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|---|---|
physical-chemistry, solubility, enthalpy
Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if $\Delta_\mathrm r G$ is favourable (i.e. negative), and $\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy. | {
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camera
Originally posted by hsu with karma: 5780 on 2012-09-12
This answer was ACCEPTED on the original site
Post score: 4
Original comments
Comment by Gazer on 2013-07-04:
Hi. I am looking to simulate swissranger_camera in Gazebo; I found that previous model use "gazebo block laser" as a camera plugin. We will need pointcould2 as an input source for PCL_Ros, which produce the image. But the "block laser plugin" only publish pointcloud1. I am wondering
Comment by Gazer on 2013-07-04:
will ROSDepthCamera suits our needs?
Comment by tasneem2000 on 2023-06-17:
i used GazeboRosOpenniKinect in my URDF to generate a 3D pointcloud but the z values are always either 4.0 or 0.0. this means it cannot see all z levels right? if yes please recommend me a way to generate a 3D pointcloud to determine the heights of the obstacles. | {
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exoplanet
Given this is the case, I am curious to know how it was formed in the first place? If I am not wrong planets are formed when the debris orbiting in the gravitational pull of a star stick together overtime gets in to the form of the planetary bodies. So, if WASP-18b was not always so close to its sun and it was formed far away in cold space, then could there be a possibility that, this exoplanet was formed in an orbit of another star and overtime it got away from its original orbit and trapped in the tug of its current sun? (Its my imagination! just bare with it :))
Also could any one point me to good documentations from where I can learn more. | {
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bandpass
the DFT is sampled in frequency while the DTFT is continuous. Sampling in one domain causes aliasing in the other when the number of samples over the duration is not sufficient. This is what drives a longer filter length than what we could achieve with the other approaches to meet similar performance. | {
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c#, performance, game, console, 2048
I won't show the game implementation here. The Main method becomes
class Program
{
static void Main()
{
IMainUserInterface userInterface = new MainUserInterface();
IGameLoop gameLoop = new GameLoop();
IGame game = new Game2048();
userInterface.StartGame(gameLoop, game);
}
} | {
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quantum-field-theory, renormalization, perturbation-theory, greens-functions, correlation-functions
Title: Renormalization conditions of the Callan-Symanzik equation Assume we have a massive $\phi^4$ theory the exact two-point correlation function is given as
$$G=\frac{iZ}{p^2-m_r^2}+\text{terms regular at } p^2=m_r^2 $$ and if I want to apply renormalized perturbation theory I find
$$G=\frac{i}{p^2-m_r^2-\Sigma(p^2)}$$
where
$$-i{\Sigma(p^2)}$$ is sum of all one particle irreducible diagrams.
Then the renormalization condition is
$$-i{\Sigma(p^2)}|_{p^2=m_r^2}=0$$ such that it will have a pole with residue 1 at $p^2=m_r^2$.
However, when we derive Callan-Symanzik equation for massless theory we define the renormalization condition as
$$G=\frac{i}{p^2} \quad\text{at }\quad p^2=-M^2$$ where $M$ is renormalization scale.
As far as I understood the first $G$ is exact, also the second one is exact and in order to match them we say $$-i{\Sigma(p^2)}|_{p^2=m_r^2}=0$$.
But the confusing part is, in the massless case the exact $G$ should be exactly | {
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astronomy, space, power, radio, satellites
As to your last question, if the satellites emit at different frequencies there should be no problem. It is not practible or even impossible to construct antennas with sufficient directional properties to distinguish the signals of the satellites. | {
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# Computing a map between two lists
What I want is a little difficult to explain. So I try to it with an example.
I have 2 lists, each of which has $(n-1)^2$ elements. Each element is an integer between 0 and $n-1$, inclusive.
For example for $n=4$:
L1 = {1,2,3,2,0,2,3,2,1}
L2 = {2,3,0,3,0,1,0,1,2}
From these lists I want to compute possible elements of L2 that correspond to an element of L1. The result is an ordered list with $n$ elements:
{{0},{2},{1,3},{0}}
This means: | {
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harmonic-oscillator, frequency
To try and justify this, consider the trajectory for simple harmonic motion
$$x = A\sin{(\omega t + \phi)} \, .$$
If we increase $t$ by $2\pi / \omega$, then
$$x_2 = A\sin \left(\omega \left(t + \frac{2\pi}{\omega} \right) + \phi \right) = A\sin{(\omega t + \phi + 2\pi)}$$
Increasing $t$ by $2\pi / \omega$ has increase the phase by $2\pi$, so the object has returned to its previous position. This time increment is then interpreted as the time period of the oscillation.
As for your second question, to understand why the oscillator behaves the way it does under driving forces and/or damping, you really need to try and solve the differential equation for the motion. If you "underdamp" the oscillator, then you do indeed obtain the relationship $\omega^{2} = \omega_{0}^{2} - \beta^{2}$ where $\beta$ is the damping ratio. | {
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javascript, html, event-handling
<a href="#" data-filter=".12"><img src="https://docs.google.com/drawings/d/1IcgefHKcMCxEs6ZUjpVhisnxLE9hZ92zZKqLyy597W4/pub?w=40&h=40" alt="image" onmouseover="this.src='https://docs.google.com/drawings/d/1Sb4sO71xOTgg3nyfJ1PmbqAC_ucTDsEqDm2CiZFyDA8/pub?w=40&h=40';"
onmouseout="this.src='https://docs.google.com/drawings/d/1IcgefHKcMCxEs6ZUjpVhisnxLE9hZ92zZKqLyy597W4/pub?w=40&h=40';"></a>
</p>
</div> You can actually do this entirely in CSS with no need for JavaScript. You just have to switch to using background-image instead of an image within the anchor and apply a unique class (or id) to each element to target them in your CSS. | {
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black-holes, white-holes
Title: Is there a way to move a black hole? Is it possible to move or transport a black hole? What about a white hole? In addition to what @Jack Mazy said about force due to gravitational attraction, there are also charged (Reissner–Nordström) black holes. I would imagine in theory you could take another charged object and place it near the black hole which would cause Coulomb force on black hole putting it in motion. | {
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python-3.x
Your code is fine now: no more global variables.
For a final note, I would suggest to add a suffix to AvailableProxies, such as AvailableProxiesProvider or AvailableProxiesManager, but that's up to you. | {
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python, beginner, algorithm, programming-challenge, pathfinding
print("found entrance at", entrance)
print("found exit at", exit)
print("[#] finished in", time.time()-entry_time, "seconds")
return entrance, exit
def solve(filename, dfs, dji):
print("[*] solving", filename)
solve_time = time.time()
print("opening image file")
try:
maze = Image.open(filename)
except:
print("unable to open file, quiting")
return
width, height = maze.size
print("size =", width, "x", height)
maze.mode = 'RGB'
maze_matrix = maze.load()
graph = generate_graph(maze_matrix, width, height)
entrance, exit = get_entrance_and_exit(maze_matrix, width, height)
if dfs:
path = solve_dfs.alg(graph, entrance[0]*width+entrance[1], exit)
save(filename, path, entrance, exit, "DFS")
if dji:
path = solve_dji.alg(graph, entrance[0]*width+entrance[1]
, exit[0]*width+ exit[1])
save(filename, path, entrance, exit, "DJIKSTRA") | {
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electromagnetism, special-relativity, electric-fields, charge
Edit:
My source - This question was asked in IIT JEE Physics 2006 as follows: As stated by @Philip, both the $E$ and induced $B$ fields are time-independent if the charge density is uniform enough. However, if the charge distribution is somehow discrete, the mentioned fields are time-dependent. For instance, if you have only one point charge moving uniformly in a circular path, the fields are time-dependent everywhere in the plane of rotation except for the center about which the charge revolves. At the center of rotation, both the direction and strength of the induced $B$ field, as well as the strength of the $E$ field, are constant, whereas the direction of $E$ is changing. | {
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javascript, html, audio, simon-says
That { ... }; thing is called an object, and the [color] part is requesting the value of the corresponding color key passed as a parameter. Objects are great at associating specific keys ("red", "green", ...) to specific values ("audoR", "audioG", ...).
Then in the blue element you could put:
<div class="pushPad blue" onclick="audoByColor['blue'].play()">
Using a switch statement
You can achieve a similar effect with a switch statement:
function getAudioByColor(color) {
switch (color) {
case "red":
return document.getElementById("audioR");
case "green":
return document.getElementById("audioG");
case "blue":
return document.getElementById("audioB");
case "yellow":
return document.getElementById("audioY");
default:
return null;
};
} | {
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quantum-mechanics, thermodynamics, special-relativity, statistical-mechanics, kinematics
Title: Obtaining the temperature from Bose-Einstein and Fermi-Dirac distribution Lets say you are given a distribution function $f(p)$ and you want to define a temperature, $T_f$, for this distribution. (I assume $\mu = 0$.)
It is then natural to define a temperature the following way:
\begin{equation}
T_f \equiv \frac{ \int d^3p \ G(p) f(p)}{\int d^3p \ f(p)},
\end{equation}
where $G(p)$ is defined by the following equation
\begin{equation}
T = \frac{ \int d^3p \ G(p) f_{eq}(p,T)}{\int d^3p \ f_{eq}(p,T)},
\end{equation}
where $f_{eq}(p,T)$ is the equilibrium thermal distribution function.
I know that if $f_{eq}$ is given by the Maxwell-Boltzmann distribution, then $$G_{MB}(p) = \frac{p^2}{3E},$$
where $E = \sqrt{p^2 + m^2}$.
What I need is to find an expression for $G(p)$ if $f_{eq}$ is the Bose-Einstein or Fermi-Dirac distribution
$$ f_{eq} = \frac{1}{e^{E(p)/T} \pm 1}.$$ | {
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• Of course, the dimension can be as big as 3, if $W_2\subset W_1$. It can't be bigger; it can be as small as 1, but no smaller; and it can be anything in between, that is, it can be 2. – Gerry Myerson May 12 '16 at 12:32
• @ Gerry Myerson what is the use of field Z/6Z here? – Arun Sharma May 12 '16 at 12:38
• Is $W_1 \bigcup W_2 = V$ true? – Arun Sharma May 12 '16 at 12:47
• Z / 6 Z isn't a field, nor does it appear in the question, Arun, and the union of two subspaces is never a vector space, unless one of the subspaces contains the other. – Gerry Myerson May 15 '16 at 12:46
Note that $$W_1 \subseteq W_1 + W_2 \subseteq V$$. So $$4 = \dim(W_1) \leq \dim(W_1 + W_2) \leq \dim(V) = 6$$. Putting in the information we know into your formula gives that:
$$\dim(W_1 + W_2) = 7 - \dim(W_1 \cap W_2).$$
Using our inequality with this gives:
$$4 \leq 7 - \dim(W_1\cap W_2) \leq 6.$$
This inequality simplifies to:
$$1 \leq \dim(W_1 \cap W_2) \leq 3.$$ | {
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general-relativity, black-holes, gravitational-redshift
Conversely, I know that for someone on the surface of the star that is collapsing to form a black hole it will appear quite different. The observer on the surface will not see anything unusual happen as they cross the event horizon and in a finite time they will reach the singularity at the center of the black hole where we do not know what will happen since general relativity breaks down in a singularity. | {
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r, text-mining
Title: Text Mining - Similarity among words to determine thresholds While trying to find how one place (Military High School) like a school is similar to another one (Military H School or Military High S), I used the LV algorithm to find the string distance and then converted them into a percentage on a calibrated scale.
stringdist(data[i,1],data[i,2],method = "lv")
Though this is admissible, I still need a more precise way to determine the threshold. The goal is to set a threshold of acceptance for a variation of a name from the standardized name, so that say if the threshold of this word is less than say 85%, it would be rejected. This exercise is part of data pre-processing. | {
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means that for a matrix to. matrix is orthogonal. This article is showing a geometric and intuitive explanation of the covariance matrix and the way it describes the shape of a data set. Compute the matrix solution of the orthogonal Procrustes problem. The inverse of an orthogonal matrix is also orthogonal which can be easily proved directly from the definition. The trace of A , denoted Tr A , is defined to be the sum of its diagonal. The adjacency matrix of an undirected simple graph is symmetric, and therefore has a complete set of real eigenvalues and an orthogonal eigenvector basis. We know the first column, [a b] T, of A is a unit vector, since all of the columns of an orthogonal. By letting p i, q. (diag(A)) ij = δ ijA ij eig(A) Eigenvalues of the matrix A vec(A) The vector-version of the matrix A (see Sec. Note: Every Square Matrix can uniquely be expressed as the sum of a symmetric matrix and skew-symmetric matrix. If A has rank n, then the first n columns of P will be an | {
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python, programming-challenge, interview-questions, roman-numerals
There is no need for the if (End == 1) special case: the general case for the algorithm handles one-character inputs just fine.
I see LetterToValue[RomanCharacterized[…]] written everywhere! Why not convert the letters to numbers early on to avoid that verbosity? (See values in my solution below.)
start is not a good variable name, since it is incremented in the loop (in contrast to End, which stays fixed). Note that next is always start + 1, so there is an opportunity to either write assert next == start + 1 or to eliminate one of the variables.
Going further, if you are always considering neighbouring elements of a sequence, it may be better to use zip() in some way. Also, the loop logic would be much simpler if you never advance by two positions. As long as the input is legal (i.e. not like IIX), you can get the same results using the function below, which I consider to be more Pythonically expressive.
def roman_to_int(roman):
""" | {
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organic-chemistry, biochemistry, chemical-biology
Complex IV (Cytochrome c Oxidase)
Cytochrome c oxidase (COX or CcO), the terminal enzyme of the electron transport chain, catalyzes the one-electron oxidations of four consecutive reduced cytochrome c molecules and the concomitant four-electron reduction of one $\ce{O2}$ molecule to yield $\ce{H2O}$:
$$\ce{4cytochrome c^2+ + 4H^ + O2 -> 4 cytochrome c3+ + 2H2O}$$
Overview: | {
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filters, convolution
Title: 2D convolution of image with filter as successive 1D convolutions I want to prove (or more precisely experiment with) the idea that a 2D convoltion as produced by the Matlab conv2() function between an image I (2D matrix) and a kernel (smaller 2D matrix) can be implemented as some 1D conv i.e. the Matlab conv() function and NOT conv2(). Of course possibly some reshapes and matrix multiply might be needed but no conv2().
And to make it clear, I am NOT refering to that kind if thing:
s1=[1,0,-1]'
s2=[1 2 1]
diff=conv2(x,y)-conv2(conv2(x,s1),s2) | {
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human-biology, circadian-rhythms
For instance the Clock and Bmal1 nuclear proteins, when present in sufficiently large amount, will dimerize and act as transcription factors on a series of genes that contain a region in their promoter element called E-box. Amongst these are the Period (Per) genes.
Per is then synthesised, and exported to the cytoplasm. Here, if in a sufficiently large amount, it can heterodimerize with the Cryptocrome (Cry) proteins: the dimer will then enter the nucleus and inhibit the transcription actions of the Clock/Bmal couple. This will in turn, block the transcription of Per, that will therefore not be able to dimerize with Cry and stop the inhibition of Clock/Bmal and so on. All of this takes approximately 24 hours.
This basic loop (and some more) is nicely schematised in this review by Reppert and Weaver
Coordination of circadian timing in mammals. - Reppert and Weaver, Nature. 2002 Aug 29;418(6901):935-41. | {
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thermodynamics, acoustics, vibrations
Title: Difference in Sound and Heat What makes sound and heat distinct ?
The answer is said to lie in the fact that the former consists of vibration in an ordered fashion while the latter is not.
But why would ordered vibration not be heat, when heat is the just the 'jingling' of molecules (ordered or disordered) ?
Shouldn't it be in this way : 'All sounds are heat while all heat is not sound' ? What makes sound and heat distinct ?
What makes them distinct is that sound is the transport of mechanical energy from one place to another in the form of mechanical longitudinal waves, whereas heat is defined as the transfer of energy from one substance to another due to temperature difference. Further explanation follows. | {
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quantum-mechanics, quantum-spin, eigenvalue, observables
So the answer to your question is yes, the standard procedure is to start with the observable $O$ as defined in the absence of spin and then construct the direct product extension to account for spin. | {
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optics, quantum-optics, decoherence
However, you didn't do that. You started the mirror off in a thermal state at room temperature. This state can be considered as a superposition of different momentum states of the mirror1, with a phase associated to each one. If you change the momentum by a small amount, the phase associated to this state in the superposition only changes by a small amount. Now, let $p_\gamma$ and $p_m$ be the original momenta of the photon and the mirror, and let $\Delta p_\gamma$ be the change in the momentum when the photon bounces off the mirror. When you send the photon towards the mirror, the original state $p_m$ (photon passes through) will end up in the same configuration as the original state $p'_m = p_m - \Delta p_\gamma$ (photon bounces off). These two states $p_m$ and $p'_m$ had nearly the same phase before you aimed the photon at the mirror, so they will interfere, and if the phase on these two states are really close, the interference will be nearly perfect. | {
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"tags": "optics, quantum-optics, decoherence",
"url": null
} |
hash, binary, crc
Title: What are the chances of hash collision given large input and small hash? I have an input of 128 bits (binary, 0s and 1s) and want to hash this input with 32 bit CRC. But I am not sure if collision rate is moderate or too high ?
Is it 2^128/2^32 = 2^98.
And does that means 1 collision after 2^98 hashes OR something else ?
I am a little bit confused with the math involved.
Is there any other 32 bit hash better (collision resistant) than CRC 32. I can not use any other hash greater than 32 bits as I have to store these values somewhere. We don't know anything at all about the distribution of your keys, so we will assume they're random.
You are guaranteed to have one collision after $2^{32}+1$ hashes, by the pigeonhole principle. | {
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"url": null
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c++, algorithm, search
Make it work for any comparable type
Turning this index-based algorithm to an iterator-based algorithm made me realize that there weren't any tricky things making use of the value of key to be more efficient and that the key was only used for comparison. Therefore, you can also make the searched type generic instead of restricting it to int. Actually, it is as simple as merely changing its signature and it works with any LessThanComparable type:
template<typename RandomAccessIt, typename LessThanComparable>
RandomAccessIt search(RandomAccessIt begin, RandomAccessIt end, LessThanComparable key); | {
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"url": null
} |
thermal-radiation
Title: In-Vacuum Cooling Via Blackbody Radiation If thermal energy is lost via blackbody radiation, and it's relative to the temperature of the body and its surface area - could you effectively cool a large body by pumping all the thermal energy into a small space, and use blackbody radiation to remove energy from the system (cooling the average large body temperature down)?
For example, you have a space craft in a vacuum with an average body temperature of 500K. Using heat pumps, you bring the average body temperature down to 200K except on a single spot where the temperature is much much higher (defined by he energy previously in the rest of the large body). This would assume you have perfect thermal insulation between the main craft and the hot spot, and no practical upper limit temperature of the 'hot spot'.
I'm thinking about possibility from a physics point of view, ignoring the difficulties in thermal isolation and pumping that much heat around. | {
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php, html, datetime, layout
//color settings for prices
$colorlow = '#6F6';
$colormid = '#09F';
$colorhigh = '#F90';
$colorspecial = '#F0F';
$colorfull = 'rgba(255,0,0,0.3)';
/* CONTENT CALENDAR
-----------------
-----------------
*/
echo '<div id="calendar">';
$count['month'] = 0;
foreach($row as $month)
{
$lastcourse = end($month['course']);
$laststart = $lastcourse['date'];
$enddate = new DateTime($laststart);
$enddate->modify('+ '.($lastcourse['length']-1).' days');
$iterate = new DateTime('01-'.date('m',strtotime($laststart)).'-'.$month['year']);
if (!isset($stored['year']) || isset($stored['year']) && $stored['year'] != $month['year'])
{ | {
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python, python-3.x
if(typ and self.typ != typ.upper().strip()):
return False
if(name and self.ref.name.upper().strip() != name.upper().strip()):
return False
return True
else:
return False
@staticmethod
def update(instance) -> None:
logger.info(f'Updating Synonyms for {instance.name}...')
typ = instance.TYP
if(typ == 'CustomSheet'):
return Synonyms.updateCustomSheet(instance)
elif(typ == 'Projektliste'):
return Synonyms.updateOtherCustomSheet(instance)
else:
logger.error(f'Cannot update synonyms. {typ} is unknown instance.') | {
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vba, winapi
lngTryAgainCtr = 0
TryAgain:
lngTryAgainCtr = lngTryAgainCtr + 1
'Change delimiter to pipe
Call SetLocaleInfo(GetUserDefaultLCID(), LOCALE_SLIST, "|")
'Check to make sure setting separator as pipe worked correctly
strBuffer = String$(85, 0)
Long1 = GetLocaleInfoEx(LOCALE_NAME_USER_DEFAULT, LOCALE_SLIST, lpLCData, 0)
strListSeparator = String$(Long1, 0)
Long1 = GetLocaleInfoEx(LOCALE_NAME_USER_DEFAULT, LOCALE_SLIST, strListSeparator, Long1)
If Instr(strListSeparator, "|") = 0 Then
If lngTryAgainCtr < 3 Then
GoTo TryAgain
Else
If GetLocaleInfoEx(LOCALE_NAME_USER_DEFAULT, LOCALE_SLIST, strListSeparator, Long1) <> 0 Then Debug.Print GetLastError
'pass part 1 error message
'need error message for if run actively instead of part of script
End If
End If | {
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take into account the composition of the remaining cards in the deck. Card deck probability calculator. In this tutorial, you'll learn some of these concepts using a deck of cards and generic poker situations. Hope this helps. One deck has twice the number of cards in the other deck with the same color ration (so one deck has 52 cards and the other has 104, both half red and half black). Since there are 26 black cards in the deck, the probability that the second card is black is 26/52 = 1/2. How to calculate the probability of drawing 27 cards (without replacement) and getting at least 1 card of each rank but this time with only 10 ranks 9 ranks- 4 of each rank 1 rank- 16 of that rank so my formula needs adjusting in other words a standard deck of cards using almost Blackjack ranks A-9 (A only = 1) and 16-10 value cards for 10,J. P(heart or club) b. Thirteen cards in each suit. Given this sampling procedure, what is the probability that exactly two of the sampled cards will be aces (4 | {
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quantum-mechanics, wavefunction, schroedinger-equation, potential, education
How does this help? Well, just as in the particle in a box, the introduction of the finite volume of space tells us that our momenta can no longer take arbitrary values, but must, instead, be quantized as
$$\textbf{p}=\frac{2\pi\hbar}{L}\textbf{n}$$
Where $\textbf{n}=\langle n_x,n_y,n_z\rangle$ is a vector of integers. From this, we easily see that the energy levels are of the form $E_{\textbf{n}}=2\pi^2\hbar^2(n_x^2+n_y^2+n_z^2)/mL^2$. This seems just like the Particle in a Box energy with an extra factor of four (this factor comes from the fact that we chose periodic as opposed to vanishing boundary conditions at the edge). This doesn't seem particularly useful until we start trying to calculate sums over momenta. Using this discretization, however, one can show the correspondance
$$\sum_{\textbf{p}}f(\textbf{p})\sim V\int\frac{\mathrm{d}^3\textbf{p}}{(2\pi\hbar)^3}f(\textbf{p})$$
You might still be skeptical. "When will this actually be used?" you might ask. So let's do an example. | {
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neural-network, perceptron
Is there any equation to predict the output of a multi-layer perceptron network other than iterating over each neuron with $w*x+b$?
Should I just tell my teacher that a logistic regression is a different case and the same does not apply to this type of neural networks?
Is the first formula correct to show that a value of a neuron is the sum product of the previous layers plus the bias? | {
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c++, reinventing-the-wheel, memory-management
const Chunk* prev_chunk(const Chunk* chunk) const noexcept {
if (!chunk || is_begin(chunk)) {
return nullptr;
}
return reinterpret_cast<const Chunk*>(
reinterpret_cast<const unsigned char*>(chunk) -
realsize(chunk->prev_size_));
}
Chunk* next_chunk(Chunk* chunk) noexcept {
if (!chunk || chunk == top_chunk_) {
return nullptr;
}
return reinterpret_cast<Chunk*>(reinterpret_cast<unsigned char*>(chunk) +
realsize(chunk));
}
const Chunk* next_chunk(const Chunk* chunk) const noexcept {
if (!chunk || chunk == top_chunk_) {
return nullptr;
}
return reinterpret_cast<const Chunk*>(
reinterpret_cast<const unsigned char*>(chunk) + realsize(chunk));
} | {
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physical-chemistry, stoichiometry, mole, density
Although I solved the question, I came across a very interesting step in the procedure of doing so:
Weight of $\pu{112 mL}$ of $\ce{H2}$ at STP is calculated as $\frac{112 \times 2}{22400}$
Does this mean density of a gaseous substance at STP can be given as
$$\text{density} = \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{\dfrac{\text{Volume(mL)}\times\text{Molar Mass}}{22400}}{\text{Volume(mL)}} = \dfrac{\text{Molar Mass}}{22400} $$
Can we say that this holds true? This is true for ideal gases at standard temperature and pressure. The Volume of an ideal gas is 22.4 L/mol at 298.15 K and 1 bar.
You can check this yourself by rearranging the ideal gas equation and solving for the molar volume at STP.
Edit: Make sure you always use correct units, preferably SI in your equations. And use them in all equations. | {
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ros, translation
import rospy
import yaml
from std_msgs.msg import String
from std_msgs.msg import Float32
from crazyswarm.msg import FullState
from geometry_msgs.msg import Pose
from geometry_msgs.msg import Twist
from geometry_msgs.msg import Vector3
with open('crazyswarm/ros_ws/src/crazyswarm/launch/crazyflies.yaml', 'r') as file:
cfs_yaml = yaml.safe_load(file) | {
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} |
cell-biology, apoptosis, autophagy
All three activate different components and caspases in the beginning but finally all activate Caspase 3, which finally activates endonucleases (which degrade the chromosomal DNA), Proteases (which degrade the proteins in the nucleus and the cytoskeleton) which leads to the degradation of the cell. In the end apoptotic bodies are formed from the cell. Additionally the mitochondria break down and release cytochrome c and ATP. Their release from the cell attracts macrophages which take up and eliminate the apoptotic bodies and in turn release cytokines which suppress an inflammatory response.
Apoptosis goes on in a ordered way and does not trigger any further reactions in neighbouring cells (except these receive the same signal). See also reference 1 and 2 for more details. | {
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• @BillDubuque Did I use the transitivity result in the argument that I posted in my answer? – Doug Spoonwood Aug 17 '17 at 18:21 | {
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c#, form, session
// Query all servers until we find a session
while (newsession.ID == null)
{
newsession = GetSession(oldsession);
}
// While lync isn't running, sleep 1 second then check again
while (IsRunning(newsession, "lync.exe") == false)
{
System.Threading.Thread.Sleep(1000);
}
return newsession;
}
//////////////////////////////////////////////////////////// Is Running
public static bool IsRunning(Session session, string processName)
{
string strcmdIn = "tasklist /S " + session.ServerName + " /FI \"SESSION eq " + session.ID + "\"";
string cmdOut = Cmd.StdOutAdmin(strcmdIn); | {
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electric-circuits, electrical-resistance, capacitance, inductance, electronics
Inductors couple current and magnetic flux, with the constitutive relationship $\Phi=LI$ (again, with extensions for the nonlinear case).
Current and charge are in some sense inherently related (you don't need a component to give them a relationship). We express their relationship by $I=\frac{dQ}{dt}$ or $Q=\int I\ dt$.
As the duals of current and charge, voltage and flux are also inherently related. We express their relationship as $V=\frac{d\Phi}{dt}$ or $\Phi = \int V\ dt$.
That means there's one combination of circuit variables that aren't coupled by one of our traditional circuit elements: charge and flux.
To fill in this gap in circuit theory, in 1971 Leon Chua proposed the definition of an additional circuit element, the memristor that can couple charge and flux. The memristor has the constitutive relationship $\Phi = M Q$. | {
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star, mass, hr-diagram
But this does not mean that there is a flaw in the H-R diagram. It is correct as it is: it displays Luminosity Vs. Color and is perfect at that. And then it happens that some extra relations appear: stars are not randomly scattered on the HR diagram, but they group in branches, and these branches are related to stages of their life. The position on each branch is related to each star's mass. This is not a general relation holding for all the diagram. Almost the same happens with size, but this is more tightly related to Luminosity, per definition. | {
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algorithm, sorting, vba, excel
'Pass 2 - write sort criteria to an empty row.
Dim numArtists As Long, last As String, counter As Long
'Need to be sorted for this pass.
Call sheet.UsedRange.Sort(sheet.Columns(1))
For row = 1 To length
'Get the artist to use as the key.
artist = sheet.Cells(row, 1).Value2
'Reset the counter if the artist changed.
If artist <> last Then
counter = 0
last = artist
End If
counter = counter + 1
'Calculate where it falls in the sort.
sheet.Cells(row, 5).Value2 = counter / artistCounts(artist)
Next row
'Now just sort on the criteria column and delete it.
Call sheet.UsedRange.Sort(sheet.Columns(5))
sheet.Columns(5).Delete
Debug.Print "FastShuffled " & length & " songs from " & (UBound(artistCounts.Keys) + 1) _
& " artists in " & Timer - startTime & " seconds."
End Sub
Sample output:
FastShuffled 30000 songs from 190 artists in 1.359375 seconds. | {
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as well to get 1 × 16 = 16. Math Lesson Plan Metric Measurements, Fractions and Number Lines. For example, 3/2, 4/3, 5/4, 6/5 are all improper fractions. Students must fill in the missing numerator or denominator to make the two fractions shown equivalent. x. Teach your students to calculate fractions of 100 using a number line model. Tim cut his pizza(the same size) into 3 equal slices and ate one of them. Problem 1. Equivalent fractions. Let us see how their values are all equal. For example, 1/3, 2/6, 3/9, 4/12 are all equivalent fractions . After you have written down and explained what it means to create equivalent fractions (MP3), I want you to create a visual model to show equivalent fractions (MP4 & MP7). For example, you are following a recipe to make a cake that requires half a cup of water. Equivalent Fractions: Problems with Solutions By Catalin David. Speaking of which, we like to look at the glass as full. Equivalent Fractions: 4. So, 3/5 = 6/10 = 9/15 = 12/20. 4th grade. | {
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ros, navigation, turtlebot-navigation, amcl, move-base
Restart roscore in between your trials, which also restarts the parameter server. This is the sledgehammer method that should always work, but might be inconvenient.
Manually remove the parameters with rosparam or with according commands in the launch file.
Use the clear_params attribute to the node tag in your launch files. This clears the private namespace of a node before launch. Depending on how you load parameters, this might not work. http://wiki.ros.org/roslaunch/XML/node
Use anonymous node name. This way the name is different for each run and you can be sure there are no 'stale' parameters. This and the previous solution don't do anything about global parameters, but I think in your case it is only about private ones.
I suggest making sure stale parameters of previous runs don't affect you any more, and then try again and see if it makes more sense.
Originally posted by demmeln with karma: 4306 on 2014-04-11
This answer was ACCEPTED on the original site
Post score: 2 | {
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homework-and-exercises, thermodynamics, entropy, chemical-potential
I know that the coefficients of $\mathrm{d}U$,$\,\mathrm{d}V$ and $\mathrm{d}N$ must match for equations $(1)$ and $(2)$ ie.
$$\left(\frac{\partial S}{\partial U}\right)_{V,N}=\frac{1}{T}\tag{A}$$
$$\left(\frac{\partial S}{\partial V}\right)_{U,N}=\frac{1}{T}P\tag{B}$$
$$\left(\frac{\partial S}{\partial N}\right)_{V,U}=-\frac{1}{T}\mu\tag{C}$$
But I simply have no idea how to show $(\mathrm{A})$, $(\mathrm{B})$ and $(\mathrm{C})$. So this means I am stuck at the very beginning and hence cannot show my attempt at providing a solution (reason for question closure). | {
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quantum-information
The Computational Complexity of Linear Optics. S Aaronson and A Arkhipov. Proceedings of the 43rd annual ACM symposium on Theory of Computing (STOC '11), pp. 333-342 . Full paper (96 pages) at arXiv:1011.3245 [quant-ph].
For a more understandable reference, try this blog entry by Aaronson. | {
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complexity-theory, qma
Then I realized it is not just that; if we ever want to compute a superposition over some artificial objects, it is almost inevitable to get your superposition with some components being non-sense encoding. There must (or better be) some way to sanitize the input, right?
In classical computing we usually can easily solve this. In quantum computing, it seems to be not a "big-deal" as well, but I just can't figure this out. | {
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java, jdbc
Title: Connection to a database, do I need to create a new one and close it everytime? I currenctly have a class which does several requests to a database with several methods (called from main method) sharing this pattern :
public int getSomething(String id, File file) {
Connection conn = null;
PreparedStatement ps = null;
ResultSet rs = null;
int lines = 0;
try {
conn = openConnection();
ps = conn.prepareStatement(SQL_QUERY.replace("?", id));
ps.setFetchSize(1500);
rs = ps.executeQuery();
lines += nnpw.writeInFile(file, rs);
} catch (Exception e) {
e.printStackTrace();
} finally {
close(rs, ps, conn);
}
return lines;
} | {
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c++, object-oriented, template-meta-programming, numerical-methods, signal-processing
/**
* @brief Method passes the filtered value into the filter
* @param input filtered input
*/
void setInput(float input)
{
input_buffer[input_index++] = input;
if (input_index == NO_INPUT_COEFFICIENTS) {
input_index = 0;
}
}
/**
* @brief Method calculates the filter.
*/
void calculate()
{
// filter implemented in the direct form
convolveInputs();
convolveOutputs();
calculateOutput();
storeOutput();
}
/**
* @brief Method returns output of the filter.
* @return filter output
*/
float getOutput() const
{
return output;
} | {
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c#, event-handling
e.Handled = true;
}
private void SelectItemInComboBox(ComboBox comboBox, object item)
{
var comboBoxItem = comboBox.ItemContainerGenerator.ContainerFromItem(item) as ComboBoxItem;
comboBoxItem.IsSelected = true;
}
I will leave it up to you as an exercise whether you can further eliminate the duplicate code from selectorFunc, and whether it would be useful. | {
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magnetic-fields, magnetic-moment
2) Magnetic Dipole as limit of current carrying loop:
"A magnetic dipole is an idealized current loop - where area goes to zero and current to infinity so that their product is finite. The closer you approximate this situation, the more the field starts to look like a dipole field - analogous with the field created when you approach a positive and negative charge closer and closer together, with the product of charge and distance (the dipole moment) constant."
(a) Why area zero and current infinity? What good is that?
(b) When positive and negative charges approach each other, the distance between them changes. So, how can the product of charge and distance be constant ?
(c) This image again contradicts my interpretation of Magnetic Dipole Moment. Let me quote the sentences from the beginning of the paragraph about the Magnetic pole representation (boldly highlighted by me): | {
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Mathematics has lots of notation!. Scientific notation (also referred to as scientific form or standard index form, or standard form in the UK) is a way of expressing numbers that are too big or too small to be conveniently written in decimal form. Use subscripts to attach a notation to \lim: \lim_{x\to 0} $$\lim_{x\to 0}$$ Nonstandard function names can be set with \operatorname{foo}(x) $\operatorname{foo}(x)$. why math symbols are used. Write the exponential form with words. What Is Domain and Range Interval Notation? Interval notation is a method used to write the domain and range of a function. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. $\begingroup$ I have shown that e^i(kx-wt) is an oscillating function with the same frequency as sin(kx - wt). This site is supported by donations to The OEIS Foundation. Latex Derivative. If you want the limits of an integral/sum/product to be specified above and below the symbol in | {
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ros, custom-message, build, catkin
Title: Message generation fails after JADE update
Hello.
This morning I apt-get update and upgrade my Linux (14.04 Xubuntu).
After this, catkin_make is not able to generate header files for a custom message of mine anymore.
It worked before and I have all the necessary dependencies and message-generations set up.
After googling around here is what I tried but what was unsuccessful:
catkin_make clean #of course
catkin_make -j1 #Some reported that it would help to build one "jobbed"
Looking for my CustomMessage works:
rosmsg show my_package/CustomMessage
When I search in the devel/include/my_package I find no headerfile.
My Error that I get while building is, because the header file was not generated:
In file included from /home/username/catkin_ws/src/my_package/src/MyProgramm.cpp:1:0:
/home/username/catkin_ws/src/my_package/src/MyProgramm.h:5:37: fatal error: my_package/CustomMessage.h: No such file or directory
#include "my_package/CustomMessage.h" | {
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Ascend H10 Hybrid Kayak, Fresh Anchovies Morrisons, Tooth And Claw Game, Work Permit News, Conditorei Coppenrath & Wiese Halal, Maar's Pizza Coupon Code, Agave Attenuata Variegata, Quicken Errors During Quicken Cloud Sync, Lifetime Payette Kayak 9'8, Ocean Reef Phone Number, | {
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"lm_q2_score": 0.8633916047011594,
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"openwebmath_score": 0.5848663449287415,
"tags": null,
"url": "http://haninhe.com/au4i3hgo/6cd83b-power-questions-maths"
} |
fourier-transform, digital-communications, noise, snr
Where $K$ is an adjustment factor depending on the peak-to-average of the constellation and how the constellation was normalized to the peak for computing EVM.
This would be affected by the portion of the noise as in the OP's figure that is within the bandwidth of the receiver after demodulation just prior to symbol decision (which is the noise bandwidth we are concerned with: the waveform bandwidth). | {
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c#, algorithm, programming-challenge, graph
it converts string values to numbers for every test case
it builds graph for every test case
it solves the problem for every test case
Those should be three different methods.
You can further simplify MaximizeValues method if you change its signature to:
public static long MaximizeValues(string[] datas, string[] allFriendships)
and call it once for every query instead.
Your code does not follow C# naming conventions. Public fields and properties should use PascalCase. Also the names themselves could be better. What is id_1? How is it different from id_2? Hard to tell.
Your algorithm is pretty hard to follow, at least for me (I am no expert in graph theory). If you add some local variables with descriptive names - that might help. For example, instead of writing:
result += (i + 1) * (long)i + sum;
you could write:
var explainWhatThatIs = (i + 1) * (long)i + sum;
result += explainWhatThatIs; | {
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special-relativity, reference-frames, collision
Title: What is the minimum energy of a photon for the reaction to occur? If a photon with sufficient energy interacts with a stationary proton, the reaction occurs
$$\gamma+p\to p+\pi^0$$
What is the minimum energy of a photon $E_\textrm{min}$ for this reaction to occur.
I've heard that the way to find minimum energy is to use reference frame in which both proton and $\pi$ meson have zero velocity after the collision. I've tried this approach and I've even gotten the energy of the photon in this reference frame as
$$E_\gamma'=\frac{m_\pi^2+2m_pm_\pi}{2(m_p+m_\pi)}c^2$$
However I'm having problems with transforming this energy back to the original reference frame. Is this the right approach to this problem? If you think geometrically
(using the strategy of the answer I referenced in the comment:
Lowest kinetic energy of particle for which reaction is possible (invariant mass)),
one need not "transform" to the COM-frame and back.
Draw an energy-momentum diagram. | {
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graph-theory, co.combinatorics
$$
\ell_j \geq \beta\, \sum_{i=0}^{j-1} \ell_{i}
$$
Now, look for the level $\ell_j$ which contains vertex $\frac{n}{3}$. That is, so $\sum_{i=0}^{j-1} \ell_i < n/3$ and $\sum_{i=0}^{j} \ell_i \geq n/3$. If this level is large, i.e., $\ell_j \geq n/6$, we are done. Otherwise, the next level has size
$$
\ell_{j+1} \geq \beta \, \sum_{i=0}^{j} \ell_{i} \geq \beta \frac{n}{3},
$$
and we are done.
While this proof looks at the number of vertices in a level rather than the number of edges (which the OP asked about), there are always at least as many edges added in step $i$ as vertices in level $i$, since each vertex must be reached by some edge. | {
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(*Project p onto plane thru {a,b,c}*)
Project[p_, {a_, b_, c_}] := Module[{v},
v = Project0[p - a, {b - a, c - a}];
Return[v + a]
];
(*Project0[] assumes one point @origin*)
Project0[p_, {b_, c_}] := Module[{nvec, d},
nvec = Normalize[b\[Cross]c];
d = nvec.p;(*dist from p to plane along nvec*)
Return[p - d nvec]
];
-
You'll get my +1 if you add example code for calling and displaying the results (as in your question). – bill s Oct 12 '13 at 13:39
You got my vote for posting what is no doubt working code. For something I thought would be straightforward, this problem was far from. (As for getting other votes, what can I say? It's a tough crowd, and one's gotta deliver.) – Daniel Lichtblau Oct 12 '13 at 23:27 | {
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human-biology, biochemistry, human-anatomy
As you see (click on the image to see full version), the fingers contain extensor muscles for movement. However, the index finger and the little finger have separate extensors, because of which they can move more independently as compared to the middle finger and ring finger, which contain a common extensor muscle called extensor digitorum. Some people lack the interconnections between these extensors, and can thus move their ring finger quite easily. One can also achieve this through practice.
Now, as to why the ring finger moves easily when moved along with another finger, have a look at another image:
The fingers are connected to brain via two nerves:
radial nerve, which connects with thumb, index finger and one side of middle finger
ulnar nerve, which connects with little, ring and the other side of middle finger | {
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it up into a number of. Figure 3: Matlab’s Statistics Toolbox To give the readers an idea of the available Matlab toolboxes, a list of widely used Matlab Toolboxes is provided in. This fractional population artifact is an unfortunate side effect of using continuous functions to model a supposedly discrete population. ! The duration of infectivity is as long as the duration of. Lesson 2 looked at the Runge-Kutta approach to solving ODEs and showed us how to use Matlab's built in function to do so. The virus model we use is basically an SIR model that updates daily based on simulated daily contacts between individuals. The Logistic Equation (Verhulst Model) Describes Population Growth DN/dt = R*N*(1-N/K )N - Population R - Maximum Population Growth Rate K - Carrying Capacity (finite Resources To Sustain The Population) (A) Write A FUNCTION N = Logistic(r,K,N0,t) That Takes In Model Parameters R And K, The Initial Condition N0, And A Array T. | {
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Single line Sudoku solver: https://rosettacode.org/wiki/Sudoku#Mathematica solve[sudoku_] := NestWhile[ Join @@ Table[ Table[ReplacePart[s, #1 -> n], {n, #2}] & @@ First@SortBy[{#, Complement[Range@9, s[[First@#]], s[[;; , Last@#]], Catenate@ Extract[Partition[s, {3, 3}], Quotient[#, 3, -2]]]} & /@ Position[s, 0, {2}], Length@Last@# &], {s, #}] &, {sudoku}, ! FreeQ[#, 0] &] | {
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python, array, numpy
alternative solution
try to vectorise as much as possible. The easiest thing to do is count the zeroes in front and at the back:
def count_zeroes(array):
return (array.cumsum(axis=1) == 0).sum(axis=1)
zeroes_front = count_zeroes(array)
and then the same for the reverse:
zeroes_back = count_zeroes(test_data[:,::-1])
The amount each row needs to roll is:
roll = (zeroes_front + zeroes_back) //2 - zeroes_front
array([ 0, -1, -2, -2])
and then you apply this roll over each row:
np.array([np.roll(row, r) for row, r in zip(test_data, roll)])
array([[0. , 0.149, 0.064, 0.736, 0. ],
[0. , 0.258, 0.979, 0.618, 0. ],
[0. , 0.786, 0.666, 0. , 0. ],
[0. , 0.782, 0.954, 0. , 0. ]])
in total:
def centre(array):
zeroes_front = count_zeroes(array)
zeroes_back = count_zeroes(array[:,::-1])
roll = (zeroes_front + zeroes_back) //2 - zeroes_front
return np.array([np.roll(row, r) for row, r in zip(array, roll)]) | {
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php, security
Specific issues I'm concerned about:
security -- is my DB connection info safe? This file is in the root directory of public content, so dbiInfo.php, with the database connection information, is not publicly accessible (I think)
security -- am I open to SQL injection attacks? I build a SQL query with string concatenation
security -- $user_table is untrusted input; is it safe? It's only used as a key to look up trusted input ...
error handling -- have I dealt with all error conditions
there are lots of versions of PHP functions -- am I using the right ones?
General issues:
following conventions
quality/readability/comments
Edit: the data is publicly available -- I'm worried about somebody getting more than read access to one of the listed tables, or any access to any other table in the DB. This:
$tables = array(
"day" => "p_day",
"month" => "p_month"
... etc. .....
);
$table = $tables[$user_table]; | {
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arduino, imu, rosserial
Comment by tonybaltovski on 2015-02-02:
Did you create the ros_lib machine that is connecting to the Arduino?
Comment by nvoltex on 2015-02-02:
I didn't understand your question. The arduino is connected to a computer running ubuntu 14.04 and I installed the ros library on the arduino IDE following the tutorial: http://wiki.ros.org/rosserial_arduino/Tutorials/Arduino%20IDE%20Setup.
Comment by tonybaltovski on 2015-02-02:
After updates, you may need to remake your ros_lib. Can you run any example sketches currently?
Comment by nvoltex on 2015-02-02:
Yes I can. In fact if I simply comment the parts related to the subscriber, I'm able to get the desired information on the topic /IMUdata (as expected). What seems to be the problem is that once I instantiate the subscriber and subscribe to the desired topic I'm unable to retrieve data from the IMU.
Comment by nvoltex on 2015-02-02: | {
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(iii) For any square matrices $X,Y$ we have $\text{tr}(XY)=\text{tr}(YX)$.
So the coefficient of $x^{n-k}$ in $\det(xI-AB)$ is $$\text{tr}((AB)^{(k)})=\text{tr}(A^{(k)}B^{(k)})$$ whereas the coefficient of $x^{n-k}$ in $\det(xI-BA)$ is $$\text{tr}((BA)^{(k)})=\text{tr}(B^{(k)}A^{(k)})$$ and these are equal. | {
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random-quantum-circuit, haar-distribution, swap-test
$$K(\Phi_1) = \int dU \, U\otimes\bar U = |m\rangle\!\langle m|,$$
with $|m\rangle\equiv\frac{1}{\sqrt d}\sum_i |i,i\rangle$ the maximally entangled state. Finally, you obtain the expression for $\int dU\, U\otimes U^\dagger$ by taking the partial transpose of this, and end up with the swap operator on the right-hand side. | {
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It's best not to split the polynomials up using $$i$$, but collect terms in powers of $$x$$.
\eqalign{f_0 = f(x) &= \left( 1+i \right) {x}^{6}+ \left( 3-3\,i \right) {x}^{4}-8\,{x}^{3}+ \left( 3+3\,i \right) {x}^{2}+1-i \cr f_1 = f'(x) &= \left( 6+6\,i \right) {x}^{5}+ \left( 12-12\,i \right) {x}^{3}-24\,{x }^{2}+ \left( 6+6\,i \right) x } The remainder of $$f_0$$ on division by $$f_1$$ is $$f_2 = f_0 - (x/6) f_1 = \left( 1-i \right) {x}^{4}-4\,{x}^{3}+ \left( 2+2\,i \right) {x}^{2}+ 1-i$$ Then take remainder of $$f_1$$ on division by $$f_2$$, etc. | {
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solar-system, neptune, triton
Do we know how much nitrogen ice is on Triton?
If all the nitrogen ice on Triton thawed, how thick would its atmosphere be?
Will Triton thaw naturally, and have a thicker atmosphere, in the future as the Sun's luminosity increases?
If so, would its atmosphere be stable over a long period of time?
In the far future, perhaps as the Sun turns into a red giant, would Triton be warm enough to have liquid water? Triton's orbit is decaying due to tidal interactions, it is predicted to reach Neptune's Roche limit in about 3.6 billion years (Chyba et al., 1989) where it would likely be disrupted, possibly forming a ring system. This timescale is shorter than the timescale for the Sun to become a red giant. | {
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\] In our case, the area is replaced by the length \(l$$ of the capacitor and the distance by the logarithm of the ratios of outer and inner radius. A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C_0. 35 Solution (details given in class): 2. apacitors and capacitance, combination of capacitors Capacitance of a conductor C = Q/V; obtain the capacitance of a parallel plate capacitor (C = ∈₀A/d) and equivalent capacitance for. Consider, a capacitor consisting of two thin conducting plates 1 and 2, each of area A held parallel to each other, at a distance d apart. Let the parallel plate capacitor be filled with two blocks of identical size and dielectric constants k_1=3. C = k ε 0 A/d. Multiple Choice (4 points each) Choose the one best answer to each of the following problems. Capacitors vary in shape and size, but the basic configuration is two conductors carrying equal but opposite charges (Figure 5. The capacitance of the parallel plate capacitor is | {
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"url": "http://mqeo.agenzialenarduzzi.it/capacitance-of-a-parallel-plate-capacitor-with-3-dielectrics.html"
} |
cc.complexity-theory, np, average-case-complexity
Title: Are there any known NP problems which are conjectured to be exponentially hard on average? ETH states that SAT cannot be solved in the worst case in subexponential time. What about average case? Are there natural problems in NP that are conjectured to be exponentially hard in the average case?
Take average case to mean average running time with uniform distribution on the inputs. It might be conjectured that the Learning Parity with Noise Problem (LPN) at constant error rate requires time $2^{n^{1-o(1)}}$. The fastest known algorithm (Blum-Kalai-Wasserman) uses time $2^{O(n/\log n)}$. | {
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electromagnetic-radiation, condensed-matter, schroedinger-equation, time-evolution, bloch-oscillation
so this is how we actually add an electromagnetic coupling to any system, just make the system invariant under $x$ dependent gauge transformation, it automatically adds the gauge coupling.
Ok, now the next step is just doing this in the lattice. First we wick rotate, than we enforce the local gauge invariance and then, we discretize, only the spatial coordinates, and leave the temporal coordinate continuous then the action is now,
\begin{equation}
S_{WF}[A]=\int d\tau \bigg[\sum_{\vec{r}}\bar{\psi}(r)(\gamma^{0}_E(\partial_0+iA^{E}_0)\psi(r)+(m_0+2R)\sum_{\vec{r}}\bar{\psi}(r)\psi(r)-\frac{1}{2}\sum_{\vec{r}i}[\bar{\psi}(r)(R-\gamma^i_E)\psi(r+\hat{i})U_{ir}+\bar{\psi}(r+\hat{i})(R+\gamma^i_E)\psi(r)U^{\dagger}_{ir} ]\bigg]
\end{equation}
where $r=(\tau,\vec{r})$, $R$ is the wilson parameter which is dropped in continuum limit and as long as it is finite, the particular value is not important and $U_{ir}=\exp(iA_{ir})$ | {
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large-hadron-collider, accelerator-physics, particle-accelerators
This is a very stupid question, but I bet I'll get some really remarkable answers. If the power output of the Large Hadron Collider were infinite (or at least a very big number), and notwithstanding a failure or limitation of the mechanical and engineering aspects of the machinery, would it eventually "explode" if the power is turned up too high? In the case of the LHC, yes the beam can do quite a lot of damage.
At full power there is something like 350MJ stored in the beam - close to a freight train, or roughly the kinetic energy of a full jumbo jet at take-off. There is a very complex safety system to dump the beam safely eventually steering the beam energy into a large block of graphite inside a much larger cooled block of metal inside a very big block of concrete. | {
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## Equations Involving Fractional Expressions
To solve equations involving fractional expressions, eliminate the denominators, for example, by multiplying each side by the least common multiple (LCM) of all denominators — although any common multiple works. Then test all the solutions to find the extraneous ones.
Example 1
Solve the equation $\dfrac{5}{x}+\dfrac{1}{x-2}=\dfrac{4}{x^{2}-4}$.
Solution
To eliminate the denominators, we multiply each side by the least common multiple of all denominators: $x,x-2,x^{2}-4$. Because $x^{2}-4=(x-2)(x+2)$, the LCM of the denominators is $x(x-2)(x+2)$ or $x(x^{2}-4)$
\begin{align*}
\begin{array}{rclr}
x(x-2)(x+2)\left(\dfrac{5}{x}+\dfrac{1}{x-2}\right) & =& x(x-2)(x+2)\dfrac{4}{x^{2}-4}\\
5(x^{2}-4)+x(x+2) & =& 4x\\
5x^{2}-20+x^{2}+2x & =& 4x &{\small \text{(expand LHS)}}\\
6x^{2}-2x-20 & =& 0&{\small \text{ (simplify)}}
\end{array}
\end{align*}
Using the quadratic formula, we find the solutions of the above equation: | {
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java, complexity
if (resourceData.size() > 0){
planResourceData.stateKey = ((TheatreResourceData) resourceData.get(0)).stateKey;
}
for (TheatreResourceData resData : resourceData){
switch (resData.type){
case TheatreResourceType.SURGEON1:
planResourceData.staff.surgeon1 = resData;
break;
case TheatreResourceType.ASSISTANT:
assistants.add(resData);
break;
case TheatreResourceType.OTHER_SURGEONS:
surgeons.add(resData);
break;
case TheatreResourceType.BOOKABLE_RESOURCE:
bookableResourceKeyToCountMap.put(Long.parseLong(resData.resourceID), resData.count);
bookableResourceKeyToCountMapOutter.put(Long.parseLong(resData.resourceID), resData.count);
break;
}
}
reverseBookableResMap.put(activityEntry.getKey(), bookableResourceKeyToCountMap); | {
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Function. The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by: The lower limit of 0 − is short notation to mean and assures the inclusion of the entire Dirac. Proven the the Heaviside function is a tempered distribution I must evaluate: $$\langle F Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I suggest you Google "Fourier Transform of the Heaviside Function" to gain some further insights - particularly as to the origin of the delta function term. Fourier transforms 1. But the Laplace transform is especially suited for the study of initial value problems, whereas the Fourier transform is appropriate. Find the Fourier transform of re(r), where e(r) is the Heaviside function. The Fourier transform is particularly well-suited to understanding problems | {
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"tags": null,
"url": "http://lotoblu.it/satw/heaviside-function-fourier-transform.html"
} |
java, algorithm, object-oriented, design-patterns, finance
/**
* @return the balance
*/
public double getBalance() {
return balance;
}
/**
* @param balance the balance to set
*/
public void setBalance(double balance) {
this.balance = balance;
}
}
OperationFactory.java class:
package operations;
public class OperationFactory {
public AccountOperations getOperation(String operation) {
if(operation.equals("1")) {
return new AccountSummary();
} else if(operation.equals("2")) {
return new AccountDeposit();
}
return null;
}
}
AccountOperations.java interface:
package operations;
import accounts.Account;
public interface AccountOperations {
public void performOperation(Account account);
}
AccountSummary.java class:
package operations;
import accounts.Account;
public class AccountSummary implements AccountOperations { | {
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# Thread: Generate range of values exponentially
1. ## Generate range of values exponentially
I would like to generate S values from A to B both included, with a logarithmic or exponential increase between the values and I don't know which one could be a good formula to use, right now I'm using:
$m=B/A$
$f[1]=A$
$f[n+1]=n*(m^{1/S})$
It works well but it doesn't let me control how fast the curve grows, just the start, end and the number of steps. Is there a better one? I thought this would be a fairly used formula but I can't find a straightforward answer anywhere.
2. Hello, Berem!
I would like to generate S values from $A$ to $B$, both included,
with a logarithmic or exponential increase between the values.
Here's how I'd approach the log function . . .
Let: . $f(x) \:=\:p\ln(x) + q$
We are given: . $\begin{array}{c}f(1) \:=\:A \\ f(5) \:=\:B\end{array}$
Since $f(1) = A$, we have: . $p\ln(1) + q \:=\:A \quad\Rightarrow\quad q \:=\:A$ | {
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"url": "http://mathhelpforum.com/calculus/109668-generate-range-values-exponentially.html"
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energy, work, potential-energy, conventions, conservative-field
How can the equation $W=-\Delta PE$ be justified/derived using the work-energy principle or otherwise? Short answer
There is no need to use the Work-Energy Theorem to justify or derive the equation $W = -\Delta PE$ for the force of gravity because the equation $W = -\Delta PE$ for the force of gravity is true by the definition of potential energy.
Long answer
The definition for a conservative vector field is as follows:
A vector field $\mathbf{v}: D \to \mathbb{R}^n$, where $D$ is an open subset of $\mathbb{R}^n$, is said to be conservative if and only if there exists a $C^1$ scalar field $\varphi$ on $D$ such that
$$\mathbf{v} = \nabla \varphi \tag{1}$$
where $\nabla \varphi$ denotes the gradient of $\varphi$. When the equation above holds, $\varphi$ is called a scalar potential for $\mathbf{v}$. (Note: I have used $D$ instead of the $U$ that Wikipedia uses to denote the open subset so that I may use $U$ to denote the potential energy function later in this answer) | {
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"url": null
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neural-network, classification, loss-function, categorical-data, activation-function
Title: Activation and Loss Function not chosen correctly when use Neural Network I have three classes for my text dataset before.
These are my classes:
0 = Cat
1 = Not Both
2 = Dog
Then I use this code:
df_result = df[df["class"] != 1]
So, now my classes are 0 and 2.
When I use neural networks, what can I choose for the Loss and Activation? Then, what should I do for choose the last Dense of neural network's model?
Before, my code like this:
model.add(Dense(2, activation = 'softmax'))
history = model.compile(loss = sparse_categorical_crossentropy,
optimizer = 'adam', metrics=['accuracy']) | {
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java, algorithm, sorting, mergesort
if (tail2.compareTo(head1) < 0) {
if (mergedRunHead == null) {
mergedRunHead = headInterval2;
mergedRunTail = headInterval2;
} else {
mergedRunTail.next = headInterval2;
headInterval2.prev = mergedRunTail;
mergedRunTail = headInterval2;
}
headInterval2 = headInterval2.next;
continue;
}
int index = findLowerBound(aux,
headInterval2.from,
headInterval2.to + 1,
head1);
Interval newInterval = new Interval(headInterval2.from,
index - 1);
headInterval2.from = index; | {
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c#, change-making-problem
We've defined how to get the coins, we've assumed that the calculation (of how many coins of a given type should be used already exists) (cfr the other answers)
All that remains is the logic inbetween these two:
private static void PrintListOfChange(decimal totalChange)
{
//Get the coins and order them by value (descending)
var coinTypes = GetCoinsForLocalCurrency()
.OrderByDescending(x => x.Value)
.ToList();
//This is where we construct our result
Dictionary<CoinType, int> neededChange = new Dictionary<CoinType, int>();
//Placeholder value for calculations
decimal remainingChange = totalChange;
foreach(var coinType in coinTypes)
{
int coinAmount = CalculateMaxCoins(coinType, remainingChange);
//Do nothing if the result was 0
if(coinAmount > 0)
{
//Add the results
neededChange.Add(coinType, coinAmount); | {
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qiskit, programming, ibm-q-experience, ibm-quantum-devices
Depending on your engagement you also have access to different levels of collaboration from simple support to joint development.
No special qiskit library for IBM Q Network but special content/documentation and education materials.
Note that between the common user and the IBM Q Network member there is also a new way to engage for academic researchers: https://www.ibm.com/blogs/research/2020/07/quantum-researcher-program/ | {
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equations, but the fourth equation is our first foray into the world of special functions, in this case Bessel functions. Neukirchy School of Mathematics and Statistics, University of St Andrews, St Andrews, UK, KY16 9SS. Maxwell’s equations are complicated in spherical coordinates. Unfortunately, analytical solutions of Initial Condition : Cðr; t ¼ 0Þ ¼ f ðrÞ (4) Correspondence concerning this article should be addressed. An equation of the sphere with radius #R# centered at the origin is. Made by faculty at the. Let the potential be V 0 on the upper hemisphere,and V 0 onthelowerhemisphere, V(R) = V 0 ˇ 2 ˇ 2 4. Then our volume element r2 sin θdθdφdr = −r2dμdφdr. ; Fotiadis, D. The separation of variables equation for deriving free surface eigenfunctions is as follows As in the previous section, Laplace's equation must be solved in cylindrical coordinates satisfying the free surface and the radiation condition. Simplifying most terms, I have: (1/r)(d/dr)(k*r*dT/dr) + q,dot = 0, where | {
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"tags": null,
"url": "http://dpdf.bainsetcuisines.fr/heat-equation-in-cylindrical-coordinates-separation-of-variables.html"
} |
homework-and-exercises, potential, field-theory
Second set of definitions. If you define $g=\frac{G M}{r^2}$, and if you define $\phi=-\frac{G M}{r}+C$ for some constant $C$, then you still have $g=\frac{d\phi}{dr}$, the physics and forces are totally unchanged, but you no longer have $\phi=-gr$. Physically, you can add any constant to a potential and the result stays unchanged. So your teacher may want to point out that this alternative definition is valid, and in this alternative definition your law does not hold.
Also note that usually, one would prefer to say $g=-\frac{G M}{r^2}$ and $g=-\frac{d \phi}{dr}$, reflecting the fact that the force $mg$ should accelerate objects down the potential. That negative sign is an important when defining potentials! | {
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"tags": "homework-and-exercises, potential, field-theory",
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image-processing, computer-vision, filtering, correlation
Title: How does Correlation for 1D Signals work? On page 2 of the slide, I don't understand the example. Like $j$ and $s$ are both on the x-axis, what does j+s mean? The filter is $(f(−1), f(0), f(1)) = (−0.5, 0, 0.5)$. How can we use it? Could anyone explain how we get $r(j)$ with numbers? For the simple $3$-tap filter given in the example, each output value $r(j)$ is computed as
$$\begin{align}r(j)&=f(-1)I(j-1)+f(0)I(j)+f(1)I(j+1)\\&=-\frac12 I(j-1)+\frac12 I(j+1)\tag{1}\end{align}$$
So each input value is replaced by the (scaled) difference between its neighboring values. | {
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"url": null
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Notice that $$x^{\underline k}=\frac{x^{\underline{k+1}}}{x-k}$$ for $k\ge 0$. If we generalize this to negative $k$, we have \begin{align*} x^{\underline{-1}}&=\frac{x^{\underline0}}{x-(-1)}=\frac1{x+1}\\\\ x^{\underline{-2}}&=\frac1{(x+1)(x+2)}\\\\ &\;\vdots\\\\ ... 3 This is really the same answer as that of Qiaochu Yuan, but I find the "binomial coefficients of x", much as I approve the notation, a bit distracting when next to ordinary binomial coefficients. One can do without them, using falling factorial powers instead: x^\underline n=x(x-1)\ldots(x-n+1), which is of course the same as n!\binom xn. Elementarily ... 3 Here is an old scicomp.SE question that answered some of your question: What are criteria to choose between finite-differences and finite-elements? In my humble opinion, FEM is the most flexible one in terms of dealing with complex geometry and complicated boundary conditions. FEM also allows the adaptive/local procedure to get higher order local ... 3 The problem | {
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"openwebmath_score": 0.9530885219573975,
"tags": null,
"url": "http://math.stackexchange.com/tags/finite-differences/hot"
} |
java, interview-questions
bidTracker.recordUserBidOnItem(new Bid(user1, 10), item1);
Bid lowBid = new Bid(user2, 5);
bidTracker.recordUserBidOnItem(lowBid, item1);
}
@Test
public void recordUserBidOnItem_shouldThrowInvalidBidException_whenBidIsSameAsCurrentlyWinningBid() throws InvalidBidException {
thrown.expect(InvalidBidException.class);
thrown.expectMessage("A bid of £10 on item { id: i1, name: item1, description: Brilliant! } is too low. It should be more than the current winning bid: £10)");
bidTracker.recordUserBidOnItem(new Bid(user1, 10), item1);
Bid sameBid = new Bid(user2, 10);
bidTracker.recordUserBidOnItem(sameBid, item1);
} | {
"domain": "codereview.stackexchange",
"id": 35523,
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"openwebmath_score": null,
"tags": "java, interview-questions",
"url": null
} |
evolution
I think it may help if I answer with an example that is widely promoted by AQA in their A2 Biology syllabus.
Stabilising Selection in human birth weight
It is harmful for an infant to be born with a very low birth weight. They are much more vulnerable to heat loss due to their high surface area to volume ratio and consequently their respiratory demands are very high. Pre-term babies (which account for 67% of low-birthweight infants(1)) are particularly susceptible to respiratory problems (lack of surfactant in the lungs), cardiac problems (Patent ductus arteriosus - the lungs are still bypassed when the umbilical cord has been cut) and dangerous intestinal problems (Necrotizing enterocolitis) amongst many other conditions can all be fatal (further information on mentioned conditions) and are reflected in high mortality rates at these low birth rates. It is therefore not beneficial to be on the extremes of birth weight. | {
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ros
[ INFO] [1342451679.062215026]: HectorSM p_use_tf_scan_transformation_: true
[ INFO] [1342451679.062233934]: HectorSM p_pub_map_odom_transform_: true
[ INFO] [1342451679.062255121]: HectorSM p_scan_subscriber_queue_size_: 5
[ INFO] [1342451679.062276856]: HectorSM p_map_pub_period_: 2.000000
[ INFO] [1342451679.062296779]: HectorSM p_update_factor_free_: 0.400000
[ INFO] [1342451679.062316553]: HectorSM p_update_factor_occupied_: 0.900000
[ INFO] [1342451679.062335595]: HectorSM p_map_update_distance_threshold_: 0.400000
[ INFO] [1342451679.062354975]: HectorSM p_map_update_angle_threshold_: 0.060000
[ INFO] [1342451679.062376253]: HectorSM p_laser_z_min_value_: -1.000000
[ INFO] [1342451679.062395483]: HectorSM p_laser_z_max_value_: 1.000000
loading model xml from ros parameter | {
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evolution, ecology, anaerobic-respiration
Which is closer to the truth? (An environment that takes extremely long to equilibriate with the atmosphere, long enough for evolution to be fast in comparison, would count as isolated for the purposes of this question since it would serve as a reservoir of life from which aerobic respiration could emerge.) In the question body it says: | {
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battery
If there is no way to measure it, how do I open this thing? I tried to open it using flat screwdriver and it leaves a dented mark. The DC 12v output may or may not be direct from the battery, if it is current limited then probably not - you will need to check the spec sheet.
The top is probably screwed down as the pv controller is under there. The screws to get access are hidden under the orange graphic with all the labels, removing it without damage depends on how strong the adhesive is.
One way is to rub your thumb over the surface to find the screw holes then just uncover those... but if they fitted plastic hole covers you may not be lucky. | {
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because it is tested & true method that will 100% work for you. It begins with setting up the variables that you will solve for. Let's plot a few more curves in the phase space. x 1 + 2x 2 = 16: x 1 + x 2 = 9: Substitution Method. See the full health analysis review. Solve a system of ordinary differential equations using lsoda from the FORTRAN library odepack. The above example is just to let you get a taste of what ODE is and how to use python to solve ODE in just a few lines of codes. solve_ivp to solve a differential equation. linalg, which offers very fast linear algebra capabilities. An example of using GEKKO is with the following differential equation with parameter k=0. An expression does not have equality. In the differential equation system, $$pS(t)$$ must be replaced by $$p(t)S(t)$$, and in this case we get a differential equation system with a term that is discontinuous. •An alternative is to use solvers for Ordinary Differential Equations (ODE) in Python, so-called ODE | {
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quantum-mechanics, schroedinger-equation
If you specify the initial conditions as $\Psi(x, 0)$ then you have effectively created a wavepacket describing your particle, so it does have a finite uncertainty in position, and of course now a finite uncertainty in momentum. You can now calculate the expectation value of position as a function of time.
Your $\Psi(x, 0)$ will probably be expressed as a linear superposition of the plane wave solutions. To calculate the superposition just Fourier transform your $\Psi(x, 0)$.
Response to comment:
In your comment you ask: | {
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plasma-physics
The reason is as follows: the electronic current increases because of edge effects due to the increase of the collection area by the expansion of the sheath. A follow up question is:
2. why does the electons sheath expand in the first place?
(I know that the collection area of a cylindrical prob is not limited by the physical area of the probe's tip).
I've read many references, but I didn't find satisfactory answers to these tow questions. Can anyone elaborate on these points please? Thanks in advance! To answer your first question, we have to remind ourselves on a very important property of the plasma: if you insert a test charge into a plasma, plasma particles of the opposite charge will form a space charge cloud around that test charge cancelling its electric field such that the test charge is not "seen" by the plasma outside of this space charge cloud. The distance over which the test charge is shielded is the Debye length which depends on plasma temperature and density. | {
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Regardless, what you want is called the highest posterior density interval (HPDI) which is unique under most conditions. This posting should answer your question.
The R package "coda" contains an algorithm to compute the HPDI from a sample. Also, SAS will compute this interval as part of any Bayesian analysis.
• The HPDI should be the smallest interval that contains 95 % of posterior probability, correct? Jun 11 '15 at 15:48
• Yes that is correct. Jun 11 '15 at 17:55
Nathan L is right that the HPDI is one reasonable answer to this question. But it seems worth pointing out that there is no single perfect answer to your question. The posterior is a probability distribution, and there is no perfect way to summarize all the information in a generic probability distribution with just two scalars. One reasonable way to describe a distribution using an interval is to give the HPDI, one way is the 5%-95% interval, and there are other possibilities. None is perfect. | {
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radiation, microscopy
Title: How does the decay of an atom look like? There is this famous image recorded by IBM using the Scanning tunneling microscope. Those dots are individual Xenon atoms. | {
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