text stringlengths 1 1.11k | source dict |
|---|---|
python, visualization
so the default value is, indeed, .95, as you guessed.
EDIT: How CI is calculated: barplot calls utils.ci() which has
seaborn/seaborn/utils.py
def ci(a, which=95, axis=None):
"""Return a percentile range from an array of values."""
p = 50 - which / 2, 50 + which / 2
return percentiles(a, p, axis) | {
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homework-and-exercises, thermodynamics, statistical-mechanics
fluidic process on our hands. The work done is then of course equal to $$W = p_{0}V$$ So, the final energy of our system is given by $$E_{f} = U + W = CT_{o} + p_{0}V$$ For an ideal gas, we have also have $$E_{f} = CT_{f}$$ and thus: | {
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javascript
Use Object.entries to get the key and the value from an object at once, instead of iterating over the keys and separately extracting the values.
const mergeObject = (customFields) => {
const mergeFields = {}
for (const [key, value] of Object.entries(customFields)} {
mergeFields[key] = transformValue(value);
}
return mergeFields
} | {
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drag, aerodynamics, lift
(As I'm typing this I just realized that the answer might be "as fast as the speed of the wind it takes to make a person lean that far".) If you wanted to get an estimate of how fast you'd need to be moving in order to lift yourself with air deflection from running, you need to establish a couple of important parameters:
Body weight $p$
Torso angle $\theta$ (defined such that $0^\circ$ would be running upright)
Speed $s$
Torso length $\ell$
Torso width $w$
Air density $\rho$ | {
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} |
quantum-mechanics, electrons, atomic-physics
It takes the Schrodinger equation to solve for the wave functions and energy levels of stationary orbits. Other solutions are not stationary. The ones that are are so called eigenstates of the energy, which is conserved. | {
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python, game, dice
Verify/validate input – When asking for input, it is often useful to verify that the user entered legal options. This can be done using an external method
Version compatible calling of parent constructor – I'm not sure if you're using Python 3 or Python 2, but a way to call the parent constructor in both version are to use super(HumanPlayer, self).__init__(name) where you replace the current class name instead of HumanPlayer.
Use var += 1 to increment variables – This looks a little neater than doubling the variable name | {
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c++
0.912665 -0.666179 -1.409592 -1.440303 1.48104 -88.842338 21.44113 -47.07769 -30.874201 7.402907 1.658714 -26.159671 -89.100513 -21.301901 22.968179 22.688747
0.483921 -0.731897 -1.305772 -1.414634 1.387132 -88.705916 22.56705 -48.006169 -30.72695 3.01525 1.773535 -23.261877 -89.057369 -21.94138 25.371108 23.26944
-0.040279 -0.812168 -1.178862 -1.38335 1.272311 -88.539071 23.943755 -49.141486 -30.546927 -2.300542
-1.279702 -1.002047 -0.878746 -1.309267 1.000844 -88.144647 27.198731 -51.825795 -30.121219 -14.640345 2.279228 -11.180994 -88.879752 -24.572118 35.256071 25.658447
-1.947199 -1.10432 -0.717115 -1.269332 0.854625
following is the publishJoint.cpp file:
#include <string>
#include <ros/ros.h>
#include <sensor_msgs/JointState.h> | {
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\begin{align} R(R-2r)&=2rR ,\\ R&=4r . \end{align}
• So nice! Really thank you. – user143993 Feb 5 at 6:46
I you perform a circular inversion w.r.t the black circle, the red circle becomes the red tangent line in the picture below, while the blue circle gets reflected to another circle, tangent to the black circle, the black line and the red line. The diameter of this new circle must be $$1$$. Therefore $$\overline{AB}=1+1=2$$ and $$\overline{AC}=1/\overline{AB}=1/2$$.
• Really thank you. – user143993 Oct 27 '18 at 11:18
Based on the figure, one can make some inspired guesswork.
Construct a rectangle $$ABCD$$ with side $$AB$$ of length $$1$$ and diagonal $$AC$$ of length $$3.$$ Extend $$AB$$ to $$E$$ so that $$B$$ between $$A$$ and $$E$$ and $$AE = 2.$$
Construct a circle of radius $$4$$ about $$A,$$ a circle of radius $$2$$ about $$E,$$ and a circle of radius $$1$$ about $$C.$$ | {
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c#, optimization, sorting, contest-problem
//assume correct input
int numOfWords = int.Parse(firstLine.Split(" ".ToCharArray())[0]);
string alphabet = firstLine.Split(" ".ToCharArray())[1].ToUpper();
for (int i = 1; i <= numOfWords; i++)
{
words.Add(Console.ReadLine());
wordsCopy.Add(words[words.Count - 1]);//preserve index
}
Console.WriteLine();
//Quicksort is unstable (meaning "equal" elements will still be swapped, e.g. "go" and "Go")
words.Sort(delegate(String x, String y)
{
int max = (x.Length > y.Length) ? y.Length : x.Length;
if (x.ToUpper().Equals(y.ToUpper()))
{
//dirty work-around to stablize the sort
return wordsCopy.IndexOf(x) < wordsCopy.IndexOf(y) ? -1 : 1;
} | {
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c#, performance, generics, reflection, interval
timeWindowIdentifier.SetValue(output, timeWindowMinutes);
datePropertyInfo.SetValue(output, currentWindowFrom);
prevTimeWindow = output;
yield return output;
}
}
}
while (nextWindowFrom <= dateTo);
}
}
private static DateTime GetPropertiesAndDictionaries<T>(DateTime dateFrom, List<T> stateModels, out PropertyInfo datePropertyInfo, out List<PropertyInfo> copyProperties, out PropertyInfo timeWindowIdentifier, out int size, out TimeWindowDictionary[] dictionaries, out int i) where T : new()
{
Type tType = typeof(T);
var propInfos = tType.GetProperties(); | {
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filters, finite-impulse-response
pass band edge: 20 kHz
stop band edge: 28 kHz
pass band ripple: 0.1dB
stop band attenuation: 40 dB
Let's say you want that filter at 4*44.1kHz instead of 4*48kHz. In Matlab you would do the following:
%% match the filter at 44.1 kHz
fs = 4*44100;
d = fdesign.lowpass('Fp,Fst,Ap,Ast',20000,28000,0.1,40,fs);
hd = design(d,'equiripple');
% convert to polyphase
h4 = 4*reshape(hd.Numerator',4,12)'; | {
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"url": null
} |
electrical-engineering, control-engineering, control-theory
$x < -\sqrt{k}$
$-\sqrt{k} < x < 0$
$0 < x < \sqrt{k}$
$x > \sqrt{k}$.
Region 1, $ x < -\sqrt{k} $:
$$ x'=-kx+x^3 < 0 $$
Region 2, $ -\sqrt{k} < x < 0 $:
$$ x'=-kx+x^3 > 0 $$
Region 3, $ 0 < x < \sqrt{k} $:
$$ x'=-kx+x^3 < 0 $$
Region 4, $ x > \sqrt{k} $:
$$ x'=-kx+x^3 > 0 $$
Summarizing the results in a "phase portrait''
Here it can be seen that the equilibrium point $x = 0$ is attractive (stable), while the equilibria $x = \pm \sqrt {k}$ are repulsive (unstable).
Hence the region of attraction is $-\sqrt{k} < x < \sqrt{k}$. | {
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a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. We also know that AB= CD as they are opposite sides in a parallelogram. Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. Geometry doesn't have to be so hard! And from the definition of a rectangle, we know that all the interior angles measure 90° and are thus congruent- and we can prove the triangle congruency using the Side-Angle-Side postulate. For example, the two triangles ΔABD and ΔDCA, in which the diagonals form corresponding sides. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). What is the formula of | {
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"url": "http://migf.com/acupuncture-in-pnmg/are-the-two-diagonals-of-a-rectangle-equal-why-68fef3"
} |
human-biology, biochemistry
Title: Help me understand my membrane potential assignment Epithelial with the ability to transport NaCl is bathed in an isotonic salt solution with these values (in mM): 136 Na, 6 K, 140 Cl.
The intracellular concentrations are: 45 Na, 100 K, 13 Cl.
The cell's resting potential is -63 mV and the temperature is 37 Celcius.
Which ions will passively diffuse over the membrane? | {
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star, stellar-evolution, metallicity
stars could form out of the gas (enriched by the supernova ejecta) in a continuing cycle. In the low-mass, isolated protogalactic clouds which probably contributed to the halo, the initial round of supernovas ejected most of the gas (including the original gas that hadn't yet formed stars) -- thus, little opportunity to form more stars (out of higher-metallicity gas) on a continuing basis. | {
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• thanks for your delicate solution! – user 1357113 Jun 18 '12 at 10:57
Another approach: $$\begin{eqnarray*} \int_{0}^{\infty} dx\, \frac{e^{-x} \sin(x)}{x} &=& \int_{0}^{\infty}dx\, \frac{e^{-x}}{x} \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!} \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \int_{0}^{\infty}dx\, x^{2k} e^{-x} \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}(2k)! \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \hspace{5ex} \textrm{(Leibniz series for \pi)}\\ &=& \frac{\pi}{4}. \end{eqnarray*}$$
• this is another magic shot! Nice job! Thanks! :-) – user 1357113 Jun 19 '12 at 5:52
• @Chris: Thanks, Chris. Another good question! – user26872 Jun 19 '12 at 7:40 | {
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fluid-dynamics, turbulence, navier-stokes
The finite-energy statistical equilibrium for any continuous field is then a zero temperature state where all modes contain an infinitesimal amount of energy, which is the partition of the initial energy. The energy cascade is the method by which a fluid tries to do the paritition, by sending energy down into short wavelength modes in a random looking way, to get closer to the statistical equilibrium state. Since this is impossible, you just get a continuous draining of energy from long-wavelength motion to short wavelength motion, and when this draining process reaches a scale-invariant steady state, we call the situation isotropic homogenous turbulence. | {
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np-hard, shortest-path
$1$ waypoint and $k$ turnstiles and $c(e) = 1$ $\in \mathrm{NP\ hard}$, maybe?
If we restrict each edge to having capacity $1$, meaning it can only be included once in the shortest path, then we can make a reduction from the NP-complete $2$-link-disjoint paths problem. See section 2.3 of Amiri et al. [2] for the full proof. The idea is that with $c(e) = 1$ we can make our graph directed by using turnstiles to establish the direction of each edge.
However, this is only true for non-planar graphs. If we restrict it to planar graphs, the question appears to be open [3]. That being said, it is known to be in P under certain constraints [4], but still unclear for the general case.
$1$ waypoint and $k$ turnstiles and $c(e) = 2$ $\in$ ??? | {
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quantum-field-theory
i) $f(c\overline{u}) = c \cdot f(\overline{u})$
ii) $f(\overline{u} + \overline{v}) = f(\overline{u}) + f(\overline{v})$.
Only, what is $\overline{u}, \overline{v}$ in the context of of the Klein-Gordon equation? I guess it is e. g. $u = \phi$ and $v = \phi'$, which would make the mapping indeed linear (, right?).
And why would an inhomogenous mapping, i. e. the right-hand side of the Klein-Gordon equation is non-zero, is a non-linear mapping? Consider ($\partial^2 +m^2)\phi = const \cdot \phi^3$. Why is this not a linear mapping?
Apparently, linearity is equivalent to have scalar field interactions, whereas non-linearity implies no interaction. Why is that so?
Thank you in advance for any hints. When we say a particular equation is linear, we mean that it follows the principle of superposition: | {
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electromagnetism, maxwell-equations, differentiation
$$\begin{matrix}
\text{E and B in a moving inertial frame} \\
\begin{align}
\vec E' &= \vec E + \vec v \times \vec B \\
\vec B' &= \vec B - {1 \over c^2} \vec v \times \vec E \\
\end{align}
\end{matrix}$$
These are the (non-relativistic) conversions of electric and magnetic fields to those in an inertial frame moving at velocity $\vec v$. They make lots of quasistatic problems trivial, and Faraday's law is no different. Applying it to the sliding bar example from above:
$$\begin{array}{c|c}
\text{The rest frame} & \text{The moving frame}\\
\hline
\vec E = 0 & \vec E' = \vec v \times \vec B \\
\vec B = +B\hat z & \vec B' = \vec B \\
\end{array}$$ | {
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c#, game, mvc, winforms, chess
bool legal = false;
Piece movedPiece = board.GetPieceAt(start);
Piece capturedPiece = board.GetPieceAt(end);
if (movedPiece != null && movedPiece.Colour == player.Colour && !start.Equals(end))
{
if (IsEnPassant(board, move) || IsCastle(board, move))
{
legal = true;
}
else if (movedPiece.CanMove(board, start, end))
{
if (capturedPiece == null || (capturedPiece.Colour != player.Colour))
{
if (!board.CheckMoveInCheck(player, move))
{
legal = true;
}
}
}
}
return legal;
} | {
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"tags": "c#, game, mvc, winforms, chess",
"url": null
} |
black-holes, metric-tensor, coordinate-systems, singularities
Which aspect of the metric in Schwarzschild coordinates indicates that the coordinates are only valid outside the event horizon? Coordinates are not sacred objects in GR. Any coordinate system is just as good as any other coordinate system. So to ask whether the Schwarzschild coordinates are valid or not is a meaningless question ${}^1$.
However it is reasonable to ask if coordinates have an intuitive meaning for some specified observer. So for example if we take an observer far from the massive object then the Schwarzschild time coordinate is the time as measured by that observer's clock, and the Schwarzschild radial coordinate is the circumference of a circle centred on the object divided by $2\pi$. Both these are intuitively meaningful measurements for our observer. | {
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"tags": "black-holes, metric-tensor, coordinate-systems, singularities",
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fluid-dynamics, flow, viscosity
Title: How viscosity and velocity relate in a pipe? Assume we have a pipe that there is a fluid stream in it. By increasing the velocity of the fluid the resistance will increase either (because of the viscosity I think). My question is how are these to parameters depending to each other and what is the formula? With hydrodynamics we normally find that at low shear rates the flow is limited by the viscosity of the liquid while at high shear rates it's limited by inertial forces and the viscosity doesn't matter. This is the case for flow in a pipe. At low flow rates the pressure drop $\Delta P$ is related to the flow rate $Q$ by the Hagen-Poiseuille equation:
$$ \Delta P = \frac{8\mu \ell}{\pi r^4} Q $$
where $\ell$ is the pipe length and $r$ is the pipe radius. So in this case the pressure drop is proportional to the viscosity $\mu$.
However at high flow rates the pressure drop is given by the Darcy-Weisbach equation:
$$ \Delta P = f_d \frac{\ell}{2r} \frac{\rho v^2}{2} $$ | {
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mathematics, cryptography, teleportation
The last detail is important here.
It means that the basis is either the eigenbasis of the $Z$ operator, or the eigenbasis of the $X$ operator — which means that when Bob wants to do the corrections, one of the two corrections which he will perform will in principle have no effect, while the other will simply interchange the two possible basis elements. | {
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quantum-mechanics, homework-and-exercises, quantum-information, quantum-computer
Title: Could anyone prove this result from Deutsch's algorithm? Does anyone know how to prove the result shown in one of the images, that $$U_f(\left|x\right>(\left|0\right>-\left|1\right>)/\sqrt{2} = (-1)^{f(x)}\left|x\right>(\left|0\right> - \left|1\right>)/\sqrt{2}~ ?$$ The desired action of $U_f$ is shown in the image of a quantum circuit. This example is taken from Nielsen and Chuang's Quantum Computation and Quantum Information. As you can see from your first scan, $U_f$ maps $\lvert x\rangle \lvert y\rangle$ to $\lvert x\rangle \lvert y \oplus f(x)\rangle,$ where the range of $f$ (a function with a 1 bit domain range) is $\{0,1\}.$ Applying it to your input state $\lvert x\rangle (\lvert 0\rangle - \lvert 1\rangle)$ we get $\lvert x\rangle (\lvert f(x)\rangle - \lvert 1\oplus f(x)\rangle),$ which if $f(x)=0$ yields $\lvert x\rangle (\lvert 0\rangle - \lvert 1 \rangle)$ and $\lvert x\rangle (\lvert 1\rangle - \lvert 0 \rangle)$ if $f(x)=1,$ both can be abbreviated as $(-1)^{f(x)} | {
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"tags": "quantum-mechanics, homework-and-exercises, quantum-information, quantum-computer",
"url": null
} |
c, queue
/*
The queue structure just stores meta info and points to lines. The member index
is there because I thought it would make the the other functions simpler since
they won't have to deal with special cases. They just treat it as if it were a
regular line. And that's ok because most functions take the line that comes before
the one they are operating on as an argument so they don't have to walk through
the list to find that node in order to update the pointer to the next;
The queue stores only elements of sizeof(void *), whatever is received by enqueue
will be sent back by dequeue, so it can store a pointer to anything and
also work with other data types of size up to sizeof(void *). | {
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homework-and-exercises, thermodynamics
Title: fusing two air containers with fixed pressures $p_1,p_2$ and temperatures $T_1, T_2$, what end-result $T$ and $p$ will be? Let's say that I have two boxes with both volumes equal $V$, filled with an air.
Air in first box have pressure $p_1$ and temperature $T_1$, likewise air in second box have pressure $p_2$ and temperature $T_2$.
Suppose this is ideal gas, then from Clapeyron equation we have:
$$ p_1 V = n_1 R T_1 $$
$$ p_2 V = n_2 R T_2 $$
Thus, after simple transformation:
$$ \frac{p_1 V} {R T_1} = n_1 $$
$$ \frac{p_2 V} {R T_2} = n_2 $$
Now, let's say that we want to fuse both boxes and remove bulkhead, so we will have instead 1 box with volume equal $2V$.
What will be temperature $T$ and pressure $p$ of air in the new box? We can use the same law as above to get:
$$ 2 p V = ( n_1 + n_2 ) R T = \left( \frac{p_1 V} {R T_1} + \frac{p_2 V} {R T_2} \right) R T $$,
thus:
$$ p = \frac { T } { 2 } \left( \frac { p_1 } { T_1 } + \frac { p_2 } { T_2 } \right) $$ | {
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} |
energy-conservation, harmonic-oscillator, oscillators
A better understanding is possible trying an explicit form for $x(t)$:
$$x(t) = a\,\sin \omega t.$$
Although you can't expect this is the right form of the limit cycle, it may give a semiquantitative idea of what will happen. I leave for you to compute the integral and to show that for $a>2$ dissipation prevails whereas for $a<2$ there is a net energy creation. A balance is obtained for $a=2$.
Therefore if you start from rest or in general from an amplitude $<2$ the oscillation will grow larger, until balance is reached. The opposite happens if you start from $a>2$: amplitude decreases until the limit cycle is reached from above. This shows stability of the limit cycle. | {
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"url": null
} |
ros, kuka, pcl, youbot, pointcloud
/home/youbot/catkin_ws/src/ira_laser_tools-master/src/laserscan_virtualizer.cpp: In member function ‘void LaserscanVirtualizer::virtual_laser_scan_parser()’:
/home/youbot/catkin_ws/src/ira_laser_tools-master/src/laserscan_virtualizer.cpp:109:13: warning: format ‘%ld’ expects argument of type ‘long int’, but argument 8 has type ‘std::vectorros::Publisher::size_type {aka unsigned int}’ [-Wformat]
Running just one laserscan_multi_merger node of this package gives me next error:
[laserscan_multi_merger-1] process has died [pid 16008, exit code -11, cmd /home/youbot/catkin_ws/devel/lib/ira_laser_tools/laserscan_multi_merger __name:=laserscan_multi_merger __log:=/home/youbot/.ros/log/6a789f88-6310-11e4-9d7c-000bab4510aa/laserscan_multi_merger-1.log].
log file: /home/youbot/.ros/log/6a789f88-6310-11e4-9d7c-000bab4510aa/laserscan_multi_merger-1*.log
all processes on machine have died, roslaunch will exit
shutting down processing monitor...
... shutting down processing monitor complete
done | {
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"tags": "ros, kuka, pcl, youbot, pointcloud",
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} |
javascript, beginner, jquery, html, calculator
$('#4sfm3').val(parseFloat($('#4sfa3').val() / 12));
$('#5sfm1').val(parseFloat($('#5sfa1').val() / 12));
$('#5sfm2').val(parseFloat($('#5sfa2').val() / 12));
return $('#5sfm3').val(parseFloat($('#5sfa3').val() / 12));
}); | {
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"tags": "javascript, beginner, jquery, html, calculator",
"url": null
} |
analog-to-digital, quantization, resolution, approximation
The OP asked for how the noise equivalence can be verified. It is obvious by inspection of the plots given that the only difference between the two is a static offset, but the equivalence can be computed simply enough for further verification by creating a floating point full scale sine wave and then creating two quantization signals by both rounding and truncating the sine wave to the closest integer. Subtract the quantized fixed-point (integer) signal from the original floating point signal to get the quantization error alone. Then compute the standard deviation of each and they will be equivalent. | {
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"url": null
} |
$$p(m,k) = \sum_{s=1}^m (-1)^{s-1} \binom{m}{s} p^{(s)}(m,k).\tag{3}$$
In particular, \eqalign{ p(4,13) &= \frac{(3181)(233437)(25281233)}{(2^6)(3)(5^4)(7^4)(17^3)(19^2)(23^2)(29)(31)(37)(41)(43)(47)} \\ &=\frac{18772910672458601}{745065802298455456100520000}\\ &\approx 2.519631234522642\times 10^{-11}. }
The small primes in the denominator were expected: they cannot exceed $mk=52$. The large primes in the numerator strongly suggest there exists no general closed form formula for $p(m,k)$.
### What does this answer mean? | {
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"tags": null,
"url": "https://stats.stackexchange.com/questions/288200/what-is-the-chance-that-someone-is-dealt-a-suit-in-the-game-of-bridge"
} |
Case 3: Eigenvalues complex
when: $\alpha$ < $\frac{-25}{8}$
critical points: $\alpha$ = $\frac{-25}{8}$, $\frac{-25}{8}$ + 1
c) will be posted below:
15
Quiz-6 / Re: Q6 TUT 5101
« on: November 17, 2018, 04:08:43 PM »
Phase portrait attached
Pages: [1] 2 | {
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} |
php, authentication, session, authorization
class Session {
public $is_logged;
public $error;
private $useragent;
public function __construct(){
$this->useragent = $_SERVER["HTTP_USER_AGENT"];
session_name(SESS_NAME);
session_set_cookie_params(COOK_TIMEOUT,COOK_PATH,COOK_DOMAIN,COOK_SECURE,COOK_HTTP);
session_start();
}
public function set_session($user){
if(isset($_SESSION['user'])){
$this->error = "Session Already Set";
return false;
}
$nonce = $this->gen_nonce();
$_SESSION['user'] = $user;
$_SESSION['nonce'] = $nonce;
$_SESSION['useragent'] = $this->useragent;
setcookie('nuo',$nonce,time()+COOK_TIMEOUT);
header("refresh:0");
} | {
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"id": 16650,
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"openwebmath_score": null,
"tags": "php, authentication, session, authorization",
"url": null
} |
javascript, react.js
Anyways, not much change here:
var SearchBar = React.createClass({
handleInputChange: function(){
this.props.onUserInput(this.refs.myInput.value, this.props.inStockOnly);
},
handleCheckboxChange: function(){
this.props.onUserInput(this.props.filterText, this.refs.myCheckbox.value);
},
render: function() {
return (
<form>
<input type="text" placeholder="Search..." value={this.props.filterText} onChange={this.handleInputChange} ref="myInput" />
<div>
<label>
<input type="checkbox" checked={this.props.inStockOnly} onChange={this.handleCheckboxChange} ref="myCheckbox" id="myCheckbox" />
<span>Only show products in stock</span>
</label>
</div>
</form>
);
}
}); | {
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"openwebmath_score": null,
"tags": "javascript, react.js",
"url": null
} |
homework-and-exercises, newtonian-mechanics
$$T\cos\alpha = mg-ma\sin\alpha, \quad T\sin\alpha = ma\cos\alpha$$
Note that instead of $ma$ from your formulae, I had to write $ma\sin\alpha$ and $ma\cos\alpha$ because the acceleration is a vector in a direction that is neither vertical nor horizontal. Multiply the first (correct) equation by $\sin\alpha$, the second one by $\sin\alpha$, and subtract them in order to eliminate $T$. You will get
$$ 0 = T(\sin\alpha\cos\alpha-\cos\alpha\sin\alpha) =\\ = mg\sin\alpha - ma\sin^2\alpha-ma\cos^2\alpha=mg\sin\alpha-ma $$
which gives you $a=g\sin\alpha$ again. In other words, you have forgotten to realize that the acceleration $a$ is in a direction tilted by $\alpha$ so if you rewrite the vectors in the horizontal and vertical components, you have to rotate it into $a\sin\alpha$ and $a\cos\alpha$ with the right signs. | {
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resonance, inductance, antennas
If the equations assume no self inductance and a velocity factor of 1, what causes the reactance ? The impedance formula you quoted assumes that the radiator is made of an ideal cylindrical conductor, ie., completely lossless idealized material that is represented by the boundary conditions: $n\cdot H=0$ and $n\times E=0$. Whatever low frequency inductance and/or capacitance it may have as a pair of metal pieces is already built in the formula.
It has a reactance because the pair of cylinders do not have a constant wave impedance as a pair of transmission lines (imagine you are pushing the far ends of parallel wires apart while keeping their feed points together) and also because the vacuum impedance ($377\Omega$) is not a proper termination at the far ends. | {
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} |
quantum-mechanics, harmonic-oscillator, conventions, density-of-states
Your suggestion of $4\pi k^2/8$ was correct for the 3D box, but here he/she is using the 3D box result and correcting using $k_i=\pi n_i/L \to k_i=n_i$.
One way of calculating the 3D density of states directly is to use integral 4.634 of Gradshteyn and Ryzhik (7th). I can supply more details if interested. Also see http://arxiv.org/abs/cond-mat/9608032. | {
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ozone, upper-atmosphere
To some extent, there most likely will always be an area over Antarctica with reduced stratospheric ozone content in late Southern Hemisphere winter. (I wrote "always". Read that as meaning for the next several millions of years.) The issues is how long that annual ozone hole exists, how large that ozone hole becomes, how deeply depleted that ozone hole is, and (most importantly) what that means for the world at large. The reason it will continue to exist to some extent is that volcanos can inject ozone-deleting chemicals directly into the stratosphere.
However, right now, the worst of the ozone-deleting chemicals in the stratosphere are chemicals created by humans. Many of those chemicals will remain in the atmosphere for decades after production of them is ceased. For example, the graph below depicts the amount of trichlorofluoromethane (aka Freon-11, and other names) in the atmosphere: | {
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fft
We can write this roughly as
$$Y_k = X_k\cdot \hat{W}_{N,k} \cdot M_N$$
For N=2 the recombination matrix is simply
$$ M_2 = \begin{bmatrix}
1 & 1\\
-1 & 1
\end{bmatrix} $$
For N = 4, we get
$$M_4 = \begin{bmatrix}
1 & 1 & 1 & 1\\
1 & -j & -1 & j\\
1 & -1 & 1 & -1\\
1 & j & -1 & -j
\end{bmatrix} $$
Note that the first column is always 1.
For the execution of the radix 2 butterfly we need 1 complex multiply and 2 complex adds (or subtracts). That's a total of 4 real multiplies and 6 real adds.
For the execution of the radix 4 butterfly we need 3 complex multiplies for the twiddle factors. The recombination matrix consists ONLY of powers of $j$ so we don't have to actually execute any complex multiplies but can achieve this by simply swapping real/imaginary parts and/or signs. Due to the specific structure of this matrix we can implement this with 8 complex adds (not 12 if we were to do this directly). That's a total of 12 real multiplies and 22 real adds. | {
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php, mvc, codeigniter
Title: Volunteer Signup Page In a previous code review, it was strongly suggested that I start using PHP Frameworks to improve security for my websites. I tried Laravel and got stuck due to the steep learning curve. Then I tried CodeIgniter and found it to be a good fit. I did a CodeIgniter tutorial and was able to pick up the basics right away, so I converted my Volunteer Signup Website to CodeIgniter.
As this is my first time working with a PHP Framework and with MVC, I want to run it by the community and make sure I am following best practices and organizing everything correctly.
What does this page do? | {
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navigation, amcl-demo.launch, plugin, amcl, pluginlib
[ INFO] [1396554687.373924581]: Using plugin "inflation_layer"
[ INFO] [1396554687.994371819]: Created local_planner base_local_planner/TrajectoryPlannerROS
[ INFO] [1396554688.069237432]: Sim period is set to 0.20
[ INFO] [1396554690.131045226]: odom received! | {
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"tags": "navigation, amcl-demo.launch, plugin, amcl, pluginlib",
"url": null
} |
computability, oracles, random-oracles
First, since it took me a minute to figure this out myself, let me formalize the difference between your question and $\mathsf{AlmostP}$; it's the order of quantifiers. $\mathsf{AlmostP} := \{L : Pr_R(L \in \mathsf{P}^R) = 1\}$, and the result you allude to is $\forall L\, L \in \mathsf{BPP} \iff Pr_R(L \in \mathsf{P}^R) = 1$. If I've understood correctly, you are asking if $Pr_R(\forall L\, L \in \mathsf{P}^R \cap \mathsf{COMP} \iff L \in \mathsf{BPP}) = Pr_R(\mathsf{P}^R \cap \mathsf{COMP} = \mathsf{BPP}) = 1$.
Consider
$p := 1-Pr_R(\mathsf{P}^R\cap \mathsf{COMP} = \mathsf{BPP}) = Pr_R(\exists L \in \mathsf{P}^R \cap \mathsf{COMP} \backslash \mathsf{BPP})$. | {
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"url": null
} |
homework-and-exercises, newtonian-mechanics, newtonian-gravity, orbital-motion, celestial-mechanics
Title: Calculate the Earth’s velocity in its orbit (assumed to be circular) around the Sun Data:
GM of the Sun: $\text{GM} = 1.327 \times 10^{20} \;\mathrm{m^3 s^{-2}}$
Radius of the Earth’s orbit: $r_{Earth} = 1.496 \times 10^{11} \;\mathrm{m}$
Orbital period of a planet in seconds: $T = \sqrt{\frac{4\pi^2}{GM}r^3}$ | {
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"url": null
} |
c#, beginner, object-oriented, winforms, playing-cards
AllAchievements[StraightAchievement.EnumCasted] = Properties.Settings.Default.GetStraight;
AllAchievements[FlushAchievement.EnumCasted] = Properties.Settings.Default.GetFlush;
AllAchievements[FullHouseAchievement.EnumCasted] = Properties.Settings.Default.GetFullHouse;
AllAchievements[FourOfAKindAchievement.EnumCasted] = Properties.Settings.Default.GetFourOfAKind;
AllAchievements[StraightFlushAchievement.EnumCasted] = Properties.Settings.Default.GetStraightFlush;
AllAchievements[RoyalFlushAchievement.EnumCasted] = Properties.Settings.Default.GetRoyalFlush; | {
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navigation, amcl
Originally posted by David Lu with karma: 10932 on 2014-10-02
This answer was ACCEPTED on the original site
Post score: 3
Original comments
Comment by Marcus on 2014-10-03:
Hey David,
thank you for your quick answer. I was already suspecting something like this, since I tried to run my simulation without AMCL and ros demands the transform from base_footprint to map. I was just not aware that it might be that simple.
Thanks,
Marcus
Comment by jamman on 2014-11-07:
Hey David, I was wondering if we can use the robot_state_publisher instead of multiple tf in Markus case ?
Comment by David Lu on 2014-11-07:
I'm not sure what you're asking. Can you open up a new question instead? | {
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Also, when you get an equation like
v2⋅96=(v2−12)⋅(96−40v2)
You should divide out by the common factor of 8 and get
v2⋅12=(v2−12)⋅(12−5v2)
before multiplying it out. | {
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"url": "https://math.stackexchange.com/questions/2599667/speed-of-two-trains-travelling-side-by-side/2599677"
} |
c++, median, constrained-templates
struct frugal_strategy
{
template<std::ranges::forward_range Range,
std::invocable<std::ranges::range_value_t<Range>> Proj,
projected_strict_weak_order<Range, Proj> Comp,
midpoint_function<Range, Proj> Midpoint>
auto operator()(Range&& values, Comp compare, Proj proj, Midpoint midpoint) const
-> median_result_t<Range, Proj, Midpoint>
requires std::invocable<inplace_strategy_rvalues_only, Range, Comp, Proj, Midpoint>
|| std::invocable<copy_strategy, Range, Comp, Proj, Midpoint>
|| std::invocable<external_strategy, Range, Comp, Proj, Midpoint>
{
if constexpr (std::invocable<inplace_strategy_rvalues_only, Range, Comp, Proj, Midpoint>) {
return inplace_strategy_rvalues_only{}(std::forward<Range>(values), compare, proj, midpoint);
} | {
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"url": null
} |
optics, waves, huygens-principle
Mathematically, we solve for the Green's function $G$ in the following equation.
$$\left(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)G(\mathbf{x},t;\mathbf{x}_0,t_0) = \delta(\mathbf{x}-\mathbf{x}_0) \delta(t-t_0)$$
With some math, this can be shown to be a spherically propagating wave.
The idea behind Green's functions is that once you know $G$, you can write any wave in roughly the following way:
$$u(\mathbf{x},t) \sim \int d\mathbf{x}^{\prime} G(\mathbf{x},t;\mathbf{x}^{\prime},0)u(\mathbf{x}^{\prime},0)$$
This is why, given the initial distribution of the wave $u(\mathbf{x}^{\prime},0)$, you can find the wave at a later time by propagating each point of the wave out with the spherical wave Green's function. For the exact form for the wave, see here https://math.dartmouth.edu/~ahb/notes/waveequation.pdf | {
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python, pandas
print(merged)
Output
B_0 B_1 ... B_248_higher_tf B_249_higher_tf
A datadate ...
0 2022-01-01 8 6 ... NaN NaN
2022-01-02 2 2 ... 4.0 4.0
2022-01-03 0 3 ... 4.0 4.0
2022-01-04 1 7 ... 4.0 4.0
2022-01-05 5 0 ... 4.0 4.0
... ... ... ... ... ...
3999 2022-02-11 7 0 ... 3.0 1.0
2022-02-12 0 5 ... 3.0 1.0
2022-02-13 8 8 ... 5.0 1.0
2022-02-14 5 8 ... 5.0 1.0
2022-02-15 1 9 ... 5.0 1.0
[184000 rows x 500 columns]
Runs in about three seconds. | {
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slam, navigation, ros-melodic, rtabmap
terminate called after throwing an instance of 'UException'
what(): [FATAL] (2020-05-05 22:47:50.222) Rtabmap.cpp:3338::process() Condition (_memory->getSignature(id) != 0) not met! [id=113]
[rtabmap/rtabmap-5] process has died [pid 1931, exit code -6, cmd /home/pi/catkin_ws/devel/lib/rtabmap_ros/rtabmap rgb/image:=/d435/color/image_raw depth/image:=/d435/aligned_depth_to_color/image_raw rgb/camera_info:=/d435/color/camera_info rgbd_image:=rgbd_image_relay left/image_rect:=/stereo_camera/left/image_rect_color right/image_rect:=/stereo_camera/right/image_rect left/camera_info:=/stereo_camera/left/camera_info right/camera_info:=/stereo_camera/right/camera_info scan:=/scan scan_cloud:=/scan_cloud user_data:=/user_data user_data_async:=/user_data_async gps/fix:=/gps/fix tag_detections:=/tag_detections odom:=/odom imu:=/imu/data move_base:=/move_base grid_map:=/map __name:=rtabmap __log:=/home/pi/.ros/log/6fd4bf78-8f58-11ea-8779-dca63207de8a/rtabmap-rtabmap-5.log]. | {
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} |
zoology, mammals
Title: Are Sloth tongues soft or rough? Are Sloth tongues soft or rough?
Dogs have soft tongues, while cats have rough tongues, so the feeling when they lick e.g. a human hand is quite different. So what about sloths?
This question came up in a recent discussion but I was unable to find any materials as google queries for sloth tongues leads to tons of cute pictures. Sloths have long, thick, sticky tongues covered in a carpet of tiny, rear-ward pointing spikes that they can pull leaves in with.
So the tongues are quite different from human tongues and likely much less "soft" to touch and more "rough".
According to scanning electron microscopy studies on the topography of a sloth's tongue, the following results were found:
The results revealed that the rostral part of the tongue presents a
round apex and covered by filiform and fungiform lingual papillae and
a ventral smooth surface. | {
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beginner, c, formatting, io
int
main(void)
{
char s1[MAXLINE];
char s2[MAXLINE];
while (get_line(s1, MAXLINE) > 0) {
expand(s1, s2);
printf("%s", s2);
}
return (0);
}
/**
* Here I have tried to write a loop equivalent to the loop seen
* previously in chapter 1. (without using && and ||,
* as specified in chapter 2 of the book, exercise 2.2).
*
* for (i = 0; i < lim-1 && (c = getchar()) != EOF && c != '\n'; ++i)
* ...
**/
int
get_line(char s[], int lim)
{
int c, i;
i = 0;
while (--lim > 0) {
c = getchar();
if (c == EOF)
break;
if (c == '\n')
break;
s[i++] = c;
}
if (c == '\n')
s[i++] = c;
s[i] = '\0';
return (i);
}
void
expand(const char s1[], char s2[])
{
int i, j, ch; | {
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• The number of conjugates of $\;\alpha\in K\;$ ( over $\;F\;$ ) is the number of different roots of the minimal polynomial of $\;\alpha\;$ in $\;F[x]\;$ , which is $\;[F(\alpha):F]=\;$ the degree of the polynomial mentioned above. – DonAntonio May 22 '17 at 22:16
• @DonAntonio That is more or less the OP's argument in the post. – Kenny Wong May 22 '17 at 22:18
• @SamY. Why do you think that ${\rm Gal}(K/F(\alpha))$ is a normal subgroup of ${\rm Gal}(K/F)$? This is true if and only if $F(\alpha)$ is a normal extension of $F$; it is not true in general. [By the way, you also wrote ${\rm Gal}(K/F(\alpha))$ and ${\rm Gal}(K/F)$ the wrong way round in your post.] – Kenny Wong May 22 '17 at 22:19
• @KennyWong Indeed. These posts that write "I don't understand this or that..." and then later it happens to be they actually do perplex me. I many times address the OP's very question and don't read the whole thing. – DonAntonio May 22 '17 at 22:20 | {
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resource-recommendations, spectroscopy, resonance, vibrations, molecules
Some science books say that the frequency of microwaves used in a
microwave oven is the 'natural frequecy' of water and resonance is the
mechanism that causes the vibration. That is not the case - it is down
to the pull of the electric and magnetic forces from the wave.
They are dealing with two halves of the same coin: microwaves do excite molecular resonances and they do that through "pull of the electric and magnetic forces" (with electric forces being the most important). So, both are true.
Separately, in the gas phase, molecular resonances are very sharp. In liquid phase, like water in a microwave, they are less sharp. As they become less and less sharp, some might choose to stop using the word resonance. | {
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magnetic-fields, path-integral, wick-rotation
The time-derivative term in path integral for Dirac fermions. When relating the path integral formulation to the canonical formulation, this term can/should be regarded as part of the definition of the inner product, so we should be relieved that its coefficient is not affected by Wick rotation.
The so-called theta term $\int F\wedge F$ in QCD in four-dimensional spacetime. Section 6.1 in reference 12 mentions an interesting connection (an analogy, at least) between theta terms and complex saddles.
The Chern-Simons term in odd-dimensional spacetime. The fact that its coefficient is not affected by Wick rotation is important in the study of chiral anomalies, because the anomaly (which is matched by the Chern-Simons term through the anomaly-inflow mechanism) can only occur in the phase of the partition function, even when the signature is euclidean. | {
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c#, .net, cryptography, validation, mvp
Title: User name and password validation in a MVP application In a WinForms application I'm doing the password validation as follows:
When the user presses the OK button after entering username and password, an event will be fired and the listener in the presenter will then do the validation with the help of the Encryption class. It will basically compare the two hash values.
User Model
class User
{
string UserID { get; set; }
string Name { get; set; }
string NIC {get;set;}
string Designation { get; set; }
string PassWord { get; set; }
List<string> Permission = new List<string>();
bool status { get; set; }
DateTime EnteredDate { get; set; }
}
Login form
public partial class frmLogin : Form , IView
{
public event EventHandler OnValidatePassword;
public string UserName
{
get { return txtUser.Text; }
}
public string Password
{
get { return txtPassword.Text; }
}
public bool Valid { get; set; } | {
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organic-chemistry
Why does the above compound produce hydrogen gas? | {
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special-relativity, energy, momentum, potential-energy, mass-energy
Title: Is the Einstein Energy-Momentum equation $E^2 = p^2c^2 + m_0^2c^4$ valid only for Free Particles? Is the energy -momentum relation
$$E^2 = p^2c^2 + m_0^2c^4$$
satisfied only by free particles or even bound particles?
Does the Energy refer to total Energy(including potential) or only (kinetic +rest mass). The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept.
The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes an entirely quantum answer I'll make this an entirely no quantum answer.
So you used to be able to have potentials exert forces. In reality you knew that the forces came in action reaction pairs so there was something somewhere else that felt an equal and opposite force. | {
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javascript, performance, ecmascript-6, quiz
]
},
{
"id": 6,
"category": "AD",
"messages": [
{
"score": "High",
"message": "You enjoy change and can go with the flow with ease!"
},
{
"score": "Medium",
"message": "You can adapt when changes are needed, but also enjoy when things stay the same and you can get into a routine."
},
{
"score": "Low",
"message": "You are not a fan of change, and prefer to stick with a steady routine."
}
]
},
{
"id": 7,
"category": "GR",
"messages": [
{
"score": "High",
"message": "You push through to accomplish your goals, no matter what life throws at you! You are well-suited to work on long-term goals."
},
{
"score": "Medium",
"message": "You see things through if you can, but sometimes a difficult situation leads you to change your goals."
},
{ | {
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C is a constant. The Surprising Details About Definite Integral Calculator That Most People Aren't Aware Of Rumors, Lies and Definite Integral Calculator . Our online … Math Help . Now, this identity successfully states that the evaluation of the triple integral in the LHS amounts to evaluating a surface integral (double integral) in the RHS. An integral of the form intf(z)dz, (1) i.e., without upper and lower limits, also called an antiderivative. Integral Calculator. Integrals - Step-By-Step. List of Indefinite Integral Formulas . where both F and ƒ are functions of x, and F is differentiable. Then you've got … (2) This result, while taught early in elementary … Learn how to find limit of function from here. the function inside the integral….) Each was made to assess distinctive attributes as objectively as possible. calculators. Fixing Integration Constants Example 3 Consider a rocket whose velocity in metres per second at time t seconds after launch is v = bt2 where b = 3ms−3. | {
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control-engineering, aerospace-engineering, pid-control
In your PID loop, you add $k_d\dot{E}$ instead of subtracting it. By subtracting, it reduces the input as you increase angular velocity. For instance, if I am $\frac{\pi}{2}$ radians off and not rotating, my input will be large. As I increase in angular velocity, my input decreases from both the decrease in error and the increase in angular velocity. Eventually, there must be 0 input and it will happen at some point before I get to my position. The UAV keeps rotating and error decreases. However, $\dot{E}$ remains at its previous level and the input becomes negative, decreasing my angular velocity. If, somehow, the angular velocity, got to 0 before achieving the setpoint, this whole process would be back at the beginning. This means you should only have to tune your gains. | {
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sql, sql-server, random, t-sql
SELECT TOP 1 @amount = I
FROM Numbers
ORDER BY CRYPT_GEN_RANDOM(4)
OPTION (MAXRECURSION 0)
First, this generates a table of numbers 1000 to 5000 using a recursive CTE, then selects one randomly using the CRYPT_GEN_RANDOM() that we discussed earlier.
To If or Not to If
Might I also suggest an alternate method of generating the @payment_way. Instead of using an If, might I suggest using a Simple Case Statement.
Personally I find this
SET @payment_way =
CASE @amount % 2
WHEN 1
THEN 'cash'
ELSE 'credit card'
END | {
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1. Fibonacci's sequence
Two consecutive Fibonacci numbers are relatively prime
This is my proof - am i correct?
Let Fn and Fn+1 be any two consecutive Fibonacci numbers and suppose there is an integer d > 1 such that d divides Fn and d divides Fn+1.
Then Fn+1 - Fn = Fn-1 will also be divisible by d (if d divides a and d divides b, then a = d*m and b = d*n for some integers m and n. Then a - b = d*m - d*n = d * (m-n), so d divides (a - b) as well). But now notice that Fn - Fn-1 = Fn-2 will also be divisible by d.
We can continue this way showing that Fn-3, Fn-4, ... , and finally F1 = 1 are all divisible by d. Certainly F1 is not divisible by d > 1. Thus we have a contradiction that invalidates the assumption. Thus it must be the case that Fn and Fn+1 are relatively prime.
Has anyone know of any other properties of the Fibonacci’s Sequence?? please
2. Originally Posted by Natasha1
Two consecutive Fibonacci numbers are relatively prime
This is my proof - am i correct? | {
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ros, sensor-msgs, range
*Warning: Limiting the max range of a range sensor without adjusting the range (e.g. max_range=0.5, range=0.567) will result in obstacles not getting removed from the range layer sensor. So be sure to clip all values > max_range to max_range.
** Assuming nonholonomic robots with a reasonable size.
Originally posted by Humpelstilzchen with karma: 1504 on 2016-12-27
This answer was ACCEPTED on the original site
Post score: 0 | {
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mechanical-engineering, materials, applied-mechanics, stresses, simulation
Is the yield stress referring to just the stress. Having trouble understanding the inputs, any clarification would be helpful. While the yield stress is a well defined point in materials such as steel or other metals, that is not the case when we talk about foams.
Here they are calling "yield stress" to the whole volumetric strain-stress curve.
So they are basically asking for a number of strain-stress curves at different strain rates.
Or you can use material 63 if strain rate effects are not important. | {
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c++, snake-game
When you do that in the other for loops as well, there will be a compile error:
head_map[i] = temp1;
board[head_map[i].row][head_map[i].column] = token;
This compile error means that your code is somewhat unusual. You used i in a loop, and usually that variable is not needed after the loop. Not so in this case.
When the for loop is finished, i will be the same as snake_length. Therefore you can replace the code with the very similar:
head_map[snake_length] = temp1;
board[head_map[snake_length].row][head_map[snake_length].column] = token; | {
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newtonian-mechanics, computational-physics, spring, simulations
Title: How to integrate over a timestep in a mass-spring model? I'm writing a simulation of a block of matter using a "mass-spring" model, where the matter is modelled as a 3D lattice of point masses, where each point is connected by springs to the (up to) 26 neighboring point masses.
At each "tick" I calculate the new force at each point mass with (pseudocode):
for each neighbor:
distance = length(vector_from_me_to_neighbor)
force_here += spring_constant * (distance - ideal_distance_from_me_to_neighbor) * normalised_vector_from_me_to_neighbor
Using verlet integration, I update the position of each point mass with:
new_pos = current_pos + velocity_approximation + (force / point_mass) * timestep * timestep * 0.5; | {
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android
Originally posted by Ben_S with karma: 2510 on 2011-11-12
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by manabu on 2011-11-13:
Thanks for your answer. Making a web service is a better approach, but developing web application with high transparency to ROS network is difficult for me. No authentication may become a problem in the future indeed. Tunneling seems to be a only way to use android.rosjava applications. | {
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Theorem is a special of the sine rule. sinA + sinB = 2 sin 1. Formulas for cos(A + B), sin(A − B), and so on are important but hard to remember. Rather than adding equations (3) and (8), all we need to do is subtract equation (3) from equation (8): cos(A For the tan(A + B) formula, I will explain that you could use sin(A + B)/cos(A + B) and that it will simplify to the form they will see in textbooks. The main idea is to create a triangle whose angle is a difference of two other angles, whose adjacent sides, out of simplicity, are both 1. 12/13 sin C = 15/65 + 48/65 = 63/65. 07. It can be rearranged to: $$\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}$$ This The cosine of a compound angle a plus b is expressed as cos (a + b) in trigonometry. In a right triangle ABC the sine of α, sin(α) is defined as the ratio betwween the side adjacent to angle α and the side opposite to the right angle (hypotenuse): cos α = b / c. The shape of the cosine curve is the same for each full rotation of the angle | {
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} |
organic-chemistry, stereochemistry, isomers
Consequently, I also counted 32 isomers (same as yours). When compared to your count, I realized your count of chain alcohols (6), ethers (5), and carbonyl compounds (3) are all correct.
However, your count of three-membered ring alcohols (7) and three-membered ring ethers (6) are incorrect. There are only 6 possible three-membered ring alcohols (one less than your count) while 7 three-membered ring ethers are possible (one more than your count; see the chart). Therefore, it might not be the miscount of meso-isomer (if this is the reason, you should count 8 possible three-membered ring ethers), but be an error on structure drawing.
Your final count of for-membered ring (4) and five-membered ring (1) ethers are also correct. Precisely, only 1 four-membered ring alcohol is possible.
Conclusively, you and I both have the same structure count but wrong counts on different kind of structures. You can compare your drawing with attached chart here and find out where you made the mistake. | {
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java, performance, thread-safety, concurrency
if (message != null && message.equals("q")) {
// it's definitely bad if scheduler decides to reschedule here
// because output lock happens only inside stop server method
// (trying to lock right after readLine procedure creates possibility of losing the result)
server.stopServer();
break;
}
}
});
thread.setDaemon(true);
thread.start();
}
handles second type of cancellation (prompt)
private void runPromptScheduler() {
scheduledExecutor.scheduleWithFixedDelay(this::displayPrompt, INITIAL_DELAY, SCHEDULER_PERIOD, TimeUnit.SECONDS);
}
private void displayPrompt() {
promptLock.lock();
try {
Scanner scanner = new Scanner(System.in);
System.out.print("continue(1), continue without prompt(2), cancel(3): ");
int readValue = scanner.nextInt(); | {
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strings, vba, formatting, vb6
Else 'only the "P" is specified
thisFormat = "#,##0"
End If
'Append the percentage sign to the format string:
thisFormat = thisFormat & "%"
Case "R", "r" 'ROUND-TRIP format (a string that can round-trip to an identical number)
'example: ?StringFormat("{0:R}", 0.0000000001141596325677345362656)
' ...returns "0.000000000114159632567735"
'convert value to a Double (chop off overflow digits):
v = CDbl(v)
Case "X", "x" 'HEX format. Formats a string as a Hexadecimal value.
'Precision specifier determines number of total digits.
'Returned string is prefixed with "&H" to specify Hex.
v = Hex(v)
precisionSpecifier = CInt(precisionString) | {
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faster-than-light, causality, wormholes
And as for entanglement if you go with a realist (and nonlocal)/quantum mechanics, it isn't one causing the other it is the spin state continuously changing (in configuration space) based on the Hamiltonian which also changes the spatial wave in configuration space. So you just have a nonlocal thing, the wave (defined on configurations) and it all evolves according to be same Schrödinger equation, it isn't one thing in one place causing another it is a wave that is already everywhere and more than everywhere it is in configuration space. And all parts of it evolve based on just the values assigned to configurations near it. So it is local in the sense that only nearby configurations affect the value or spin at a given configuration. But the object itself isn't a thing with properties in real space and in particular that means there isn't a wave at each point in spacetime so no obvious connection to wormholes. | {
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android, database
Title: Storing icons in Room Database I want to store some 512x512 icons for my Android app in a Room database. This is so each item in a list can have all the text data along with the image data stored in a single record.
The answers to this question https://stackoverflow.com/questions/46337519/how-insert-image-in-room-persistence-library indicate this is not a good practice, however the images I'm saving aren't even 1 kilobyte.
Under these circumstances would it be acceptable to store images in a database like this?
I want the database to come preloaded with certain default list items, but the users will also be able to add their own items with custom icons. Given that this is the case it may be better to just store URIs in the database but I would like some insight on this.
would it be acceptable to store images in a database like this? | {
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java, parsing, serialization
static Class<?> extractTypeParameter(TypeParser<?> typeParser) {
return extractTypeParameter(typeParser.getClass(), typeParser);
}
/*
* Pass the original TypeParser as well, to be able to print
* it in error message.
*/
private static Class<?> extractTypeParameter(Class<?> c, TypeParser<?> typeParser) {
if(c == null){
String message = "Can not find parametirized type in %s !?!?!? Must be a bug somewhere...";
throw new IllegalArgumentException(String.format(message, typeParser));
}
for (Type t : c.getGenericInterfaces()){
if(t instanceof ParameterizedType){
ParameterizedType type = (ParameterizedType) t;
if(TypeParser.class.equals(type.getRawType())){
return (Class<?>) type.getActualTypeArguments()[0];
}
}
}
return extractTypeParameter(c.getSuperclass(), typeParser);
}
} | {
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information-theory, coding-theory
Title: Problems with the proof of Huffman Optimality in Cover's book There is a some question that arise from the proof of Lemma 5.8.1 of Cover's book on information theory that confuse me.
First question is why he assumes that we can "Consider an optimal code $C_m$. Is he assuming that we are encoding a finite number of words so that
$\sum p_i l_i$ must have a minimum value? Here I give you the relevant snapshot.
Second, there is an observation made on this notes that was also done in my class before proving the theory of optimality of huffman codes, that is,
observe that earlier results allow us to restrict our attention to
instantaneously decodeable codes | {
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php, beginner, object-oriented, classes
A part from that the Validator should be able to tell what validation rules passed an what didn't:
$validator->passed('SmallerThenSixtyFour'); //true || false
A smart person would ofcource refractor the rules to:
$validator = new Validator(
new MaxSizeRule(63),
new MinSizeRule(0)
);
If writing new all over the place becomes a pain in the a$$. We could use a FactoryPattern that creates the validator for us given some custom validation rule language:
ValidatorFactory::make('max:63,min:0,unique');
Laravel does a pretty good job here. Or if you want to go all the way with validation, why not look into how symfony did it, they implemented the JSR303 Bean Validation specification (143 pages of pure fun).
To sum up
Keep every little piece of code stupid. The less it knows, the better. Use interfaces, or better. Reuse interfaces. Think abstract, and once in a while, go all the way.
Some other small remarks: look into the php-fig standard. | {
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java, swing, gui
/**
* Convenience method for setting the bounds of a window after its
* preferred, centred bounds have been determined. The window will be
* packed, which comes with the side-effect of setting it displayable.
* @param window the window whose bounds are to be set
* @return {@code true} if the window's bounds were changed as a result of
* this operation
* @see #getSpaceCentredBounds(java.awt.Dimension)
* @see Window#pack()
*/
public boolean setSpaceCentredBounds(Window window) {
return setBounds(window, mwb, ReferenceRegion.AVAILABLE_SPACE);
} | {
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python, python-3.x, async-await, configuration
Title: Managing python application configuration in a key-value store For a project we're working on, we need a central place to store the configurations for various applications (they're all headless services running in docker containers so local configuration files or command line parameters aren't going to cut it in production)
We've chosen to use Consul as the central key-value store and since most of the modules are written in Python I've created a config wrapper to interact with it. It's making use of the python-consul SDK for that purpose.
There are two main modes of operation:
On initialization we load the current configuration values synchronously.
After that a background monitoring job gets kicked off which executes a callback whenever a key changes. | {
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"tags": "python, python-3.x, async-await, configuration",
"url": null
} |
javascript, jquery, plugin
// Add a callback event
if (typeof callback === "function") {
// Execute the callback
callback.call(this);
// Unbind the event handlers
$this.unbind( transitionEnd );
}
});
}
// Check if delay exists or if it"s a callback
if (!delay || typeof delay === "function") {
// If it"s a callback, move it to callback so we can call it later
callback = delay;
// Run the animation (without delay)
run();
} else {
// Start a counter so we can delay the animation if required
setTimeout( run, delay );
}
});
};
})(jQuery, window, document); Awesome code, | {
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python, performance, regex, csv, pandas
Frequencies -- 1325.0427 1339.0678 1353.0061
Frequencies -- 1369.0614 1408.5258 1433.0543
Frequencies -- 1452.4148 1454.6319 1500.4304
Frequencies -- 1511.2305 1517.2562 1552.9189
Frequencies -- 1560.5313 1636.2290 1640.1732
Frequencies -- 1664.8747 1681.5566 1703.2026
Frequencies -- 1770.2627 3058.4143 3122.3743
Frequencies -- 3147.1828 3192.5897 3199.1398
Frequencies -- 3211.0676 3222.0033 3236.3394
Frequencies -- 3262.2119 3556.7997 3862.4791 | {
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biochemistry, yeast, pathogenesis, microbiome, alcohol
Title: The effects of auto-brewery syndrome Why are some people affected so differently by auto-brewery syndrome differently if the syndrome seems to be caused by the single organism saccharomyces cerevisiae? It is known that the syndrome has been used to waive DUI charges — like most recently for this New Yorker whose blood alcohol level was 0.4 but she had none of the usual symptoms of intoxication, even though she had ingested alcohol earlier that day. She also carried out most of her daily activities normally, even with high levels of alcohol in her system.
On the other hand, there was another case where the carrier experience the effects of auto-brewery syndrome differently: | {
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# Trigonometryfinding the impedance in rectangular and polar form
#### MFletch
##### New member
I dont fully understand how to work out the impedance from the given equation (5j-5)x(11j-11)/(5j-5)+(11j-11). Any help would be greatly appreciated. Thanks.
The answer needs to be in rectangular and polar form.
#### DavidCampen
##### Member
The numerator factors into 55$$(j-1)^2$$ and the denominator to 16(j-1); from here it should be simple to calculate the rectangular form.
#### MFletch
##### New member
Thank you, however, How would that convert to polar and Rec. form?
#### DavidCampen
##### Member
$$\frac{55{(j-1)}^{2}}{16(j-1)} = ?$$
Hint, cancel like terms and you will have the rectangular form.
so the rectangular form will be
$$\frac{55}{16}(j-1)$$
Last edited:
#### MFletch
##### New member
great thank you I've got it!
What would be the best way to find the admittance of Z = (1/5j-1) + (1/2j+6) + (1/4j) in rectangular and polar form?
#### DavidCampen | {
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computability, terminology
See if you can show that all these definitions are equivalent. You might want to look up the technique of "dovetailing".
What about the digits of $e$? We need to phrase the problem of computing the digits of $e$ in our framework. The set of digits of $e$ is the set of pairs $(k,d)$, where $d$ is the $k$th decimal digit of $e=2.71828\ldots$, i.e. $\{(0,2),(1,7),(2,1),(3,8),(4,2),(5,8),\ldots\}$. Since there is an algorithm computing the $k$th digit of $e$ (e.g. using the Taylor series $e = \sum_{k=0}^\infty 1/k!$), the set of digits of $e$ (under this encoding) is computable. | {
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thermodynamics
$$\mathbf v(\mathbf x,\mathbf p) = g(\mathbf x,\mathbf p),$$
for some function $g$.
Suppose that $\mathbf x\in\mathbb R^n$ and that $\mathbf v\in\mathbb R^m$; then among the $n+2m$ variables $\mathbf x,\mathbf v,\mathbf p$ there are $m+1$ constraints: $m$ coming from $g$ and 1 coming from $f$. Therefore you only have $n-1$ independent variables. The ones that you can then take as independent among the set $\mathbf x,\mathbf v,\mathbf p$ usually depends on how well-behaved the functions you get are. In general you get a (possibly smooth) submanifold of $\mathbb R^{n+2m}$, and the choice correspond to the fact that your atlas should have well-behaved charts (think of the sphere in $\mathbb R^3$, where you can't give a global chart, but you have to patch at least two charts together). | {
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bioinformatics, rna-sequencing
e was for England and in that lab's strain list the daf-2 allele was the 1,370th on their list. So daf-2(e1370) designation means that the strain carries a lf mutation in the daf-2 gene, and the specific allele is one that was isolated in Cambridge, UK. I believe that allele is temperature-sensitive (ts) dauer-constitutive. | {
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history-of-chemistry, vsepr-theory
Title: What are actually the differences between Sidgwick & Powell theory and Nyholm & Gillespie Theory? The famous VSEPR theory was proposed by first Sidgwick & Powell. It was modified then by Nyholm & Gillespie.
I was reading Inorganic Chemistry by J.D.Lee where he discusses very briefly the theory of Sidgwick & Powell; I couldn't somewhat conceive the language. So, I googled it but to my misfortune I could nowhere get more info other than this:
The idea of a correlation between molecular geometry and number of valence electrons (both shared and unshared) was originally proposed in 1939 by Ryutaro Tsuchida in Japan, and was _independently presented in a Bakerian Lecture in 1940 by Nevil Sidgwick and Herbert Powell of the University of Oxford. In 1957, Ronald Gillespie and Ronald Sydney Nyholm of University College London refined this concept into a more detailed theory, capable of choosing between various alternative geometries [1] | {
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electromagnetism, atmospheric-science, earth
source | {
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c#, object-oriented, game, winforms, minesweeper
this.MineFieldTable.RowStyles.Add(new System.Windows.Forms.RowStyle(System.Windows.Forms.SizeType.Absolute, 20F));
this.MineFieldTable.RowStyles.Add(new System.Windows.Forms.RowStyle(System.Windows.Forms.SizeType.Absolute, 20F));
this.MineFieldTable.RowStyles.Add(new System.Windows.Forms.RowStyle(System.Windows.Forms.SizeType.Absolute, 20F));
this.MineFieldTable.RowStyles.Add(new System.Windows.Forms.RowStyle(System.Windows.Forms.SizeType.Absolute, 20F));
this.MineFieldTable.RowStyles.Add(new System.Windows.Forms.RowStyle(System.Windows.Forms.SizeType.Absolute, 20F));
this.MineFieldTable.Size = new System.Drawing.Size(210, 210);
this.MineFieldTable.TabIndex = 4;
this.MineFieldTable.Visible = false;
//
// TB_mineCount
//
this.TB_mineCount.Location = new System.Drawing.Point(100, 25);
this.TB_mineCount.MaxLength = 3; | {
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Example: [0.3 0.2 0.1]
Example: 'green'
Marker size, specified as a positive value in points.
Example: 10
## Output Arguments
collapse all
One or more implicit function line objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific line. For a list of properties, see ImplicitFunctionLine Properties.
### Topics
#### Mathematical Modeling with Symbolic Math Toolbox
Get examples and videos | {
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angular-momentum, rotational-kinematics
There are two types of changes to consider. a) Change in rotation speed and b) Change in rotation direction. I think the first is kind of trivial to understand, but the second is a little more tricky. If you have vector that is riding on a rotating body, which direction does the vector change? You need the right hand rule to figure it out as it is a cross product.
Look at:
http://en.wikipedia.org/wiki/Rotating_reference_frame
and
http://www.envsci.rutgers.edu/~broccoli/dynamics_lectures/lect_06_dyn12_mom_eq_rot.pdf
Hope this helps. | {
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newtonian-mechanics, rotational-dynamics, work, torque
Also, Torque = angular acceleration * moment of inertia.
Would this mean that the same force can do a different amount of work on an object depending on where it is applied?
For example, consider a rod floating in space. If it is pushed by force F at its center of mass, there will be no torque. However, if it is pushed near one of its ends, there will be the same acceleration of the center of mass as before, plus some torque. Yes, you are right! Only when a force is applied purely through the center of mass it results in the body gaining a linear action with no rotational components.When any force is applied at a distance from the center of mass, it results in the body gaining the linear acceleration mentioned above plus an angular acceleration which depends on the moment arm (perpendicular distance to the point of application). In all, yes, the work done by a force does depend on its point of application. (and of course the time interval through which it acts.) | {
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philosophy, artificial-consciousness, self-awareness
Dialog
Playing to win
Detecting honesty or dishonesty
Now consider how similar or different these mental activities are when we compare self-directed or externally directed attention.
One can talk to one's self or talk to another
One can play both sides of a chess game or play against another
One can scrutinize one's own motives or those of another | {
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ros, octomap-mapping
If you succeeded in doing all these procedures, then you can render octomap with Marker Arrays on Rviz.
This is just my case, if your are using other sensors, modify corresponding topics according to your needs. Good luck ! | {
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machine-learning, reinforcement-learning, discount-factor, actor-critic-methods
Why do the weights updates depend on the discount factor via $I$?
It seems that the more we get closer to the end of the episode, the less we value our newest experience $\delta$.
This seems odd to me. I thought discounting in the recursive formula of $\delta$ itself is enough.
Why does the weights update become less significant as the episode progresses?
Note this is not eligibility traces, as those are discussed separately, later in the same episode. This "decay" of later values is a direct consequence of the episodic formula for the objective function for REINFORCE:
$$J(\theta) = v_{\pi_\theta}(s_0)$$
That is, the expected return from the first state of the episode. This is equation 13.4 in the book edition that you linked in the question.
In other words, if there is any discounting, we care less about rewards seen later in the episode. We mainly care about how well the agent will do from its starting position. | {
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quantum-mechanics, angular-momentum, symmetry, schroedinger-equation, coordinate-systems
The separated equations for $R(\rho)$, $P(\phi)$ and $Z(z)$ are all second order; this is why there are two solutions for $P$ and $Z$ for each $m^{2}$ or $k_{3}^{2}$, respectively. There should also be two solutions to the radial equation you found for $R(\rho)$; the two solutions are proportional to $J_{m}(k_{1}\rho)$ and $N_{m}(k_{1}\rho)$—where $k_{1}^{2}+k_{3}^{2}=k^{2}$—in terms of Bessel Functions of the first and second kind. However, the Bessel Function of the second kind (or Neumann function) is not normalizable in the vicinity of the origin, so it does not get an acceptable solution. This leaves you with just the $J_{m}$ solution that you found.
Since $\rho=R$ must be a root of the Bessel Function, $k_{1}$ is quantized as well. $k_{1}R$ has to be one of the $z_{m,j}$, where $z_{m,j}$ is the $j$th root $J_{m}(z_{m,j})=0$ of the Bessel Function $J_{m}$. This makes the quantized energy level of the system | {
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# How could we find the largest number in the sequence $\sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$?
How to find the largest number in the sequence$$\sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$$
I am interested in a "calculus-free" approach. Thanks,
-
If you square all of them, the largest of the squares will correspond to the largest of the square roots. – Ｊ. Ｍ. Nov 8 '11 at 10:45
@GerryMyerson Shame on me :) I did not read the question carefully enough. – Sasha Nov 8 '11 at 13:42
The $n$-th term in the sequence is $n\sqrt{51-n}=\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\le n\le 50$), does $n^2(51-n)$ become the largest?
If you want to avoid calculus, you could use the AM-GM inequality: if $x,\,y,\,z\ge 0$, then $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz},$$ with equality if and only if $x=y=z$. | {
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} |
ros, orocos
Title: Do rosservice calls from rtt_ros respect orocos operation thread spec?
Say I have an orocos component that has an operation such as,
this->addOperation("do_work", &MyWorker::doWork, this, RTT::ClientThread).doc("Does work");
Where when called, this blocks the callers thread and does not block the task context thread in OROCOS.
I'm unable to find anywhere in the documentation for rtt_ros_integration whether or not ROSServiceService calls respect the orocos operation spec on whether to block the client thread of the task context thread.
Originally posted by jlack on ROS Answers with karma: 78 on 2017-07-30
Post score: 1 | {
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ros, catkin, library, cmake
Title: export a prebuilt library in catkin
I am wondering how can I export an already built library within a package to make it available to other catkin packages.
Before with rosbuild this could be done by by using the cpp tag in the manifest. For instance:
<export>
<cpp lflags="-Wl,-rpath,${prefix}/lib -L${prefix}/lib -lmylib" cflags="-I${prefix}/include"/>
</export>
As far as I understand this functionality has been replaced by the catkin_package() command. However, I have not found a clear way to make a prebuilt library within a catkin package visible to other catkin packages.
I have tried setting the prebuilt library as an imported library, but then when I do cmake it does not set it as a target visible by other packages. In CMakeLists.txt I have:
...
catkin_package(LIBRARIES mylib)
...
add_library(mylib STATIC IMPORTED)
set_target_properties(mylib PROPERTIES IMPORTED_LOCATION /path/to/mylib)
Any suggestions? | {
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