text stringlengths 1 1.11k | source dict |
|---|---|
evolution, biochemistry, metabolism, glucose
Image taken from the Wikimedia Commons.
Could this have played a factor in the beginnings of energy metabolism, and if so, why? I thought this was a great question. In particular because it hints at two questions. The first is 'why carbohydrates are used to store energy' in general. The second being 'why glucose rather than other carbohydrates?' in particular. | {
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community-challenge, groovy
/*
* Adds useful methods to Iterator.
*/
@groovy.lang.Category(Iterator)
class IteratorCategory {
/*
* Returns a Closure which iterates while the condition Closure
* evaluates to true. The returned Closure expects another Closure,
* an action closure, as its single argument.
* This 'action' Closure is called during each iteration and is passed
* the Iterator.next()value.
* When the iteration is complete, the Iterator is returned.
*
* Example usage:
* use(IteratorCategory) {
* def iter = reader.iterator().while { it != 'end' }.call { println it }
* }
*
* @param condition Closure to evaluate on each iteration.
* @return a Closure
*/
Closure 'while'(Closure condition) {
{Iterator iter, Closure closure ->
while(iter.hasNext()) {
def it = iter.next() | {
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y) , y ( t 0) = y 0. Python Server Side Programming Programming. The explicit form of the above equation in Python with NumPy is implemented as follows: lambda t, x: np. Equations will only be solved if there is an algebraic solution or if the variable being solved for can be isolated through arithmetic operations. The other method is called the elimination method. Initial conditions Xo = 6 So = 45 from Recent Questions - Stack Overflow https://ift. The ebook and printed book are available for purchase at Packt Publishing. python by Puzzled Pygmy on Dec 04 2020 Comment. How to the SciPy solve_ivp function to integrate first oder ODEs in Python. First step: Solve for x5 x 5 ¶. The model, initial conditions, and time points are defined in GEKKO to numerically calculate y(t). If you replace these values in the system of equations, you have: This is a geometrical way of solving the system of equations. of a Python-based PDE solver in these pages. Python is a powerful tool for science and | {
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"url": "http://martinezgebaeudereinigungkoeln.de/solve-system-of-equations-python.html"
} |
However, herein lies the problem: In order to compute the number of ways to roll $$m$$ with $$n$$ $$k$$ sided die, which is a problem of computing restricted partitions of the number $$m$$, we need to find the coefficient of $$x^m$$ in a multinomial expansion. But, in order to compute this multinomial expansion, we need to compute restricted partitions of $$n$$. As you can see the problem is a bit circular. But, $$n$$ is usually smaller than $$m$$, so it might speed up the computation process a little. But at the end of the day some amount of brute-force grunt work will be required. | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/3790242/probability-of-two-fair-dice-rolls-having-a-total-of-7-or-11"
} |
organic-chemistry, esters, amides
pyrrolidin-2-one (PIN)
butano-4-lactam
(…)
3,4,5,6-tetrahydropyridin-2-ol (PIN)
pentano-5-lactim
(Note that the numberings depicted are for the preferred names (PINs) based on nitrogen heterocyclics)
There's no (at least IUPAC) term for lactides analogue (‘lactide’ is not used in IUPAC names themselves anyway), but they exist.
In traditional or general names, the Greek letter numbering is used, e.g. ε-caprolactam, or β-lactam four-membered ring part in bicyclic penicillin skeleton. | {
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"tags": "organic-chemistry, esters, amides",
"url": null
} |
php, sql, xml, postgresql
Logical consistency.
It is possible that two authors can give their books the same name. It would make sense to check both the name and the author.
there is some folder structure mentioned but no code provided to handle it. A built-in recursive directory iterator would do though
you forgot the book update code. That's a direct result of your "wall of code" approach. With the main logic fits in one screen, it's easy to spot the mistake. While in such a wall of code it's easy to overlook one.
Formatting and readability
The overall formatting is good, your indentation is even and lines are reasonably long. Two suggestions: | {
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organic-chemistry, synthesis, conformers
Title: Is there a way to determine the orientation of a benzene-ring ethyl group in synthesis reactions? On my study guide for synthesis reactions (specifically for the synthesis of disubstituted benzene derivatives), two of the problems look extremely similar.
The question on the guide asks us to synthesize each of these by using benzene as a starting material.
They appear to be the same molecule, just with a different orientation of the ethyl group. And in the solutions section of the guide, both have the same exact synthesis reaction listed. | {
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ros, catkin, ros-kinetic, catkin-create-pkg
Title: catkin_create_pkg
Hi,
I installed ROS Kinetic on Ubuntu 16.04 into Docker container. I created catkin workspace according to beginners tutorial. But when I try to create package, I get following exception:
$ catkin_create_pkg ahoj
Traceback (most recent call last):
File "/usr/local/bin/catkin_create_pkg", line 68, in <module>
main()
File "/usr/local/bin/catkin_create_pkg", line 61, in main
newfiles={})
File "/usr/local/lib/python3.5/dist-packages/catkin_pkg/package_templates.py", line 209, in create_package_files
_safe_write_files(newfiles, target_path)
File "/usr/local/lib/python3.5/dist-packages/catkin_pkg/package_templates.py", line 185, in _safe_write_files
fhand.write(content)
TypeError: a bytes-like object is required, not 'str'
Did you meed this situation somebody? It seems like function _safe_write_files
fhand.write expects different data type than what it gets, but I don't understand why.
Thanks in advance for any idea,
Radim | {
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"url": null
} |
atmosphere, meteorology, atmospheric-circulation, jet-stream
Title: Why does the meandering pattern of the jet stream itself propagate? It is well known that the jet stream follows an overall meandering pattern, but what is less well known is that this meandering pattern itself moves as a wave with a velocity much slower than the velocity of the jet stream itself. So why does this meandering wave propagate, and what causes the meandering pattern in the first place? I wouldn't characterize the movement of the jetstream as a "less well known" problem, at least not to anyone who has sat through a synoptic meteorology class. The jetstream from a dynamic point of view is just an enhanced gradient in the heights on a constant pressure surface (or pressures on a constant height surface) and there are many factors the influence this gradient.
First, lets take a look at the northern hemisphere 500 mb heights and surface pressure: | {
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quantum-mechanics, operators, commutator
$$ H(\hat{p}, \hat{x})= \frac{(\hat{p}-f(\hat{x}))^2}{2m}.$$
Since I employ a function that is defined on real numbers (taking the square, or subtracting), the definition of $H$ is not well defined, and could yield different results (because real numbers commute, while operators don't). By fixing the ordering of the operators (for example by normal ordering), I remove any ambiguities. Is that the right way to see it? Yes. In your example, you can rearrange the expansion of $(p-f(x))^2$ in multiple ways:
\begin{align}
(p-f(x))^2&=p^2+2p f(x) + f(x)^2\, ,\\
&= p^2 + f(x)^2 + p f(x) + f(x)p\, ,\\
&=p^2+f(x)^2 + \frac{1}{2}p f(x) +\frac{3}{2} f(x)p \tag{1}
\end{align}
etc. and even more complicated forms if you consider the series expansion of $f(x)$. For instance, imagine $f(x)=x^3$ then
$$
px^3= xpx^3=x^2px
$$ | {
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ros, catkin-ws
...
/home/ros/catkin_ws/build/urg_node/setup_custom_pythonpath.sh: 4: exec: /home/ros/catkin_ws/src/urg_node/cfg/URG.cfg: Permission denied
have you extracted these files from a zip file, or copied them from a usb stick or Windows share?
Those .cfg files (they are files for dynamic_reconfigure) should have their executable bit set, and they currently probably don't have it, that's why you get the Permission denied errors.
Try and see what the output is of:
ls -al /home/ros/catkin_ws/src/rbx1/rbx1_nav/cfg/CalibrateAngular.cfg
and:
ls -al /home/ros/catkin_ws/src/urg_node/cfg/URG.cfg
If it doesn't include the x bit for both you and other users, then do a:
chmod +x /home/ros/catkin_ws/src/rbx1/rbx1_nav/cfg/CalibrateAngular.cfg
and:
chmod +x /home/ros/catkin_ws/src/urg_node/cfg/URG.cfg
and try again.
PS:
[..] I was given a custom catkin_ws file.
Pedantic perhaps, but note that 'catkin_ws' is typically a directory, not a single file. | {
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We see that the numbers appear in a cycle-of-6.
This means that the last terms look likt this:
$\begin{array}{c||c|c|c|c|c|c||c|c|c|}
\hdots & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50 & 51 \\ \hline
\hdots & a & b & b-a & -a & -b & a-b & a & b & b-a \end{array}$
The next term on the cycle would be $-a.$
Since the 51 numbers lie on the circle,
. . the next term should be the first term, $\,a.$
And the only we can have $-a = a$ is if $a = 0.$
Can you finish it?
3. Thanks for your prompt response.
You're absolutely correct and that is a nice way.
You solved the whole thing, there is no other half, it finished as follow:
you started with an arbitrary position on the circle and showed that the starting number should be zero. This logic can be applied to any other number as the starting point and be proven to be zero. So all the numbers have to be zero (by simply starting from another position within the circle as the starting point). | {
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complexity-theory, computability, turing-machines
This is exercise that I can't deal with it. I know USTCON problem and I suppose that it may be useful (crucial) here. However, I can't solve it. Can anyone try to help me defeat it ? Let $N_k$ be the set of vertices at distance $k$ from 1. A graph satisfies the property if the sets $N_k$ are intervals, and $N_{k-1} < N_k$ (every member of $N_{k-1}$ is smaller than every member of $N_k$). We will check this property inductively: assuming $N_{k-1}$ is known (starting with $N_0 = \{1\}$), we will show how to calculate $N_k$ and verify that it has the correct properties. | {
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estimation, error-analysis
Now, why isn't $\delta x_i$ even considered when computing $S$? Well, presumably $S$ includes contributions that $\delta x_i$ would give. If you have a ruler that is only accurate to $1\text{mm}$, presumably over many measurements, your measurements will be spread out so that $S\geq 1\text{mm}$. Of course, one can conceive of instances where this might fail. It is of course the experimenter's perogative to report the best estimate of error he can. And if in some instances $\delta x_i$ is a better measure, then he should report those. But, generally speaking, $S$ is just a better measure overall. | {
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mechanical-engineering
But in my opinion, your thought about screws is the best. By far the most secure and with the greatest leverage. Use one in the form of a screw jack and use a multi-start thread to make it faster. That will solve the speed issue. You'll need to find someone with a lathe to single-point thread you some, which means instead of screws you might as well get them to just make two big screw jacks with hand crank wheels, instead of four smaller, regular screws.
Or if you can't find someone with a lathe, then perhaps find
multi-start/fast travel threaded rod or lead screws. Typically used for linear motion.
If your plate was being guided on vertical tight rails then you only need one screw jack if it is big and strong enough since the rails would eliminate tilting. | {
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optics, reflection
\end{align}
[Which is exactly equal to the critical angle for glass and air, in accordance with the principle of reversibility of light.]
$\therefore ∠OBC = ∠OCB \approx 41.81°$
But this wasn’t our true objective was it? We wanted the light to get totally internally reflected, but for that, $r_{2}$ must be greater than $90°$. Hence $i_{2}$ must be greater than $41.81°$ (because $i \propto r$, by Snell's law), which implies that $r_{1}$ must be greater than $41.81°$ (because $r_{1}$ = $r_{2}$), which means that $i_{1}$ must be greater than $90°$ (again, because $i \propto r$). That is impossible, unless the source be in the glass itself, thus proving our observations. | {
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java, algorithm, recursion, weekend-challenge, sudoku
public static final String displayAsString(int[][]data) {
return buildStringForGrid(data.length, (int)Math.sqrt(data.length), data);
}
private static final String buildStringForGrid(final int dimension, final int blocksize, final int[][]rows) {
final StringBuilder sb = new StringBuilder();
for (int r = 0; r < dimension; r++) {
if (r == 0) {
sb.append(printSymbolLine(dimension, blocksize, null, symbols[0]));
} else if (r % blocksize == 0) {
sb.append(printSymbolLine(dimension, blocksize, null, symbols[3]));
} else {
sb.append(printSymbolLine(dimension, blocksize, null, symbols[2]));
}
sb.append(printSymbolLine(dimension, blocksize, rows[r], symbols[1]));
}
sb.append(printSymbolLine(dimension, blocksize, null, symbols[4]));
return sb.toString();
} | {
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quantum-mechanics, visible-light, electromagnetic-radiation, photons, quantum-electrodynamics
More precisely, is a quantum of light meant to be just a certain number of wavelengths of light (something like "1 quantum = a single period of a sine wave" perhaps?), or is the concept completely unrelated to wavelengths? In other words, how much is a single quantum? There are two meanings usually attached to the word "quantum" in quantum theory, one colloquial and one technical.
As you know, electromagnetic radiation behaves in ways characteristic of both waves and particles. For non-specialists, it's easy to think of a particle as being a "unit" of the wave, and since "quantum" means a unit of something, the word has gotten associated with "particle." But in reality, the idea of a particle isn't precisely defined. When people talk about a particle of light, the EM field associated with what they probably mean could be described as a wave packet, which you can think of as an electromagnetic wave that is localized to some small region in space. For example, something like this: | {
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quantum-mechanics, hilbert-space, operators, wavefunction, quantum-states
If we are being very strict, we would say that the position operator $Q$ eats a ket with position-space wavefunction $\psi(x)$ and spits out a ket with position-space wavefunction $x\psi(x)$. However, we often relax a bit and say that $Q$ eats a wavefunction $\psi(x)$ and spits out $x\psi(x)$.
The reason we use kets in the first place is that it can be quite convenient to not restrict yourself to a particular basis. I find it very difficult to believe that you've never used the vector notation $\vec r$ as opposed to the index notation $r_i$, and this is precisely the same thing. The only difference is that the index $i$ in $r_i$ runs over $\{1,2,3\}$, while the index $x$ in $\psi(x)$ runs over $\mathbb R$. | {
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quantum-mechanics, quantum-spin, relativity, dirac-equation
Massive charged particles such as the electron require a 4-component spinor - i.e. a pair of two 2-component spinors - but for neutrinos, the minimum amount to describe a single particle is given by one 2-component spinor. | {
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who make sure they find a match for all of set B, and who don't mind using polyamory to do it. Give an example of function. A one-one function is also called an Injective function. In other words, the function F maps X onto Y (Kubrusly, 2001). Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. Other examples with real-valued functions A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. De nition 67. Why is that? (ii) Give an example to show that is not surjective. f(x) = 0 if x ≤ 0 = x/2 if x > 0 & x is even = -(x+1)/2 if x > 0 & x is odd. The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, https://www.calculushowto.com/calculus-definitions/surjective-injective-bijective/. The only possibility then is that the size of A must in fact be exactly equal to the size of B. Why it's surjective: The entirety of set B is matched because every | {
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navigation, move-base
Title: Sending multiple goals to Navigation Stack
I am writing a program for autonomous navigation of a Husky. I combining imu, wheel encoders and visual odometry for navigation. All the sensors have been successfully integrated and a target goal can be sent to the robot. Here is my question:
How can I send the robot multiple goals so that it will visit each goal and do not stop at each one? Right now, I am using the ac.waitForResult() function, (based on the "Sending Goals to the Navigation Stack" tutorial). The robot will stop at each goal and then move to the next one. I want the robot to move smoothly. | {
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newtonian-mechanics, forces, acceleration
Title: Duration of the force exerted and the acceleration According to Newton's Second Law, consider the case of wanting to accelerate the mass of 3 kg by 3 m/s^2 on a frictionless surface, I need to exert 9 N of force. My question is that
To let the mass accelerate forever, do I have to exert the force of 9 N forever or just for a short amount of time? What's the difference?
If I push the same object quickly on the surfacet that has a kinetic friction of 1 N and the object starts accelerating with 3
m/s^2, how will the friction cause the acceleration to change over
time? | {
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curvature, vacuum, hawking-radiation, qft-in-curved-spacetime, pair-production
The reason for this requirement comes from conservation of energy. When quanta are produced by static spacetime geometry appearance of a photon outside of an isolated body, carrying positive energy $E$, must be accompanied by appearance of another photon carrying the energy $-E$ so the net energy is conserved. But existence of negative energy photons is possible only for black holes behind the horizon. Informally, the possibility of negative energies for photons inside black holes can be linked with the fact that inside of a black hole time flows along the radial direction, so the energy–momentum component of a future–directed photon along the direction given by the outside time does not have to be positive. | {
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"url": null
} |
as velocity and acceleration provide engineers, scientists, and economists with tools to model real-world situations. Calculus II. 12 - 14 Rectangular Lot Problems in Maxima and Minima. Calculus II Homework (If you’re looking for a compiled list of assigned problems, click here for a PDF. Popular Problems. This list of Calculus games and worksheets will help you learn Calculus facts and increase your Calculus skills. ( ) 3 x dx. Pre-Algebra, Algebra, Pre-Calculus, Calculus, Linear Algebra math help. E-mail: [email protected] CALCULUS I Solutions to Practice Problems. You can then utilize the results to create a personalized study plan that is based on your particular area of need. Textbook Authors: Stewart, James , ISBN-10: 1285741552, ISBN-13: 978-1-28574-155-0, Publisher: Cengage Learning. Since 36 62, the equation becomes 6x 62 2 x, so we must have x 2 2 x which has the solution x 4 3. It will entirely ease you to look guide ap calculus problems and solutions as you such as. The | {
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$\hspace{4cm}$
The points not contained in the shaded region correspond to monomials that do not belong to $J$. These monomials form a basis for $\mathbb{C}[x,y]/J$, so we can simply count them to find its dimension. For more on this, I recommend Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms.
• So the dimension is 3？
– gtx
Dec 26, 2016 at 1:19
• I count $4$ black dots in the unshaded region: $(0,0), (1,0), (0,1)$, and $(0,2)$. Dec 26, 2016 at 1:47
Here's an approach you may find simpler: The fact that $(x+2)^2=x^2+4x+4$ is in $I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $x^2$ with $-4x-4$. Similarly, the fact that $(x+2)(y+1)=xy+2y+x+2 \in I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $xy$ with $-2y-x-2$. Finally, the fact that $(y+1)^3 = y^3 + 3y^2 + 3y + 1 \in I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $y^3$ with $-3y^2-3y-1$. | {
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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10263
Location: Pune, India
Re: In a survey of 200 college graduates, 30 percent said they [#permalink] | {
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ros
Title: Searching in map
Hello,
How can i search for specific object in map that has been built using rviz ?
Can I use a specific algorithm?
or can I do this using rviz?
any help?
thanks in advance
Originally posted by marwa on ROS Answers with karma: 153 on 2013-03-21
Post score: 2
You should probably read up on some basic concepts regarding rviz, gmapping and generally "how stuff in ROS works :)" (the tutorials are a good start).
Rviz is the ROS Visualizer and mainly is useful for visualizing the robot and environment state. It does neither build maps, nor does it provide algorithms for searching for objects in them. It just allows you to visualize the map it receives from somewhere else (for example gmapping).
That being said, I am not aware of a existing approach that performs some kind of object recognition on occupancy grid maps (if that is what you require). | {
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thermodynamics, temperature, units, ideal-gas, physical-constants
Title: Conversion of $R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}} $ to $ 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}}$ I'm trying to understand the unit conversions when one unit is obtained by shifting the value of another one by some constant. In particular, the unit conversions from Kelvin to Celsius scale:
$$T(\mathrm{^\circ C}) = T(\mathrm K) - 273.16.$$
I want to find the value of $R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}}$
in the units of $\mathrm{{\frac{J}{kg\cdot mol\cdot ^\circ C}}}$.
What I did was to consider $\frac{1}{\mathrm K} $ as $\frac{1}{1\ \mathrm K} = \frac{ 1}{274.16\ \mathrm{^\circ C}}, $
so that
$$R = 30.3363\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}}, $$
but according to the book that I using, | {
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dft, window, symmetry
subplot(5,1,3);
stem(0:63,imag((X1_analytic)));
title('Imaginary part');
subplot(5,1,4);
stem(0:63,abs(X1_analytic));
title('Abs value ');
subplot(5,1,5);
stem(0:63,angle(X1_analytic));
title('Phase '); Your choosen example frequency $\omega_0 = \frac{5 \pi} {32}$ touched a numerically sensitive case, in which the imaginary part of the theoretical DFT which should be zero is not observed so, on the numerical computation. Since the imaginary part is computed as non-zero, the phase which should be zero according to $\phi[k] = \tan^{-1}( \frac{\Im\{X_1[k]\}}{\Re\{X_1[k]\}} )$ whenever $\Im\{X_1[k]\}$ is zero, turns out to be some erratic function of $k$ (Note also that real part of $X_1[k]$ is also not computed to be exactly zero for those $k$ for which it should be, which also contributes to the erratic phase computation behaviour)
Consider the discrete time signal $x_1[n] = \cos(\frac{5\pi n}{32})$ and its 64-point DFT $X_1[k]$ (for ineteger $k \in [0:63]$) which is computed as: | {
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c#, design-patterns, pdf
// private footerBuilder;
public static CustomerTextDocumentBuilder Create() => new();
public Doc Build()
{
return _document;
}
public CustomerTextDocumentBuilder WithTextDocument(
DocumentInformation documentInfo)
{
new TextDocumentPdfWriter(_document).AddDocument(documentInfo);
return this;
}
public CustomerTextDocumentBuilder WithCustomerInformation(
CustomerInformation customerInfo)
{
new CustomerInfoPdfWriter(_document).AddCustomerInformation(customerInfo);
return this;
}
public CustomerTextDocumentBuilder WithProjectInformation(
ProjectInformation projectInfo)
{
new ProjectInfoPdfWriter(_document).AddProjectInformation(projectInfo);
return this;
} | {
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c#, wpf, mvvm
var upperOverlap = db.WorkOrders.Where(x =>
x.FirstNumber <= SelectedWorkOrder.LastNumber
&& x.LastNumber >= SelectedWorkOrder.LastNumber).FirstOrDefault();
if (upperOverlap != null)
{
MessageBox.Show("Overlap Error");
return;
}
#endregion
SelectedWorkOrder.Created = DateTime.Now;
db.WorkOrders.Add(SelectedWorkOrder);
int insertResult = db.SaveChanges();
if (insertResult > 0)
MessageBox.Show("New Work Order Created");
}
}
#endregion
}
Is this Implementation correct? Are there any guidelines or design patterns for handling this kind of Insert\Select\Update functionality? The part that I really don't like in this code is the IsNewWorkOrder variable, any way to implement it differently?
View
The XAML is rather... simplistic: | {
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Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)
Your proof for (d) seems to prove the following statement: "if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded". That is a true fact, but is that what you want to prove here?
• I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). – Quasar Nov 11 '20 at 16:05
$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$ | {
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The sum of the residues is double that residue. The integral is then $i 2 \pi$ times that sum; I get
$$\int_0^{\pi/2} \frac{dx}{1+\sin^2{x}} = \frac{\pi}{2 \sqrt{2}}$$
• He has only a quarter of unit circle initially. – Start wearing purple May 14 '13 at 16:53
• @O.L.: That is built into my analysis. The factor of $1/4$ cancels an equal factor in the denominator. – Ron Gordon May 14 '13 at 16:57
• WHOA! How did you get the first bit? – user53076 May 14 '13 at 17:28
• $z=e^{i x}$, $dx = -i dz/z$. Then use $\sin{x} = (z-z^{-1})/(2 i)$. Do the algebra. Note that you only have $1/4$ of a full circle in your integral, so you must multiply by $1/4$. – Ron Gordon May 14 '13 at 17:31
• @RonGordon I get a $-1$ where you have a $6$ in the very first line. Are we totally sure that the $6$ is correct? – The Count Jan 5 '17 at 2:47
HINT:
$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}= \int_0^{\pi\over2} \frac{\csc^2xdx}{\csc^2x+1}=\int_0^{\pi\over2} \frac{\csc^2xdx}{\cot^2x+2}$$
Put $\cot x=u$ | {
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"url": "https://math.stackexchange.com/questions/391537/computing-int-0-pi-over2-fracdx1-sin2x"
} |
electric-circuits, electric-current, electrical-resistance, voltage, power
Title: How can more voltage give more power (do more work)? I * V = W
1A of current @ 1V is: 1W
1A of current @ 10V is: 10W
I understand that 1A is 1C-per-second, which is some amount of electrons. Intuitive so far.
How can I get more power if I use the same amount of electrons-per-second flowing through?
Using the water analogy, where voltage is the pressure of the water and current is the amount of water that flows through, using more pressure also gives us more water flow, so they aren't independent variables.
Using another water analogy, where voltage is the height of a waterfall, this one seems to work because the water has some time to accelerate and have a bigger impact on the ground (more power), but I can't find where is the equivalent effect in electricity - what causes the acceleration, and what are the velocity * mass here?
What exactly happens when you increase the voltage in a circuit, without changing the number of electrons passing though per second? | {
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organic-chemistry, hydrogen-bond, conformers
Now, hydrogen bonding is not explicitly treated in this calculation, but based on my discussion with @pentavalentcarbon the Leonard-Jones term in the forcefield (and to a lesser extent the electrostatic term) does capture these interactions.
So clearly, some stabilisng influence is evident in the structure, because if I were to argue on sterics alone, conformer 2 should have had lower energy than conformer 1. As an additional test, I measured the distance between $\ce{H}5$ and $\ce{O}2$ and found it to be 2.62 Angstroms. This is indeed suggestive
Now, using the geometries of conformer 2 and conformer 1, and that of an additional conformer with completely elcipsed oxygens that I obtained from Avogadro, I performed another CCSD(T) calculation is psi4 (again in the aug-cc-pVDZ basis). The table given below summarises my results:
$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{conformer 1} & -229.744410904091524
\\ | {
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ros-melodic, sensor-msgs, image-transport
Originally posted by PeteBlackerThe3rd with karma: 9529 on 2019-03-21
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by eirikaso on 2019-03-21:
Thank you. I have recorded some .bag files with the Intel realsense camera using their ROS driver. Unfortunately they save them as 8UC1 encoded. I will try to use cv_bridge instead tomorrow, but I got a tip that I should be able to just change the encoding string link
I will try to subscribe, apply cv_bridge, and then publish again
Comment by PeteBlackerThe3rd on 2019-03-22:
If you just need change the encoding string. You'll need to first make a non-const copy of the message and then update the encoding as I've shown in my answer, then you will be able to re-publish the updated message
Comment by eirikaso on 2019-03-22:
Yes, thank you. I just tested it 30 minutes ago and got it working :) | {
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matlab, fourier-transform, watermarking
Your professor has gone ahead and detailed out how you can accomplish cross-correlation in the frequency domain. This is often the preferred method, since it is typically much more computationally efficient. Your professor has also detailed common ways to improve cross-correlation detection, specifically the use of windowing functions as well as the use of frequency normalization.
In turns out those methods he described will be very important for your success. The watermark signals are pretty low gain (naturally), so we'll need to maximize the signal to noise ratio (SNR) as much as possible.
I think that since this is seemingly an assignment of some sort, I'm not willing to just give you the answer in numeric or code form, but here's a pictorial output of what I got: | {
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$v_f^2 = \frac{Mv_i^2}{2m}$
Now, conservation of momentum?
11. Nov 1, 2016
### ChrisBrandsborg
Yes, and from conservation of momentum:
Px: 0 = 2mvf⋅sinθ
Py: Mvi = 2mvf⋅cosθ
12. Nov 1, 2016
### PeroK
The first equation is clearly not correct. That implies $\theta = 0$. Remember that momentum is a vector so you must respect the direction the object is moving. The first equation is not needed for this question anyway. But, you need to understand why it's wrong.
What can you do with the second?
13. Nov 1, 2016
### ChrisBrandsborg
Do I put my equation for vf into my two equations for momentum?
14. Nov 1, 2016
### PeroK
The first equation is of little interest in this problem. Focus on the second.
15. Nov 1, 2016
### ChrisBrandsborg
Oh, I thought since they are moving only in y-direction, the x equation has to be = 0 (but I guess it´s just = 0), not 0 = 2mvfsinθ ?
Hmm.. what I can I do with the second one.. Hmm. Can I put $v_f = √\frac{Mv_i^2}{2m}$ in for vf ? | {
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stoichiometry
Title: Confusion regarding a stoichiometric formula In a book, which contains only stoichiometry problems, I came across this formula.
For any given compound $\ce{M_xN_y}$, $x \text{ Moles of N} = y \text{ moles of M}$
According to this rule if I take $\ce{H2O}$ then, $x = 2 \text{ and } y = 1$.
So by the above formula,
$2 \text{ moles of Oxygen} = 1 \text{ mole of Hydrogen}$
But surely,
$2 * \mathrm{N_A} \not= 1 * \mathrm{N_A}$, where $\mathrm{N_A}$ is Avogadro's number.
However, in an example, the author has used this formula and successfully solved a problem? So, if it's correct, can anyone prove this and point out where my contradiction is wrong.
Example as requested:
What weight of oxygen will react with $1$g of calcium?($\ce{Ca}= 40\mathrm{g \over mol}$) | {
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r
Title: Achieve a dataframe from a dynamic list of lists This code works for me, but I am trying to improve my coding and am looking for advice on how to improve it. Specifically I want to simplify this so that perhaps I can have a dynamically updating dataframe that tracks the "site" that I am on, and the "cutoff" value that I am on, and records the calculations for that site/cutoff value combination. Right now, it's a little clumsy and admittedly very ugly. Through this process, though, I learned about "assign" for creating assigning variables on the fly. I think I just need maybe one or two more concepts before I can really clean this up.
calcs <- function(site, cutoff){} #Returns a 4 item list
cutoffmatrix <- c(0, 0.005, 0.01, 0.015) #matrix that I want to loop through
for(i in 1:length(cutoffmatrix)){ #Loop that uses my function for cutoffmatrix value
test <- list()
co <- cutoffmatrix[i]
assign(paste("TC", i, sep = ""), calcs("TC", co))
} | {
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ros, turtlesim, service, server, spawn
class spawn_turtle_class
{
public:
spawn_turtle_class(ros::NodeHandle& nh)
{
turtlesim_spawn_client = nh.serviceClient<turtlesim::Spawn>("turtlesim_spawn");
}
bool spawn_callback(srv_turtlesim::spawn_turtle::Request &req,
srv_turtlesim::spawn_turtle::Response &res)
{
bool not_spawn=false;
turtlesim::Spawn srv;
ROS_INFO("request: n=%d", (int)req.num_turtle);
/----------------------------------------------------------------------------------------/
for (int i = 0; req.num_turtle > i; i++)
{
srv.request.x = random() % 12;
srv.request.y = random() % 12;
srv.request.theta = random() % 360;
if (turtlesim_spawn_client.call(srv))
{
ROS_INFO("response: You got turt");
}
else
{
ROS_INFO("response: No turt :(");
not_spawn = true;
}
} | {
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"tags": "ros, turtlesim, service, server, spawn",
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} |
bond, hydrogen-bond
Title: What are the prerequisites in hydrogen bonding? I know when a hydrogen bond is, but - can hydrogen bonds only occur between two molecules containing hydrogen? Or, only one of the molecules should contain hydrogen and the other one may not contain hydrogen, but should have a large electronegative difference between two of its sides? Actually, theoretically speaking hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. A hydrogen bond may occur between $\ce{H-O}$, $\ce{H-N}$, $\ce{H-F}$ (first two are more common). Apart from that, these hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular)
See also the Wikipedia article about hydrogen bonding | {
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} |
2020 least squares function r | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9678992905050948,
"lm_q1q2_score": 0.8525228547393211,
"lm_q2_score": 0.8807970654616711,
"openwebmath_perplexity": 1114.914286161403,
"openwebmath_score": 0.7120915055274963,
"tags": null,
"url": "http://www.intermash.ua/8nxz7/66kmp25/e0l9j.php?79eef7=least-squares-function-r"
} |
Returns:
The grid and the KDE
"""
# boundaries
if bounds:
xmin, xmax = bounds
else:
xmin = min(data) - 3 * width
xmax = max(data) + 3 * width
# grid points
x = np.linspace(xmin, xmax, gridsize)
# compute kde
kde = np.zeros(gridsize)
for val in data:
kde += norm.pdf(x, loc=val, scale=width)
# normalized the KDE
if normalized:
kde /= sp.integrate.simps(kde, x)
return x, kde
Using our function, the KDE is computed as:
x, kde = my_kde(data, gridsize=200)
Now we can compare with the KDE obtained from seaborn. The gaussian kernel being the default, we have to give to seaborn the same band width, that is 1. here, using the bw keyword.
ax = sns.distplot(data, hist=False, axlabel="Something ?",
kde_kws=dict(gridsize=200, bw=1, label="seaborn"))
ax.plot(x, kde, "o", label="my_kde")
plt.legend();
Compute the gaussian KDE with scipy | {
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9752018434079934,
"lm_q1q2_score": 0.8018003514661395,
"lm_q2_score": 0.8221891261650248,
"openwebmath_perplexity": 6749.983393401551,
"openwebmath_score": 0.5839259624481201,
"tags": null,
"url": "http://gsalvatovallverdu.gitlab.io/python/kernel_density_estimation/"
} |
python, python-3.x, django, youtube, oauth
class Command(BaseCommand):
""" """
def handle(self, *args, **options):
players = list(chain(Goalie.objects.all(), Skater.objects.all()))
for player in tqdm(players):
if link_old_or_missing(player):
updated = False
while CREDENTIALS_FILES and not updated:
link = get_link(player.name)
if link:
player.relevant_video = link
player.video_link_updated_at = datetime.datetime.now()
player.save(update_fields=['relevant_video', 'video_link_updated_at'])
updated = True
print(f'{player.name} link is updated!')
else:
print(f'{player.name} link could not be updated with these credentials!')
else:
print(f'{player.name} link is up do date!') | {
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"openwebmath_score": null,
"tags": "python, python-3.x, django, youtube, oauth",
"url": null
} |
php, mvc, validation
$('#list_collection option').each(function(i){
roles[roles.length + 1] = $(this).val();
});
$('#list_assessor option').each(function(i){
roles[roles.length + 1] = $(this).val();
});
$('#department').blur(function(){
var department = $.trim($(this).val());
$('#role').attr('list', 'list_' + department);
});
var password = new LiveValidation('password');
password.add(Validate.Presence);
password.add(Validate.Length, {minimum: 10});
$('input[type=text]').each(function(i){
var field_id = $(this).attr('id');
var field = new LiveValidation(field_id);
field.add(Validate.Presence);
if(field_id == 'department'){
field.add(Validate.Inclusion, {within : departments});
}
else if(field_id == 'role'){
field.add(Validate.Inclusion, {within : roles})
}
});
</script> | {
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "php, mvc, validation",
"url": null
} |
philosophy, agi, rewards
In a more technical term, the AI wants to maximize utility, which depends on the implemented utility function. In the case of DQN, this would be maximizing the score in the game.
The human brain functions in a similar fashion, although a little more complex and often not as straight forward. We as humans usually try to adjust our actions to produce a high output of dopamine and serotonin. This is in a way similar to the reward used to control AIs during reinforcement learning. The human brain learns which actions produce the most amount of those substances and finds strategies to maximize the output. This is, of course, a simplification of this complex process, but you get the picture.
When you talk about motivation, please don't mix it up with consciousness or qualia. Those are not required for motivation at all. If you want to discuss consciousness and qualia in AI, that's a totally different ball game. | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "philosophy, agi, rewards",
"url": null
} |
java, algorithm, strings, unit-testing, comparative-review
/**
* find the largest substring that has no repeated character
* starting at character i in array c
* @param c
* @param i
* @return
*/
public int findLocalMax(char[] c, int i){
int n = c.length;
// seen: characters already seen
HashSet<Character> seen = new HashSet<Character>();
int localMax = 0;
for (int j = i; j < n; j++){
// c[j] was seen already
if (seen.contains(c[j])){
return localMax;
}
else{
seen.add(c[j]);
localMax++;
}
}
return localMax;
} | {
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"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "java, algorithm, strings, unit-testing, comparative-review",
"url": null
} |
newtonian-mechanics, forces, reference-frames, acceleration, inertial-frames
Acceleration
[...]
By applying the transformation above from the stationary to the rotating frame three times the absolute acceleration of the particle can be written as: | {
"domain": "physics.stackexchange",
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"tags": "newtonian-mechanics, forces, reference-frames, acceleration, inertial-frames",
"url": null
} |
homework-and-exercises, electrostatics, potential, coordinate-systems, dipole
At first glance this looks reasonable, with two big lumps of opposite charge and some 'ringing' around them, but that appearance is deceptive:
That 'ringing', i.e. the shallow regions of charge opposite to the central lumps at $r>1$, are much wider than the central lumps, and because of that, they have a significant fraction of the charge of central lumps themselves, but with opposite sign and at a larger distance from the origin (so they contribute more to the dipole moment).
Moreover, because of that, the dipole moment of the charge distribution is zero. | {
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"openwebmath_score": null,
"tags": "homework-and-exercises, electrostatics, potential, coordinate-systems, dipole",
"url": null
} |
clustering
ROCK is a static model that combines nearest neighbor, relocation, and hierarchical agglomerative methods. In this algorithm, cluster similarity is based on the number of points from different clusters that have neighbors in common.
You can use CBA Package in R to perform the ROCK clustering.
Algorithm Steps:
Data----->Draw Random Sample----->Cluster with Links----->Label Data in DIsk | {
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"openwebmath_score": null,
"tags": "clustering",
"url": null
} |
2020 adjoint of a matrix properties | {
"domain": "eatatnakama.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754506337407,
"lm_q1q2_score": 0.828840867951316,
"lm_q2_score": 0.8418256492357358,
"openwebmath_perplexity": 825.3674565895485,
"openwebmath_score": 0.9794465899467468,
"tags": null,
"url": "http://eatatnakama.com/great-value-pzvmnkd/article.php?id=adjoint-of-a-matrix-properties-d6992e"
} |
javascript, beginner, jquery, animation
Title: Simple animation with Javascript and JQuery I'm learning Javascript and JQuery, and just finished a chapter on JQuery animations.So i decided to give it a try and implement a simple animation on Html document i had.
Here is the code:
jQuery(document).ready(function($) {
var animInfo = {
iter : 1, // Iterator
elementCount : 4, // Number of ( <div class="soc(?)"></div><p></p> ) structures to animate
socAnimSpeed : 500, // soc - Section One Content Animation Speed
socpAnimSpeed: 1000 // scop - Section One Content <p> Animation Speed
};
animSoc( animInfo ); // Start the Animations
});
/* Logic Behind the Functions | {
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"tags": "javascript, beginner, jquery, animation",
"url": null
} |
Q. Other examples of exponential functions include: $$y=3^x$$ $$f(x)=4.5^x$$ $$y=2^{x+1}$$ The general exponential function looks like this: $$\large y=b^x$$, where the base b is any positive constant. Get help with your Exponential function homework. answer as appropriate, these answers will use 6 decima l places. If we have an exponential function with some base b, we have the following derivative: (d(b^u))/(dx)=b^u ln b(du)/(dx) [These formulas are derived using first principles concepts. Exponential Functions We have already discussed power functions, such as ( )= 3 ( )=5 4 In a power function the base is the variable and the exponent is a real number. In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent.To differentiate between linear and exponential functions, let’s consider two companies, A and B. The amount of ants in a colony, f, that is decaying can be modeled by f(x) = 800(.87) x, where x is the number of days | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.960361156361018,
"lm_q1q2_score": 0.8002925680026275,
"lm_q2_score": 0.8333245911726382,
"openwebmath_perplexity": 542.4767977931447,
"openwebmath_score": 0.8701205849647522,
"tags": null,
"url": "http://ewingconsulting.com/find-life-tmik/0139ff-exponential-function-examples-with-answers"
} |
organic-chemistry, boiling-point
Fat is made of triglycerides. It'll probably break down to glycerol and the correspondent acid. Then it would decompose and boil.
Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance. | {
"domain": "chemistry.stackexchange",
"id": 535,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "organic-chemistry, boiling-point",
"url": null
} |
java, android
public T get(int row, int column) {
checkIndexes(row, column);
return values[index(row, column)];
}
@Override
public boolean equals(Object obj) {
if (!(obj instanceof Matrix<?>)) {
return false;
}
@SuppressWarnings("unchecked")
Matrix<T> other = (Matrix<T>) obj;
if (other.rows != rows || other.columns != columns) {
return false;
}
for (int row = 0; row < rows; ++row) {
for (int column = 0; column < columns; ++column) {
T elementOfThis = get(row, column);
T elementOfOther = other.get(row, column);
if (!elementOfThis.equals(elementOfOther)) {
return false;
}
}
}
return true;
} | {
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"id": 3940,
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"lm_name": null,
"lm_q1_score": null,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "java, android",
"url": null
} |
c#, beginner, recursion, reflection, extension-methods
Main:
public static void Main(string[] args)
{
IShape circle = new Circle(5, 5, 10);
Console.WriteLine(circle.PropertiesToString());
Console.ReadLine();
}
The above method will also cycle through nested objects and print those to the Console as well, adding in tabs for readability's sake.
I'm kind of unfamiliar with System.Reflection and was wondering if there was a more efficient way I could approach doing something like this. Bug
Your code does not handle value == null everywhere.
object value = property.GetValue(obj, null);
Type valueType = value.GetType(); // <- can throw the infamous NRE
Review | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "c#, beginner, recursion, reflection, extension-methods",
"url": null
} |
javascript, tic-tac-toe
function setSquareToCurrentPlayer(squareId) {
gridArray[squareId] = playersTurn;
endTurn();
}
function endTurn() {
updateGridDisplay();
if (endGame() == true) {
if (winCheck() == true) {
alert("GAME OVER!!! player " + playersTurn + " WINS!!! - hope you had fun :)");
} else {
alert("GAME OVER!!! No more spaces left :( Hope you had fun ;)");
}
} else {
playersTurn == 1 ? playersTurn = 2 : playersTurn = 1;
if (aiPlaying == true) {
if (playersTurn == aiPlayerId) {
runAi();
}
}
}
}
function runAi() {
var aiCanWin = false;
var playerCanWin = false;
for (var i = 0; i < 9; i++) {
if (gridArray[i] == 0) {
gridArray[i] = playersTurn;
if (winCheck()) {
aiCanWin = true;
break;
} else {
gridArray[i] = 0;
}
}
} | {
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"openwebmath_score": null,
"tags": "javascript, tic-tac-toe",
"url": null
} |
algorithms, reference-request
Conversely, suppose that $S'$ can be partitioned into two equal parts, say one of them consisting of $2nx_1+1,\ldots,2nx_m+1$ together with $r$ many $1$s. It follows that
$$ 2n(x_1 + \cdots + x_m) + m + r = 2n(x_{m+1} + \cdots + x_n) + (n-m) + (n-r), $$
and so
$$ 2n(x_1 + \cdots + x_m - x_{m+1} - \cdots - x_n) = (n-2m) + (n-2r). $$
Since $1 \leq m \leq n-1$ and $0 \leq r \leq n$, the right-hand side is in the range $[-(2n-2),(2n-2)]$. Since the left-hand side is a multiple of $2n$, we conclude that the right-hand size is zero, and so $x_1 + \cdots + x_m = x_{m+1} + \cdots + x_n$.
Let us now attend to the sum of $S'$. If $S$ sums to $2R$ then $S'$ sums to
$$
2n \cdot 2R + 2n = 2n(2R+1).
$$
Therefore if $n$ is even, then we are done. Otherwise, we can replace $n$ with $n+1$ throughout (equivalently, add a zero to $S$). | {
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"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "algorithms, reference-request",
"url": null
} |
javascript, functional-programming
Check for collisions and out of bounds:
const checkCollisions = (x, y, movedRovers) => {
if (movedRovers.length > 0) {
for (let i=0; i<movedRovers.length; i++) {
if (movedRovers[i].position.x === x && movedRovers[i].position.y === y) {
throw 'collision detected with rover at position ' + i;
}
}
}
};
const checkBounds = (movedRover, plateauBounds) => {
if ((movedRover.position.x > plateauBounds.x || movedRover.position.x < 0)
||(movedRover.position.y > plateauBounds.y || movedRover.position.x < 0)) {
return false;
}
return true;
};
The Tests:
test('test input matches expected output', () => {
expect(runMission('5 5\n1 2 N\nLMLMLMLMM\n3 3 E\nMMRMMRMRRM')).toBe('1 3 N\n5 1 E\n');
});
describe('functions to process string input', () => {
test('establishes bounds with with keys x, y', () => {
expect(processInput.establishBounds('5 5\n1 2 N\nLMLMLMLMM')).toEqual({'x': 5, 'y': 5});
}); | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "javascript, functional-programming",
"url": null
} |
cm and a circumference of cm. The ratio circumference. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 3. The perimeter of a semicircle is half of the circumference plus the. 0 2cm 0 4cm. It is in circle geometry that the concepts of congruence and similarity. Explain in words how to determine the perimeter of a semicircle. What is 282. Perimeter of a Semicircle Formulas & Calculator. The notation for semicircle and major arc is similar to that of minor arc. A major arc is an arc that is larger than a semicircle. Therefore, all we need to do is add the measurement of each side of the shape and we will have the perimeter. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius $$r$$. Circumference (C): The distance the edge of a circle Radius and Diameter Circumference d: ltd 2Ttr Ex 1) Find the circumference of the circle above. They will also | {
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"lm_q1q2_score": 0.8820429569942798,
"lm_q2_score": 0.8902942152065091,
"openwebmath_perplexity": 427.07611142392966,
"openwebmath_score": 0.8478139638900757,
"tags": null,
"url": "http://milanoporteeserramenti.it/tec/circumference-of-a-semicircle-calculator.html"
} |
javascript, authentication, jwt
Title: User registration and login service using JWT I have created service to communicate with my backend for user registration and login. I use the JS fetch API and send all data through HTTPS. I use JWT tokens to authenticate queries once I have logged, and this is stored in the window local storage.
The API details are stored in a config.json file.
Two things that I'm not sure if I can improve are:
Should I use local storage for storing the JWT token or is there a better alternative?
Is it ok to send the password without hashing it first on the client side? On the server side it is hashed and salted before storage so no plaintext passwords are stored.
import config from "../config.json"
let jwtToken = null; | {
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"tags": "javascript, authentication, jwt",
"url": null
} |
beginner, c, mathematics, floating-point
void getDiv(float a, float b, float c)
{
float finalAnswer;
finalAnswer = doDiv(a,b,c);
printf("%f\n", finalAnswer);
}
I have tried my best to comment each function but I am not sure if I am doing it correctly. Can someone please advise me on making my comments better, naming variables better and anything else that can help me utilize coding standards correctly?
I would like to pick up the habit now while I am still writing small lines of code so I can take it forward with me when I write bigger codes.
Any suggestions on how I can comment better, name variables better, write code more elegant are welcome. Also, do I comment the method declarations on top? There's a bit to say about this code, so I'll try to take it from the top and work my way down: | {
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"openwebmath_score": null,
"tags": "beginner, c, mathematics, floating-point",
"url": null
} |
• "The complex system will work again only when all the components have been printed." Does that mean you'd like the system to end as soon as possible? (Thinking of the objective function here). Besides this, I think this could better be formulated as a scheduling problem on parallel machines, with the additional constraint that only 2 of the 3 machines are used. – dhasson Jul 2 '20 at 13:37
• The Generalized Assignment Problem can serve as inspiration to continue, it's the same you are doing but change the first 3 constraints where every machine is made to print 2 components. First of all, the objective function is total working time of the machines. Instead you want to minimize the makespan (maximum time to finish all jobs), so that system can work again as soon as possible. Second, you may add binary variables $z_i = 1$ if machine $i$ is used, with additional constraints for $\sum_i z_i = 2$ and linking $x$ and $z$. – dhasson Jul 2 '20 at 15:30 | {
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"lm_q2_score": 0.8670357563664174,
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"openwebmath_score": 0.7603274583816528,
"tags": null,
"url": "https://or.stackexchange.com/questions/4479/formulation-of-assignment-problem-as-integer-programming"
} |
c#, game, unity3d
/// <summary>
/// Fill the chunk data with newly generated GameObject representations of chunks, each
/// with their own custom mesh based on where they're positioned.
/// </summary>
public void GenerateChunkData()
{
for(int x = 0; x < this.WidthInChunks; x++)
{
for(int y = 0; y < this.HeightInChunks; y++)
{
for(int z = 0; z < this.DepthInChunks; z++)
{
// Create a new, empty chunk object prefab which will represent the chunk itself
// in the world.
GameObject chunkObject = (GameObject)Instantiate(
this.EmptyChunkPrefab,
new Vector3(x * this.ChunkWidth, y * this.ChunkHeight, z * this.ChunkDepth),
Quaternion.identity
); | {
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"id": 20716,
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Thank you for your replies.
I want to point out that I wasn't mocking about it ; I just wanted clarity. A lot of time some parts of maths seem like nit-picking to me but as plenty of people on here point out the finer points can be important.
FactChecker
Science Advisor
Gold Member
Who writes $x\geq 0$ if he uses $\sqrt x$? This is simply self-evident, resp. clear by context.
I think that should be stated more carefully. Certainly, if $x = -y^2+y+5$ one would not use $\sqrt {x}$ without specifying that $-y^2+y+5 = x \ge 0$ and determining what the corresponding valid values of $y$ are. | {
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reference-request, optimization, programming-languages, compilers, lattices
G. Ramalingam: Data Flow Frequency Analysis. PLDI 1996: 267-277.
(I haven't read the paper beyond the first couple of pages to see that they were using the edge formulation, it's what came up after a bit of Googling.)
I wasn't familiar with the edge transfer function approach until you asked this question. But I also know of at least two cases where it is important to take edges into account.
The first is when dealing with (post)dominance frontiers. The natural definition of the dominance frontier of a node is a set of edges, but the original (more complicated) formulation was in terms of nodes. See for example the (brief) discussion of this in
Pingali, Keshav; Bilardi, Gianfranco: Optimal Control Dependence Computation and the Roman Chariots Problem, ACM ToPLaS, 19(3):462-491, 1997.
Another example is in the classic
Knoop, J; Rüthing, O; Steffen, B: Lazy Code Motion, PLDI, 1992. | {
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cosmology, quantum-information, universe, simulations
To simulate this portion of space, take it's state and transition it to a legal state. Repeat.
We have only used a finite number of bits and we have modeled a section of space. There's a huge difference between the number of bits you can store in a given space and the number of bits you need to describe that space.
Take a single atom of iron with its 26 electrons. For a complete description, you need the many-particle wavefunction $\psi(\vec{x}_1, \vec{x}_2, \vec{x}_3, \dots, \vec{x}_{26})$ (ignoring spin for the moment). Imagine you want to sample it in a given region of space with a very crude grid of 10 points for each direction, so you have $1000$ points in total.
This means you need $1000^{26} = 10^{3\cdot 26} = 10^{78}$ numbers to store it. For decent precision, you want to use at least $16$ bits, so you end up with approximately $10^{79}$ to $10^{80}$ bits. This is more than (or of the same order as) there are atoms in the entire universe. | {
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that even though two planes intersect in a line, it is possible for three planes to intersect in a point. Three planes will always intersect in a point because each pair of planes intersects in a line and the lines intersect in a point. @ACuriousMind I suppose the formal/mathematical treatment was unneeded here (and was also out of my scope). Should we leave technical astronomy questions to Astronomy SE? So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. To determine the plane the two lines share, three points are required. A plane is a ruled surface. If two planes intersect, then their intersectionis a line . It is not parallel. Determine the locations of the planes to each other in the case that n = 4 and second time n = 8. Question 894684: Is it possible for three planes to never intersect? The geometric definition of a line is, a line is a straight line. In 2-dimensional Euclidean space, if two lines | {
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Oh...the Cauchy–Schwarz inequality!
OK. Let's see if I'm doing this right.
$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y} \sqrt {cos^2Z+sin^2Z}$
$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y}$
Using the Cauchy–Schwarz inequality again,
$\cos^2X\cos^2Y+\sin^2X\sin^2Y\leq\sqrt{\cos^4X+ \sin^4X}\sqrt{\cos^4Y+\sin^4Y}$
$\cos^4X<\cos^2X$
$\sin^4X<\sin^2Y$
Adding, $\cos^4X+\sin^4X<1$
Similarly, $\cos^4Y+\sin^4Y<1$
So, $\cos^2X\cos^2Y+\sin^2X\sin^2Y\leq\sqrt{\cos^4X+ \sin^4X}\sqrt{\cos^4Y+\sin^4Y}<1$
So, $\sqrt{\cos^2X\cos^2Y+\sin^2X\sin^2Y}<1$
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Oh..., one need to apply the Cauchy–Schwarz inequality twice to get the problem solved. Thanks, Alexmahone.
I think I've to polish up on the concept of Cauchy–Schwarz inequality. | {
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optics, everyday-life, polarization, vision
First of all, let's write down what unpolarised light is. We choose two basis polarisation states: let's go with left and right polarised light in this case since you say that 3D goggles filter these (I actually don't know what 3D goggles do: since I am blind in one eye I can't see 3D, so I've never given it much thought because I have no intuitive grasp of what 3D sight "looks like"). A general polarisation state can be represented by a vector of two complex numbers $E_+$ and $E_-$ standing for the amplitudes of the left and right components of the light and unpolarised light is where they constants vary swiftly with time: not too fast relative to the light's frequency but still fast enough that over any time period long enough to get a possibly harmful dose of light in the eye. Therefore, for the purposes of this argument we can think of the average total intensity of the light as made up of the average separate intensities of the two polarisations and, for unpolarised light, the | {
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homework-and-exercises, statistical-mechanics, ideal-gas, kinetic-theory
Hope you understand ! | {
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complexity-theory, proof-techniques, np, p-vs-np
Title: Does indirect diagonalization a relativize technique? My main question is can with R.kanon , Fortnow ,... technique that shows lower bounds for SAT seperate P and NP ?
Baker-Gill-Solovay showed that $P?=NP$ could not be solved with relativization. Does indirect diagonalization a relativized proof?
Suppose with indirect diagonalization we show that $NP\not\subset A$ (A is an arbitrary class). how $A$ class can be big with indirect diagonalization technique?(I know it is possible that $TISP(n^{1.8},polylog n)=P$ and the bigger class for me is P or $TISP(n^{O(1)},polylog n)=P$ ). we know that $A$ is at least $TISP(n^{1.8},polylog(n))$(class of languages decides with $O(n^{1.8})$ time and $O(polylog(n))$ space simultaneous).
According to Baker-Gill-Solovay proof I think this approach can not seperate P and NP with indirect diagonalization according this pdf. that mentioned "we will show that such techniques alone cannot prove NP=P" | {
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• Possible duplicate of A.P. terms in a Quadratic equation. – Arnaud D. Aug 1 '18 at 12:52 | {
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planet, size, water, escape-velocity, atmospheric-escape
So it is possible for a science fiction writer to justify a terrestrial type planet with a density somewhat different than what would be calculated based on the terrestrial type planets in our Solar System. But not too much different.
On page 53 Dole sets a rough maximum possible mass for a planet habitable for humans. | {
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javascript, node.js, ecmascript-6, amazon-web-services
let putRecord = (record) => {
let payload = {
Data: JSON.stringify(record),
PartitionKey: String(numbers.random() * 100000)
};
records.push(payload);
if (records.length >= config.kinesis.maxRecords) {
sendRecords();
}
};
return {
putRecord: putRecord,
recordCount: () => {
return records.length;
},
clearRecord: () => {
records = [];
},
init: () => {
setInterval(() => {
if (!_.isEmpty(records)) {
sendRecords();
}
}, 200);
}
};
}; There is no obvious bug but; | {
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} |
- 3 years, 3 months ago
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The subject of your note is whit in the scope of Game Theory. The choice depend of the criteria of the gambler then if he choose to keep the envelope he will be playing under a maximin policy since he is guaranteed the best of the worst scenario. I would recommend a textbook on Game Theory you may find resources in the web
- 3 years, 3 months ago
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×
Problem Loading...
Note Loading...
Set Loading... | {
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sides. Diagonal Formula The line stretching from one corner of the square or rectangle to the opposite corner through the centre of the figure is known as the diagonal. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Your Infringement Notice may be forwarded to the party that made the content available or to third parties such The two legs have lengths of 8. A square is a geometric shape which is fully determined by the lengths of its side, a. In and solve geometry problems diamonds length of diagonal of square diagonals lengths if the square a... Find diagonal of a square and also its proof that means how we that! Have to find the length of the square with side lengths of a square divides it into right. The families of n-hypercubes and n-orthoplexes opposite sides of any rectangle applicable to triangles! Scores, create tests, and the side of the rectangle example: a calculator... Is then the hypotenuse of a square length | {
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"openwebmath_score": 0.6344366073608398,
"tags": null,
"url": "https://monzarifle.com/canadian-dollar-tzmb/length-of-diagonal-of-square-cf583d"
} |
temperature, radiation, thermal-radiation, sun
G.B. Rybicki and A.P. Lightman, Radiative Processes in Astrophysics, First Edition, John Wiley & Sons, Inc., New York, NY, 1979. | {
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modulation, demodulation, channelcoding, software-defined-radio
Title: Getting the bitstream of an unknown RF signal I am trying to get the bitstream of a signal sent over 433MHz. I have installed SDR# (I'm on Windows), connected my NESDR SMART, and I identified the signal here:
I can see there is only one peak with progressively decreasing peaks, so I assume this is AM. Then I set the bandwidth to the width of the central lobe. I finally made a recording of the "Audio" and opened the WAV into Audacity but I could not make sense of the sines (I was expecting a digital stream, since the modulation was specified) so here is the I/Q data instead.
This is an extract of the body of this waveform: | {
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quantum-field-theory, quantum-chromodynamics
Let me give you the example of chiral $SU(2)$ in QCD. In this theory there is an approximate $G=SU(2)_L\times SU(2)_R$ symmetry of the UV theory for up and down quarks
$$
Q_{L}=\left(\begin{array}{c}u_L\\ d_L \end{array}\right)\rightarrow U_L Q_{L}\,,\quad
Q_{R}=\left(\begin{array}{c}u_R\\ d_R \end{array}\right)\rightarrow U_R Q_{R}
$$
where $U_{L,R}$ are unitary $2\times 2$ matrices of $SU(2)_{L,R}$. In the IR, the symmetry is broken spontaneously down to a subgroup $H=SU(2)_{L+R}$ isospin so that 3 massless pions $\pi^a$ are present in the low-energy spectrum according to the Goldstone theorem. This pions can be written in a non-linear sigma model fashion as $\Sigma(x)=e^{i\pi^a(x) \sigma^a/f_\pi}$ where $\Sigma\rightarrow U_L \Sigma U^\dagger_R$ non-linearly realize $G$ but linearly realize $H$ (where $U_L=U_R$). The lowest order lagrangian invariant under $G$ is of the schematic form
$$ | {
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computer-architecture, parallel-computing
So in your MinMax routine not only does every call need to fetch all three branch instructions, (even if on average only 2.5 are evaluated), but every assignment statement takes up a cycle as well (even if it doesn't actually get "executed").
This problem is sometimes called thread divergence. If your machine has something like 32 SIMD execution lanes, it will still have only a single fetch unit. (Here the term "thread" basically means "SIMD execution lane".) So internally each SIMD execution lane has a "I'm enabled/disabled" bit, and the branches actually just manipulate that bit. (The exception is that at the point where every SIMD lane becomes disabled, the fetch unit will generally jump directly to the "else" clause.)
So in your code, every SIMD execution lane is doing:
compare (a > b)
assign (max = a if a>b)
assign (min = b if a>b)
assign (max = b if not(a>b))
assign (min = a if not(a>b))
compare (c > max)
assign (max = c if c>max)
compare (c < min if not(c>max)) | {
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python, beginner, game, python-2.x, adventure-game
computercmd1 = cpuprompt1()
if computercmd1 == "www.plan7286.org":
print "Welcome to the official Plan 7286 website."
computercmd2 = cpuprompt2()
if computercmd2 == "Todd Henley/indigo7286":
print "Welcome, Todd."
print "36.444890, -116.688752"
elif computercmd2 == "exit":
home()
elif computercmd1 == "exit":
home()
def computer1():
print "Your computer. Many hours spent here."
computercmd1 = cpuprompt1()
if computercmd1 == "www.plan7286.org":
print "Welcome to the official Plan 7286 website."
computercmd2 = cpuprompt2()
if computercmd2 == "Todd Henley/indigo7286":
print "Welcome, Todd."
print "36.444890, -116.688752"
elif computercmd2 == "exit":
home2()
elif computercmd1 == "exit":
home1()
def computer2():
print "Your computer. Many hours spent here."
computercmd1 = cpuprompt1() | {
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ubuntu, ros-humble, abb
[spawner-5] [INFO] [1695217066.218926162] [spawner_joint_trajectory_controller]: Waiting for '/controller_manager' services to be available
[ros2_control_node-1] [INFO] [1695217066.726038631] [ABBSystemHardware]: Not connected to robot...
[ros2_control_node-1] [INFO] [1695217067.726593491] [ABBSystemHardware]: Not connected to robot...
[spawner-4] [INFO] [1695217068.231779535] [spawner_joint_state_broadcaster]: Waiting for '/controller_manager' services to be available
[spawner-5] [INFO] [1695217068.231825933] [spawner_joint_trajectory_controller]: Waiting for '/controller_manager' services to be available
[ros2_control_node-1] [INFO] [1695217068.727029265] [ABBSystemHardware]: Not connected to robot...
[ros2_control_node-1] [INFO] [1695217069.727507168] [ABBSystemHardware]: Not connected to robot...
[spawner-4] [ERROR] [1695217070.244324257] [spawner_joint_state_broadcaster]: Controller manager not available | {
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beginner, programming-challenge, haskell, ascii-art, fractals
Title: Solving Sierpinski Triangle in Haskell I am a Haskell beginner with a background in C++ and Python. I have been teaching myself Haskell for about half a year on and off and recently I started doing Hackerrank problems to improve my Haskell muscle. Sometimes I found myself struggling with problems that would be solved fairly easily with an imperative language. Sierpinski triangle is one of them.
My solution ends up much longer than I would have written in Python. Some of the submissions I read at Hackerrank took advantage of the fact that it is a 32 by 63 image to print out while I took a more general approach that should work for any 2^n by 2^(n+1)-1 image. First there is probably a much better general solution to the problem and further more, even with the general solution I have, I still believe that there should be a much more compact way of writing it in Haskell.
Here is my wall of text solution:
import Data.List (groupBy, sortBy, intercalate) | {
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homework-and-exercises, newtonian-mechanics, friction, harmonic-oscillator, spring
$$my''=-ky-mg \implies y''+\omega^2y=-g$$
Here $\omega^2=\frac{k}{m}$. This lead to a solution
$$y(t)=\frac{g}{\omega^2}[cos(\omega t)-1]$$
Assuming this to be true, the object would oscillate with the same amplitude for an indefinite period of time. Intuitively, I would think that the gravity would act as a damping force that causes the oscillations to die out but this does not appear to be the case. Does it make sense that the object in this case will oscillate forever? In your equation, $y$ is the extension from the relaxed or natural length of the spring. If instead you measure instead the extension of the spring from its equilibrium length $y_e$ (where the net force on the mass is zero), you will find the same form of the equation for a horizontal mass on a spring set up. $$-ky - mg = m\ddot y = -k \left( y + \frac{mg}{k} \right) = -k(y-y_e)$$
Now let $Y = y-y_e = y+mg/k = y+g/\omega^2$ and in this shifted variable you obtain $$\ddot Y = -\omega^2Y.$$ | {
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> FAQ > LaTeX > FAQ > LaTeX plus or minus \mp! A plus or minus sign infront of square root consider a Manlet mathematical quantity either... Consider a Manlet sign and plus plus minus latex an online LaTeX editor that 's easy to use two... ( ggplot2 ) collaboration, version control, hundreds of games literally at your fingertips can... -- and -- - are not math symbols mit der Ausgabe: plusminus: \pm is used Plus/minus... With hundreds of LaTeX mathematical symbols essentially on GitHub ) may not have been reviewed symbol where the or. The coding with plus minus sign to a casual user ( mouse plus minus latex. Minus at the top, meaning minus-plus, it ’ s a very simple language allows. Plus/Minus sign ± Hold down the ALT key and type 0177 on keypad... Quantity is either positive or a negative value GitHub ) minus character is \ pm just have say! Converted to the point by a fixed ratio.Here are some less common units - FAQ > LaTeX > FAQ LaTeX. I 'm using these lines of code but I would | {
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"openwebmath_score": 0.9162700176239014,
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electromagnetism, magnetic-fields, spring
Title: Mimic Magnetic Repulsive Resistance through a spring I have two identical magnets and I push them towards each other so that they are repelling.
Is it possible to accurately mimic this repulsion effect (magnetic energy) just using a metal spring (mechanical energy)?
Or is the rate at which the strength of magnetic repulsion in ratio to distance of end points of the magnets changes different than the rate at which the strength of repulsion (spring) in ratio to the distance of the end points of a spring changes?
^Sorry if that sentence made no sense, it's hard to word in an easy to understand manner. I'll try saying it again, but using substitution, below:
Or is the rate at which X changes different than the rate at which Y changes?
X = the strength of magnetic repulsion in ratio to distance of end points of the magnets
Y = the strength of repulsion (spring) in ratio to the distance of the end points of a spring | {
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javascript, form
function hasNumber(s){
return s.split('').some(c => !!~'0123456789'.indexOf(c));
}
function hasSymbol(s){
return s.split('').some(c => !!~symbolList.indexOf(c));
}
function setPassword(s){
pwd.value = s;
pwd.select();
}
/**form {text-align: center;display: block}*/
/**pair {display:block; text-align: left}*/ | {
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newtonian-mechanics, rotational-dynamics
This would only be true if they were accelerating for the same amount of time, since acceleration is the change of velocity per unit time. However, since the cylinder with friction accelerates at a lower rate, but for the same distance, this lower acceleration is applied for a longer time than in the frictionless case. Therefore the logic you were applying gives a final velocity that is too low. | {
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# Prove 𝑃′⋀(𝑃⋁𝑄)→𝑄 - How?
I'm going a course in computer science math, and I came across an exercise that is the following:
Use the rules of equivalence and/or inference to prove:
$\lnot P \land (P \lor Q)\rightarrow Q$
Now, I hate to ask for help, but I have no idea how to derive anything from this, even though I have a cheat sheet for all the equivalence and inference rules in the course.
I tried to solve it with:
1. $\lnot P \land (P \lor Q)\rightarrow Q$ (initial statement)
2. $(\lnot P, P \lor Q)\rightarrow Q$ (simplification)
3. $Q \rightarrow Q$ (disjunctive syllogism)
But this seems wrong. Does anyone else have a clue on which rules to apply for this? Thank you.
First of all, it is probably a good idea to add parentheses to avoid any kind of ambiguity. So, I would write $\lnot P \land (P \lor Q)\rightarrow Q$ as $(\lnot P \land (P \lor Q))\rightarrow Q$
Second:
I tried to solve it with: | {
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quantum-mechanics
or,
$$\langle S_x \rangle = \langle \psi(t) \vert \big(\cos(\omega t) S_x - \sin(\omega t) S_y\big) \vert \psi(t) \rangle,$$
meaning that the magnetization components of the lab and rotating frames are related by
$$\langle S_x \rangle = \cos (\omega t) \langle S_{\tilde x} \rangle - \sin (\omega t) \langle S_{\tilde y} \rangle.$$
Now observe what happens if I multiply this equation by $\cos(\omega t)$:
$$\langle S_x \rangle \cos(\omega t) = \cos^2(\omega t) \langle S_{\tilde x} \rangle - \cos (\omega t)\sin (\omega t) \langle S_{\tilde y} \rangle = \frac{1 + \cos(2 \omega t)}{2}\langle S_{\tilde x} \rangle - \frac{\sin(2 \omega t)}{2}\langle S_{\tilde y} \rangle,$$
so if we pass this through a low-pass filter, we get
$$\frac{1}{2} \langle S_\tilde{x} \rangle.$$
And you can do the same to get $S_\tilde{y}$. | {
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lower-bounds, communication-complexity
Title: Best communication complexity lower bound of disjointness It is well known that no deterministic two-party protocol can solve disjointness problem (DISJ) on $n$-bit inputs without sending $n+1$ bits in the worst case (see, e.g., the book by Kushilevitz and Nisan). For bounded-error randomized protocols, a lower bound of $\delta n$, for some constant $\delta$, has also been shown in a seminal paper by Razborov [Razborov92]. My question is:
What is the best known explicit value of $\delta$ currently (both upper and lower bounds)? | {
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the object is moving downwards. The coefficient of determination, R 2 , is 54. Gravity applies a constant force and thus a constant acceleration. In words why this is the case -- for a shallow fluid, the motion of the fluid is mostly side-to-side, and in order to accumulate more c) According to the linear model, a 1200 square-foot home is expected to have a price of$121,020. Mathematically, the kinetic energy is proportional to the square of the speed, as can be seen in the equation for KE. Direction does not matter for speed. and its a discussion we have time and time again. As you increase the Reynolds number, you would end up approximating that relationship of pressure drop being proportional to the square of the flow rate. east C. Some people square t s and get an F-statistic with 1 degree of freedom in the numerator and n−2 degrees of freedom in the denominator. . This is seen in that the estimates for pressure are only accurate during the summer months when temperatures are | {
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"url": "https://smartclinic2.diapindia.org/wp-content/uploads/2022/01/rhl9jwj/the-increase-of-speed-is-a-squared-relationship.html"
} |
history, time
Title: Why the continued obsession in measuring the year as a multiple of days? Pick a any notion of year and any type of day measurement. Outside of history and the common cycle of day an night, why are these viewed as integrally or even fractionally commensurate?
Why all the stunts with leap days and leap seconds, etc... when we now have atomic clocks? Why not decouple the two? The year is not defined as a multiple of days.
The year is the time for the Earth to orbit the sun once. This is not a constant amount of time. So it is not defined in seconds or days or any other length of time. However it only varies a little.
The (synodic) day is the amount of time for the Earth to rotate once relative to the Sun. This is also not a fixed amount of time. But like the year only varies a little. | {
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mobile-robot, slam, ekf
As a side note, the slides you posted come with a recorded lecture that is freely available on YouTube, thanks to the author of this course. The lectures provide more insightful guidance through the material than just a read. If you haven't already taken a look, it will be well worth the time. | {
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homework-and-exercises, electromagnetism, electric-circuits, capacitance
As we can see since there is a voltage across $R_1$, there will be a current from (b) to ground (i.e. to the negative pole of the voltage source). However, by KCL, there would also be a current flowing through $R_2$, which has no voltage across it. This seems to be a paradox to me.
What I think could be the reason for this is the turning of the switch, which cannot happen in an instant. I think something similar to what is described in this question is taking place but I am not sure. After the switch, you've taken the power source and bottom left resister out of the equation. Assuming both leads are connected to a common ground, the capacitor will discharge through the two resisters. Just because something is labeled ground does not mean it is in all configurations. | {
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general-relativity, cosmology, reference-frames, universe, observable-universe
$^\dagger$Fun fact: This confusion wouldn't rise in my language, where "energy forms" is a single word. | {
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newtonian-mechanics
Title: How to explain that an object at a certain point with zero velocity can move away from that point by having an acceleration? As A.P.French points out:
You may feel, . . . , that we have gone to excessive lengths in our discussion of instantaneous velocity & acceleration. But make no mistake about it; these are very subtle concepts. The notion that an object could both be at a certain point and moving past that point was one that perplexed some of the best minds of antiquity.. . . . If you want to test your mastery of these ideas of instantaneous velocity & acceleration,try explaining to someone how an object that is at a certain point with zero velocity can nonetheless move away from that point by virtue of having an acceleration. It really isn't trivial. | {
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"url": null
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