq Analysis After reading some of the forum posts in Biostar and SeqAnswers I find it very confusing whether to filter out the duplicate reads from aligned files or not. As far I understand it's very difficult to distinguish between highly expressed genes and duplicate reads and we may lose important information during the filtration process.
So, is it really necessary to remove the duplicates in differential expression analysis using RNA-seq data? For normal RNA-seq PCR duplicates are normally kept in, but the duplication rate can be used as a quality control: The higher the duplication rate, the lower the quality. For expression analysis, it is probably best to discard high duplication rate samples, rather than deduplicate them. | {
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However, it looks 'hairy' when I try with larger derivatives, like, for the function:
$$(13x^2-5x+8)^{\frac{1}{2}}$$ we do: $$\frac{d(13x^2-5x+8)^{\frac{1}{2}}}{dx} = \frac{d\color{Green}{(13x^2-5x+8)}^{\frac{1}{2}}}{d\color{Green}{(13x^2-5x+8)}}\frac{d(13x^2-5x+8)}{dx} = \frac{1}{2\sqrt{\color{Green}{13x^2-5x+8}}}(26x -5)$$ but it's really better for me to do like this, instead of doing the bla bla bla of changing variables and stuff. But I'm afraid my teacher does not accept this. Is this notation/way of doing good for you guys?
One more example: $$\frac{d}{dx}\sqrt{(\sin(7x+\ln(5x)))} =$$ | {
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python, sorting, pandas
So I created a function (not finalized, but the idea)
def sort_df_values(df, column_names, ascending=False, fixed_categories=None):
if fixed_categories is not None and fixed_categories:
original_positions = {name: position for position, name in enumerate(df.index.values) if name in fixed_categories}
original_positions = dict(sorted(original_positions.items(), key=operator.itemgetter(1), reverse=True))
excluded = df.loc[fixed_categories]
included = [name for name in list(df.index) if name not in fixed_categories]
new_df = df.loc[included].sort_values(column_names, ascending=ascending)
result = pd.concat([new_df, excluded])
new_index_values = list(result.index.values) | {
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java, beginner, gui, calculator
} else {
try { // standard "=" operation
memory.add(new BigDecimal(numValue));
numValue = "";
result = opEqual(memory, operation);
mainDisplay.setText(result
.setScale(prec, RoundingMode.HALF_DOWN)
.stripTrailingZeros().toPlainString());
operation.add(0, "0");
memory.removeAll(memory);
secDisplay.setText("");
sideTextArea
.append("\u00BB" + mainDisplay.getText() + "\n");
// if user will try to get result, after unnecessary usage
// of operation button(+,-,etc. | {
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"openwebmath_score": null,
"tags": "java, beginner, gui, calculator",
"url": null
} |
ros, catkin, package.xml, metapackage, buildfarm
Title: How is the info in the package.xml of a metapackage used?
My guess is that one of the usage is to let buildfarm know which pkgs in the given metapkg to be built. | {
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fft, fourier-transform, frequency-spectrum, python, physical-units
Euler's identity shows the OP's example best. From Euler's we see that:
$$cos(\omega t) = \frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$
If we scale the OP's result by $N=1000$, we get $\frac{1}{2}$ which is exactly the coeffiicents of each the exponential terms within the signal. The Fourier Transform itself decomposes arbitrary signals into each of its $e^{j \omega t}$ terms, either as a continuous function for waveforms of infinite duration without repetition, or as discrete terms when repetition exists. The DFT is similar when we consider the finite duration sampled time domain signal will have the same mathematical result in the Fourier Transform as a signal that repeats over that same interval extending to infinity in time, with the distinction that the frequency domain result will be continuous but will be zero every where except at the non-zero DFT results. | {
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which is option (4) in the problem.
Once again, the limit should converge relatively quickly, so let’s check. The claimed value is 1.81093; taking $$n=5$$ the sum is $$0.390476+0.373913+0.36+0.348148+0.337931\approx1.810468$$. That looks pretty good.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {
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gazebo, rviz, urdf, xacro, robot
You can get the fixed version here.
Originally posted by Martin Günther with karma: 11816 on 2011-10-09
This answer was ACCEPTED on the original site
Post score: 3
Original comments
Comment by Martin Günther on 2011-10-10:
Yes, the collision model can be different from the visual model. However, in your case the collision model was offset quite a large distance from the visual, and so the wheels wouldn't touch the ground. You can visualize this in RViz: select "Robot model" and check the box "collision enabled".
Comment by ZdenekM on 2011-10-10:
Thanks for your time! Your answer helps me a lot. So, the visual and collision model must be the same? I read somewhere that collision model can be simpler than visual so I thought they can differ. | {
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Establishing the bijectivity of $Q$ and defining the geometric program precisely seem a bit long for a MathOverflow post. I'll post a more detailed description to ArXiv and put a link here when I'm done.
-
I look forward to see how the inversion can be expressed as a geometric problem. – mpiktas Nov 26 '15 at 16:20
(Edited to fix a bug.)
I think the following bijection will do what you want.
For $1\leq i,j\leq n-1$, define $$r_{ij}=\log(p_{ij}/p_i)$$
Given the $r_{ij}$ and the marginals, we can recover the $p_{ij}$ as follows:
$p_{ij}=p_i \exp(r_{ij})$ for $1\leq i,j \leq n-1$.
$p_{in}=p_i - \sum_{j=1}^{n-1}p_{ij}$ for $1 \leq i \leq n-1$.
$p_{nj}=q_j - \sum_{i=1}^{n-1}p_{ij}$ for $1 \leq j \leq n$.
Note that we do not use the fact that $\sum p_i=\sum q_j=1$, only that the individual $p_i$ and $q_j$ are known and non-zero. For example, we could apply this transformation to doubly stochastic matrices with non-zero entries. (In that case, $p_i=q_j=1$ for all $i$ and $j$.) | {
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"tags": null,
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gel-electrophoresis
Quality of EtBr. I've never actually had this happen, but your stock may be bad for some reason. Ask another lab for a ul or two to easily control for this.
Quality of DNA. Obviously, your sample must actually have DNA in it, and not be degraded. You can check amount and purity easily with a spectrophotometer, but it will not tell you if the DNA has been digested into its component nucleotides (that is usually checked with a gel).
Amount of gel. The crucial variable is obviously not total amount of EtBr, but concentration. Less gel, brighter bands - but you seem to be using a standard gel amount.
Quality of gel. Again, never had this happen, but try making the gel with someone else's reagents. | {
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Problems 1-3; Unit 9 Test:Volume and Surface Area, Friday, Monday May 1, 4. Solve problems requiring the calculation of the dimensions as well as the surface area of cylindrical shapes abstractly and in context. Eight basketballs are put into a holding. Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in. 5 in the prism? 3 2 A desk is 2 feet wide and 4 feet long. The difference between their surface areas is 160 p. Find the lateral surface area and the total surface area of a cuboid whose dimensions are 26 m,14 m and 6. SOLID CUBE:A solid cube is the part of the space enclosed by six faces of the cube. A painter uses a roller to paint a wall. 3) Calculate the surface area of composite shapes. In Activity 3, students they solve for the surface area of pyramids and cones. What is the volume of each freezie? 2. ^3\) is to be constructed. Volume = 1/3 | {
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ros, osx
Title: After so many years of ROS, why hasn't Mac OSX been stabilized?
After so many years of ROS, why hasn't Mac OSX been stabilized?
Originally posted by shyamalschandra on ROS Answers with karma: 73 on 2012-12-02
Post score: 5
Maintaining a huge software system for many platforms is hard and it gets even harder because ROS packages are not created by only one group of developers but by many institutions all over the world, most often in an academic environment. To provide full support for OSX, every developer who is releasing packages would need to port his code on OSX besides the normal Ubuntu platforms. Since there are not even OSX virtual machines, this would require to own a Mac. I don't thing this is feasible.
But ROS is open source and if you are missing something, you can volunteer for maintaining ROS packages for OSX. I'm sure the community would be really happy about that. | {
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To talk more about the rotated ellipse, such as how I found the tilted axes, would take us too far afield. But if this made you curious, here is a thorough textbook explanation from Libretexts.
But let’s do what I suggested, and try taking $$x=\frac{\sqrt{2}}{2}$$. First, we find y: $$3x^2+2xy+y^2=2\\ 3\left(\frac{\sqrt{2}}{2}\right)^2+2\left(\frac{\sqrt{2}}{2}\right)y+y^2=2\\ \frac{3}{2}+\sqrt{2}y+y^2=2\\ y^2+\sqrt{2}y-\frac{1}{2}=0\\ 2y^2+2\sqrt{2}y-1=0\\ y=\frac{-2\sqrt{2}\pm\sqrt{16}}{4}=\pm 1-\frac{\sqrt{2}}{2}$$ So the two points we get are $$(0.707,-1.707)$$ and $$(0.707,0.293)$$.
Now, using the formula we got for the derivative, we get $$y’=\frac{-(3x+y)}{x+y}\\ =\frac{-(3\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2})}{\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2}}\\ =\frac{-(\sqrt{2}\pm 1)}{\pm 1}=-\sqrt{2}-1,\sqrt{2}-1$$ These are the same as the slopes of the axes, $$y=\left(-1\pm\sqrt{2}\right)x$$. | {
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classical-mechanics, notation, action, calculus, variational-calculus
Title: The $\delta$ notation in Goldstein's Classical Mechanics on the calculus of variation In Goldstein's classical mechanics (page 36) he introduces the basics of the calculus of variation and uses it to effectively the Euler-Lagrange equations. However, there is a step in which the $\delta$ notation is defined:
$$\delta y\equiv \left(\frac{\partial y}{\partial \alpha}\right)\text d\alpha,$$
in which $\alpha$ is the parameter used in the path modification:
$$y(\alpha,x)=y(0,x)+\alpha\eta(x),$$
$x$ is effectively a generalised time parameter. Both of these definitions are fine, however this notation is then introduced into the action integral:
$$\frac{\text dJ}{\text d\alpha}=\int^{x_2}_{x_1}\left( \frac{\partial f}{\partial y} - \frac{\text d}{\text dx}\frac{\partial f}{\partial \dot y}\right)\frac{\partial y}{\partial \alpha}\text dx,$$
which becomes: | {
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## Use in physics
The radian is widely used in physics
Physics
Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.Physics is one of the oldest academic...
when angular measurements are required. For example, angular velocity
Angular velocity
In physics, the angular velocity is a vector quantity which specifies the angular speed of an object and the axis about which the object is rotating. The SI unit of angular velocity is radians per second, although it may be measured in other units such as degrees per second, revolutions per...
is typically measured in radians per second (rad/s). One revolution per second is equal to 2π radians per second. | {
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robotic-arm, manipulator, trajectory
I think that for let understand my doubts, I have to first say how I thought to tackle the problem, even if it is clearly the wrong way.
My idea for solving this proble was to find the maximum velocity, which is:
$V_{max}=\frac{T}{4}$
and from it derive the time $T$, which in my opinion was the minimum motion of time because it is referred to the maximum velocity. So, I thought the minumum time velocity to be:
$T = \frac{4V_{max}}{A}$
but this is obviusly wrong. Infact, when I went looking for the solution that my professor proposed for this exercise, the correct way was the following: | {
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machine-learning, deep-learning, supervised-learning, gan
Even the random distribution that the fake images follow will have some pattern. Hence convergence will be faster.
You can control the output of generator at test time by giving label for image you want to generate. | {
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roslaunch
Title: pausing inside .launch file?
Is there any way to add a sleep (delay or pause) to a .launch file between statements?
In my current code I need to kill several default PR2 controllers, change some rosparams, then reload these same controllers. However, this all happens so fast in my launch file that it seems to cause problems. | {
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java, numbers-to-words, divide-and-conquer
@Override
public ILangNumber multiply(ILangNumber number) {
long mult= getNumber()*number.getNumber();
return createNumber(mult);
}
@Override
public ILangNumber divide(ILangNumber number) {
long divide= getNumber()/number.getNumber();
return createNumber(divide);
}
@Override
public ILangNumber pow(ILangNumber exponent) {
long pow= (long)Math.pow(getNumber(), exponent.getNumber());
return createNumber(pow);
}
@Override
public long getNumber() {
return number;
}
abstract protected String parseNumber(long number);
abstract protected ILangNumber createNumber(long number);
@Override
public String toString(){
return parseNumber;
}
}
Implementation
package numberTranslator;
public class SpanishNumber extends AbstractLangNumber{ | {
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"url": null
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lagrangian-formalism, linear-systems
identical to \eqref{A-12}.In other words, since $\:A^{\boldsymbol{-}1}\:$ is also orthogonal it leaves the inner product of two vectors invariant, so in \eqref{B-14} we could replace any inner product $\:\left(A^{\boldsymbol{-}1}\mathbf{x}\right)\boldsymbol{\cdot}\left(A^{\boldsymbol{-}1}\mathbf{y}\right)\vphantom{\dfrac{a}{b}}\:$ by | {
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We know $$\overline{DEF} = 99x$$ but that doesn't help us find the three digits we need in order to calculate $$\overline{FED}$$.
Let's start by taking a step back, what's $$99x$$ equal to?
$99x = \overline{ABC} - \overline{CBA}$
$99x = (100A + 10B + C) - (100C + 10B + A)$
We can use this to find the separate digits, since each digit in $$\overline{DEF}$$ is equal in some way to the corresponding digits in the two previous numbers. In other words:
$D = A - C$
$E = B - B = 0$
$F = C - A$
So now $$\overline{DEF} = 100(A-C) + 10(0) + (C-A)$$
We already said that $$x = A - C$$ so we'll substitute that in
$\overline{DEF} = 100x - x = 99x$
We've now just proved that we still have the same number, so we'll stop using $$x$$ for now and switch back to using the three digits.
$\overline{DEF} = 100(A - C) + 10(0) + (C - A)$ | {
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perl, linux
# tofix: could figure out beforehand what the largest filename is, and then position the file-size category accordingly.
# tofix: also, the header should be either singular or plural, not both.
# tofix: sorting week1 .. week12 does not work correctly.
# tofix: add: type(eg .pm) and line_quantity.
########################################################################### | {
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ros, rosserial, publisher
class SensorStatePublisher
{
public:
SensorStatePublisher()
{
state_listener = nh.subscribe<range_msgs::SensorStates>("/range_data",100000, &SensorStatePublisher::sensor_msg_callback, this);
ir_1_publisher = nh.advertise<sensor_msgs::Range>("/ir_1_data", 100000);
ir_2_publisher = nh.advertise<sensor_msgs::Range>("/ir_2_data", 100000);
ir_3_publisher = nh.advertise<sensor_msgs::Range>("/ir_3_data", 100000);
ir_4_publisher = nh.advertise<sensor_msgs::Range>("/ir_4_data", 100000);
ir_5_publisher = nh.advertise<sensor_msgs::Range>("/ir_5_data", 100000);
ir_6_publisher = nh.advertise<sensor_msgs::Range>("/ir_6_data", 100000);
ir_7_publisher = nh.advertise<sensor_msgs::Range>("/ir_7_data", 100000);
} | {
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beginner, php, html
Title: Methods To Generate an HTML String using an Array in PHP Problem
Method getQuoteHTML generates an HTML string using an array returned from getQuoteParams method. They are methods of a large class, UpdateStocks.
Would you be so kind and help me to replace them with faster, easier or more efficient methods, if possible?
getQuoteHTML:
/**
*
* @return a large HTML string of current quote section
*/
public static function getQuoteHTML($a){
$i=UpdateStocks::getQuoteParams();
$bt='<a href="#" class="s18 ro tx-1 b119 r100 t-21 p-2 br-5 mv-3" onclick="J.s({d:this});return false;" title="'.$a["symbol"].' latest quote"> Quote: '.date('l, d F Y \⏱ H:i T',microtime(true)).'</a>';
$hs='';
foreach($i as $k=>$p){
$h='';
if(isset($p["id"])&&!empty($a[$p["ky"]])){
if(preg_match('/(^s-*)/i',$p["id"])==1){
$r=rand(20,99);
$h=$h.'<p id="'.$p["id"].'">'; | {
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performance, beginner, vba, excel
On that subject: I see getBook and I think "That's a method to get a workbook object? Maybe a boolean to denote whether a workbook should be retrieved?". Wait, what do you mean that's the name of the active workbook? Why on earth wouldn't you call it activeWorkbookName!
Continuing in the same vein: If something is a Worksheet object then that should be obvious from its' name. Rather than data, which could be meant to represent literally anything , how about wsData? just 2 characters makes it so much clearer what's going on here. wsLo, wsWa, wsOh, wsAt, the abbreviations are still painful to keep track of, but at least when they pop up in the code later, I'm going to know roughly what they are. | {
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slam, navigation, rostopic
/camera/depth_registered/image_raw/theora/parameter_updates
/camera/depth_registered/points
/camera/depth_registered/sw_registered/camera_info
/camera/depth_registered/sw_registered/image_rect
/camera/depth_registered/sw_registered/image_rect/compressed
/camera/depth_registered/sw_registered/image_rect/compressed/parameter_descriptions
/camera/depth_registered/sw_registered/image_rect/compressed/parameter_updates
/camera/depth_registered/sw_registered/image_rect/compressedDepth
/camera/depth_registered/sw_registered/image_rect/compressedDepth/parameter_descriptions
/camera/depth_registered/sw_registered/image_rect/compressedDepth/parameter_updates
/camera/depth_registered/sw_registered/image_rect/theora
/camera/depth_registered/sw_registered/image_rect/theora/parameter_descriptions
/camera/depth_registered/sw_registered/image_rect/theora/parameter_updates
/camera/depth_registered/sw_registered/image_rect_raw
/camera/depth_registered/sw_registered/image_rect_raw/compressed | {
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26/05/2011В В· The method of convolution is a great technique for finding the probability density function (pdf) of the sum of two independent random variables. We state the convolution formula in the continuous case as well as discussing the thought process. Some examples are provided to demonstrate the technique and are followed by an exercise. Improved approximation of the sum of random vectors by the skew normal distribution Christiansen, Marcus C. and Loperfido, Nicola, Journal of Applied Probability, 2014; Approximation of partial sums of arbitrary i.i.d. random variables and the precision of the usual exponential upper bound Hahn, Marjorie G. and Klass, Michael J., The Annals of | {
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ros, moveit, follow-joint-trajectory, ros-kinetic, movegroup
Title: FollowJointTrajectoryAction feasibility checking
MoveIt seems to do some sanity checking before executing trajectories and I was wondering if using the follow_joint_trajectory action gets you those same checks. I sent a bad trajectory that didn't start from the current pose today in simulation and expected an error but instead I got a instantaneous jump in the robots simulated state (very bad). I was expecting an error such as this but just got straight up execution.
Is this expected behavior?
Should I perhaps wrap my trajectory with a MoveGroup::Plan and then execute via MoveGroup::execute(const Plan &plan) call? Do I get other checks this way as well like collisions, joint limits, etc.?
Edit: In this particular case I am using a move group on top of industrial_robot_simulator
Originally posted by BrettHemes on ROS Answers with karma: 525 on 2018-10-16
Post score: 1 | {
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complexity-theory, closure-properties
After understanding the definition of closed under left/right polynomial composition, I think I was able to solve items (a), (b), (c) and (f). However, I was not able to solve items (d) and (e).
My solution for item (a):
Closed Under Left Polynomial Composition: consider an arbitrary $f(n) \in C$ and an arbitrary polynomial $p(n)$. Then, $f(n)$ is of the form $n^{k'}$, for some $k' > 0$ and therefore $p(f(n))$ is a polynomial. Let $k''$ be the degree of the polynomial $p(f(n))$. Take $g(n) = n^{k''} \in C$ and we have $p(f(n)) = O(g(n))$.
Closed Under Right Polynomial Composition: same reasoning.
My solution for item (b):
Not Closed Under Left Polynomial Composition: consider as a counterexample $f(n) = n \in C$ and $p(n) = n^2$. Then, $p(f(n)) = n^2$. For every $g(n) = k n \in C$ we have $O(g(n)) = O(n)$. Since $n^2 \neq O(n)$ we conclude.
Not Closed Under Right Polynomial Composition: the previous counterexample applies.
My solution for item (c): | {
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• How did you involve the derivative to calculate the sum? – user19405892 Jul 7 '16 at 0:41
• @user19405892: $\sum_kkq^k=q\frac{\mathrm d}{\mathrm dq}\sum_kq^k$. – joriki Jul 7 '16 at 2:33
• @user19405892 are you familiar with the derivative? Just remember that $kq^{k-1} = \frac{d}{dk} q^k$ (it is just the power rule). Multiply both sides by $q$ and you get that $kq^k = k\frac{d}{dk} q^k$. All that is left to do is take $q$ and $\frac{d}{dq}$ outside the sum (taking the derivative out isn't always valid, but it is here because polynomials are nice functions in this regard) – Brevan Ellefsen Jul 7 '16 at 3:44 | {
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with at least 5 vertices.viii. With 4 vertices (labelled 1,2,3,4), there are 4 2 How many of these are not isomorphic as unlabelled graphs? By continuing you agree to the use of cookies. But still confused between the isomorphic and non-isomorphic $\endgroup$ – YOUSEFY Oct 21 '16 at 17:01 A bipartitie graph where every vertex has degree 3. iv. Copyright © 2021 Elsevier B.V. or its licensors or contributors. Their degree sequences are (2,2,2,2) and (1,2,2,3). In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. A bipartitie graph where every vertex has degree 5.vii. Therefore, a large class of graphs are non-isomorphic and Q-cospectral to their partial transpose, when number of vertices is less then 8. An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. These can be used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). | {
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"url": "http://jitsurei.net/boys-twin-xdjeaej/0ad894-non-isomorphic-graphs-with-8-vertices"
} |
quantum-mechanics, solid-state-physics, harmonic-oscillator
Title: Kronecker delta in the harmonic lattice potential energy This is a rather mathematical question regarding a derivation of the harmonic lattice energy $\Phi^{ham}$ in a book about the quantum theory of solids. Specifically, I found the third equivalence in Eqn. (1) below is quite bizarre:
$$\begin{aligned}
\Phi^{\mathrm{harm}} &=\Phi_{0}+\frac{1}{4} \sum_{i \kappa, j \nu} \sum_{\alpha \beta}\left(u_{i \kappa, \alpha}-u_{j \nu, \alpha}\right) \phi_{\alpha \beta}\left(\boldsymbol{R}_{i}+\boldsymbol{d}_{\kappa}-\boldsymbol{R}_{j}-\boldsymbol{d}_{\nu}\right)\left(u_{i \kappa, \beta}-u_{j \nu, \beta}\right) \\
&=\Phi_{0}+\frac{1}{2} \sum_{i j} \sum_{\kappa \alpha, \nu \beta} \phi_{\alpha \beta}\left(\boldsymbol{R}_{i}+\boldsymbol{d}_{\kappa}-\boldsymbol{R}_{j}-\boldsymbol{d}_{\nu}\right)\left(u_{i \kappa, \alpha} u_{i \kappa, \beta}-u_{i \kappa, \alpha} u_{j \nu, \beta}\right) \\ | {
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1, 2, 3 -- 2 is magenta, 3 is magenta
1, 3, 2 -- 3 is magenta, 2 is magenta
2, 1, 3 -- 1 is red, case is excluded
2. 3. 1 -- 3 is magenta, 2 is red
3. 1. 2 -- 1 is red, case is excluded
3. 2. 1 -- 2 is red, case is excluded
Of the three cases where the first of the colored balls is magenta, two of them have the other colored ball magenta too.
We could get through the case analysis a bit faster if we had started by observing that $$P(\text{second ball magenta}\mid\text{first ball magenta})= P(\text{second ball red}\mid\text{first ball red})$$ by symmetry, and then both of these are the same as the (unconditional) probability that the two colored balls have the same color.
The two colored balls have the same color exactly when the ball you pick to determine colors is either the highest numbered between the three balls-to-be-picked or the lowest numbered between the three balls-to-be-picked. And this happens, of course, in $2/3$ of all cases. | {
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} |
antarctica, continent, history
Rather inaccessible.
Finally, it's worth notice that the Internet is flooded by ridiculous conspiration theories about the lack of information about Antarctica. Whiles being entertaining to watch (with a couple of beers or so), there is not any truth in them, but the phenomina gives in insight in how myths and lies are created and blooming on the internet and how important it is to stay critical about unreliable sources. The reality is that Antarctica is very difficult and expensive to access and, thankfully, its sensitive nature is well protected by international agreements. | {
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So all the perfect square cells will be open
those are 1,4,9,16,25,36,49,64,81,100
selected
1 vote
The initial state of every lock = locked
On every NON-SQUARE numbered door, the number of wanders changing the state will be EVEN. And hence the door will return to locked position.
E.g. Door No.$8$: Warders $1, 2, 4$ and $8$ will act.
Door No.$30$: Warders $1,2,3,5,6,10,15,30$ i.e. a total of $8$ warders will act.
On every SQUARE numbered door, the number of wanders changing the state will be ODD. And hence the door will be in unlocked position.
E.g. Door No.$16$: Warders $1, 2, 4, 8$ and $16$ will act.
Door No.$36$: Warders $1,2,3,4,6,9,12,18,36$ i.e. a total of $9$ warders will act.
So door No. $\color{gold}{1, 4, 9, 16, 25, 36, 49, 64, 84}$ and $\color{gold}{100}$ will be in Open position and rest in Locked position.
Concept used: Perfect squares have odd number of factors and others have even number of factors.
edited
0
Why square nos. have even number of factors | {
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c++, boost
Now:
Exception: Expected <factor> at ""
Intermission: superficial changes to here: Live On Coliru
In the printer there's a lot of repetition (apply_visitor(*this...) and it's slightly less than readable due to operator(). My preference is to relay to a call or apply function
Also in the printer, don't hardcode the output stream. One day(TM) you will want to format to a string. Or std::cerr, or a file
Combining these notes on the printer: Live On Coliru
struct printer {
std::ostream& _os;
template <typename T> std::ostream& operator()(T const& v) const
{ return call(v); }
private:
std::ostream& call(expression const& ast) const {
return boost::apply_visitor(*this, ast.expr);
}
std::ostream& call(binary_operator const& expr) const {
_os << "op:" << expr.op << "(";
call(expr.lhs) << ", ";
return call(expr.rhs) << ')';
} | {
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ros, bash, setup.bash, macos-mountain-lion, osx
Originally posted by Kevin with karma: 2962 on 2013-06-14
This answer was ACCEPTED on the original site
Post score: 0 | {
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} |
homework-and-exercises, friction
Title: Friction between two stacked blocks I was studying friction when I came across a question in my book, A block of mass 2kg is stacked on top of a block of mass 4kg. A force of 20N is acting at an angle of 37 degrees on the system. Find the accelerations of both the blocks if it is known that the coefficients of static and kinetic friction between the two blocks are 0.3 and 0.2 respectively and there is no friction between the ground and the 4kg block.
I started by calculating the acceleration of the lower block (caused by friction) when I got confused so as to take 0.3 or 0.2 (static or kinetic) as the coefficient of the frictional force pushing the lower block.
Can someone pls throw some light on this...... Assume that there is no slipping between the blocks and try finding the value of static friction, if it is more than limiting value of static friction then blocks will slip and kinetic friction would act. | {
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Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last | {
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"url": "https://s12762.gridserver.com/1gb40i0t/43d397-master-electrician-test"
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ros, rosconsole
Originally posted by thebyohazard on ROS Answers with karma: 3562 on 2022-02-21
Post score: 1
How do I make the output of the rosconsole command line utility show in the desired format?
Looking at the source, I don't believe you can.
That script uses a rospy.Subscriber to subscribe to the /rosout topic topic (here), and uses a relatively simple callback to print incoming messages in a fixed format (here):
def _print(self, msg):
print('[ {} ] [\033[1m{}\033[21m]: {}'.format(
self._level_string_map[msg.level], msg.name, msg.msg))
if self._verbose:
stamp_sec = msg.header.stamp.to_sec()
stamp_tz = datetime.fromtimestamp(stamp_sec, tzlocal())
print(' [{} ({:.6f})] [{}]: {}:{}'.format(
stamp_tz, stamp_sec, msg.function, msg.file, msg.line)) | {
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expansion, redshift, observable-universe, cosmological-inflation, hubble-constant
In other words, the distance to an object at redshift $z$ when it emitted the light we see, is a factor (1+z) smaller than it is today (disregarding the galaxies' relatively small peculiar velocities).
The most distant galaxy
The current galaxy redshift record holder is, as you say, GN-z11 with $z=11.1$ (Oesch et al. 2016). Solving the equation above yields a current distance of
$$
d_\mathrm{phys,now}=32.2\,\mathrm{Glyr},
$$
(a Glyr is a billion lightyears), and hence the distance between GN-z11 and the Milky Way was, when it emitted its light some 400 Myr after the Big Bang
$$
d_\mathrm{phys,then}=\frac{32.2\,\mathrm{Glyr}}{1+11.1}=2.66\,\mathrm{Glyr}.
$$
As a small curiosity I can tell you, that GN-z11 recede from us with a velocity of roughly $4c$ when it emitted its light, while today it "only" recedes at $2.2c$. And yes, that's allowed.
The cosmic microwave background | {
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c++, object-oriented, opengl
void GLProgram::DeleteUniform(const std::string& identifier)
{
unsigned int i = 0;
for (Uniform* u : m_Uniforms)
{
if (u->GetName() == identifier)
m_Uniforms.erase(m_Uniforms.begin() + i);
i++;
}
m_UniformLocations = GetUniformLocations();
}
int* GLProgram::GetUniformLocations()
{
if (m_UniformLocations != nullptr) free(m_UniformLocations);
unsigned int count = m_Uniforms.size();
int* ptr = (int*)malloc(count * sizeof(int));
for (unsigned int i = 0; i < count; i++)
{
GLCall(*(ptr + i) = glGetUniformLocation(m_RendererID, m_Uniforms[i]->GetName().c_str()));
ASSERT(*(ptr + i) != -1);
}
return ptr;
}
void GLProgram::ParseUniform(Uniform* uniform)
{
unsigned int locationOffset = 0;
for (Uniform* u : m_Uniforms)
{
if (u == uniform)
break;
locationOffset++;
} | {
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ions, elements, electronegativity, ionization-energy
A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases. | {
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beginner, c, tic-tac-toe, ai, curses
///////////////////////////////////////////////////////////////////////////////////
// End AI Logic ///////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////
static int spaces_left(const char playable_spaces[NUM_SPACES]) {
// Returns 0 if no spaces left
int hits = 0;
for (int k = 0; k < NUM_SPACES; k++)
if (playable_spaces[k] == ' ')
hits++;
return hits;
} | {
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} |
chirality, symmetry, group-theory
Title: Example of an achiral molecule without a plane of symmetry or inversion center? Some websites, such as this textbook, mention how generally a plane of symmetry or inversion center in a molecule is enough to consider it achiral. Nonetheless, this source clarifies
but if these symmetry elements are absent the molecule should be checked carefully for an S$_n$ axis before it is assumed to be chiral.
Furthermore, Wikipedia notes the definition of chirality as "a molecule which has no S$_n$ axis for any value of n is a chiral molecule."
Thus, are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation? Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?
edit: This post does cover some definitions which is helpful, but the point of this question was to see an actual example. Part 1 | {
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python, performance, python-3.x, programming-challenge, balanced-delimiters
Explanation -
Scan from left to right, make sure count["("] >= count[")"].
Then scan from right to left, make sure count["("] <= count[")"].
Here are some inputs/outputs -
print(remove_invalid_parentheses("()())()"))
>>> ["()()()", "(())()"]
print(remove_invalid_parentheses("(a)())()"))
>>> ["(a)()()", "(a())()"]
print(remove_invalid_parentheses(")("))
>>> [""]
Here are the times taken for each output -
%timeit remove_invalid_parentheses("()())()")
6.54 µs ± 176 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit remove_invalid_parentheses("(a)())()")
8.43 µs ± 1.29 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit remove_invalid_parentheses(")(")
3.69 µs ± 67.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) | {
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python, performance, statistics
len_exp = len(expenditure)
window = sorted(expenditure[:d])
windex = window.index
winsort = window.sort
if d % 2 == 0:
m1 = d // 2
m0 = m1 - 1
ai = 0
for zi in range(d, len_exp):
median_x2 = window[m0] + window[m1]
a, z = expenditure[ai], expenditure[zi]
ai += 1
if z >= median_x2:
count += 1
# Note: do not write `window =` anywhere, or you invalidate
# windex and winsort.
window[windex(a)] = z
winsort()
else:
raise NotImplementedError("I haven't written this code.")
return count
if __name__ == "__main__":
from timeit import timeit
from random import randrange as rr
results =[line.strip().split() for line in """
replsort 9.730
replsort2 11.538
linsert 17.954
insbisect 10.970
noparity 10.574
orig 11.801
""".splitlines() if line.strip()] | {
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homework-and-exercises, newtonian-mechanics, acceleration, harmonic-oscillator, spring
I know how to solve it using the law of energy conservation but as a challenge I wanted to find the solution without it.
So from the 2nd Newton's law it's obvious that a = k*x/m. But I don't know how to integrate it in order to find velocity since a = f(x(t)).
How would I solve this without using the law of energy conservation? I believe you could use some differential calculus.
A helpful formula is a = d(v^2 / 2)/dx, derived via a = dv/dt = dv/dx * dx/dt
$$a = {-kx \over m}$$
$$a = {\operatorname{d}\!({1 \over 2}v^2)\over\operatorname{d}\!x}$$
$$\therefore -{1 \over 2}v^2 = \int {k \over m} x dx = {k \over 2m}x^2+c$$
When $x=A$, $v=0$: $\therefore c = +{kA^2 \over 2m}$
$$\therefore v = +\sqrt{{-kx^2+kA^2 \over m}}$$
When $x=0$, $v=\sqrt{{kA^2 \over m}}$
This is the same result as that you would get using energy.
(I've assumed that the force causing the spring to retract is F = -kx. | {
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neural-networks, mathematical-foundations
By contrast, there is a universal approximation theorem that says that, for sufficiently nice nonlinear activation functions, any function can be approximated well by using enough neurons.
The universal approximation theorem was not discovered for many years after when neural networks were first invented, so it was not a motivating factor in their invention. Early neural network research was mainly inspired by biological neurons (the kind in your brain) and control theory.
While the universal approximation theorem says that a sufficently large neural network has the potential to approximate any function well, the actual reason why the standard method of training neural networks (stochastic gradient descent backpropagation) performs so well in practice is still poorly understood and an active area of research. | {
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(* A model of the form a + b*o + c*t + d*t^2 *)
model = Fit[data2, {1, o, t, t^2}, {o, t}]
0.941482 + 1.00358 o + 0.00545877 t + 0.0000743324 t^2
This model can be used as:
model./{o->value1, t->value2}
Plot[{model /. o -> 30, model /. o -> 31}, {t, 80, 200}, Epilog -> Point@Cases[data2, {30, t_, f_} :> {t, f}]]
Blue for $\theta=30$, along with the corresponding data points
Orange for $\theta=31$ | {
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ros, lsd-slam, camera, usb-cam, ros-indigo
slam.launch
<launch>
<!-- Activate the color webcam. -->
<node name="usb_cam" pkg="usb_cam" type="usb_cam_node" output="screen" >
<param name="video_device" value="/dev/video0" />
<param name="image_width" value="640" />
<param name="image_height" value="480" />
<param name="pixel_format" value="yuyv" />
<param name="camera_frame_id" value="usb_cam" />
<param name="io_method" value="mmap" />
</node>
<!-- Convert the color webcam's output to mono. -->
<node name="to_mono_node1" pkg="image_proc" type="image_proc" ns="usb_cam" />
<!-- Display the mono stream. -->
<node name="image_view" pkg="image_view" type="image_view" respawn="false" output="screen">
<remap from="image" to="/usb_cam/image_mono" />
<param name="autosize" value="true" />
</node>
</launch> | {
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python, git
raw = str(vers).strip('-')
rawsplit = raw.split('-')
st = rawsplit[0] | {
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quantum-mechanics, particle-physics, neutrinos, weak-interaction
Here, then, you monitor in t the oscillating mean energy of the combination of the two energy eigenstates 1,2. But t differences amount to L differences, so... | {
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} |
We use the notation $X_1 \cong X_2$ to mean that the spaces $X_1$ and $X_2$ are homeomorphic. Then we have:
$C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R}) \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$
Thus $C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$. This completes the proof that $C_p(L_\tau)$ is topologically the $\Sigma$-product of $\tau$ many copies of the real lines.
____________________________________________________________________
Looking at the $\Sigma$-Product
Understanding the function space $C_p(L_\tau)$ is now reduced to the problem of understanding a $\Sigma$-product of copies of the real lines. Most of the facts about $\Sigma$-products that we need have already been proved in previous blog posts. | {
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quantum-field-theory, lagrangian-formalism, standard-model
\begin{equation}
\langle 0 | F[\phi(x)] | 0 \rangle = \int \mathscr{D}\phi(x) F[x] e^{i S[\phi(x)]}
\end{equation}
Using some manner of equivalent of the fundamental theorem of calculus, we have that the path integral of this total derivative is zero :
\begin{equation}
\int \mathscr{D}\phi(x) \frac{\delta}{\delta \phi(x)} \left[ F[x] e^{i S[\phi(x)]} \right] = 0
\end{equation}
And therefore, we get that
\begin{equation}
\int \mathscr{D}\phi(x) e^{i S[\phi(x)]} \left[ \frac{\delta F[x]}{\delta \phi(x)} + i F[x] \frac{\delta S}{\delta \phi(x)} \right] = 0
\end{equation}
This leads to the Schwinger-Dyson equation :
\begin{equation}
\left\langle \frac{\delta F[x]}{\delta \phi(x)} \right\rangle = - i \left\langle F[x] \frac{\delta S}{\delta \phi(x)} \right\rangle
\end{equation}
In particular, if we consider the raw quantity for $F = 1$, we simply get
\begin{equation}
\left\langle \frac{\delta S}{\delta \phi(x)} \right\rangle = 0
\end{equation} | {
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c#, .net, rabbitmq
public interface IRabbitConnectionFactory : IDisposable
{
/// <summary>
/// Username to use when authenticating to the server.
/// </summary>
string UserName { get; }
/// <summary>
/// Password to use when authenticating to the server.
/// </summary>
string Password { get; }
/// <summary>
/// Virtual host to access during this connection.
/// </summary>
string VirtualHost { get; }
IConnection CurrentConnection { get; }
IModel CurrentChannel { get; }
IConnection CreateConnection();
IModel CreateModel();
} IRabbitConnectionFactory
Single or Multiple interfaces
As far as I can see this interface (in some degree) resembles/mimics the RabbitMq's IConnectionFactory
For me this mixture of properties (like UserName, Password, etc.) and methods (like CreateConnection) feels a bit odd | {
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quantum-mechanics, unitarity
We have: $A|\phi_n\rangle=n|\phi_n\rangle$
In Schrodinger picture:
$|\psi(t)\rangle =\sum_n c_n(t_0)e^{-\frac{iE_n(t-t_0)}{\hbar}}|\phi_n\rangle$
$|\psi(t)\rangle =\sum_n c_n(t)|\phi_n\rangle$
Here we can see that the state vector is time-dependent while the basis kets are not. So I believe, that I correctly translate mathematically the initial part of the statement.
Than in the Heisenberg picture at some arbitrary time t:
$|\psi(t)\rangle_H=\sum_n c_n(t_0)e^{-\frac{iE_n(t-t_0)}{\hbar}}|\phi_n\rangle$
$|\psi(t)\rangle_H=\sum_n c_n(t_0)|\phi_n(t)\rangle$
But at the same time when we move from the Schroedinger picture to the Heisenberg one we do:
$|\psi(t)\rangle_H=U(t,t_0)^\dagger |\psi(t)\rangle=U^{\dagger} U |\psi(t_0)\rangle$.
So I don't know how to show the 2nd part of the statement.
But at the same time when we move from the Schröedinger picture to the Heisenberg one we do:
$|ψ(t)⟩_H=U(t,t_0)^†|ψ(t)⟩=U^†U|ψ(t_0)⟩$.
So I don't know how to show the 2nd part of the statement. | {
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# Math Help - Eigenvectors!
1. ## Eigenvectors!
(i) Determine all the eigenvalues of A
(ii) For each eigenvalue of A, find the set of eigenvectors corresponding to
A = ( 1 2 )
( 3 2 )
I found eigenvalues of A to be 4 and -1.
I also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1
BUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1
I'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?
Here's how i calculate eigenvector for =4
A-4I = (-3 2)
(3 -2)
then (-3 2 |0)
(3 -2 |0)
and i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)
and for =-1
A-(-1)I = (2 3 )
(2 3 )
then (2 3 |0)
(2 3 |0)
and i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)
2. You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine! | {
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turing-machines, computability, artificial-intelligence, philosophy
Title: Why can't humans translate all of their abilities into an algorithm? Will that hold true in the future? I know that an algorithm can't decide whether another algorithm halts on an input or not (a Turing machine can't decide whether another Turing machine will halt on an input). But I, as an human, can: I mean you just don't end up in your activity as a programmer with a looping algorithm and you don't know why!
To do this, you generally analyze the (somewhat) natual language of the code you wrote. Considering this, it comes to no surprise to me that for a machine this is hard to do: if I had to write a convicing chatbot (also a problem in the realm of natural language) in a programming language with traditional techniques it would turn out to be pretty bad. But why is that? Is there any term or concept that clearly differentiates between what humans are able to do, what machines are able to do and why is there exists this distinction? | {
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star, classification, stellar-classification
Title: Dataset for machine learning MK stellar classification I would like to create program for automatic Morgan-Keenan stellar classification using machine learning. For that, I need dataset of stars with known absolute magnitude, temperature and luminosity type (0, Ia, Ib, ..., VII). I found some datasets, (e. g. kaggle.com), however they contain only few hundreds stars and not all star types are represented.
Is there any large enough dataset (at least 1000 stars) in which all types of stars are represented, from hypergiants to dwarfs, and with all the information mentioned? That I know there is the XHIP catalog via VizieR and you can enter a range in UMag (or B- and V-band) for example of -20 .. 20. Check the box SpType and Tc to get also the spectral type and temperature (double check the literature for the quantities you need) | {
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special-relativity, speed-of-light
Title: A doubt about the classic problem of measuring the speed of light in only one direction Why can't two clocks synchronized with signals SLOWER than the speed of light, measure the speed of light in one direction?
I know that it is not possible to measure the speed of light in only one direction, and I understand why two clocks synchronized with light signals are useless. But I realize that I can't explain why two clocks synchronized with signals slower than light don't work either. Because neither the speed of light, nor the speed of a turtle in one direction can be measured without arbitrary assumptions about the synchronism of clocks at the point of departure and point of arrival. | {
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homework-and-exercises, electromagnetism, point-particles
as all the time dependence is contained (through $\vec r(t)$) in the delta function. Now, for $\delta(y-z)$, we have the result that
$$\frac{\partial}{\partial y} \delta(y-z) = -\frac{\partial}{\partial z} \delta(y-z)$$
We can use this here to get:
$$\frac{\partial \rho}{\partial t} = -q\frac{d\vec{r}}{dt}\cdot\vec\nabla\delta^3(\vec x - \vec r(t))$$
The divergence of the current density is:
$$\vec\nabla\cdot\vec J = \vec\nabla\cdot\left[q\frac{d\vec{r}}{dt}\delta^3(\vec x - \vec r(t))\right]$$
$$ = q\frac{d\vec{r}}{dt}\cdot\vec\nabla\delta^3(\vec x - \vec r(t))$$
where the field-like dependence on position is once again contained only in the delta function. | {
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a matching M is. The problem is to find a path through a graph in which non-negative weights are associated with the arcs. The length "($)of a path$= '(,'*,…,', is defined as "$=-. The complexity of. The graph given in the test case is shown as : The shortest paths for the 3 queries are :: The direct Path is shortest with weight 5: There is no way of reaching node 1 from node 3. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. These values become important when calculating the. Intro to Networks Notes Weighted graphs and networks Graphs that have a number associated with each edge-Shortest path E. The algorithm is driven by a priority queue of nodes, ordered by their cost. Create Maximum Occurring Character in a String C Program code to find maximum occurring character in a string using Loop. The algorithm must run in O(V+E) *We cannot edit the Bellman-Ford run on the algorithm. 2013/2014 Consider a | {
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python, outlier, grid-search, isolation-forest
X_train = np.round(X_train,2)*100
X_train = X_train.astype(int)
X_test = np.round(X_test,2)*100
X_test = X_test.astype(int)
'y'
yeah = StandardScaler().fit(y_train)
y_train = yeah.transform(y_train)
y_test = yeah.transform(y_test)
y_train = np.round(y_train,2)*100
y_train = y_train.astype(int)
y_test = np.round(y_test,2)*100
y_test = y_test.astype(int)
I chose the IForrest because its fast, has pretty good results and can handle huge data sets (i currently only use a chunk of the data for testing). Setting Up the GridSearchCV:
clf = IForest(random_state=47, behaviour='new',
n_jobs=-1)
param_grid = {'n_estimators': [20,40,70,100],
'max_samples': [10,20,40,60],
'contamination': [0.1, 0.01, 0.001],
'max_features': [5,15,30],
'bootstrap': [True, False]}
fbeta = make_scorer(fbeta_score,
average = 'micro',
needs_proba=True,
beta=1) | {
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reductions, np-hard, np, space-complexity, polynomial-time-reductions
I am unsure about the second item of proof. The confusion is about where I am saying SAT is in PSPACE-hard and also in NP and using that to prove the inclusion. Is the logic correct? If yes, Is there a clearer way to write it? "SAT is PSPACE-hard. Since SAT $\in$ NP $\implies \forall L \in$ PSPACE $\leq_{P}$ SAT $\implies$ PSPACE $\subseteq$ NP" is indeed confusing.
What is much clearer is "SAT is PSPACE-HARD $\implies (\forall L \in$ PSPACE, $L\leq_{P}$ SAT) $\implies$ ($\forall L \in$ PSPACE, $L$ is in NP) $\implies$ (PSPACE $\subseteq$ NP)".
The first implication is according to the definition of PSPACE-HARD.
The second implication holds since NP is closed under polynomial-time many-one reduction and SAT is in NP.
The third implication is a tautology. | {
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statistical-mechanics, hamiltonian-formalism, phase-space, complex-systems, ergodicity
Now if a point (system) gets to visit the entire volume of the system, doesn't that imply that the microstate changes its energy, and isn't that in contradiction to the Liouville theorem
No, because, sometimes implicitly, by "entire volume" it's meant "entire accessible volume". So, for instance, the orbit is not required to (nor could it) visit states with different energies to be considered ergodic.
Edit: What does it mean for a system to spend time in a region of phase space?
If a trajectory $\vec{x}(t)$ in the phase space $S$ of a classical system traverses a region $R \subset S$, the length of the time interval $t\in(t_\mathrm{in},t_\mathrm{out})$ between going in and out of $S$ can be called the time spent in this region. | {
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java, algorithm, taxicab-geometry
// The hole is where the piece used to be.
return new FifteenBoard(childIndexToPiece, childDistance, index);
}
The full code is at Github here. There are some changes I made that are incompatible with the original, such as changing the distanceTo(Board) function into a distance() function. I ran a simple speed test that tested the generation of new boards and retrieving the distance of each new board. The new format is about 10-20% faster than the old one, but I didn't test it rigorously. | {
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quantum-mechanics, homework-and-exercises, quantum-information, hilbert-space, linear-algebra
Title: These two operators commute...but their eigenvectors aren't all the same. Why? The Hamiltonian
$$H = \left[ \begin{array}{cccc}
a & 0 & 0 & -b \\
0 & 0 & -b & 0\\
0 & -b & 0 & 0\\
-b & 0 & 0 & -a \end{array} \right] $$
commutes with the qubit exchange operator
$$ P = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array} \right] $$ | {
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slam, navigation, ros-melodic, rtabmap
log file: /home/pi/.ros/log/6fd4bf78-8f58-11ea-8779-dca63207de8a/rtabmap-rtabmap-5*.log | {
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life
Reproduction is not only cell division obviously. Following this same definition one would not consider a human to be a living things but only to be a collection of living (and non-living) things. It is important to understand that a spermatozoid is just one phase of a life-cycle. This phase yield to the next phase. That's it. It seems weird to say that a kid is not alive just because he cannot reproduce. It seems weird to say that a grandmother is not alive because it cannot reproduce. You can say however that a kid is a living thing that cannot reproduce and at another moment of its life cycle it will be able to reproduce (assuming it will survive to this age). One should not think of spermatozoids as something totally detached from the human phase as we know it. These things just form a cycle and it seems to me miseleading to say that a part of this cycle is not alive. Saying such thing would yield someone to think that two living things create non-living things that by fusion will | {
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schmidt-decomposition
Title: How to write Schmidt decomposition for pure tripartite state? Supposing a pure tripartite state, i wrote schmidt decomposition as follows
$\sum_{}^{r}p_{ijk}|i\rangle|j\rangle|k\rangle$
Did i write it correctly?
Like pure bipartite states If r>1 then the state is entangled? Assuming the Schmidt decomposition exists, it would be written as:
$$\sum_{i=1}^rp_i\left|i_A\right\rangle\left|i_C\right\rangle\left|i_B\right\rangle$$
With $\left\{\left|i_A\right\rangle\right\}_i$, $\left\{\left|i_B\right\rangle\right\}_i$ and $\left\{\left|i_C\right\rangle\right\}_i$ being orthonormal basis of the respective Hilbert space.
Writing it like this allows to have a single index, and thus to define the Schmidt number $r$. However, you have to be careful, in that the Schmidt decomposition doesn't always exist for more than two systems. This paper gives necessary and sufficient conditions for it to exist for higher number of systems. | {
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c++, file, error-handling, logging, macros
But that’s just one of dozens of techniques you can use to AVOID the use of macros… which is a much more useful skill to have in 2020.
In other words, all this has brought us back to the original point I wasn’t going to belabour. The best macro is the one you don’t write.
Questions
Are there any apparent issues with how I implemented the macro?
Yes, it is needlessly long and complex. Even for a non macro function, this is unnecessarily long and complex. It should be refactored into smaller functions for each of the four different behaviours.
Can there be any improvements made to it?
The best way to write a macro is: don’t.
I can’t conceive of why anyone would want to write a macro in 2020. Macros were a dirty hack when they were first created back in the 1970s. There may be a few very rare cases where you still need them, but by and large, if you can possibly solve a problem without macro, then THAT is the skill you should be exercising.
Is the intent clear on its usage? | {
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algorithms, complexity-theory, np-complete, reductions, traveling-salesman
However, I'm not able to find any significance for TSP problems with unit edge costs. Is option 1 merely a trick, or is there more to it? OP here.
Papadimitriou and Yannakakis have shown that it is possible to approximate the TSP problem 1 in polynomial time within accuracy $ \frac{7}{6} $. This guarantee has been further improved by Bläser and Shankar Ram to $ \frac{65}{56} $. However, no matter how good those results are, they are still approximations, and not an optimal solution. Thus, option 1 is incorrect.
The DP algorithm doesn't make any assumptions on the edge costs, so option 2 is incorrect.
If all edge weights are negative, then deleting a vertex and all of its incident edges can certainly increase the minimum sum because in effect, that edge weight is now added to the previous minimum. Thus, option 3 is incorrect. | {
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experimental-physics, atomic-physics, history, physical-chemistry, elements
Title: How is the number of electrons in an atom found? I was wondering, what type of experiments were held to identify the number of electrons in an atom?
(For example, how do we say that carbon has 6 electrons and magnesium 12.) I would like someone to give me a link to a video of this experiment in comment form or answer form, because it would be more clear me if I saw the experiment rather if someone explained it to me. As John Rennie answered it very clearly, I would like to add some more details too.
See, around early 1900 the idea of atoms was floating around the scientists' heads. At first everything was theory, but these things happened: | {
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thermodynamics, special-relativity, mass
Consequently, when you shift energy from potential to kinetic in the box, the observed mass from the outside doesn't change.
Instead, the reason that the object at absolute 0 has an "increasing entropy" is purely due to thermal energy transfer with its surroundings, which will appear as energy -- hence mass -- going into the box. Measuring the mass very carefully will, as John Rennie says in a comment to your question, reveal a change in the total internal energy, but not anything deep or fundamental about the entropy contained within it: it could have gotten that mass because its temperature increased, but it could also have gotten that mass in the form of positronium (electron-positron pairs) which are a tenth of a microsecond away from exploding into gamma rays. You don't really know just by looking at the mass of the box which has happened. | {
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javascript, strings, unit-testing, regex
return r;
});
return t;
}
function _dFlag(chr, pos, search, replace, obj) {
let r = '';
if (replace) {
let ind = search.indexOf(chr);
if (replace.length >= ind) {
r = replace.charAt(ind);
}
}
return r;
}
function _sFlag(chr, pos, search, replace, obj, escaped) {
let escapedChrDeets = _charsNeedEscaping(chr);
let escapedChr = escapedChrDeets.text;
let searchRegExp = new RegExp(escapedChr, 'y');
if (escaped) {
pos = search.indexOf(chr);
}
searchRegExp.lastIndex = pos;
let searchMatch = search.match(searchRegExp);
let r = '';
if (searchMatch) {
let searchChr = searchMatch[0];
if (searchChr in obj) {
r = replace.charAt(obj[searchChr]);
if (obj[searchChr]+1 === searchMatch.index) {
r = '';
}
} else {
let replacementIndex = searchMatch.index; | {
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c#, object-oriented, linked-list, console, inheritance
{
Console.ForegroundColor = ConsoleColor.Red;
base.RemoveFirst();
}
public override void RemoveLast()
{
Console.ForegroundColor = ConsoleColor.DarkRed;
base.RemoveLast();
}
public override void Remove(int e)
{
Console.ForegroundColor = ConsoleColor.DarkMagenta;
base.Remove(e);
} | {
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homework-and-exercises, classical-mechanics, spring
This yields the effective spring rate as
$$ F = k_{eff} x = \left( \frac{k_1 k_2 (a+b)^2}{a^2 k_1 + b^2 k_2} \right) x $$
Summary:
When the springs are in parallel with equal extensions the effective spring rate is the sum of the two springs. When the lever is free to rotate then the effective spring rate is less. | {
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mathematical-physics, dirac-delta-distributions
$$
\int dx~\delta(-x) f(x) = \int dy~\delta(y) f(-y)
$$
By definition of $\delta(y)$,value of this integral is $f(0)$, the same as the left-hand side. So (**) is satisfied.
The equation $\delta(x) = \delta(-x)$ is thus consequence of the definition of $\delta(x)$, it is not independent assumption.
Your function $\xi$ may actually obey the second statement too (and thus be symmetric in that sense), even though the $\Delta$-dependent expression after the limit sign does not. This is similar for other approximations of delta distribution; the approximation may not have properties of $\delta$ (such as symmetry), but the limit does. | {
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python, linear-regression, feature-engineering
to
id Male Female
0 1 0
1 0 1
This can easily be done with pandas:
import pandas as pd
data = pd.DataFrame({'Sex': ['Male', 'Female']})
print(pd.get_dummies(data))
will result in:
Sex_Male Sex_Female
0 1 0
1 0 1
Using categorical data for groupby operations
This is an additional usecase but in your case it is not necessary to use this but if you feel so, you can try implementing this as well
You could use the mean for each category over past (known events).
Say you have a DataFrame with the last known mean prices for cities:
prices = pd.DataFrame({
'city': ['A', 'A', 'A', 'B', 'B', 'C'],
'price': [1, 1, 1, 2, 2, 3],
})
mean_price = prices.groupby('city').mean()
data = pd.DataFrame({'city': ['A', 'B', 'C', 'A', 'B', 'A']})
print(data.merge(mean_price, on='city', how='left')) | {
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vba, excel
Else: Stop
End If
I had to reread this a bunch of times and I'm still not sure if I'm doing all the checking you are. You aren't consistent with how you're checking for a numeric value, you use a worksheet function and the vba function, there's probably no need to use the worksheet function if your data is standardized.
There's also no need to check if something like isempty = true as the result of isempty is either true or false. | {
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radio-telescope, black-hole, radio-astronomy
Notice the Y-shaped configuration of the current EHT array sites in user rugk's question. More sites in a narrow slice of the Earth's circumference allow increased capabilities but if a particular sector is competently (capability) covered it's a duplication of resources and an increase of processing time compared to a perfect location that fills in a hole. A site that is near the edge is useful to increase the observation window, but one too far away to be able to simultaneously observe with a sufficient number of other site isn't anywhere near as useful. Wherever the site is it must be capable of operating at the necessary frequencies and have enough sensitivity to gather useful data at that distance. It's a tall order. | {
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At this point I would make a sign chart. First, I need "critical points." These are values which (1) the fraction equals 0, and (2) the fraction is undefined. I find (1) by setting the numerator equal to zero:
\begin{aligned}
(x - 2)^2 &= 0 \\
x - 2 &= 0 \\
x &= 2
\end{aligned}
I find (2) by setting the denominator equal to zero: x = 0.
Then I draw my sign chart like this:
Code:
und 0
----------+---------------+-------------
0 2
(und = undefined) | {
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waves, acoustics, greens-functions
The solution to this problem is well-known:
$$\hat{G}(x,\omega|\xi) = -\frac{i}{2}\frac{c}{\omega} e^{-i\frac{\omega}{c} \lvert x - \xi \rvert}. $$
The dipole solution
Then, for the multipole we place two lower-order poles next to each other with opposite polarity. In particular, for the dipole we assume the space-time source-function is given as $\tfrac{\partial \delta(x-\xi)}{\partial x}\delta(t)$, i.e., the spatial derivative of the delta function. We find the dipole solution by a integration of the Green's function with this new source function. Extending it for the $n$-pole ($n=0$ for monopole, $n=1$ for the dipole, ...) solution:
$$ \hat{G}_\text{$n$-pole}(x,\omega|\xi) = \int \hat{G}(x,\omega|x')\frac{\partial^n \delta(x'-\xi)}{\partial x'^n} \text{d}\! x' = (-1)^n\frac{\partial^n \hat{G}(x,\omega|\xi)}{\partial \xi^n}. $$
Carrying this out, I find for the dipole (note: $\partial \lvert x-\xi \rvert/\partial \xi = -\text{sign}(x-\xi)$): | {
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gazebo
Title: What is the future of Gazebo on a sans-OpenGL macOS?
Since Apple deprecated OpenGL and OpenCL for the current OS already, it is not unlikely that the next version will remove support for these technologies altogether. Apple wants developers to use Metal.
My question is if there is a way forward for Gazebo if this becomes a reality.
Is there e.g. a way to make use of Metal or keep using a OpenGL version that is implemented in user land and/or has an abstraction layer on top of metal?
I think it is important to know for people who rely on macOS and think about working with Gazebo/Ignition in the future. | {
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complicated: $$-3 + 4 \log(2) + \pi \bigl(1 + \log(\pi/2) - \gamma - 4 \log\Gamma(3/4) \bigr).$$ This requires evaluation of $\log(1)/1 - \log(3)/3 + \log(5)/5 - \log(7)/7 + - \cdots$, which can be achieved by differentiating the functional equation for the Dirichlet L-function $L(s,\chi_4) = 1 - 3^{-s} + 5^{-s} - 7^{-s} + - \cdots$ and evaluating at $s=1$. | {
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"url": "https://mathoverflow.net/questions/184787/parity-of-lfloor-1-x-y-rfloor-not-equally-distributed/184791"
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