text stringlengths 1 1.11k | source dict |
|---|---|
$$y=2x^2$$ has derivative $$y'=4x$$. So the derivative will be positive for all $$x>0$$. This means the tangent line will intersect the x-axis at some point $$x_a for a given x.
This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $$x which is made up of only area under the parabola, and a region $$x>x_a$$ where area is from the area between the parabola and the tangent to $$(3,18)$$.
The equation of a tangent line at $$(x_0,y_0)$$ is $$\frac{y-f(x_0)}{x-x_0}=f'(x_0)$$
Or : $$y=f'(x_0)(x-x_0)+f(x_0)$$
The x intercept happens where $$y=0$$.
Requiring $$y=0$$ implies an x intercept of $$x_c=\frac{-f(x_0)}{f'(x_0)}+x_0$$
So from the above arguments with $$x_0=3$$:
$$A=\int_0^{x_c}2x^2dx+\int_{x_c}^32x^2-(12x-18)dx$$
But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle. | {
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red-giant
However, the spectrum of light that is being scattered is very different in the case of a red giant. The solar spectrum peaks at about 500 nm and is about a factor of two less intense at both 400 nm and 700 nm. A red giant has a spectrum that peaks at around 900 nm (in the infrared), and the flux is about 100 times lower at 400 nm and two times lower at 700 nm (which is why they are called red giants).
If Rayleigh scattering was all that was going on, and the total flux incident at the top of the atmosphere was the same, then the scattered spectrum from the red giant illumination would be quite different. The overall amount of scattered red light would be about the same as in the solar case, but the amount of scattered blue light would be reduced by about a factor of 50. The net effect would be that the sky was much darker, and rather than being dominated by blue light, would actually have a redder spectrum (what colour this would be perceived as, I'm not sure). | {
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semiconductor-physics
Title: Why does the depletion layer of a p-n junction diode decrease in width upon forward biasing? The current in the direction of intrinsic diffusion increases, as the potential barrier decreases. I'm okay with that, but I feel that the dip in potential barrier has nothing to do with the width of the depletion region. The absence of free carriers in the depletion region (DR) produces charge separation. This is because the n-type material in the DR contains positively charged donor dopant atoms with no electrons to neutralize their charge and the p-type material in the DR contains negatively charged acceptor dopant atoms with no holes to neutralize their charge. This separation of positive charge to the n-type part of the DR and negative charge to the p-type part of the DR creates an electric field in the DR. The line integral of this electric field (in volts) tells you the potential energy (in electron-volts) that an electron or hole gains or loses in passing from one side of the | {
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javascript, jquery
"want" / "did"
"want again" / "did" (active)
"want" (active) / "done"
"want again" / "done" (active)
"want again" (active) / "done again"
"want again" / "done again" (active)
The path is one of these two options. In the first path, done is clicked first. In the other, want is clicked first.
1--2-----5--6
1--3--4--5--6
Acknowledging the state machine, look at the "want"/"want again" button. The text will always say "I want to do it again" after the done button has clicked. Essentially we can boil it down to this:
$('.want-text').toggle(done <= 0)
$('.want-again-text').toggle(done > 0) | {
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optimization, heuristics
Of course, you can combine these methods. There may be many other variations you could try. I would suggest that you read about methods for optimization, brainstorm other similar methods, and then try them all to see which work best.
Also, you might want to try a simulation: choose $f_t$ that you believe to be representative and where you can evaluate them as many times as desired to find the true maximum and their shape, and then simulate each of these strategies to see how well they work. You might also want to visualize the shape of the $f_t$ and what these strategies are doing (what points they are visiting), in case that helps you refine them or choose suitable constants that are appropriate for your particular application domain. | {
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} |
javascript, jquery, playing-cards
const showGameButtons = (cardDealt) => {
if (cardDealt) {
$("#btnDeal").hide();
$("#btnHit").show();
$("#btnStand").show();
//document.getElementById("btnDeal").disabled = true;
//document.getElementById("btnHit").disabled = false;
//document.getElementById("btnStand").disabled = false;
} else {
$("#btnDeal").show();
$("#btnHit").hide();
$("#btnStand").hide();
//document.getElementById("btnDeal").disabled = false;
//document.getElementById("btnHit").disabled = true;
//document.getElementById("btnStand").disabled = true;
}
if (player.isWinner === true) {
document.getElementById("containerDealer").classList.remove("winner");
document.getElementById("containerPlayer").classList.add("winner"); | {
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homework-and-exercises, lagrangian-formalism, field-theory, variational-principle, klein-gordon-equation
\begin{equation}
\delta S=\int\mathrm{d}^dx\left(\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial_{\mu}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\mu}\partial_{\nu}\delta\phi+\cdots\right),
\end{equation}
where the $\cdots$ terms appear if you have higher derivatives involved (there also may be an annoying factor of $2$ somewhere in that last line since partial derivatives commute, and we don't want to overcount). We can also simply overcome the difficulty of using the above equation by just directly finding $\delta\mathcal{L}$ by finding the first order variation of $\mathcal{L}$ with respect to $\phi\to\phi+\delta\phi$. This gives
\begin{equation}
\begin{gathered}
\mathcal{L}+\delta\mathcal{L}=\frac{1}{2}(\phi+\delta\phi)\partial^2\left(\phi+\delta\phi\right)+\frac{1}{2}m^2\left(\phi+\delta\phi\right)^2\\ | {
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python, performance, programming-challenge, recursion, matrix
You probably do not need to understand the process of this deep math so good. But maybe someone Python experienced can point me to some basic performance upgrade potential the code has.
Questions:
What I am curious about. Is .append inefficient? I heard list comprehensions are better than append.
Would not (0 <= dimensions_coordinate < len(arr)) changed to len(arr) <= dimensions_coordinate or dimensions_coordinate < 0) boost the code?
Are there performance differences between == and is?
Is dimensions = len(coordinates)... if curr_dim == dimensions:... slower than if curr_dim == len(coordinates)?
if you understood the math do you see a way to do it iterative? Because I heard recursions are slower in python and theoretical informatics says "Everything recursive can be iterative"
The whole code:
matrix is a N-dimensional matrix
coordinates of the cell is a N-length tuple
distance is the reach of the neighbourhood | {
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Since $V$ is open in the product $X \times Y$, choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$. With $p \in B$, there exists some $j$ such that $t_j \in B$. First, $(t,t_j) \in V$. Since $t \in U_j$, $(t,t_j) \in U$. Thus $U \cap V \ne \varnothing$. This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$
Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.
Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal. | {
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photons, electrons
Title: Electron-positron annihilation Photon pair production in electron-positron annihilation is necessary to conserve the linear momentum. In such discussions the electron-positron pair collide with equal and opposite momentum, if they collide without the above condition then only a single photon can conserve the momentum (or two or more photons with different energies).
Another question is why more than two photons are not produced in the symmetric collision of electron-positron. The problem is that you need to conserve both energy and momentum. This requires at least two photons that are not quite colinear. A single photon would carry too much momentum since $p=\frac{E}{c}$ for a single photon; two photons at nearly right angles provide a mechanism for conserving energy and momentum. | {
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java, beginner, reflection, factory-method
Title: Factory Design Pattern in Java I am currently attempting to understand the fundamentals of factory methods in Java and I found a lot of exampels where a pattern like this has been used.
I got some exercises about sorting algorithms where one task is to write a factory class for all existing algorithms but with the requirement that newly added algorithmes can be integrated via "plug and play". That phrase itched me.By using the switch case solution I would be unable to add Algorithms without rewriting the factory class so I came up with this:
SorterFactory.java
public final class SorterFactory
{
private SorterFactory()
{
} | {
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c++, beginner
But the more usual check is
cin // If used in a boolean context (like an if) it will be converted
// to a boolean based on the call to good.
if (cin && grade >= 0 && grade <= 100)
But even this is not usual. Usually you check the result of the read operation by placing the read into the if statement.
if (cin >> grade && grade >= 0 && grade <= 100)
Note: the operator>> returns a a reference to the stream, so this is functionally the same as the line above. But it imparts more information.
What you are saying here is: if the read works and the grade is in the range [0..100] we succeeded.
Looking at your dispatch. You have a structure that is the same every time you call.
vector<tuple<int, char>> fork {
tuple<int, char> {90, 'A'},
tuple<int, char> {80, 'B'},
tuple<int, char> {70, 'C'},
tuple<int, char> {60, 'D'},
tuple<int, char> {0, 'E'},
}; | {
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quantum-field-theory, fourier-transform, education, anticommutator
\ \delta_{\bf{k}\bf{k'}}\delta_{\sigma\sigma'}
=
\\
\frac{1}{V} \sum_{k} e^{i {\bf k} \cdot ({\bf r} - {\bf r}^{'}) }
\delta_{\sigma\sigma^{'}}
=
\delta({\bf r} - {\bf r}^{'}) \ \delta_{\sigma\sigma^{'}}
$$
Moreover, if we have electron pair operators, such that:
$$
\hat{\varphi}^{+}({\bf R}) = \int \varphi ({\bf r})
\ \psi^{+}_{\uparrow} \big( {\bf R + \frac{r}{2}} \big)
\ \psi^{+}_{\downarrow} \big( {\bf R - \frac{r}{2}} \big)
\ d^{3}r
$$
where ${\bf R}=(1/2)({\bf r_{1}+r_{2}})$ and ${\bf r}={\bf r_{2}-r_{1}}$, are the center of mass location and distance between the electrons, respectively; Can the commutation relation:
$$
[\hat{\varphi}({\bf R}),\hat{\varphi}^{+}({\bf R}^{'})]
$$
be determine in the same fashion? Since I agree with how you went from the Fourier space to real space commutation relations, I will apply the same steps to the electron pair operator you defined.
\begin{align} | {
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batch
Title: Copying files into Dropbox folder from command line I made the following code.
Since it is a Batch, any kind of problems can appear.
Please review it and point any problems when copying files into Dropbox!
Its intent is that you will copy a file into an Dropbox folder.
Its syntax is dropbox <files-to-dropbox> <dropbox-folder-to-copy-to>.
@echo off
set dropbox=%userprofile%\dropbox
if exist %dropbox% (if not exist %dropbox%\%2 mkdir %dropbox%\%2) else (echo Get Dropbox for PC before using this!)
if exist %dropbox%\%2 copy %1 %dropbox%\%2
@echo on
It would have been cool to use the dropbox tag as well, right? Your script will be a lot easier to read if you expand the long statements to multiple lines:
@echo off
set dropbox=%userprofile%\dropbox
if exist %dropbox% (
if not exist %dropbox%\%2 mkdir %dropbox%\%2
) else (
echo Get Dropbox for PC before using this!
)
if exist %dropbox%\%2 copy %1 %dropbox%\%2
@echo on | {
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First, to get a better look at how the 'fill' option worked, use the 'XData' and 'YData' options in imtransform to force some additional space around the output image.
R = makeresampler({'cubic','nearest'},'fill');
Bf = imtransform(A,T,R,'XData',[-49 500],'YData',[-49 400],...
'FillValues',orange);
figure, imshow(Bf);
Now, try the 'replicate' method (no need to specify fill values in this case).
R = makeresampler({'cubic','nearest'},'replicate');
Br = imtransform(A,T,R,'XData',[-49 500],'YData', [-49 400]);
figure, imshow(Br);
And try the 'bound' method.
R = makeresampler({'cubic','nearest'}, 'bound');
Bb = imtransform(A,T,R,'XData',[-49 500],'YData',[-49 400],...
'FillValues',orange);
figure, imshow(Bb); | {
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python, homework, tic-tac-toe
which is really bad style. It will for example also catch stuff like keyboard interrupts. Since you want to catch errors in the conversion what you probably want is except ValueError:.
And lastly not really an error but: if the user inputs anything other than n on the "play again?"-prompt it'll restart.
I also feel like the runGame method is way too large - I'll see if I can come up with a clearer solution and post it up here if I can.
EDIT:
I've tried my hand at refactoring you're code:
import os
class Board():
"""Represents the game-board"""
def __init__(self):
self.board = [i for i in range(10)]
self._win_combinations = [
(1, 2, 3),
(4, 5, 6),
(7, 8, 9),
(1, 5, 9),
(3, 5, 7),
(1, 4, 7),
(2, 5, 8),
(3, 6, 9)]
self.game_over = False | {
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The first one is trivial: it is a diagonal matrix, so $\begin{bmatrix}1&0\\0&-1\end{bmatrix}^4 = \begin{bmatrix}1^4&0\\0&(-1)^4\end{bmatrix} = I_2$.
The middle one is eliminated: it has duplicated column, so it is singular, so its fourth power is also singular, but the identity matrix is not singular.
Let $R_\theta$ denote the rotational matrix of angle $\theta$.
The thing with rotational matrices is that $R_\varphi R_\psi = R_{\varphi + \psi}$, so $(R_\theta)^4 = R_{4\theta}$.
Then, the rest of the matrices are $R_{-45^\circ}$, $R_{60^\circ}$, $R_{90^\circ}$.
Therefore, their fourth powers are respectively $R_{-180^\circ}$, $R_{240^\circ}$, $R_{360^\circ}$, and only the last one is the identity matrix.
Therefore the answer is the first and the fifth one.
A more algebraic approach: $A^4=I$ means $A$ is root of the polynomial $$f(X):=X^4-1=(X+1)(X-1)(X+i)(X-i).$$ | {
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python, python-3.x, json, web-scraping, email
def process_page(number):
link = "https://www.yell.com/customerneeds/sendenquiry/sendtoone/"
payload = {'message':'Testing whether this email really works.','senderPostcode':'GL51 0EX','enquiryTimeframe':'withinOneMonth','senderFirstName':'mth','senderLastName':'iqbal','senderEmail':'habibrahman@gmail.com','senderEmailConfirm':'habibrahman@gmail.com','uniqueAdId':number,'channel':'desktop','ccSender':'true','marketing':'on'}
headers = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/59.0.3071.109 Safari/537.36'}
response = requests.post(link, data = payload, headers = headers)
items = response.json()
item = items['emailToCustomerUUID']
print(item)
main_page(main_url)
These are the four email IDs I've parsed from that webpage:
1. 468145be-0ac3-4ff0-adf5-aceed10b7e5e
2. f750fa18-72c9-44d1-afaa-475778dd8b47
3. d104eb93-1f35-4bdc-ad67-47b13ea67e55
4. 90d2db3b-6144-4266-a86c-aa39b4f99d9a | {
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newtonian-mechanics, reference-frames, acceleration, centripetal-force, centrifugal-force
Imagine you're in a car. You take a sharp left. Approximate the turn as perfectly circular. Even though you end up turning left, your body feels as though it "wants" to go right, outwards. After all, if you're leaned against the door, you will feel the door exert a force on you. The force would be the normal force, providing the necessary centripetal force to keep going in that circle (also friction, but we can ignore that for this purpose). The conclusion is that $F_C = N$ | {
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microbiology, bacteriology, metabolism, respiration
Yes, there are. They are called aerotolerant anaerobes as shown in this figure on tube 5. In the thioglycollate broth, growth is seen throughout the length of the tube
Legend
Obligate aerobe
Obligate anaerobe
Facultative anaerobe
Microaerophilic
Aerotolerant anaerobes
Are there any which can survive?
Obligate anaerobes wouldn't survive in aerobic conditions. Hence they are named so.
Such a bacterium would be read false positive for oxidizers in such a case because it would ferment glucose even in aerobic conditions.
The thioglycollate broth shown here is used to differentiate bacteria into the above said criteria. The Hugh-Liefson test does'nt do that. You can not conclude that a bacteria is aerobic or anaerobic using that test. The H-L test, is interpreted in two ways.
Fermentative - When both the closed tube and open tube show a colour change
Oxidative - Colour change only in the open tube | {
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homework-and-exercises, fluid-dynamics, navier-stokes
Title: Navier-Stokes system I have to study this system which name is Navier-Stokes. Can you explain please what means that $p$, $u$ and $(u \cdot \nabla)u$. What represents in reality? Tell me please, how should I read the factor: $(u \cdot \nabla)u$? "$u$ multiplied with gradient applied to $u$ " ?
$ (N-S)\begin{cases} -\mu \Delta u +(u \cdot \nabla)u+\nabla{p}=f &\mbox{in } \Omega, \\
\mbox{div }u=0 & \mbox{in } \Omega,\\
u_{\mid{\Gamma}}=0.
\end{cases} $ | {
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general-relativity, mathematical-physics, category-theory
Namely these tensor multiplets are no longer encoded by a Cartan-connection with values in an ordinary group (the Poincare group), but they are encoded by higher Cartan connections with values in higher categorical groups!
This is a long and fascinating story, which does not fit into this comment box here. To get started you might try these lecture notes here
https://ncatlab.org/schreiber/show/Structure+Theory+for+Higher+WZW+Terms
or some of the links provided there. | {
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For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006
This result looks for me like $ln(4)$, but it doesn't help much, since the solution is supposed to be purely analytical. How can I find a sum?
-
As for convergence, we can prove that $f(n)\le \log_2(n+1)$, and the series $\sum_{n\ge 1}\frac{\ln n}{n^2}$ converges. – TZakrevskiy Mar 16 '14 at 9:30
For a graph of $f(n)$, see OEIS A000120 – Henry Mar 16 '14 at 9:33
$f(n)$ is max$\{r\in\mathbb N:2^r | \binom{2n}n\}$ – ajotatxe Mar 16 '14 at 9:48
@user2425 Would you clarify your comment, pls? Probably I'm failing to understand it. f(10)=2. But I can't get 2 from your formula: binomial(20,10) gives 184756 and... "|" is logical or? – wf34 Mar 16 '14 at 10:10
@wf34 it means that $2^r$ divides ${2n\choose n}$. – TZakrevskiy Mar 16 '14 at 10:46
## 1 Answer | {
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c++, strings, reinventing-the-wheel
//reserves space for n chars and copies old elements to new space.
void reserve(size_t n);
void resize(size_t n, char c);
void resize(size_t n) { resize(n, ' '); }
const char* c_str() const noexcept
{
return m_contents;
}
//inserts n elements with value c starting at index pos
my_string& insert(size_t pos, size_t n, char c);
//inserts C-style string s at index pos
my_string& insert(size_t pos, const char* s);
//erases count elemenets starting at index
my_string& erase(size_t index, size_t count);
//erases pos if it is found in m_contents
iterator erase(const_iterator pos);
void pop_back();
void push_back(char c);
my_string& operator+=(const my_string& rhs);
~my_string();
};
my_string.cpp
my_string::my_string()
:m_contents{ nullptr }, m_size{ 0 }, m_space{ 0 }{} | {
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Now take four points ${(x_j,y_j \sqrt{b})}$, ${j=1,\dots,4}$ in ${S}$ in general position (so that the octuplet ${(x_1,y_1\sqrt{b},\dots,x_4,y_4\sqrt{b})}$ avoids any pre-specified hypersurface in ${{\bf C}^8}$); this can be done if ${S}$ is dense. (If one wished, one could re-use the three previous points ${(0,0), (1,0), (a,\sqrt{b})}$ to be three of these four points, although this ultimately makes little difference to the analysis.) If ${(x,y\sqrt{b})}$ is any point in ${S}$, then the distances ${r_j}$ from ${(x,y\sqrt{b})}$ to ${(x_j,y_j\sqrt{b})}$ are rationals that obey the equations
$\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2$
for ${j=1,\dots,4}$, and thus determine a rational point in the affine complex variety ${V = V_{b,x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4} \subset {\bf C}^5}$ defined as
$\displaystyle V := \{ (x,y,r_1,r_2,r_3,r_4) \in {\bf C}^6:$
$\displaystyle (x - x_j)^2 + b (y-y_j)^2 = r_j^2 \hbox{ for } j=1,\dots,4 \}.$ | {
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"url": "https://terrytao.wordpress.com/category/mathematics/mathnt/page/2/"
} |
java, optimization, multithreading, thread-safety, atomic
// .. scheduledexecutors service code to start the background thread
// call the service and get the data and then parse
// the response.
private void callServiceURL() {
String url = "url";
RestTemplate restTemplate = new RestTemplate();
String response = restTemplate.getForObject(url, String.class);
parseResponse(response);
} | {
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catkin, cmake
Now: Will my_fourth_package which depends on my_third_package be able to "see" headers and stuff of my_first_package?
Originally posted by Wolf on ROS Answers with karma: 7555 on 2014-09-12
Post score: 1
No, since the third package does not depend on the second it also does not get the transitive dependency on the first package.
It might work if all packages are installed in the same location and the include folders are all the same after installation. But you should not rely on that. In an isolated build it should not work without adding the dependency explicitly.
Originally posted by Dirk Thomas with karma: 16276 on 2014-09-12
This answer was ACCEPTED on the original site
Post score: 1 | {
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vapor-pressure
B. $\pu{0.5M \ce{NaCl}}$. Sodium chloride dissociates to two ions. Thus there are $0.5 \times \pu{2 M}$ particles in the solvent.
C. $\pu{2M}$ ethylene glycol. Again, this doesn't dissociate, so there are $\pu{2M}$ particles.
D. $\pu{2M \ce{LiBr}}$. Dissociates, resulting in $2 \times \pu{2 M}$ particles.
Thus, listing the concentration of particles in the solvent, you see where the answer comes from:
D: $\pu{4 M}$
C: $\pu{2 M}$
B: $\pu{1 M}$
A: $\pu{0.5 M}$ | {
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# Finding all values of a and b which make this piecewise function continuous
Find all values of a and b that make this piecewise function continuous.
$$f(x)=\begin{cases} (x - π)^2-1&\text{if }x < π\\ b &\text{if } x= π\\ 2\cos(x)+a &\text{if } x > π\end{cases}$$
I tried making the first two expressions equal to one another and then finally the second expression equal to the last expression. I substituted $$π$$ into x for all expressions to find a and b.
Is this the right approach? If not please correct me as I don't understand the ins and outs of piecewise functions that well yet.
First pair of expressions:
$$(π - π)^2-1 = b$$
$$b = -1$$
Second pair of expressions:
$$b =2\cos(π)+a$$
$$-1 = 2\cos(π)+a$$
$$-1 = -2 +a$$
$$a = 1$$
The values of a and b that make $$f(x)$$ continuous are $$a = 1$$ and $$b=-1$$.
• Looks good to me! – Matti P. Aug 12 at 6:43 | {
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climate, climate-change, human-influence, natural-conditions
Tucker Carson says 1:18 "The essence of science is extreme skepticism" - that just ain't so. Skeptisism doesn't equal scientific discovery. Bart Simpson is a skeptic - I don't think we should listen to his views on science. Skeptisism is important to science but it is absolutely NOT the essence of science. The essence of science is asking the right question and working out an answer through evidence and testing and peer review. Peer review doesn't mean denial. For Peer review to be meaningful, evidence or solid counter argument must be presented - not just empty skepticism. Skepticism that ignores the evidence is about as valuable as a restaurant that doesn't serve any food. Skeptics must respect and understand the evidence if they are going to provide intelligent skeptical arguments on why the evidence might mean something else. If they don't do that, it's just noise. Intelligent skepticism that respects the evidence and data is always welcome. Uneducated skepticism is a pointless and | {
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python, python-3.x, automation
define_correction(pwcoords)
# Found power button! Just do a quick check on possible previous games
if wasinlbb.execute(find_and_click=True, frequency=4, duration=4):
lveprvslbb.execute(find_and_click=True, frequency=4, duration=5)
# Now we are ready to accept invites!
# Get the invitation!!
while not accinv.execute(find_and_click=True, frequency=2):
continue
# now we can set node_5:
coordinates, is_radiant = get_node_5()
coordinates = adjusted_coords(coordinates)
# Click the correct spot
node_5 = CoordNode(node_name="Node 5: found right spot",
coords=coordinates)
time.sleep(4) # just an assurance, we're running on slow computers
while not node_5.execute():
continue | {
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discrete-signals, linear-systems, stability
Title: Oscillations in a first-order discrete-time linear systems According to my understanding, continuous first order systems do not exhibit oscillatory behavior because a first order system has only one energy storage element and thus oscillation is not possible. I was simulating the below difference equation for a first order discrete system:
$$y[n] = \tfrac{1}{10} u[n] - \tfrac{7}{10} y[n-1]$$
where $u[n]$ is a unit step function. As can be seen, this equation describes a first order difference equation. However, when plotting $y[n]$, I can see that it oscillates at the start and settles next:
However, when I simulate the following difference equation then this oscillation disappears:
$$y[n] = \tfrac{1}{10} u[n] + \tfrac{7}{10} y[n-1]$$ | {
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autonomous-vehicles, social, legal
These questions seem to not really have an answer, yet self-driving cars are almost ready. What are they doing about all of this? As far as I know, there is still a huge debate about this topic. I would say, that the main rule for every self-driving car is to avoid a crash if possible.
The question one should always ask is, in what situation would a crash really happen, and would a human react differently?
My answer is no. The point is, a human might try to avoid the child (3) but it would be out of instinct rather than "consideration". The driver might even harm others in this situation.
Since a self-driving car will normally follow the rules a critical situation will most of the time arise due to the other person, not the car. So I believe it is best to protect the driver at all costs.
The dog vs child problem could be solved via advanced animal recognition (human vs no human)
Regarding no (2): Too small = no human ==> car will ignore it. | {
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ros-melodic
Title: Failure to find libgpgme.so while building rosbag_storage
I'm building on FreeBSD from source, using catkin_make_isolated.
When I'm building rosbag_storage the linker fails to find libgpgme.so with the error ld: cannot find -lgpgme.
I have installed the libgpgme package and confirm libgpgme.so is in /usr/local/lib. Apparently, this is not a default search path (on FreeBSD only?) for libraries to link, however.
I added link_directories(/usr/local/lib) at the top of the CMakeList.txt file for rosbag_storage. The linker now finds the library and rosbag_storage builds just fine.
Is there a better way to do this than to modify the CMakeList.txt? I'm trying to keep the FreeBSD build process as close as possible to the Ubuntu process, so the fewer manual modifications I have to make, the better.
Originally posted by broomstick on ROS Answers with karma: 111 on 2020-01-22
Post score: 0 | {
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"tags": "ros-melodic",
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molecular-biology, dna, dna-sequencing, history, restriction-enzymes
Title: How was Restriction Site of EcoRI sequenced? The sequence of restriction site of EcoRI - GAATTC was identified in the early 1970s, before Sanger Sequencing was invented.(1977)
How was the restriction site of EcoRI sequenced ? The first determination of a recognition site for a restriction endonulease was reported in:
Kelly & Smith (1970) A restriction enzyme from Hemophilus influenzae II. Base sequence of the recognition site. J Mol. Biol. 51: 393-409 | {
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cc.complexity-theory, time-complexity, boolean-functions, monotone
I was Googling around a fair amount, but couldn't find any answer.
Apologies for the elementary question. Corollary 3.5 of [BHR84] shows that the problem is coNP-complete.
[BHR84] P. A. Bloniarz, H. B. Hunt, III, and D. J. Rosenkrantz. Algebraic structures with hard equivalence and minimization problems. Journal of the ACM, 31(4):879–904, Oct. 1984. http://dx.doi.org/10.1145/1634.1639 | {
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javascript, algorithm, ecmascript-6
arrayToCheck.forEach(function(element) {
if (!arrayWithoutDuplicates.includes(element)) {
arrayWithoutDuplicates.push(element);
}
});
return arrayWithoutDuplicates;
}
// --------------------------------------------------
// --------- JUST USAGE-DEMO ------------------------
// -- ... not that important, folks! ;) -------------
var arr = [0, 1, 2, 3, 4, 5, 6, 7, 1, 2, 7, 8];
var arr2 = ['a', 'b', 'c', 'd', 'e', 'a', 'b'];
function testCreateArrayWithoutDuplicates(arrayWithDuplicates) {
let unique = createArrayWithoutDuplicates(arrayWithDuplicates);
console.log('With duplicates: %o', arrayWithDuplicates);
console.log('Without duplicates: %o', unique);
console.log('Array with duplicates: %s', arrayWithDuplicates.length);
console.log('Array without duplicates: %s', unique.length);
}
testCreateArrayWithoutDuplicates(arr);
console.log('\n');
testCreateArrayWithoutDuplicates(arr2); | {
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My attempt:
i) I took the derivative of f, and found the stationary points by equating it to 0. I noticed that the only stationary point was at 0 itself, thus the graph may only change gradient at this point. I substituted for 1/2 and found that the gradient was positive, so the interval [0, 1/2] is increasing.
ii) I worked out the width of each partition is $\frac{1}{2n}$, and the height of a given partition 'k' is $f(\frac{k}{2n})$. My sum then looked something like this:
$\sum^n_{k=0}{\frac{1}{\sqrt{1 - \frac{k^2}{4n^2}}}$
But the integral is $\frac{1}{2n}\sum_{k=0}^n\frac{1}{\sqrt{1- \frac{k^2}{4n^2}}}= \sum_{k=0}^n\frac{1}{2n\sqrt{1-\frac{k^2}{4n^2}}}$ $= \sum_{k=0}^n\frac{1}{\sqrt{4n^2(1- \frac{k^2}{4n^2}})}$
This looks different to what's in part iii). Is it incorrect? Also, how do I solve part iii)? Thanks.
4. Oh! I see what you did! I must have forgotten to multiply by width! >_<
Now that I understand that bit, how do I calculate the limit? | {
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"url": "http://mathhelpforum.com/calculus/159291-integration-using-riemann-sums.html"
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java, benchmarking
runNormally(): Maybe rename to to: runMeasurements() or similar?
runMeasureTime(): Maybe rename to measureMethodRuntime() or similar?
getParamValue(): Maybe rename to getFirstOrDefault() or similar?
Someone with more expierence implementing anotations than me should probably answer your questions about the annotation implementation, so I'll not comment on that. | {
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spacetime, string-theory, sigma-models
The non-linear $\sigma$ model view: One may view string theory as being a theory essentially defined by the stringy scattering amplitude as the sum over compact 2d manifolds on which certain conformal field theories live. All you need is a conformal field theory with non-anomalous Weyl symmetry on the manifold and the definition of the stringy amplitude, and you've got a string theory. Turns out that at least a certain subset of these theories - superconformal, $\mathcal{N}=2$ theories - are often naturally expressed in terms of an action of fields whose target space is a ten-dimensional manifold. | {
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special-relativity, time-dilation, doppler-effect
Does the doppler effect take into account who is the proper time or is the rest frequency based solely on the velocity of the two frames, irregardless of "proper time"? Because for example I would think that in the twin paradox, both twins would see each other's signals as red shifted since they are moving away from each other. But how could this be the case if their time dilation aren't symmetrical due to only one being in an inertial reference frame, considering that the relativistic doppler shift is originally based off of Lorentz Transformations / Time Dilation in the form of
λ0=(c+u)T0
Is this T0 a different proper time than in the normal time dilation? The flaw is in this part of your argument: | {
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Now we optimize the number of processors using a preprocessing technique. Assuming we started with $$c \cdot n/ \log n$$ processors for some constant $$c$$ (2 is enough here). Let us start by completing the array into a power of two as above and cut the array into chunks, each of size $$\log n$$. We will have at most $$O(n / \log n)$$ of them. Each processor can find the minimum in a chunk sequentially in $$O(\log n)$$ time. Let us build a new array $$A'$$ of length $$O(n / \log n)$$, where the processor responsible for the $$i$$-th chunk writes the minimum in the $$i$$-th position of the array. We end up with an array of size $$O(n / \log n)$$ and with $$c \cdot n / \log n$$ processors we can proceed as above in $$O(\log(n/\log n) = O(\log n)$$ steps.
In total we get an algorithm with $$O(\log n)$$ time and $$O(n / \log n)$$ processors. | {
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models, climate-models, data-analysis
Title: Kalman - Data assimilation with external model I'm working with a distributed snow model which simulates Snow Depth and Snow Cover area. It takes in input some atmospherics variables : Temperature / Precipitation / Wind speed ...
I want to assimilate the output : Snow Depth retrieving remote sensing data ( I have snow cover indices for each 10 days ) and by using Kalman Filter. But some ambiguities that prevent me to do it :
I implemented equation (2a) and (2b), but I have no idea how to implement the model operator M quoted in equation (3)... If you have any idea or document that describes carefully this operator please provide it to me. $M$ is the forward model you are using. It is the dynamical model used to solve whatever discrete equation evolves the field ($d?/dt=$).
The best way to start with data assimilation and understand its nomenclature is to read Ide et al. (1997). In their work, they explain it as: | {
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scikit-learn, regression, visualization, cross-validation
That said, you're fitting and predicting the model on each fold twice by doing this. You could use return_estimator=True in cross_validate to retrieve the fitted models for each fold, or use the predictions from cross_val_predict to generate the scores manually. (Either way though, you'd need to use the splitter object to slice to the right fold, which might be a little finicky.) | {
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fft, python, power-spectral-density, window-functions, segmentation
to
fft_segment[i] = np.abs(np.fft.fft(windowed_segment,fft_size)) # fft automatically pads if n<nfft
If I do that I get the following instead. There's still a scale mismatch (a factor of 2?), but the shape is much closer. | {
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special-relativity, spacetime, velocity, coordinate-systems, inertial-frames
Given that an object is moving at a velocity $\vec{v}$ I can't think of any other way to construct a 4$^{th}$ component with the same dimensions that transforms covariantly other than
$$ V^\mu \equiv \frac{dx^\mu}{dx^\tau} $$ | {
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quantum-field-theory, hilbert-space, normalization
$$
\langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}.
$$
This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do). I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another. This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states. Words of wisdom would be appreciated.] Might as well collect my comments, most deleted, in this memo answer. | {
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special-relativity
$$
which is simplified to
$$
\frac{\partial\tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial\tau}{\partial t}=0\tag{3}
$$
I have read and know how to go from equation (1) to equation (2) using differentials and partial derivatives, but recently, I found a forum thread which stated that what Einstein means by "Making $x'$ infinitely small" is to take a Taylor Series of the components of equation (1) and reducing $x'$ to $0$. Yet, I am not sure of how to do that. Can someone help me? It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).
So we know a priori that $\tau$ is of the form:
$$
\tau\left(x',y,z,t\right) ~=~ Ax' + By + Cz + Dt + E
\,,$$
and
$$
\tau\left(0,0,0,t\right) ~=~ Dt+E | {
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the string. Image result for what is lagrange method In mathematical optimization, the method of Lagrange multipliers (named after Joseph Louis Lagrange) is a strategy for finding the local maxima and minima of a function subject to equality constraints. Please enter the necessary parameter values, and then click 'Calculate'. Get more help from Chegg. Suppose the perimeter of a rectangle is to be 100 units. com and understand syllabus for college, adding and subtracting rational expressions and plenty of other math topics. Lagrange Multipliers with Two Variables. Note that the unconstrained minimum of is at , where. (Last Updated On: October 2, 2019)Lagrange multipliers for functions of two variables. Euler 賞識 Lagrange，在1766年和 d'Alembert 一起推薦 Lagrange 為（柏林科學院）Euler 的繼承人。 在他一生浩瀚的工作中，最為所有數學家熟知的發明就是 Lagrange multiplier（拉格朗日乘數）或 Lagrange multiplier method，這是一個求極值的方法。. Use Lagrange multipliers to find three positive numbers whose sum is 190 and whose product is maximum. LaGrange County | {
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lo.logic, soft-question, ho.history-overview, intuition, function
Once you have different variables with the same name, you also need to prevent capture. The simplest example of capture is that applying the function $\lambda x y. x$ (return first argument) to $y$ must not give $\lambda y. y$ (the identity function), but $\lambda y'. y$ (a constant function).
What's worse is that the counterexamples to naive algorithms are hard to construct when you know the problem already, let alone when you don't. Bugs in almost correct algorithms often lie undetected for years. I hear that even good students typically fail to come up (on their own) with the correct definition of capture-avoiding substitution. In fact, PhD students (me included) and professors aren't exempt from these problem.
That's one reason why some (including one of the best textbooks on programming languages, Types and Programming Languages by Benjamin Pierce) recommend nameless representation (not quite combinatory logic, even though it has been used, but rather deBrujin indexes). | {
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biochemistry, metabolism
1. phosphagen system (very fast, but quickly deprived)
2. glycolysis (fast, but slower deprived)
3. oxidative system (slow, 'impossible' to deprive)
When ATP is used, for example for muscle contraction, ADP is formed. The formed ADP can be rephosphorylated to ATP by the enzym creatine kinase, which also requires creatinephosphate (the phosphate attached to the creatine provides the phosphate group for ADP to form ATP). Additionally, ATP can be formed out of 2 ADP molecules by adenylate kinase. The phosphagen system is largely driven by the law of mass action, i.e. the direction of the enzymatic reaction is controlled by the concentration of the products and substrates. In other words; when ATP is low, and ADP is high, the generation of ATP is favored and vice versa. The energy from the phosphagen system is depleted rather quickly tho. | {
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In the example we are discussing, the substitution "really is" let $u=\arcsin x$. So $u$ naturally lives in the interval $[-\pi/2,\pi/2]$, and therefore $\cos u$ is non-negative.
Hint: first integrate by parts to get,
$$\int\sqrt{1-x^2}\,\mathrm{d}x=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x.$$
Now attempt the trigonometric substitution.
To answer your question: yes, it is safe and doesn't really matter. When you do trig. substitution in your first year calculus course, you are always assuming that $\cos$ is positive as a result you can do:
$$\sqrt{\cos^2 x} = \cos x$$
and not have any problems. Also, take into account what @spencer said; whatever your final answer, you can just find it's derivative and prove yourself right or wrong. | {
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so the trapezium rule gives a method of estimating integrals. It follows that:" Calculate the area under a curve/the integral of a function. Evaluate integral of 0 to 1 ( dx/(1+x^2)) using trapezoidal rule with h = 0. Another way to look at it is that trapezoidal rule approximates the integrand by a first order polynomial and then integrating the polynomial over interval of integration as illustrated in the figure. SAS Macro to Calculate AUC /***** AREA. 1 comment: anon2315 April 19, 2018 at 9:31 PM. In this program when we will execute this program we will see that it will ask for the limit of value after that it will ask for subintervals after that you will get the result by Trapezoidal Rule. both for equally and unequally spaced points, is the trapezoidal rule. Some examples of calculus books that define the trapezoidal rule to be what is actually the composite trapezoidal rule are:. the integral can be approximated by the area of a trapezium. • Fill in the boxes at the top of this | {
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caffeinemachine
Well-known member
MHB Math Scholar
I just want to add one other approach, which I believe is worth learning and it is very easy to apply otherwise. Define a map $f:R^2\to R$ by writing $f(x, y) = (x-1)^2 + y^2$. This map takes the value $0$ only at $(x, y) = (1, 0)$. Thus $G=f^{-1}(R\setminus \{0\})$. Since $f$ is continuous, it suffices to show that $R\setminus \{0\}$ is open. This is easy to show, and certainly easier that showing $G$ is open. | {
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logic, first-order-logic, notation
Title: What does :: indicate in all-quantifier predicates? Can somebody explain to me the meaning of $::$ in the following predicate?
$$(\forall i : P(i) : Q(i)) \equiv (\forall i :: P(i) \implies Q(i))$$ "::" is ":" followed by another ":", i.e., two separators.
"$\forall i :: P(i) \implies Q(i)$" is the same as "$\forall i : : P(i) \implies Q(i)$", i.e, there is no premise/second term. Or you can consider the second term as $\text{true}$ or $\top$.
This situation is similar to the slice notation for lists in Python. For a list A, all four notations A[::3], A[0::3], A[:len(A):3] and A[0:len(A):3] mean the same thing. | {
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telecommunication, waves
$E_0 = \sqrt{ \frac{2 I}{c \epsilon_0} }$
where $c$ is the speed of light and $\epsilon_0$ is the dielectric permeability of vacuum (source)
There are several missing pieces in the specs you describe.
To sum up the specs: frequency 10GHz, input power 10-100W, amplitude 200-400V, distance 10cm
The missing parts: | {
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quantum-mechanics, quantum-interpretations, quantum-entanglement, faster-than-light, epr-experiment
spread their wave function. But doing this entangles the electron with the screen with the first slit in it and effectively erases its entanglement with the photon, unless the momentum of the screen with the first slit can be measured to sufficient accuracy after the electron passes through it. But interference will only be seen if the momentum of the screen cannot be measured to sufficient accuracy without compromising the corresponding uncertainty in the screen's position. Assuming that the level of this uncertainty cannot be controlled at will, the appearance of interference cannot be turned on and off by a distant photon/screen measurer. | {
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lagrangian-formalism, dirac-delta-distributions, variational-calculus, functional-derivatives, density-functional-theory
cf. e.g. my Phys.SE answer here.
Then it is straightforward to check that the 2nd functional derivative is symmetric for a sufficiently smooth functional.
Example:
If $$S[\phi,\psi]~=~\int\!d^nx~{\cal L}(\phi(x),\psi(x),\partial\phi(x),\partial\psi(x),x)$$ is a local functional of first order, then the 1st functional derivative is the Euler-Lagrange (EL) formula
$$ \frac{\delta S}{\delta\phi(x)}~=~\frac{\partial{\cal L}(x)}{\partial\phi(x)}-\frac{d}{dx^{\mu}}\frac{\partial{\cal L}(x)}{\partial\phi_{,\mu}(x)}, $$
and then the 2nd functional derivative becomes symmetric:
$$\begin{align}&\frac{\delta^2 S}{\delta\psi(y)\delta\phi(x)}\cr
~=~&\frac{\partial^2{\cal L}(x)}{\partial\psi(x)\partial\phi(x)}\delta^n(x\!-\!y)+\frac{\partial^2{\cal L}(x)}{\partial\psi_{,\nu}(x)\partial\phi(x)}\frac{d}{dx^{\nu}}\delta^n(x\!-\!y)\cr | {
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homework-and-exercises, special-relativity, time
Title: Again on simultaneity - Einstein train This concerns the version in which a light is flashed at the center of a moving train and while the people onboard see the flashes arrive at the two ends simultaneously, the people on the platform see the flash arrive first at the rear end and later at the front end.
Imagine that for the guy onboard the flashes arrive both at noon and the light is so powerful that it burns the hand of the watches that are fixed at the ends of the train exactly at noon position.
What exactly sees the guy on the platform? Obviously he must see the flash burning the hands in the same position, but at different times. This seems to imply that the time flows with different velocities at the two ends of the train w.r.t. the platform, but this cannot true since both of them travel together at the same velocity. | {
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Is there a field which is the union of finitely many proper subfields?
Is there a field which is the union of finitely many proper subfields? | {
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aqueous-solution, solubility, precipitation
In essence, a crystallisation is the formation of a crystalline solid (i.e. not amorphous), for instance leaving a beaker of $\ce{NaCl}$ in water out for a long time will eventually grow crystals as the water evaporates.
Precipitation can be thought of a something 'crashing out', for instance if you have certain kinds of amines, you can add $\ce{HCl}$ in diethyl ether to precipitate out the $\ce{HCl}$ salt of the amine as they are not soluble.
There are of course in-between cases. Attempts at 'crystallisation' can yield amorphous solids if the conditions are not right, and, equally, precipitation can sometimes give crystalline solids.
The choice of words is therefore always ambiguous, but in general the IUPAC nomenclature should be followed where appropriate (eg. you can never crystallise to give an amorphous solid). | {
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rownames(Z) <- paste0("p", 1:m)
Z <- as.data.table(t(Z))
#
# Compute the probability of each vector.
#
Z[, pi := exp(rowSums(Z))]
Y[, Sum := rowSums(Y)]
X <- cbind(X, Y, Z)
1-sum(X\$pi) # Should be near zero, within floating point roundoff error
#
# Summarize by group sum.
#
H <- X[, .(Probability = sum(pi)), keyby=Sum]
H.mod <- rbindlist(lapply(2:40, function(b)
H[, .(Probability = sum(Probability), Base=b),
keyby=(Sum - choose(m+1,2)) %% b])) | {
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# Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$
I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results. | {
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slam, navigation, 3d-slam, rtabmap, rtab
They are used when Grid/NormalsSegmentation=true:
rtabmap --params | grep Grid/NormalsSegmentation
Param: Grid/NormalsSegmentation = "true" [Segment ground from obstacles using point normals, otherwise a fast passthrough is used.]
Then to know more about this approach, look at the code here: https://github.com/introlab/rtabmap/blob/a4761ad397ce195c254f65be10f7e8e2810facd0/corelib/include/rtabmap/core/impl/util3d_mapping.hpp#L54
So specifically for your question, the cluster parameters are used to cluster the point clouds into multiple surfaces. This can also filter some noise.
Some examples of results:
http://wiki.ros.org/rtabmap_ros#rtabmap_ros.2Fobstacles_detection
http://wiki.ros.org/rtabmap_ros/Tutorials/StereoOutdoorNavigation#Global_costmap
http://official-rtab-map-forum.206.s1.nabble.com/Proj-map-min-obstacles-height-tp2866p2918.html
https://github.com/introlab/rtabmap_ros/issues/304#issuecomment-484716536 | {
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algorithms, algorithm-analysis, probability-theory
Old editions of standard textbooks are usually easy to get cheap. You might try Sheldon M. Ross; A First Course in Probability, a 5th edition from 1997 can be found for less than \$5 at Abe Books (or Powell's, or Zubal). Another old standard is Jay L Devore; Probably and Statistics for Engineering and the Sciences, with similarly low prices for old used editions. (This is the secret of the textbook industry: they churn out new editions with slightly different homework problems every couple of years, so that the students need to buy new editions for way too much money. The result is that if you want a book but don't need it for the homework assignments you can get it for next to nothing.)
Finally: there are the recommendations from James M. Cargal's page of Recommended Books in the Mathematical Sciences. I've used his recommendations as a starting point for several investigations into areas of mathematics where I wanted remedial help. | {
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"openwebmath_score": null,
"tags": "algorithms, algorithm-analysis, probability-theory",
"url": null
} |
ros, eigen3, movegroup
Title: movegroup eigen error when doing pick
ubuntu 14.04 trusty
ros indigo
as soon as my pick place node reaches pick the movegroup launch terminal gives this error :
[ INFO] [1410435960.615041253]: Planning attempt 1 of at most 1
move_group: /usr/include/eigen3/Eigen/src/Core/DenseStorage.h:78: Eigen::internal::plain_array<T, Size, MatrixOrArrayOptions, 16>::plain_array() [with T = double; int Size = 16; int MatrixOrArrayOptions = 0]: Assertion `(reinterpret_cast<size_t>(eigen_unaligned_array_assert_workaround_gcc47(array)) & 0xf) == 0 && "this assertion is explained here: " "http://eigen.tuxfamily.org/dox-devel/group__TopicUnalignedArrayAssert.html" " **** READ THIS WEB PAGE !!! ****"' failed.
[move_group-1] process has died [pid 6922, exit code -6, cmd /opt/ros/indigo/lib/moveit_ros_move_group/move_group __name:=move_group __log:=/home/jaysin/.ros/log/d4d874c0-39a8-11e4-a0cf-28e347742258/move_group-1.log]. | {
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speed-of-light, equivalence-principle
Title: The Equivalence Principle approaching the speed of light If, in local system, an accelerated system is completely physically equivalent to a system inside a gravitational field.
Acceleration results in an increase in speed. Right?
What happens when the accelerated "elevator" accelerates to the point where its speed hits (or exceeds?) the speed of light?
Does it stop at the speed of light?
Is that a possible way to distinguish between the two different forces? Or does something else happen? Accelerating a body beyond $c$ with a force
To accelerate the body you can use a force $\vec{F}$ and the Newton's equation in your lab frame (for experts: the very non-covariant form with relativistic mass $m$):
$$\vec{F} = \frac{dp}{dt} = m \vec{a} + \frac{dm}{dt} \vec{v}$$
But for a body with rest mass $m_0$ (i.e. the one measured at rest) and velocity $v$ we have $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ | {
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c++, multithreading
void worker(int thread_index){
while(End != state.load()){ /* Until the pool is stopped */
{ /* Wait until main thread triggers a task */
unique_lock<mutex> my_lock(state_mutex);
synchroniser.wait(my_lock,[this](){
return (Idle != state.load());
});
}
if(End != state.load()){
worker_function((*target_buffers), thread_index);/* do the work */
{ /* signal that work is done! */
unique_lock<mutex> my_lock(state_mutex);
++threads_ready; /* increase "done counter" */
}
synchroniser.notify_all(); /* Notify main thread that this thread is finsished */
{ /* Wait until main thread is closing the iteration */
unique_lock<mutex> my_lock(state_mutex);
synchroniser.wait(my_lock,[this](){
return (Start != state.load());
});
} | {
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=& \frac{\mathrm{trace}(Q\Lambda Q^T)}{n} \\ =& \frac{\mathrm{trace}(Q^TQ\Lambda)}{n} \\ =& \frac{\mathrm{trace}(\Lambda)}{n} \\ =& \frac{\lambda_1+\ldots+\lambda_i+\ldots+\lambda_n}{n} \\ \\ \geq& \sqrt[n\,]{\lambda_1\cdot\ldots\cdot\lambda_i\cdot\ldots\cdot\lambda_n} \\ =& \sqrt[n]{\det(\Lambda)} \\ =& \sqrt[n]{\det(Q^TQ\Lambda)} \\ =& \sqrt[n]{\det(Q\Lambda Q^T)} \\ =& \sqrt[n]{\det(A)} \end{align} | {
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"openwebmath_score": 0.9995983242988586,
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"url": "https://math.stackexchange.com/questions/552780/is-tran-geq-nn-deta-for-a-symmetric-positive-definite-matrix-a-in-m"
} |
javascript, jquery, performance, html, dom
Title: Efficient string concatenations with DOM I form DOM nodes as strings and append them to DOM tree like below using jQuery.
var dom = '<div><div style="display: inline-block">first name</div>'
'<div style="display: inline-block">last name</div></div>';
$("#contacts").append(dom);
The above code is a small sample. In most of my cases dom will hold big strings.
When I recently read about JS performance tutorials, I saw this post. It mentioned that this way of string concatenation is not a good practice. It mentioned the use of .join() instead of concatenation. That seems like an old post, so which one is efficient in these days? Strings are immutable. It means that when you allocate memory while creating a string, you are not able to reallocate it.
So, this code:
var a = 'a';
a = a + 'bc'; | {
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# Velocity At Earth's Equator
darshanpatel
## Homework Statement
Earth rotates in on axis that goes through both the North and South Poles. It makes one complete revolution in 24 hours. If the distance from the axis to any location on the equator is 3960 miles, find the linear speed (in miles per hour) of a location on the equator.
-None-
## The Attempt at a Solution
I used the formula V=rw to find the linear velocity, and for 'r' I used 3960, and 'w' is used 2pi/24 for making one revolution per hour, so I simplified that down to pi/12 as angular velocity.
V=3960 x (pi/12 radians/hour)≈ 1036.73 miles/hour
Is this correct? It got checked in school, but I think I was off in the decimals or something. Is there something I did wrong or?
Thanks for any help!
Homework Helper
## Homework Statement | {
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python, numpy
np.meshgrid and np.mgrid (or np.ogrid) can also be used to form the X,Y grid.
fn(*np.meshgrid(grid_x, grid_y))
fn(*np.ogrid[-Grid_side:Grid_side:N_pix*1j, -Grid_side:Grid_side:N_pix*1j])
Look up their docs, and practice. Practice with simple things like grid_x[:,None]+grid_y as well. | {
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aqueous-solution, solubility
In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.
Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :) | {
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# Matrix of $T$ over $P_2$
I'm having a bit of trouble following the logic my professor used to construct the matrix of a given transformation $T$ in class, and was wondering if anyone could share any further intuition or insight.
Given $P_2$, the set of polynomials of degree $\leq2$, define a linear transformation $T:P_2\rightarrow \mathbb{R}^3$ such that $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$.
The matrix of $T$ with respect to the basis $\mathcal{B}=\left\{1,x,x^2\right\}\\$
is given by $T(1)=\begin{pmatrix} 1\\1\\1\end{pmatrix}$, $T(x)=\begin{pmatrix} 0\\1\\2\end{pmatrix}$, and $T(x^2)=\begin{pmatrix} 0\\1\\4\end{pmatrix}$,
i.e. $\,A=\begin{pmatrix}1&0&0\\1&1&1\\1&2&4\end{pmatrix}$.
I get the whole business of the coefficients on this particular transformation being the "trivial relation" such that $c_1=c_m=0$, and that $ImT=\mathbb{R}^{3}$ such that $T$ is an isomorphism. | {
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• Thanks for your answer. You already know the limit exists and you know what it equals to ($\frac{n(n+1)}{2}$ was not my first guess). Finding what a limit equals to and proving the solution is correct by checking it satisfies the definition of the limit are two completely different exercises. (Similar to the difference between the P and NP classes in the theory of computational complexity.) I am aware I gave correct solution along with the limit, maybe that is why we did not understand each other. Sorry about it. Jul 26, 2013 at 20:01
• @DavidČepelík Personally, I think that if you prove the limit without using your super-powered tool, it doesn't matter if you used it to find the value of the limit, since the validity of your answer then does not depend on the validity of the method you applied. Jul 29, 2013 at 12:57 | {
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"url": "https://math.stackexchange.com/questions/451799/how-to-show-lim-x-to-1-fracx-x2-dots-xn-nx-1-fracnn"
} |
permutation by the factorial of the number of objects that are identical. There are methods for calculating permutations, and it's important to understand the difference between a set with and without repetition. Number of types to choose from (n) Number of times chosen (r) Permutations: Calculator ; Formula ; Simple online calculator to find the number of permutations with n possibilities, taken r times. If we reduce the number of elements by two, the number of permutations reduces thirty times. For example, consider string ABC. After choosing, say, number "14" we can't choose it again. Or you can have a PIN code that has the … This post deals with methods to generate all possible permutations in Python, of a given set of elements.We consider numeric elements in an array here and do not consider repetition of the same elements. Permutations with Repetition. In this formula, n is the number of items you have to choose from, and r is how many items you need to choose, in a situation | {
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"url": "https://centront.com/u1h5e8n7/d14994-permutations-with-repetition"
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time-complexity
With these lines:
if getline(p) ==# getline(q)
exe q . 'd_'
endif
Then :Time DelDup reports that the command took more than 8s on average.
In case it matters, here's how the :Time command is defined:
com! -count=1 -nargs=+ -complete=command Time call s:time(<q-args>, <count>) | {
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ros, slam, navigation, rtabmap-odometry, rtabmap-ros
CMake was supposed to be a fix / better to "configure" scripts lol. Not getting there any time soon I guess :)
Comment by Kendolph on 2020-09-12:
Well basically nothing changed with the new 'rtabmap_ros' :) I fixed the 'rostopic' this is the output.
topic rate min_delta max_delta std_dev window
========================================================================================
/stereo_odometry 5.006 0.184 0.2092 0.007313 21
/stereo_camera/left/image_rect_color 5.001 0.1925 0.2051 0.003302 21
/stereo_camera/left/camera_info 5.0 0.1952 0.2059 0.002057 20
/stereo_camera/right/image_rect 5.0 0.1823 0.2169 0.007288 20
/stereo_camera/right/camera_info 5.0 0.1959 0.2036 0.001369 20 | {
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particle-physics, vacuum, virtual-particles
Furry-picture Feynman diagrams depicting the one-loop QED corrections to the bound-muon g factor: (a) the electric and (b) magnetic loop vacuum polarization corrections, and (c) the selfenergy wave function and (d) self-energy vertex correction terms. A double external line represents a muonic Coulomb-Dirac wave function, and the wave line terminated by a triangle stands for the interaction with the external magnetic field. An internal wave line represents a photon propagator, and an internal double line depicts a Coulomb-Dirac muon propagator; in vacuum polarization loops, it may also represent an electron-positron propagator. | {
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beginner, c, tic-tac-toe, ai, curses
int can_winning_move = check_for_winning_move(playable_spaces, ai_side);
// Flow through the decision making logic applying the functions and checking for a fart
int thinking = 1;
int picked_space;
while(thinking){
if(playable_spaces[4] == ' '){
if(!(ai_fart(chance_to_fart_center))){
picked_space = 4;
thinking = 0;
break;
}
}
if(can_winning_move){
if(!(ai_fart(chance_to_fart_big_move))){
picked_space = can_winning_move;
thinking = 0;
}else{
picked_space = pick_random_space(playable_spaces);
thinking = 0;
}
}else if(can_block_opponent){
if(!(ai_fart(chance_to_fart_big_move))){
picked_space = can_block_opponent;
thinking = 0;
}else{
picked_space = pick_random_space(playable_spaces); | {
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c#, game, wpf, library
public void Unfill() => _pixel.Fill = null;
#endregion
#region Properties
public static Canvas Canvas { get; set; }
public static PixelType PixelType { get; set; }
public static Tuple<double,double> Corrective { get; set; }
public static int Size { get; set; }
public Brush GetFillColor => _pixel.Fill;
public int I { get; set; }
public int J { get; set; }
public bool IsFood { get; set; }
#endregion
}
Snake.cs
public class Snake
{
#region Fields
private GameField _gameField;
private readonly DispatcherTimer _dispatcherTimer;
#endregion
#region Ctors | {
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electrical-engineering, heat-transfer
Title: The concept of flux I find it difficult to understand the concept of flux and field.I have searched the internet and refered a few books.From what I understand, generally flux can be defined as "some quantity per unit of some other quantity" E.g: "the force of attraction at a point in space per unit charge placed at that point is called electric flux" , "the rate of heat transfer per unit area is called heat flux in conduction".
Please suggest whether I am right or wrong.
From what I understand, generally flux can be defined as "some quantity per unit of some other quantity". | {
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for $$a = u^2+3u+2$$
$$\frac {da}{du} = 2u+3$$
$$\frac {da}{2u+3} =du$$
Last edited: Oct 28, 2012
2. Oct 28, 2012
### Saitama
No, this is not correct.
A better approach would be to convert the denominator in a perfect square. For example:
$$x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}$$
This could be then evaluated using the standard integral ∫dx/(x^2+a^2) where a is a constant.
3. Oct 28, 2012
### Millennial
Another approach :
The denominator has the quadratic $x^2+3x+2$, which is equal to $(x+2)(x+1)$. Now, we can use partial fractions:
$$\frac{1}{x^2+3x+2}=\frac{1}{(x+2)(x+1)}=\frac{A}{x+2}+\frac{B}{x+1}$$
Would you solve that and then integrate?
4. Oct 28, 2012
### lurflurf
hint
$$d \log(\frac{p}{q})=\frac{q dp-p dq}{p q}$$
5. Oct 28, 2012
### kuskus94
Wait, shouldn't it have to be in a form like $$\frac{1}{x(x-2)^2}=\frac{A}{x}+\frac{B}{(x-2)^2}+\frac{C}{x-2}$$
I still don't get it. Is the "d" coefficient or the symbol of differentiation? | {
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ros, rosserial, rosserial-client
The values of the variables there look a little odd to me, but I'm not exactly sure what they should look like. offset seems correct (13, which matches 'std_msgs::String'). message_type seems corrupted though- it's a jumble of characters with elements of 'std_msgs' and 'String' embedded in it. I'm not sure if that indicates actual corruption, or if the debugger just isn't interpreting it correctly.
I'm really at a loss- any advice would be appreciated!
Originally posted by danep on ROS Answers with karma: 197 on 2013-01-18
Post score: 0 | {
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c++, array, c++20
*(m_begin + offset) is more conventionally written m_begin[offset].
We could really use a constructor for creating a MDSpan<const T> from MDSpan<T> where T is non-const.
To make Iterator fully functional, it needs a public default constructor and the crucial members for its traits to work:
public:
using value_type = T;
using pointer = T*;
using reference = T&;
using difference_type = std::ptrdiff_t;
using iterator_category = std::bidirectional_iterator_tag;
It shouldn't be too hard to make it satisfy the requirements for a random access iterator, which can improve its utility with standard algorithms.
The span class lacks cbegin()/cend(). I mention that because iterator should convert to const_iterator but not vice versa, which can be slightly tricky if we template the iterator to avoid writing it out twice.
Reverse iterator functions would also be good to have, and trivially implemented using std::make_reverse_iterator(). | {
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c++, performance, multithreading
void pause_thread(thread_id_t tid);
void start_thread(thread_id_t tid);
void kill_thread(thread_id_t tid);
void queue_task(thread_id_t tid, const Thread_Task& task);
void wait_for_thread(thread_id_t tid);
bool is_thread_busy(thread_id_t tid);
std::vector<Thread_Context> _thread_contexts;
};
thread_server.cpp:
#include "thread_server.h"
void work(thread_id_t tid, std::vector<Thread_Context>* pcontexts) {
auto& contexts = *pcontexts;
while (!contexts[tid].kill) {
while (contexts[tid].pause);
auto cmd = contexts[tid].task_queue.front();
if (cmd.has_value()) {
cmd.value()();
contexts[tid].task_queue.pop();
}
}
contexts[tid].dead = true;
}
Thread_Server::Thread_Server() {
}
Thread_Server::~Thread_Server() {
for (int i = 0; i < _thread_contexts.size(); i++) {
wait_for_thread(i);
_thread_contexts[i].kill = true;
}
} | {
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Let $f\in R[x]$ be primitive and, when considered in $F[x]$, irreducible. Show that $f$ is also irreducible in $R[x]$. This does not seem to require that $f$ be primitive. Suppose that $f$ were reducible in $R[x]$, say $f(x)=g(x)h(x)$ with $g,h\in R[x]$ of positive degree. Then such a factorization also exists in $F[x]$ because $g$ and $h$ can be considered as elements of $F[x]$ if we replace each coefficient $a$ by the fraction $a/1$ (the unit element $1$ exists because $R$ is a UFD, but it is not essential: the fraction $ar/r$ for arbitrary non-zero $r\in R$ would suffice). So if $f$ is reducible in $R[x]$ then it must be reducible in $F[x]$, and the contrapositive is the statement that we want. ### Herstein 3.11.13 (corollary to Theorem 3.11.1): Let $F$ be a field. Show that $F[x_1,\ldots,x_n]$ is a unique factorization domain. Theorem 3.11.1 and its corollary state that if $R$ is a UFD then $R[x_1,\ldots,x_n]$ is also a UFD. Therefore we must show here that a field $F$ is indeed a | {
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"url": "http://bergknoff.com/journal/topics-in-algebra-chapter-3-section-11b"
} |
electromagnetism, electromagnetic-radiation
\mathrm{KA}\boldsymbol{=}\Vert\mathbf{R}\Vert\boldsymbol{=}R\boldsymbol{=}c\, \Delta t\boldsymbol{=}c\left(t\boldsymbol{-}t_{\mathrm{ret}}\right)
\tag{q-01}\label{q-01}
\end{equation}
On the other hand the triangle side $\:\rm KL\:$ is the straight segment along which the charge $\:q\:$ travels with speed $\:\upsilon\:$ from the retarded time moment $\:t_{\mathrm{ret}}\:$ to the present time moment $\:t\:$ so
\begin{equation}
\mathrm{KL}\boldsymbol{=}\upsilon\left(t\boldsymbol{-}t_{\mathrm{ret}}\right)
\tag{q-02}\label{q-02}
\end{equation}
Now, the aforementioned triangle similarity is valid since
\begin{equation}
\dfrac{\mathrm{KL}}{\mathrm{KA}}\boldsymbol{=}\dfrac{\upsilon\left(t\boldsymbol{-}t_{\mathrm{ret}}\right) }{c\left(t\boldsymbol{-}t_{\mathrm{ret}}\right)}\boldsymbol{=}\dfrac{\upsilon}{c}\boldsymbol{=}\dfrac{\beta}{1}\boldsymbol{=}\dfrac{\Vert\boldsymbol{\beta}\Vert}{\Vert\mathbf{n}\Vert}
\tag{q-03}\label{q-03}
\end{equation} | {
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"tags": "electromagnetism, electromagnetic-radiation",
"url": null
} |
terminology
Title: What's the difference between constitutive laws and governing equations? I'm studying about the finite element method in a class but I don't come from a civil engineering background. Anyways, it hasn't been made clear to me what the difference between constitutive laws and governing equations are. To me they both relate physical quantities with one another. A constitutive law is generally an algebraic relation which tells you the coefficients of a differential equation, while the governing equations are the differential equations themselves.
For example, if I have a metal piston on top of a gas, I can write down the equation of motion for the piston
$$m \ddot X - PA = 0$$ | {
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"tags": "terminology",
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javascript, angular.js
var app = angular.module('store-panel', []);
app.controller('ProductController', function($scope, productService) {
this.sendToCart = function(product){
var productNew = product;
var alreadyInCart = productService.getProducts();
var existInCart = jQuery.inArray(productNew, alreadyInCart);
function checkExist() {
if (existInCart === -1 ){
productService.addProduct(productNew);
}
}
checkExist();
console.log(alreadyInCart);
};
}); | {
"domain": "codereview.stackexchange",
"id": 10598,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "javascript, angular.js",
"url": null
} |
c#, mvvm
if (item.Links != null && item.Links.Any())
feedItem.Link = item.Links.First().Uri;
}
feedData.Items.Add(feedItem);
}
You should always enclose a scope with {} braces, even if it's just for a single instruction. Also I would extract the if...else part into its own method, to reduce nesting; I think you could have a SetFeedItemLinkAndContent(SyndicationItem item, FeedItem feedItem, SyndicationFormat format) method, which would make the loop look like this:
foreach (SyndicationItem item in feed.Items)
{
FeedItem feedItem = new FeedItem();
if (item.Title != null && !string.IsNullOrEmpty(item.Title.Text))
{
feedItem.Title = item.Title.Text;
}
if (item.PublishedDate != null)
{
feedItem.PubDate = item.PublishedDate.DateTime;
}
if (item.Authors != null && item.Authors.Any())
{
feedItem.Author = item.Authors.First().Name;
}
SetFeedItemLinkAndContent(item, feedItem, feed.SourceFormat); | {
"domain": "codereview.stackexchange",
"id": 7911,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "c#, mvvm",
"url": null
} |
filters, audio, filter-design, equalization, c++
Feel free to ask if I forgot something that might help to understand my problem.
Thanks for your time.
PS : Sorry if this post is not very specific, but I want to clear that out before posting specifics problems or code. Simply put, you need a bank of passband filters. You feed your signal through each of the filters, and sum up the outputs from the filters.
Designing the filters is where the fun comes in.
First off, assuming this is just audio (music or the like) then there's no need of special filters. You can use the simplest and fastest and not worry too much about phase and ripple.
You will want to use IIR filters, as they require the least amount of calculation - since you have 28 bands and presumably are working in stereo, you will want to reduce the load on the CPU else the computer won't be able to do much else. IIR uses less CPU time than FIR filters. | {
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"tags": "filters, audio, filter-design, equalization, c++",
"url": null
} |
quantum-mechanics, mathematical-physics, space, differential-geometry
Then you build a prequantization line bundle, which is a hermitean line bundle with hermitean connection, such that its curvature is proportional to the symplectic form (the usual normalization ${i\over 2\pi}R = \omega$, where $R$ -- curvature, $\omega$ -- symplectic structure), so that symplectic form is just the first Chern class of this line bundle. You do not need polarization for this step. But you actually need the first Chern class of the bundle to be integer, which can be considered as a general case (which corresponds in the compact case to the condition that the phase space has integer volume in the units of $(2\pi\hbar)^{dim/2}$). | {
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"tags": "quantum-mechanics, mathematical-physics, space, differential-geometry",
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} |
dr\\ &= 1 - \exp(-\alpha^2) ~~ \text{for} ~\alpha \geq 0. \end{align*} giving $[1-2Q(\alpha)]^2 \leq 1 - \exp(-\alpha^2)$ for $\alpha \geq 0$, or, equivalently, $\exp(-\alpha^2) \leq 4Q(\alpha) - 4Q^2(\alpha)$. But, since $4Q^2(\alpha) > 0$ for all $\alpha$, we get that $$Q(\alpha) > \frac{1}{4}\exp(-\alpha^2) ~ \text{for}~ \alpha \geq 0$$ which is a lower bound of the form desired. But, as mentioned earlier, this is a very loose bound. | {
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9879462211935646,
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"openwebmath_perplexity": 148.10285287388086,
"openwebmath_score": 0.9909734725952148,
"tags": null,
"url": "http://math.stackexchange.com/questions/236708/lower-bound-on-px-lambda-where-x-is-gaussian"
} |
kinematics, velocity, differentiation, calculus
Title: What's the difference between average velocity and instantaneous velocity? Suppose the distance $x$ varies with time as:
$$x = 490t^2.$$
We have to calculate the velocity at $t = 10\ \mathrm s$.
My question is that why can't we just put $t = 10$ in the equation $$x = 490t^2$$ which gives us total distance covered by the body and then divide it by 10 (since $t = 10\ \mathrm s$) which will give us the velocity, like this:-
$$v~=~\frac{490 \times 10 \times 10}{10} ~=~ 4900\ \frac{\mathrm{m}}{\mathrm{s}}$$
Why we should use differentiation, like this:
$$
\begin{array}{rl}
x & = 490t^2 \\
\\
v & = \mathrm dx/\mathrm dt \\
& = \mathrm d(490t^2)/\mathrm dt \\
& = 490 \times 2 \times t \\
& = 490 \times 2 \times 10 \\
& = 9800\, \frac{\mathrm{m}}{\mathrm{s}}
\end{array}
$$
Which not only creates confusion but also gives different answer.
Any help is highly appreciated. Your question is legitimate and I don't understand why it got downvoted. | {
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"openwebmath_score": null,
"tags": "kinematics, velocity, differentiation, calculus",
"url": null
} |
meteorology, tropical-cyclone, atmospheric-circulation, troposphere
Title: Relation between direction of shear and tropical cyclone formation In an earlier question/answer Why are there no cyclones in the southern Atlantic basin there was a point raised about direction of shear(abetting/retarding) cyclone formation. It appears westerly shear hinders the formation of tropical cyclones and easterly shear enhances it. This should mean that cyclone formation is only allowed over monsoonal regions. Is that accurate ? From this article - Tropical wave dynamics and Heatinduced
steady response. | {
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"tags": "meteorology, tropical-cyclone, atmospheric-circulation, troposphere",
"url": null
} |
java, mvc, http
// get pagination details
paginationList = testimonialService.getPaginationDetails(pageNum, perPage);
// pass list of testimonials to the request object
request.setAttribute("paginationList", paginationList);
request.setAttribute("testimonialList", testimonialList);
request.setAttribute("perPage", perPage); | {
"domain": "codereview.stackexchange",
"id": 8596,
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"openwebmath_score": null,
"tags": "java, mvc, http",
"url": null
} |
thermodynamics, superfluidity
Title: Lev Landau's proof of molecular motion at absolute zero? I am reading the book "The man they wouldn't let die" by Alexander Dorozynski, which is an biography of Russian theoretical physicist Lev Landau, and I have encountered this passage
Absolute zero is considered as the lowest possible temperature for any element: $-273.16^{\circ}$ C. It was long assumed that at absolute zero all molecular motion would cease; Landau was one of the first to prove this assumption incorrect.
What proof is the writer talking about here? Does anyone know about the (corresponding) original paper? It's nothing else than Landau's 1941 two-fluid model of superfluids (similar to helium-4) that won him the 1962 Nobel prize in physics.
Landau, L. D., The Theory of Superfluidity of Helium II, J. Phys. 5, 71 (1941) | {
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"tags": "thermodynamics, superfluidity",
"url": null
} |
black-hole, mass, high-energy-astrophysics, pole
.
This question is presented here after it was asked on the space stackexchange. I want to leave it there because it has some relevant discussion to this question overall. The article you've read is not quite accurate/correct. A more correct pictue is as follows: | {
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"tags": "black-hole, mass, high-energy-astrophysics, pole",
"url": null
} |
rosbag
Title: Extract data from .bag without messages built
Hi, I am looking for a way to extract data from .bag files without messages built. Is this possible?, I have some .bag files but I don't have the messages built in my system so I can't see what types they have and I can't extract data from this bags files following the tutorials steps.
However, when I run "rosrun rqt_bag rqt_bag" and open my bag file, I can see the data published (data value).
So I think that this is possible but I can't find out information about how to do this. | {
"domain": "robotics.stackexchange",
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"tags": "rosbag",
"url": null
} |
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