text stringlengths 1 1.11k | source dict |
|---|---|
organic-chemistry, reaction-mechanism, aromatic-compounds, amines
Title: Predicting product of a Hofmann elimination with a cyclic amine What is the final product of this reaction?
Here's what I have:
Does this look correct? I'm concerned because I'm used to seeing the nitrogen removed from the product (in the cases of primary amines). With excess of methylamine, I wouldn't exclude additional methylation, yielding a trimethylanilinium iodide.
However, the reaction should stop there since a further hydrogen abstraction from a $\beta$-position isn't possible. | {
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mean is [2(65)+73]/3. It is challenging to achieve high weighted efficiency with low-power microinverters, typically because these devices are required to be low cost. How to Calculate Weighted Average Price Per Share Calculating your weighted average price per share can help you assess the performance of an investment that was made in several transactions. …Let's consider an academic course as our example. What is the uncertainty of the weighted average? What's the correct procedure to find the uncertainty of the average?. Weighted average definition, a mean that is computed with extra weight given to one or more elements of the sample. Another Tip: If number of boys and girls is same, i. May 01, 2012 · PowerPivot Weighted Average Measure Compared to Non-Weighted Average. The weighted average (or weighted mean, as statisticians like to call it) is easy to compute in SAS by using either PROC MEANS or PROC UNIVARIATE. If trim is non-zero, a symmetrically trimmed mean is computed with a | {
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clustering, apache-spark, k-means
How can I get the table that I put above? Basically I just want to extract the same fields and add a column with the cluster calculated by K-Mwans to each row...
Many thanks! This is well answered in this earlier question:
https://stackoverflow.com/q/31447141/1060350
Beware that Spark k-means is slow.
If your data fits into main memory (i.e. a few gigabyte, which means billions of vectors!) then other tools such as ELKI that don't have the cluster overhead will be much faster. Use spark only for preprocessing the data, if you e.g. have several TB of jsons, and you need to first extract the numbers out of the JSONs, then here is where Spark shines.
Once your data is then vectors, use ELKI instead, it's much faster. | {
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inflection points from extrema for differentiable functions f(x). Also, how can you tell where there is an inflection point if you're only given the graph of the first derivative? The first derivative is f′(x)=3x2−12x+9, sothesecondderivativeisf″(x)=6x−12. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points. For example, the graph of the differentiable function has an inflection point at (x, f(x)) if and only if its first derivative, f', has an isolated extremum at x. And the inflection point is at x = −2/15. Lets begin by finding our first derivative. Given f(x) = x 3, find the inflection point(s). Example: Determine the inflection point for the given function f(x) = x 4 – 24x 2 +11. In all of the examples seen so far, the first derivative is zero at a point of inflection but this is not always the case. Now, if there's a point of inflection, it will be a solution of $$y'' = 0$$. Purely to be | {
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with latest contests, videos, internships and jobs!. Question: Vector Calculus - How to expand Vector Identities directly in Maple? Tags are words are used to describe and categorize your content. freeman and Co. Stating that the normal of a vector (x,y) is (-y,x) is repeating the question, and not a proof. home depot questions and answers , calculus swokowski 6th edition solution manual pdf , gate physics question paper 2013 , aeg santo user manual , kubota l2900 manual , fujifilm finepix hs25exr manual focus , manual lancer 2008 , volvo 850 gl fuel line repair manual , 1995 acura legend spark plug manual , solution manual of computer system. Heath and Company, ISBN 0-669-35335-3. Directional derivative and gradient examples by Duane Q. Unlock your Thomas' Calculus PDF (Profound Dynamic Fulfillment) today. VECTOR AND METRIC PROPERTIES of Rn 171 22. It is ideal for students with a solid background in single-variable calculus who are capable of thinking in more general terms about the | {
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Using ListPlot3D[data, Mesh -> {0, Length[data]}] gives
Edit:
Another option is to use (from http://mathgis.blogspot.com/2009/02/howto-display-2d-plot-in-3d.html)
to3d[plot_,height_,opacity_]:=Module[{newplot}, newplot =
First@Graphics[plot];newplot=N@newplot /. {x_?AtomQ,y_?AtomQ}->{x,y, height} ;
newplot /. GraphicsComplex[xx__]->{Opacity[opacity], GraphicsComplex[xx]}];
Then
Show[Graphics3D /@
MapIndexed[to3d[ListPlot[#1, Joined -> True], First@#2, 0] &, data],
BoxRatios -> {2, 1, 2}]
-
He wants lines/contours, not a surface. More or less like the MATLAB picture he has... – Ｊ. Ｍ. Jan 18 '12 at 14:54
You can set the lines with the Mesh option. But, I'll suggest another approach, also. – Eli Lansey Jan 18 '12 at 14:59
Something like this modification of Eli's post
ListPlot3D[data, Mesh -> {0, Length[data]}, PlotStyle -> Opacity[0]] | {
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A red ace and two black cards that are not aces: Choose one of the two red aces and two of the $$24$$ black cards that are not aces.
$$\binom{2}{1}\binom{24}{2}$$
Thus, the number of favorable cases is $$\binom{2}{1}\binom{24}{2} + \binom{2}{1}\binom{24}{1}\binom{24}{1} + \binom{2}{1}\binom{24}{2}$$ so the probability of drawing at least two black cards and exactly one ace when three cards are drawn randomly from a standard deck is $$\frac{\dbinom{2}{1}\dbinom{24}{2} + \dbinom{2}{1}\dbinom{24}{1}\dbinom{24}{1} + \dbinom{2}{1}\dbinom{24}{2}}{\dbinom{52}{3}}$$
You counted each case in which a black ace and two black non-aces were drawn twice, once for each way of designating one of the black non-aces as the black non-ace. Notice that $$\color{red}{\binom{2}{1}}\binom{2}{1}\binom{24}{2} + \binom{2}{1}\binom{24}{1}\binom{24}{1} + \binom{2}{1}\binom{24}{2} = \color{red}{2808}$$ | {
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} |
quantum-mechanics, hilbert-space, operators, quantum-optics
\cos\theta & e^{i\varphi}\sin\theta \\ -e^{-i\varphi}\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \hat{a}_0 \\ \hat{a}_1 \end{pmatrix},
\tag{1}\label{eq1}
\end{equation}
where, for example, the selection of parameters $\varphi = 0$, $\theta = \pi/2$ would give us a balanced beam splitter of real coefficients. For more information about the math behind beam splitters, one can check the book by Christopher C. Gerrry and Peter L. Knight titled Introductory Quantum Optics, in particular section 6.2 - Quantum mechanics of beam splitters. | {
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quantum-mechanics
For the case of antiparallel spins, we get a total spin of $s = 0$. Thus, the magnetic quantum number is $m_s = 0$. This corresponds to the singlet state.
It is given by
$|0,0\rangle = \frac{1}{\sqrt{2}}(\xi^+ (1) \xi^- (2) - \xi^- (1) \xi^+ (2))$.
So far, all seems clear to me, but I can't find a proof that the singlet and triplet states of Helium are eigenstates of $\hat S^2$ and $\hat S_z$.
Could anyone of you help me, please? So just thinking about Spin and not the helium atom for the moment,
$$S=S_1+S_2, \quad S^2=S_1^2+2S_1\cdot S_2+S_2^2, \quad S_z=S_{1z}+S_{2z}$$
We can also use that $S_+=S_x+iS_y$ and $S_-=S_x-iS_y$ that
$$S_1\cdot S_2=S_{1z}S_{2z}+\frac{1}{2}\left(S_{1+}S_{2-}+S_{1-}S_{2+}\right)$$
Anyway what will be the old spin basis will be common eigenvectors of the commuting observables $$\{S_1^2, S_2^2,S_{1z},S_{2z}\}$$ | {
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species-identification, entomology, lepidoptera
I tried to label the pictures as best I could, you may have to look closely. You can only see a tiny portion of the moth wing in the first picture, below.
The yellow thing was quite sticky but not as hard as bubble gum, as it stopped the moth from flying away. It was frantically flapping its wings trying to get away from some ants, which seemed to be very attracted to the yellow thing. At first I thought the swarm of ants were eating the dead moth, but the moth was alive and the ants were more interested in the yellow sticky thing. I got a thin stick and pulled away some of the sticky stuff and the moth was free enough to crawl away, taking the yellow sticky thing with it, leaving a thin trail behind, like when you stretch out bubble gum. It seemed to be very sticky to the moth and my stick, and very tensile, because unlike a spider's web, I was able to draw it out more when I tried to pull it away from the moth to free it. | {
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electromagnetic-radiation, astrophysics, astronomy, spectroscopy
photons leaving stellar surfaces from being emitted as electrons rejoin the nuclei of cooled plasma
cosmic ray and magnetic field interactions
black body radiations (brown dwarfs, dead stars, non-nuclear reaction bodies)
left over microwave cosmic background radiation from big bang
radio waves from pulsars | {
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} |
homework-and-exercises, forces, fluid-dynamics, pressure, fluid-statics
But, I think my reasoning is wrong here. Because for question 10, 7.5 = 1.5 * 5, so 1.5* 200 should be 300kPa. The answer they have is 250kPa. Can someone help me out here?
For question 11, why is the answer 600kPa? Since 500kPa is 5x 100kPa, but then 1N is 1/5 of 5N, should it not be 100kPa (the original) times 5 then divided by 5 to give 100kPa, ie, no change?
Incidentally, the questions says "7.5 N of PRESSURE". Is that even correct? Shouldn't it be 7.5 N of FORCE? In these questions we are told that the mass of gas is fixed. We must assume that the temperature of the gas is also constant while the pressure and volume are varied. Thus we apply Boyle's Law which says that pressure x volume is constant for a fixed mass of gas at constant temperature. | {
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quantum-mechanics
We're interested in the shift in the ground state, that is the $n=1, l=0$ (and therefore $m=0$) state of the hydrogen atom. We write $\phi^{(0)}_n$ for the $n$-th $l=0$ eigenstate of $H_0$. By direct computation, the first order energy shift vanishes. The second order energy shift is given by
$$E^{(2)}_{1}=\sum_{m, m\neq 1}\frac{|\langle\phi_m^{(0)}|H_1|\phi_1^{(0)}\rangle|^2}{E_1^{(0)}-E_m^{(0)}}$$
This should be compared with OP's eq. $(1)$. However this can be written in a different form which can be more suited to our uses. The first order correction to the ground state wavefunction is
$$\phi^{(1)}_1= \sum_{m, m\neq 1}\frac{\langle\phi_m^{(0)}|H_1|\phi_1^{(0)}\rangle}{E_1^{(0)}-E_m^{(0)}}\phi_m^{(0)}$$
so that the corrected wavefunction (to first order) would be $\phi^{(0)}_1+\phi^{(1)}_1$. Therefore one can also write the second order energy shift as
$$E^{(2)}_{1}=\langle\phi_1^{(0)}|H_1|\phi_1^{(1)}\rangle$$ | {
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# Find the correlation coefficient
In studying the relation between the two variables $x$ and $y$ , if the equation of the regression line of $y$ on $x$ was $$y=0.421x+0.67$$ and the equation of the regression line of $x$ on $y$ was $$x=1.58y+3.9$$ \Find\ \ The linear correlation coefficient between $x$ and $y$ My solution is $$r= \pm\sqrt{0.421\times 1.58}= \pm0.8155$$ Does my solution correct or i would not take the negative value into account ?
• How did you arrive at that solution? – Jon Bown May 22 '18 at 22:40
• From the formula of the regression coefficient if we multiply the two equations of the regression coefficient in the two cases we will get the square of the formula of the linear correlation coefficient @JonBown – Hussien Mohamed May 22 '18 at 22:43
You are right, but you can find the sign too.
The correlation coefficient of $x$ and $y$ is $\dfrac{\operatorname{cov}(x,y)}{\sqrt{\operatorname{var}(x)\operatorname{var}(y)}}$. | {
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algorithms, graphs, clique
Title: Understanding CLIQUE structure I am working on the following problem:
Recall the definition of a complete graph Kn is a graph with n vertices such that every vertex
is connected to every other vertex. Recall also that a clique is a complete subset of some graph.
The graph coloring problem consists of assigning a color to each of the vertices of a graph such
that adjacent vertices have different colors and the total number of colors used is minimized. We
define the chromatic number of a graph G to be this minimum number of colors required to color
graph G. Prove that the chromatic number of a graph G is no less than the size of a maximal
clique of G.
So far, I have been thinking about the problem and came up with the following: | {
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Third generation: . $v\cdot A^2 \;=\;(0.4,0.6)\begin{pmatrix}0.7 & 0.3 \\ 0.5 & 0.5\end{pmatrix}^2 \;=\; (0.616, 0.384)$
. . Working women: 61.6%
(c) In the long run what percentage of French women will work?
We want the "steady state" vector.
This is a vector $u \:=\:(p,q)$ such that: . $u\cdot A \:=\:u$
So we have: . $(p,q)\begin{pmatrix}0.7&0.3\\0.5&0.5\end{pmatrix} \;=\;(p,q) \quad\Rightarrow\quad \begin{Bmatrix}0.7p + 0.5q \:=\:p \\ 0.3p + 0.5q \:=\:q \end{Bmatrix}$
Then we have: . $\begin{array}{ccc}\text{-}0.3p + 0.5q &=& 0 \\ 0.3p - 0.5q &=& 0 \end{array}\quad\hdots$ .
These two equatons are always equivalent.
The second equation is always: . $p + q \:=\:1$
Solve the system: . $\begin{array}{ccc}0.3p - 0.5p &=&0 \\ p + q &=& 1 \end{array} \quad\Rightarrow\quad (p,q) \:=\:(0.625, 0.375)$
Eventually, 62.5% of the women will be working.
3. 10ks sooo verry much!!! | {
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general-relativity, lagrangian-formalism, metric-tensor, variational-calculus
Either the exact form of $V(B^2)$ is known and you can simply take the variation with respect to the $B^2$-dependency in that potential.
More general, you can use use the chain rule as follows
\begin{equation}
\frac{\delta V(B^2)}{\delta B^2} \frac{\delta B^2}{\delta g^{\mu \nu}}
\end{equation}
The same strategy can be used to vary to $\delta B^\mu$
The chain rule for functional derivatives actually implies an integral, but I am not sure whether this is needed for the variation or not:
\begin{equation}
\frac{\delta F[X]}{\delta Y(r)} = \int dz \frac{\delta F[X]}{\delta X(z)} \frac{\delta X(z)}{\delta Y(r)}
\end{equation} | {
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classical-mechanics, energy-conservation
Since there is no net force, the ball is displaced without accelerating; thus all the work goes into the potential rather than kinetic energy: $W = U = mgh.$ Next Galileo releases the ball, and acting now solely under the preexisting force of gravity, it rolls down the incline. During this stage gravity does work on the ball. Again overall energy is conserved, the work representing a transfer from potential to kinetic energy. Finally, when the ball rolls back up an incline to the original height, energy is still conserved. Here work is done against gravity; it represents a transfer back from kinetic to potential energy. | {
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cameras, usb, stereo-vision
Title: Interfacing high-resolution image sensors with ARM Board I'm working on a project requiring HD (Stereo) Video Processing. Most of High Resolution (5MP+) Sensors use MIPI-CSI interface.
I managed to get a board with an Exynos5 SoC. The SoC Itself has 2 MIPI-CSI2 interfaces, the problem is that the pins to those interfaces are not exposed and It's (almost) impossible to reach them. So I decided to use the USB3.0 Buses.
The problem is when I get to Significant bandwidth (~5.36 Gibibits/s per sensor), I don't think USB3.0 will work out. Bandwidth = Colordepth*ColorChannels*PixelCount*FPS but this could be solved with a Compressed stream (via a coprocessor)
I was thinking that Cypress' CYUSB306X chip was a good candidate for the job, but one of the problems is that I can't do BGA Soldering by hand nor have been able to find a BGA Soldering Service in Switzerland.
Any Ideas on other interfaces I could implement or other coprocessors with MIPI-CSI2 Interface? | {
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c, file
Title: Counting the number of lines in a file I'm easing back into C after several months of using Python and Matlab. Can anyone give me some pointers on how to improve my code's efficiency and readability?
#include <stdio.h>
#define MAX_ARG 2
int count( char FNAME[] );
int main( int argc, char *argv[]) {
if( argc > MAX_ARG ) {
printf("Too many arguments ( %d ) supplied.\n", argc);
return 0;
}
int FILE_LENGTH = 0;
FILE_LENGTH = count( argv[1] );
printf("The length of %s is: %d lines.\n", argv[1], FILE_LENGTH);
return 0;
} | {
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c, homework, ai, connect-four
/* Function check_diagRD() */
bool check_diagRD(int board[][BOARD_SIZE_VERT], int last_move)
{
int r; //row index
int c; //column index
for (r = 0; r < 3; r++)
{
for (c = 0; c < 4; c++)
{
if (board[c][r] != 0 &&
board[c][r] == board[c+1][r+1] &&
board[c][r] == board[c+2][r+2] &&
board[c][r] == board[c+3][r+3])
{
return true;
}
else continue;
}
}
return false;
}
/* Function check_diagRU() */
bool check_diagRU(int board[][BOARD_SIZE_VERT], int last_move)
{
int r; //row index
int c; //column index | {
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logic, prolog
For an arbitrary (pure) Prolog program, the end result is that we have a collection of rules with a list of premises which consist of atomic formulas entailing a single atomic formula. All connectives can be eliminated. If you don't like this, you can easily go from here to the program corresponding to a series of axioms of the form $\forall\vec{X}.(p_1(\vec{X})\land\cdots\land p_n(\vec{X}))\to q(\vec{X})$, which are Horn clauses, or curry this to get rid of $\land$ producing $\forall\vec{X}.p_1(\vec{X})\to(\cdots(p_n(\vec{X})\to q(\vec{X})\cdots)$. This would mean you only need to understand universal quantification and implication, but even then neither ever occurs in a negative position, so you don't need their full generality which is part of why they can be eliminated. If you want to inject classical logic unnecessarily, you can use material implication and write $\forall\vec{X}.\neg p_1(\vec{X})\lor\cdots\lor\neg p_n(\vec X)\lor q(\vec X)$, and, instead of Cut below, we'd talk | {
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Aug 29, 2018
#4
+21848
+4
A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies.
What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?
Formula:
$$\begin{array}{|rcll|} \hline \gcd(a,b,c) &=& \gcd( \gcd(a,b), c ) \\ \gcd(a,b ) &=& \gcd (a-b,b ) & a \ge b \\ \gcd(a,a) &=& a & a \ge 0 \\ \gcd(a,b) &=& \gcd(b,a) \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ && \gcd(30,24) \\ && = \gcd(30-24,24) \\ && = \gcd(6,24) \\ && = \gcd(24,6) \\ && = \gcd(24-6,6) \\ && = \gcd(18,6) \\ && = \gcd(18-6,6) \\ && = \gcd(12,6) \\ && = \gcd(12-6,6) \\ && = \gcd(6,6) \\ && = 6 \\\\ \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ &=& \gcd( 6, 12 ) \\ &=& \gcd( 12, 6 ) \\ &=& \gcd( 12-6, 6 ) \\ &=& \gcd( 6, 6 ) \\ &=& 6 \\ \hline \end{array}$$
Aug 30, 2018
edited by heureka Aug 30, 2018 | {
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"url": "https://web2.0calc.com/questions/what-is-a-number-12-24-and-30-have-in-common"
} |
-
Well, given the second result, the number $23$ should be seen as the least integer $n$ satisfying ${n \choose 2} \geq 253$. I think that the fact that the least such number ($23$) actually attains this bound is a coincidence, but I'm not sure. (What would happen if a year has $366$ or $367$ days?) – TMM Mar 12 '13 at 13:14
The first problem: $23$ comes as the answer to "the smallest integer $n$ such that $n!\binom{365}{n} / {365^n} < \frac12$". Then the $253$ comes as $\binom{23}{2}$.
The second problem: $253$ comes as the answer to "the smallest integer $m$ such that $\left(\frac{364}{365}\right)^m < \frac12$."
So the first $253$ is necessarily a number of the form $\binom{n}{2}$ (a triangular number), while the second $253$ can be any integer whatsoever. In that sense, it is a coincidence that they happen to be exactly equal (that the answer to the second problem happens to be a triangular number). | {
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1. ### Math
Let ABCD be a parallelogram. Let M be the midpoint of AB and N be the midpoint of AD. Diagonal BD intersects CM and CN at P and Q, respectively. Find PQ/BD.
2. ### Finding the midpoint
Find the midpoint of the line segment joining the points P1 and P2 P1=(3,2) P2=(-5,7) The midpoint is=
3. ### Geometry
J is the midpoint of Gk, H is the midpoint of GJ and I is the midpoint of Hj. Suppose GK=64. Find the measure of HJ
4. ### Math
If D is the midpoint of AC, is the midpoint of AB,and BD = 12 cm, what is the length of AB ? Would AB=24 Cm?Thanks
1. ### Geometry
Geometric Proofs.... Write a justification for each step Given: AB = EF, B is the midpoint of Line AC and E is the midpoint of Line DF. 1. B is the midpoint of Line AC and E is the midpoint of Line DF. 2. Line AB is congruent to
2. ### Maths | {
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"url": "https://www.jiskha.com/questions/595824/if-xc-cy-then-c-is-the-midpoint-of-xy"
} |
java
Setters have a specific and well defined functionality
The intended purpose of setter methods is to discard whatever existing value the object had in the specific field and replace it with the provided value. Your setter method breaks this contract as it adds values to the exits-list instead. Thus the setter should be named something else, such as addExits(...). | {
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python, database
#upsert and merge perform the same task, having the same end result.
#The difference is that the former is optimised to work on data where usually little new rows are added
#while the latter is optimised in the case the majority of the data dealt with will be added, not already existing
def upsert(self, table, values, where):
self.update(table, values, where);
if not self.affected_rows():
self.insert(table, [values] + [where]);
def merge(self, table, values, where):
try:
self.insert(table, [values] + [where]);
except self._module.DatabaseError as e:
#TODO: Check error in case it's not due to a PK/Unique violation
self.update(table, values, where);
#Returns a row, instead of simply a list of 1. Inspired by Wordpress
def get_row(self, table, columns = None, where = None, op = "AND"):
r = self.select(table, columns, where, op);
return r[0] if r else None; | {
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quantum-mechanics, hilbert-space, heisenberg-uncertainty-principle, quantum-interpretations, superposition
Title: Are superposition and uncertainty principles logically dependent? If we assume superposition and define an Hilbert space with canonical commutation relations we can derive uncertainty relations. So it seems the uncertainty principle isn't required, or should be called the "canonical commutators principle".
On the other direction, if we would like to start assuming the uncertainty relations, it would be tricky even to give them a precise meaning, not having yet a phase space.
So is it safe to assert that uncertainty is called a principle maybe for some historical reason but it's hard to give it that role from a logical point of view? Yes, uncertainty relations are a theorem in nowadays approach to quantum mechanics, and the word principle is the remnant of a different perspective at some stage of the construction of the conceptual framework of QM. | {
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quantum-mechanics, hilbert-space, quantum-states
States are continuous functionals of the C*-algebra, associating to each observable its expected value in that particular system's configuration.
Now, every C*-algebra can be represented as an algebra of (bounded) operators acting on a Hilbert space. Such representations can be constructed starting from each state $\omega$ by a procedure called GNS conctruction. In the GNS construction built on $\omega$, the latter is identified with a ray $[\psi_\omega]$ in the corresponding Hilbert space $\mathscr{H}_\omega$. Every other ray $[\psi]$ also defines a (unique) state for the C*-algebra, as well as density matrices of the form
$$\sum_{i\in\mathbb{N}} \rho_i \lvert \psi_i\rangle\langle \psi_i\rvert$$
with $\sum_{i\in\mathbb{N}} \rho_i=1$. | {
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fourier-transform, hilbert-transform, proof
$$\hat{F}(f) = -j\text{sgn}(f) \cdot \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c}}{2} + \frac{\Pi(f+\nu_{c})|}{2} ~\right] $$
$$ \hat{F}(f) = \frac{1}{\left|a\right|} \left[~\frac{\Pi(f-\nu_{c})}{2j} + \frac{-\Pi(f+\nu_{c})|}{2j} ~\right] ~~ \text{Remember} -j = \frac{1}{j}$$
which is same as
$$ \hat{F}(f) = \frac{\Pi(f)}{\left|a\right|} \ast \left[~\frac{1}{2j} \delta(f-\nu) + \frac{-1}{2j} \delta(f+\nu) ~\right] $$
Inverse Fourier Transformation of this gives you
$$\hat{f}(t) = \text{sinc}(at)\sin{\left(2\pi\nu_{c}t\right)}$$
Hope this helps! | {
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special-relativity, inertial-frames, lorentz-symmetry
For $i=1$, I end up with a boost $B_i$ of the form:
$$B_i=
\left(
\begin{matrix}
\cosh \theta & -sinh \theta & 0 & 0\\
sinh \theta & cosh \theta & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{matrix}\right)
$$
when I should end up with:
$$B_i=
\left(
\begin{matrix}
\cosh \theta & sinh \theta & 0 & 0\\
sinh \theta & cosh \theta & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{matrix}\right)
$$
Edit: The columns of this matrix are correct because the matrix satisfies Lorentz transformation condition $\eta = B_1 ^T \eta B_1$ (where $\eta$ is the minkowski metric tensor) while my boost matrix doesn't.
Question : Since taking the square might make me lose/add a negative sign, is there a better way to obtain $e_0'$ and $e_1'$ with all the necessary negative signs in its components? Right now I would just be using trial an error, until the boost matrix I obtain satisfies $\eta = B_1 ^T \eta B_1$ | {
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homework-and-exercises, thermodynamics, statistical-mechanics, condensed-matter, solid-state-physics
Edit 2:
nephente, are you saying then that this is the correct partition function?
$$Z = \sum_{\mu_s}{e^{-\beta E_s}} = {N_s \choose N} \sum_{m = 0}^N {N\choose m} e^{-\beta N E_0} e^{-\beta m (E_1-E_0)} = {N_s \choose N} e^{-\beta N E_0} (1 + e^{-\beta m (E_1-E_0)})^N$$
$$ Z = {N_s \choose N} (e^{-\beta E_0} + e^{\beta (m-1)E_0}e^{-\beta m E_1})^N$$
Also, would you mind explaining why we need to sum on $m$ some more? The number of particles with energy $E_1$ is $0\leq m \leq N$.
There are $N_S\choose N$ possible ways to arrange $N$ particles on $N_S$ sites.
There are $N \choose m $ ways to chose $m$ out of $N$ particles to occupy the upper level $E_1$. The other $N-m$ automatically have to occupy the low-energy state $E_0$, so there are no further combinatorics involed. | {
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php, html, mysqli
<label>Username</label>
<input type="text" name="username"><br>
<label>Password</label>
<input type="password" name="password"><br>
<label>Re-enter Password</label>
<input type="password" name="confirm_password"><br>
<button type="submit">Create Account</button>
</form>
</center>
</body>
</html> | {
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python, python-3.x, gui, pyqt
'Upload labels from csv, hdf',
'Save changes to csv or hdf',
'Save changes to txt files with Yolo format',
'Open a folder from the last saved point or open a new one containing '
'photos and add them to the photo list',
'Add a video and convert it to .png frames and add them to the photo list',
'Activate editor mode',
'Delete all selections(checked items)', 'Delete all labels in the current working folder',
'Display settings', 'Display help']
tips = [f'Press ⌘⇧{key}: ' + tip for key, tip in zip(keys, tips)]
key_shorts = [f'Ctrl+Shift+{key}' for key in keys]
check_status = [False, False, False, False, False, False, False, True, False, False, False, False]
assert len(names) == len(icons) == len(methods) == len(tips) == len(key_shorts)
for name, icon, method, tip, key, check in zip(names, icons, methods, tips, key_shorts, check_status): | {
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of heads and…. The number of ways you can have 7 heads in 10 flips is (10 choose 7). A box contains ten apples, three of which are bruised. There are 10 coin flips total. Write the word or phrase that best completes each statement or answers the question. In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head. For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,? What is the probability of tossing a coin 5 times and getting 2 tails and 3 heads in that order?. Tossing a fair coin 10 times, the probability of getting exactly 5 heads (in. In a poll 37% of the people polled answered yes to the question are you in favor of the death penalt. In the end it all comes to a 50/50 somy question is: if i toss a coin and get four heads in a row, does the fifth toss has a 50/50 chance of landing heads/tails. If we throw the coin three times, the possible results are: Three heads: | {
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} |
human-biology, physiology, human-physiology, health
It clearly shows the anti-inflammatory effects and increased leukocytosis of the breath work (intermittent hypoxia) and also shows that blood pH correlates to the respiratory rate, more precisely blood CO2 levels as expected. Those are normal blood gas dynamics, nothing woo woo pH stuff and no serious scientist would attribute the health effects of the method to the temporary blood pH fluctuations seen from a short hyperventilation phase, especially not when far more dominant and better documented effects on hormonal adaptations can be discussed instead. Some practitioners get quite enthusiastic about urine pH measurements, which are VERY high after the breath work, however all this indicates is normal kidney function in response to temporary respiratory alkalosis, nothing fancy at all. The body increases bicarbonate outlet via the kidneys (renal compensation) when blood CO2 gets too low (hyperventilation) to maintain blood pH homeostasis. the pH strips measure this bicarbonate. | {
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matlab, sampling, convolution, impulse-response, speech
See the attached plots. In the last plot, around 20 s, there appears the anechoic signal:
ETC plot of IR:
When performing convolution, why do I need to divide the convolved signal with sampling frequency as I do in my code (sig_conv_meas = cconv(y_anech_resampled,y_IR)/fsIR)?
Thank you! I have no access to your audio files so I've downloaded:
IR from here (mono/r1_omni.wav) - it's a really long one
Anechoic recording from here (operatic-voice/mono/singing.wav)
Resampled voice signals:
Final convolved signal: | {
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cosmology, cosmic-microwave-background, doppler-effect
The motion of the Earth around the Sun is a modest 30 km/s, but even this doppler shifts inferred temperatures by a factor $\simeq v/c$ and needs to be taken out of the CMB analysis. A larger effect is the motion of the Sun with respect to the Galaxy/local group and with respect to the Hubble flow around us, which is 10 times bigger. This makes one hemisphere of the CMB about 1 part in a thousand hotter than the hemisphere the Earth is receding from.
To make the CMB visible we need the blackbody radiation temperature to be $>3000\ K$, requiring a blueshift of a factor 1100 from the current $2.7\ K$. This puts the peak of the Planck function at a wavelength of 966nm, outside the visible range, but the Wien tail of the distribution should be comfortably visible.
Using the relativistic redshift formula:
$$ \frac{c + v}{c-v} = 1100^2$$
gives a required velocity of $0.99999835c$. | {
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javascript
'elementBase' : function() {
var processElement = function(element, callback) {
if (typeof (element.length) !== 'undefined') {
var length = element.length;
for ( var i = 0; i < length; i++) {
var singleElement = element[i];
callback(singleElement);
}
} else {
callback(element);
}
};
return {
'processElement' : processElement
};
},
/* To show or hide element. Call displayElement.hideElement to hide and displayElement.showElement to show element.
Default display class is block to show element but one can pass display class name as third parameter in displayElement.showElement */
'displayElement' : {
'previousDisplayStatus' : {},
'defaultDisplayClass' : 'block', | {
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require two!: the set of all possible linear combinations of its row vectors x, y, z −2! Linear equation system Using Determinants understand the equivalence between a system of equations having 2 is! Occurring situations the subject of this is to notice that systems of differential equations is not subject. This chapter = −2 notice that systems of differential equations can arise quite easily naturally., an augmented matrix, or the fundamental matrix solution values of x, y = 3, z −2... Enter coefficients of your system into the input fields it exists! ( x ( t \! Be consistent example 1: solve the equation by the matrix valued function \ ( (. Vector equation, and a 2 x+b 2 y+c 2 = 0 { th }... 3, z = −2 graph of pair of linear equations is not the subject this... A case, the pair of linear equation with the formula and find the inverse of span! X, y = 3, z = −2 equations can arise easily... To sketch the graph of pair of linear equations is not the subject of this is to notice systems. | {
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"tags": null,
"url": "http://pobarabanu.com.ua/414yquca/elnfm.php?57d14b=system-of-linear-equations-matrix-conditions"
} |
entanglement
$$|\Psi_{\pm}\rangle\propto|g r\rangle \pm |r g\rangle.$$
However, he further mentioned that this is partially evidenced by the oscillations in the probability of detecting 0 and 1 atom in the Rydberg state. Nevertheless, such oscillations are not sufficient to confirm whether or not we have created an entangled state. To demonstrate entanglement, we need to measure the relative phase between the components, not just the overall populations. To achieve this in the experiment, we can introduce a differential phase shift to one atom relative to the other by applying an additional laser field (via the AC Stark effect). If this phase shift is applied when the atoms are in an entangled state, it transforms the system to a state like
$$|\Psi_+\rangle \propto|g r\rangle + e^{i \phi}|r g\rangle.$$ | {
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quality-engineering, computer-engineering
the problem exists as long as possible and then "writing software" to fix it. This is no doubt a painful learning experience. The project manager is unlikely to have the same problem again as (one can guess) he probably now works somewhere else, and Qualcomm will no doubt take greater notice of their QC department. | {
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graphs, time-complexity, graph-traversal, dag, topological-ordering
Then, if $v^*$ maximizes $d(v^*)$, the sought path is:
$$
\underbrace{
v^*, \pi(v^*), \pi(\pi(v^*)), \pi(\pi(\pi(v^*))), \dots}_{d(v^*)+1\mbox{ times}}
$$
The whole algorithm can also be formulated in terms of boxes, doing away with the graph $G$. The main observation is that if box $b_1$ stacks into box $b_2$ then the volume of $b_1$ must be smaller than the volume of $b_2$. As a consequence, it suffices to consider boxes in non-decreasing order of volume. | {
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c#, animation
pictureBox10.Image = Properties.Resources.image2;
break;
case 5:
pictureBox3.Image = Properties.Resources.image5;
pictureBox18.Image = Properties.Resources.image1;
pictureBox18.Image = Properties.Resources.image1;
pictureBox17.Image = Properties.Resources.image2;
break;
case 6:
pictureBox3.Image = Properties.Resources.image6;
break;
case 7:
pictureBox3.Image = Properties.Resources.image7;
break;
case 8:
pictureBox3.Image = Properties.Resources.image8;
break;
} | {
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thermodynamics, navier-stokes
Zero velocity, $\vec{v}=0$,
Zero viscosity, $\eta=0$,
I end up with the following heat equation
$$
\begin{align}
\frac{\partial (\rho c_v T)}{\partial t} = \left( \lambda T_{x} \right)_{x}
+ \left( \lambda T_{y} \right)_{y}
+ \left( \lambda T_{z} \right)_{z},
\end{align}
$$
and not
$$
\begin{align}
\frac{\partial (\rho c_p T)}{\partial t} = \left( \lambda T_{x} \right)_{x}
+ \left( \lambda T_{y} \right)_{y}
+ \left( \lambda T_{z} \right)_{z}.
\end{align}
$$
as given in most literature, see Wikipedia.
Any hints are appreciated, thanks! Let me quote Landau & Lifshitz: Fluid Mechanics, $\S 50$, page 188, $3^{rd}$ed: | {
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is a subset of the domain of the second function. It is best to first sketch the graphs of the 2 parts of this function on the same graph. This is unnecessary and I hope you will agree with me, after seeing the next five videos, that it is equally as hard/easy as evaluating a function for a numerical value for x. I first need to plug in function h into function g then simplify to get a new function. 3.3 DERIVATIVES OF COMPOSITE FUNCTIONS: THE CHAIN RULE1 3.3 Derivatives of Composite Functions: The Chain Rule In this section we want to nd the derivative of a composite function f(g(x)) where f(x) and g(x) are two di erentiable functions. Let's take a few values. We use cookies to improve your experience on our site and to show you relevant advertising. Function becomes the input for another way to master Piecewise and composite functions of composite functions ihsankhairir... ) and g ( x ) from known functions f and g ( x ) is one in which output! Straightforward process, and you will | {
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electromagnetism, magnetic-fields, magnetic-moment
Title: Is the total magnetic moment of an object equal to the sum of the magnetic moments of the parts that compose it? For example, given a spherical shell with charge $Q$ rotating on its axis, can I think of it as the composition of many concentric coils and integrate the infinitesimal magnetic moments $d\vec \mu$ to calculate the total $\vec \mu$? If so, why is that? It seems so simple but I'm having a hard time justifying it. You can formalise the approach generally. What is unambiguously defined is the current density $j$. By looking at the multipole expansion, the dipole moment is:
$$
m = \frac12\int x\times j d^3x
$$
Technically, it should be the tensor by analogy with electrostatic:
$$
m^{(2)} = \int x\otimes j d^3x
$$
However, due to the conservation of charge $\nabla\cdot j= 0$, $m^{(2)}$ is antisymmetric, so you can identify it as vector with the cross product. | {
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electrostatics, electric-fields, charge, potential, vectors
We are comparing work on green path and red one. Lets draw two circles with centre at $dq$, one of arbitrary radius $r$ and the second with slightly larger radius $r+dr$. They crosses green path with $AB$ and red path with $CD$. You may have noticed, that $CD = dr$, while $AB$ is longer, since it isn't parallel to radius going through $A$ to $O$. Considering $dr << r$, the work done on $AB$ is $E_x \cdot AB = E\cos\theta \cdot AB = E \cdot AO = E \cdot dr$ which is the work on $CD$ ($\theta$ denotes $\measuredangle BAO$). So both paths can be divided into intervals of the same work, thus full work on them is the same as well. | {
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Example: For the system of the previous example
It is easy to see that A=L U. Also det(A)=det(U)=-1
Pivoting and Scaling in Gaussian Elimination
At each stage of the elimination process given above, we assumed the appropriate pivot element . To remove this assumption, begin each step of the elimination process by switching rows to put a non zero element in the pivot position. If none such exists, then the matrix must be singular, contrary to assumption.
It is not enough, however, to just ask that pivot element be nonzero. Nonzero but very small pivot element will yield gross errors in further calculation and to guard against this and propagation of rounding errors, we introduce pivoting strategies.
Definition: (Partial Pivoting). For in the Gaussian elimination process at stage k, let
Let i be the smallest row index, , for which the maximum is attained. If then switch rows k and i in A and b, and proceed with step k of the elimination process. All the multipliers will now satisfy | {
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classical-mechanics, hamiltonian-formalism, coordinate-systems, phase-space, poisson-brackets
V.I. Arnold, Mathematical Methods of Classical Mechanics, 2nd eds., 1989; See $\S$44E and footnote 76 on p. 241.
--
$^1$ Although Ref. 1 and Ref. 2 don't bother to mention this explicitly, it is implicitly assumed that the map (1) is a sufficiently smooth bijection, e.g., a diffeomorphism [which depends smoothly on the time parameter $t$]. Similar smoothness conditions are implicitly assumed about $H$, $K$, and $F$. | {
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ubuntu, ros-fuerte, ubuntu-precise, openi-tracker, nite
cd ~/Downloads/NITE-Bin-Dev-Linux-x64-v1.5.2.21
sudo ./install.sh
Originally posted by isura with karma: 403 on 2013-01-13
This answer was ACCEPTED on the original site
Post score: 11
Original comments
Comment by Nash on 2013-03-06:
Thanks!, your suggestion of reinstalling Nite v1.5.2.21 worked!
Comment by ctguell on 2013-09-12:
I Did exactly what you suggestd but i still have the same problem, do you have any idea why?
Comment by isura on 2014-01-19:
Was your problem resolved @ctguell?
Comment by Gloria@HK on 2014-07-03:
I come across with the same problem with hydro,12.04. Now it works!! Thanks~ | {
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java, object-oriented, tic-tac-toe
Choose good variable names – Most of the time you've chosen good variables name, but then something like temp or c occurs. A better alternative for the first one here would be currentPlayer
Reconsider some of the function names – This a minor, but to me some of the functions are not intuitive. Like the checkXxxx, what do you check for? A row? or a winning row? or ??? And promptTurn would be better with something like readNextMove. Likewise with isAllowed, what is allowed? A better name (and close to standard naming) would be isEmpty | {
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formal-languages, regular-languages, finite-automata
Title: Index of a language and its reversal The index of a language is the number of Myhill-Nerode classes that it has. It is also equal to the number of states in the minimal DFA for the language. What is an example of a language that has a different index from its reversal? The $k$th letter equals $a$, for fixed $k$, assuming a two letter alphabet.
This generic answer is made precise in the comment by @AntonTrunov below. | {
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machine-learning, comparison, terminology, ai-field
Is one a subset of another?
Is one a tool used to build a system for the other?
What are their differences and why are they significant? Machine learning has been defined by many people in multiple (often similar) ways [1, 2]. One definition says that machine learning (ML) is the field of study that gives computers the ability to learn without being explicitly programmed.
Given the above definition, we might say that machine learning is geared towards problems for which we have (lots of) data (experience), from which a program can learn and can get better at a task.
Artificial intelligence has many more aspects, where machines may not get better at tasks by learning from data, but may exhibit intelligence through rules (e.g. expert systems like Mycin), logic or algorithms, e.g. path-finding).
The book Artificial Intelligence: A Modern Approach shows more research fields of AI, like Constraint Satisfaction Problems, Probabilistic Reasoning or Philosophical Foundations. | {
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spectrogram, audio-processing, time-frequency, voice, feature-extraction
Superimpose "clean" recordings of your gong, clang, … randomized on randomized snippets from your conversation library. Keep these snippets short.
This has, in addition to solving the privacy chicken/egg problem, the added advantage that you can create years worth of audio computationally, while actually getting labeled data for your real environment will require someone to actually record the events. For example, if you need to detect the door bell, and roughly every 2h someone rings, and to make your detector work reasonably well, you need some 100 door ringings while different conversations are going on – well, that's going to drag out your project a bit, won't it.
Optimize a system that represents these "mixture" snippets with as few coefficients as possible, and makes a detection based on these coefficients (instead of on the snippet itself).
The thing would roughly look like this:
Audio snippet N "condensed" M "detection | {
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multiclass-classification, class-imbalance, smote
Title: Why class weight is outperforming oversampling? I am applying both class_weight and oversampling (SMOTE) techniques on a multiclass classification problem and getting better results when using the class_weight technique. Could someone please explain what could be the cause of this difference? You should not expect class_weight parameters and SMOTE to give the exact same results because they are different methods.
Class weights directly modify the loss function by giving more (or less) penalty to the classes with more (or less) weight. In effect, one is basically sacrificing some ability to predict the lower weight class (the majority class for unbalanced datasets) by purposely biasing the model to favor more accurate predictions of the higher weighted class (the minority class). | {
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java, session, hibernate
Transaction tx = session.beginTransaction();
amd.savesomething(object);
bmd.savesomething(object2);
tx.commit();
session.close();
}
I would like to know if my coding here is good enough or if there is a better method to produce the same result. Instead of working with Hibernate sessions, I encourage you to use the JPA API. I think this is the most common way to work with Hibernate now. (http://www.theserverside.com/news/2240186700/The-JPA-20-EntityManager-vs-the-Hibernate-Session-Which-one-to-use)
For the transaction aspects and the initialization of your DAOs, I also think that a solution like Spring (http://projects.spring.io/spring-framework/) would help to simplify the code of your application.
Your code would look more like the following (with annotations stuff):
public class MyService {
@Resource
private AModelDAO aDao;
@Transactional
public void something(object) {
// no need to manage transactions or sessions here
aDao.save(object);
}
} | {
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general-relativity, black-holes, event-horizon, textbook-erratum, black-hole-thermodynamics
$$\nabla^a(\chi^b\chi_b)=-2\kappa\chi^a. \tag{Wald 12.5.2}\label{Wald}$$
However, Wald later claims (below equation 12.5.15) that \eqref{Wald} implies $\nabla^a(\chi^b\chi_b)$ is not zero.
Question: Is Wald not contradicting himself by essentially using $\nabla^a(\chi^b\chi_b)=0$ to imply $\nabla^a(\chi^b\chi_b)\ne 0$ on the null surface? The quantity $\chi^2$ is constant along the horizon, which means that $v(\chi^2)=0$ for any vector $v$ in the tangent space to the horizon. This can also be written as $v^a\nabla_a\chi^2=0$, and the vector $\nabla^a\chi^2$ points along the normal direction. | {
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Last edited:
#### anemone
##### MHB POTW Director
Staff member
we know $0<\dfrac {k}{n}<1$
using $0<\dfrac {1}{2}<\dfrac {2}{3}<\dfrac {3}{4}<--<\dfrac {3995}{3996}<\dfrac {3996}{3997}<\dfrac {3997}{3998}<1$
$2\times \dfrac {m}{3996}<\dfrac {k_0}{n_0}<2\times \dfrac{m+1}{3998}<1$
take $n=n_0 ,\,\, and \,\, k=k_0$
now it is easy to see that the smallese value of n =3997
Hi Albert,
Thanks for participating and I assume you meant $$\displaystyle \dfrac {3985}{3986}<\dfrac {3986}{3987}<\dfrac {3987}{3988}$$ rather than $$\displaystyle \dfrac {3995}{3996}<\dfrac {3996}{3997}<\dfrac {3997}{3998}$$, is that true?
If that's true, according to your reasoning we would arrive to $$\displaystyle \dfrac {3985}{1993}<\dfrac {2(3986)}{3987}<\dfrac {3987}{1994}$$ and I don't think this helps much as what the original question stated is it wants the difference between the two numerators on both fractions to be 1 but $3987-3985=2$. Hmm...What do you think, Albert?
#### Opalg | {
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thermodynamics, molecules
How can we represent the disappearance of surface tension on the right side? Are molecules being stripped from the surface by collisions? Because the molecules inside the compressed liquid are still bonded right?
What would you change in this crude diagram to make it more representative of what really happens?
To give a bit more context, I'm more of a fluid mechanics guy modeling inject. So I might not be using the exact terms that I should. The terminology I use is within the context of this diagram: | {
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newtonian-gravity, atmospheric-science, fluid-statics
Title: If you released a book from the thermosphere or mesosphere would the book float slightly? If you released a book from the thermosphere or mesosphere would the book float slightly?
you are further from Earth's surface compared to outer space where
there is no gravity | {
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neural-networks, deep-learning, classification, image-recognition
For dual classification and naive approach, you will again have a problem in case there are some insects that can solely be identified from features at a specific angle, if so you can then discard other angles (e.g. ignore moth face if moth back is the distinguishing feature) | {
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cosmology, gravity, space-expansion, big-bang
Space was expanding, but this was not "overcoming" gravity. In fact, the expansion of space meant that the universe was cooling down, which assisted gravity. Like a pencil balanced on its point, the universe was in a state of unstable equilibrium, which became more unstable as it cooled.
As the universe expanded and cooled, fundamental particles combined into protons and neutrons, and then into atoms (almost all of which were hydrogen and helium atoms). This took several hundred thousand years. The very small deviations from absolute symmetry were then enough to trigger the collapse of the cooling atoms into gravitationally bound clouds, and then into the first stars and galaxies. But this process was very slow, and the first stars (called Population III stars) took hundreds of millions of years to form. | {
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structural-engineering, steel, rivets
Rivets were heated to 950°F - 1050°F, handbooks say. This is not hot enough for an austenitic transition, so it must only have been for causing thermal expansion...
Putting all of this together means that rivets start out with the material properties of standard structural steel and aren't heated enough to change those properties. | {
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beginner, c, linked-list
Before this return, if current->next is NULL, the memory could not be allocated. That is a surprising situation since the function's name promises that it adds something to the list, yet in this case it doesn't add anything. Even worse, there is no efficient way for the caller to test whether something was added or not. Therefore, it should be an error if memory allocation fails.
There are many programs that wrap malloc in a simple fail-fast xmalloc function:
void *xmalloc(size_t n)
{
void *ptr = malloc(n);
if (ptr == NULL) {
fprintf(stderr, "xmalloc: %s\n", strerror(errno));
exit(1);
}
return ptr;
}
Continuing with the rest of addAtEnd:
printf("Cannot add to NULL.");
exit(1); | {
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c#, strings, extension-methods
for(int i = 0; i < 5; i++)
{
Console.WriteLine(q.Next(3));
}
Output:
abc
def
ghi
j // 1 character
// empty string
This has significantly whittled down the complexity of the syntax when using the StringQueue. All you need to do now is specify the source string and tell it how big your chunks should be.
Further improvement
There's another feature you could add here. I'm unsure if this is needed for your current use case, but it might be nicer to use in certain scenarios where you break a string in many chunks at the same time.
By implementing method chaining and out parameters, you could create a syntax that allows you to very quickly define your chunks.
public StringQueue Next(int count, out string result)
{
result = Next(count);
return this;
} | {
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"lm_q1q2_score": null,
"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "c#, strings, extension-methods",
"url": null
} |
python, time-series, forecasting, statsmodels
TypeError:seasonal_decompose() got an unexpected keyword argument 'freq'
How do I handle the number of hours in a time series model? Do I need to convert it to a different format before using it as the index? Also, the sales column is continuous numeric value, do I need to round it off?
Thanks in advance!
The frequency parameter of statsmodels’ seasonal_decompose() method has been deprecated and replaced with the period parameter. Please use period in place of frequency.
Since the data you provided is hourly, the period should be 24. The period determines how often the cycle repeats in the seasonal component. For example, with monthly data, the period would usually be 12. With hourly, it could be 24 (daily) or 168 (weekly). This is something that you should know about your data. Or you can try alternative plausible values when doing EDA (other examples 7-daily, 12-monthly, 52-weekly). | {
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star, mass
Details
The "mean molecular weight" $\mu$ is used in calculations to relate pressure to density (and temperature) - the so-called "equation of state". An example would be the perfect gas law, which can be written
$$ P = n k_B T\ , $$
where $P$ is pressure, $T$ is temperature and $n$ is the nuber density of particles - i.e. the number of particles per cubic metre.
Note that $n$ includes ALL isolated, free particles - it does not matter whether they are protons, electrons, ions, atoms or molecules; all free particles contribute equally to the kinetic gas pressure. But note that a single atom, consisting of a nucleus with electrons is treated as a single free particle for the purposes of working out the gas pressure.
However, we often wish to express pressure in terms of mass density, since if we know the mass and size of the gaseous object, then we have a direct measure of that rather than the particle number density. | {
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haskell, parsing, scheme
_ -> throwError $ BadSpecialForm "ill-formed case expression: " form
eval env (List [Atom "set!", Atom var, form]) = eval env form >>= setVar env var
eval env (List [Atom "define", Atom var, form]) = eval env form >>= defineVar env var
eval env (List (Atom "define" : List (Atom var : params) : body)) =
makeNormalFunc env params body >>= defineVar env var
eval env (List (Atom "define" : DottedList (Atom var : params) varargs : body)) =
makeVarArgs varargs env params body >>= defineVar env var
eval env (List (Atom "lambda" : List params : body)) =
makeNormalFunc env params body
eval env (List (Atom "lambda" : DottedList params varargs : body)) =
makeVarArgs varargs env params body
eval env (List (Atom "lambda" : varargs@(Atom _) : body)) =
makeVarArgs varargs env [] body
eval env (List (function : args)) = do
func <- eval env function
argVals <- mapM (eval env) args
apply func argVals
eval env (List elems) = return $ List elems | {
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c#
That doesn't make sense.
The usual approach would be to define a _nextScheduledTick
in terms of wall clock time, and then frequent pulse()
will eventually notice the clock is slightly after
that scheduled time, triggering a tick and updating the schedule.
Up in Start I think this is the part I didn't initially notice:
_timestamp = DateTime.Now.Ticks; | {
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variational-principle
$$\frac{d}{d\alpha}|_{\alpha=0} F[y_0+\alpha \delta y]=0\:.\tag{1}$$
The reason for this maybe apparently cumbersome definition relies upon the analogous fact in standard calculus: if $f=f(x) \in \mathbb{R}$ is $C^1$ with $x \in \mathbb{R}^n$, then $x_0$ is an extremal point if and only if $$\frac{d}{d\alpha}|_{\alpha =0} f(x_0 + \alpha \delta x)=0 \quad \mbox{for every $\delta x \in \mathbb{R}^n$}\tag{2}$$
In fact, (2) is equivalent to
$$\delta x \cdot \nabla f(x_0)=0 \quad \mbox{for every $\eta \in \mathbb{R}^n$}$$
which means, because $\delta x$ is arbitrary,
$$\nabla f(x_0)=0$$
which is nothing but the standard definition of extremal point $x_0$ for the function $f: \mathbb{R}^n \to \mathbb{R}$.
Coming back to (1), if $F(y) = \int_a^b f(y(t),y'(t))dt$, the equation can be rephrased to
$$\frac{d}{d\alpha}|_{\alpha=0}\int_a^b f(y_0+ \alpha \delta y, y_0' + \alpha \delta y') dt=0,$$
that is | {
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machine-learning, tensorflow, supervised-learning, beginner
EDIT: More Information
At first i am providing the answer, which would be the "Yes" or "No" on Valid, but when i provide something like:
Row;ID;Name;Start;End;
1;1;John Doe;2015-01-01;2017-01-01;
2;1;Jonh Doe;2016-02-03;2016-06-04;
3;1;Jonh Doe;2017-06-02;2018-04-01;
4;1;Jonh Doe;1990-01-01;2017-07-01; | {
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electricity, charge
Also, although conductors when left on their own will act to spread charge out across their structure through current flow, it is still possible to polarise them (i.e. create imbalances of charge) using an external means. One such method is called Electrostatic Induction. In the picture below, moving a positively charged object near to the conductor in the electroscope pulls the electrons toward the top end, and as a result causes a local net negative charge here, and a local net positive charge at the bottom. | {
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heat
Title: Is it 11% hotter today than it was yesterday? Yesterday the temperature outside was 0.5 °C. Today, the temperature is 30 °C. 30 is 5300% more than 0.5, but today is obviously not 5300% hotter than yesterday.
In Fahrenheit, the temperatures are 33 °F and 86 °F, respectively. 160% hotter sounds more reasonable, but this argument uses the same logic as Celsius, just on a different scale.
Converting these temperatures to Kelvin, we get 273.70 K and 303.15 K, respectively. Since Kelvin is an absolute scale of temperature, can we correctly say that today is 11% hotter than yesterday? | {
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quantum-mechanics, black-holes, event-horizon
Does that sound correct? Unfortunately, your desire to avoid philosophical considerations is on a losing wicket. Specifically, the ontology or nature of the underlying reality is a philosophical issue. The baseline view of science, known in quantum physics as the Copenhagen interpretation, is that you simply ignore such issues as scientifically meaningless and concentrate on the ability of your maths to predict experimental results.
A particular feature of Relativity is that the overall causal structure of objects and events is immutable and therefore, as far as the theory is concerned, is objectively real. What is subjective for each reference frame are details like exact masses, momenta, time periods, distances and directions. | {
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gazebo-11
Title: Why are Gazebo textures and colors mixed up with background windows or black?
Hi,
I'm running Gazebo 11.5.1 and ROS Galactic on Ubuntu 20.
When I load the house model in gazebo, I get the following.
The textures and colors are not right and mixed with the visual of the other software windows that I'm running.
Any ideas on what could be causing the issue?
More details:
When I start an empty gazebo world and add objects to it, they are loaded just fine.
If I save the model and load it, the objects have weird appearance again.
If I add a new instance of an object, it appears the exact same as the one in the model with weird texture and color.
Tested it it with a fresh Gazebo build from source too but the problem is still there. This is really blocking because I'm testing a visual slam algorithm so the scene appearance is critical.
Any help would be appreciated!
Thanks,
Jas | {
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elasticity, continuum-mechanics, stress-strain
Alternatively, you could find the rotation tensor $\omega$ and integrate the system directly since the sum of the two is the complete (linearized) gradient. In 2D finding the rotation tensor is equivalent to solve the Cauchy-Riemann equations. | {
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newtonian-mechanics, forces, computational-physics, discrete
ax.set_xlim((-4, 4))
ax.set_ylim((-4, 4))
ax.set_zlim((-4, 4))
ax.plot(x[:,0], x[:,1], x[:,2], 'r-')
ax.plot(x[0,0], x[0,1], x[0,2], 'bo')
ax.plot(x[-1,0], x[-1,1], x[-1,2], 'go')
plt.show()
Starting point $[3, 0, 0]$ and direction vector $[0, 1, 1]$
Starting point $[3, 0, 0]$ and direction vector $[0, 1, 2]$
Starting point $[3, 0, 0]$ and direction vector $[0, 1, 3]$ | {
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c++, array, tree
node& data2 = data1; // create a reference.
// data2 is just another name for data1.
// manipulate data2 and data1 is affected.
node* data3 = &data1; // data3 is a pointer
// It points at the address of an object. You
// can manipulate the object being pointed
// at via the pointer.
//
// This is closer to your Java/C# types as this
// value can be NULL (meaning points at nothing)
// You can also hold pointers to dynamically
// allocated data (unlike Java/C# these must be
// manually deletes (as such we don't use them
// much and prefer to use smart pointers for
// memory management). | {
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javascript, html5, canvas, snake-game
as
//2 properly named array indexes for x and y
var X = 0;
var Y = 1;
//vectors for each direction
var vectors = {
right : { x : 1 , y : 0 },
left : { x : -1 , y : 0 },
up : { x : 0 , y : -1 },
down : { x : 0 , y : 1 }
}
function updatePosition( direction ){
var vector = vectors( direction );
if( vector ){
nextPosition[X] += vector.x;
nextPosition[Y] += vector.y;
}
else{
throw "Invalid direction: " + direction
}
}
The advantages here are:
If you wanted to play snake with 8 directions, you could
No silent failure if an invalid direction is passed along
The following code gives me the creeps:
function launchFullscreen(element) {
if(element.requestFullscreen) {
element.requestFullscreen();
} else if(element.mozRequestFullScreen) {
element.mozRequestFullScreen();
} else if(element.webkitRequestFullscreen) {
element.webkitRequestFullscreen();
} else if(element.msRequestFullscreen) {
element.msRequestFullscreen();
}
} | {
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statistical-mechanics, differential-equations, statistics, brownian-motion, stochastic-processes
EDIT: As pointed out Alexander in his comment, "Langevin equation is about stochastic processes. Averages of stochastic processes are (usually) with respect to different noise realizations (all taken from same distribution). In different language - tracing out the ensemble of stochastic environment." In other words, averaging is done in fact over some parameter independent of time. It is like averaging over a random initial phase $\varphi_0$ in my last example with a periodic external force $F_0\cdot\sin(\omega t + \varphi_0)$. Thus, it is not averaging the equation or its solution over $t\le\tau\le (t+T)$. Although mathematically it is possible, I do not fully understand how it may naturally arise in physical calculations and what meaningful conclusion one can draw from such averaged things. | {
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quantum-mechanics, potential, schroedinger-equation, boundary-conditions
Title: Infinite square well separation of variables issue When solving the time independent Schrodinger equation for the infinite square well in 2 or 3 dimensions (I'll use 2 dimensions for brevity), we use separation of variables, first assuming that our solution is of the form $\psi(x,y)=f(x)g(y)$ and we get to the point where we have that
$$\frac{1}{f}\frac{\partial^2f}{\partial x^2}+\frac{1}{g}\frac{\partial^2g}{\partial y^2}=\frac{-2m}{\hbar}E$$
My problems begin at this point. Clearly the two terms on the LHS of the above equation must be equal to a constant. Thus we have that
$$\frac{1}{f}\frac{\partial^2f}{\partial x^2}=C_x \,\,\,\,\,\,\,\frac{1}{g}\frac{\partial^2g}{\partial y^2}=C_y$$
But this is never how the problem is actually solved. Instead, it is always assumed that we actually have
$$\frac{1}{f}\frac{\partial^2f}{\partial x^2}=-k^2_x \,\,\,\,\,\,\,\frac{1}{g}\frac{\partial^2g}{\partial y^2}=-k^2_y$$ | {
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c#, parsing, regex
Some people, when confronted with a problem, think “I know, I'll use regular expressions.” Now they have two problems.
Missing Edge Case(s)
There is a case I noticed you didn't cover. C# allows unicode escape sequences. Literals can have \xnnnn escape sequences, while all C# code can handle \unnnn and \Unnnnnnnn escape sequences. The preprocessor of the compiler transforms these latter two into characters, but only in the following cases:
identifiers
character literals
regular string literals
This means \u2215 which represents a slash / is not found by your regex.
I'm also not sure whether you can track inline comments in interpolated strings. Should be verified. | {
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c#
Title: Generate the name of a computer from its properties I am writing a small application to generate the domain name of computers, from their known properties, and business rules.
My Asset class is:
public enum AssetSite
{
Rotterdam,
Sydney,
}
public class Asset
{
/// <summary>
/// Gets the Asset Tag of the asset.
/// </summary>
public string AssetTag { get;}
/// <summary>
/// Gets the site to which the asset belongs.
/// </summary>
public AssetSite AssetSite { get; }
/// <summary>
/// Gets a value indicating whether the asset is a laptop.
/// </summary>
public bool IsLaptop { get; } | {
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$\displaystyle \int_{-\infty}^{\infty} L^x_t f(x)dx=\int_0^tf(X)d[X]^c.$ (2)
By eliminating the delta function, the right hand side has been transformed into a well-defined expression. In fact, it is now the left side of the identity that is a problem, since the local time was only defined up to probability one at each level x. Ignoring this issue for the moment, recall the version of Ito’s lemma for general non-continuous semimartingales,
\displaystyle \begin{aligned} f(X_t)=& f(X_0)+\int_0^t f^{\prime}(X_-)dX+\frac12A_t\\ &\quad+\sum_{s\le t}\left(\Delta f(X_s)-f^\prime(X_{s-})\Delta X_s\right). \end{aligned} (3)
where ${A_t=\int_0^t f^{\prime\prime}(X)d[X]^c}$. Equation (2) allows us to express this quadratic variation term using local times,
$\displaystyle A_t=\int_{-\infty}^{\infty} L^x_t f^{\prime\prime}(x)dx.$ | {
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Betsy DeVos Stands for the Rights of Victims September 13 2017 by Patrice Lee Onwuka
NFL protests need to take a knee September 8 2017 by Patrice Lee Onwuka
Economic Agenda Will Do More for Black People Than Statue Removal August 28 2017 by Patrice Lee Onwuka
Government Can Hate A Name, But Still Must Respect It August 14 2017 by Patrice Lee Onwuka
The Google Diversity Memo Should Start the Conversation — Not End It August 8 2017 by Patrice Lee Onwuka
Entrepreneurship Is the Real Black Girl Magic August 7 2017 by Patrice Lee Onwuka
Serena Williams Swings But Misses With Black Women’s Equal Pay Day August 4 2017 by Patrice Lee Onwuka
New York's crackdown on dog-walking licenses is only hurting low-income workers July 21 2017 by Patrice Lee Onwuka
Business should back Larry Hogan's veto of paid sick leave July 18 2017 by Patrice Lee Onwuka
Dispelling the White House pay gap myth July 10 2017 by Patrice Lee Onwuka | {
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classical-mechanics, wavefunction, schroedinger-equation, probability, born-rule
then pushes the conjecture that this analogy must be complete. That is, he assumes that just as the necessity of wave optics follows from the breakdown of ray optics for short paths of large curvature (interference and diffraction around obstacles smaller than or comparable to the wavelength), in exactly the same way the breakdown of classical mechanics, also for short paths of large curvature, calls for a wave mechanics based on equi-action wavefronts. He then infers the simplest "wave mechanics" on configuration space that has as "ray limit" the usual classical mechanics. | {
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} |
python, algorithm, graph
def __str__(self):
return f"Name: {self.name}, total cost: {self.total_cost}, connections: {[(node.name, cost) for node, cost in self.connections]}"
__repr__ = __str__ | {
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python, strings, file
This solution is a bit too much for such a task which can be easily handled on 'MS Word'.
How can such a task be accomplished in an easy way?
This solution is a bit too much for such a task
Not even close. Welcome to the Natural Language Processing space :)
In order to properly compare the English texts we at least need to apply the following things:
make the comparison case-insensitive
remove punctuation and retain alphabetic words only
remove stop words
lemmatize words
We can make use of the awesome nltk Python library to help us tokenize, remove the stop words and lemmatize. Here is something more or less generic that works for HTML documents (you can modify the "download content" part if you are working with markdown or other document types):
from bs4 import BeautifulSoup
from nltk import word_tokenize
from nltk.stem import WordNetLemmatizer
from nltk.corpus import stopwords
import requests
ENGLISH_STOPS = set(stopwords.words('english')) | {
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aerodynamics, viscosity, vortex, lift, potential-flow
... in real life, the way that nature insures the that the flow will leave smoothly at the trailing edge, that is, the mechanism that nature uses to choose the flow... is that the viscous boundary layer remains attached all the way to the trailing edge. Nature enforces the Kutta condition by means of friction. If there were no boundary layer (i.e. no friction), there would be no physical mechanism in the real world to achieve the Kutta condition.
He chooses to explain that nature found a way to enforce the Kutta condition. I prefer to think of it the other way around -- the Kutta condition is a mathematical construction we use to enforce nature in our mathematical approximation. | {
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1. The inverse of the stereographic projection is the embedding of $\mathbb{R}^n$ in $\mathbb{S}^n$. The inverse of the stereographic projection (which is already continuous) is $$i'(x_1,\ldots,x_n)=\left(\frac{-1+x_1^2+\cdots+x_n^2}{1+x_1^2+\cdots+x_n^2},\frac{2x_1}{1+x_1^2+\cdots+x_n^2},\ldots,\frac{2x_n}{1+x_1^2+\cdots+x_n^2}\right)$$ so it is also continuous. Thus, $i'$ is a homeomorphism of $\mathbb{R}^n$ onto its image, $\mathbb{S}^n\setminus(1,0,\ldots,0)$, which is open and dense in $\mathbb{S}^n$, and whose complement is the singleton $(1,0,\ldots,0)$. Moreover, $\mathbb{S}^n$ is compact. By 5., there exists an homeomorphism between $\mathbb{S}^n$ and $(\mathbb{R}^n)^*$. | {
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"url": "https://math.stackexchange.com/questions/1300027/question-on-one-point-compactification"
} |
energy, capacitance, inductance
For such a system, the capacitance matrix is a 2-by-2 symmetric matrix with $C_{11} = C_1-C_{12}$, $C_{22} = C_2-C_{12}$ and $C_{12}<0$ ($C_{11}$ and $C_{22}$ are the total capacitances from each conductor to the other conductors). Such a matrix is further positive definite if and only if its trace and determinant are positive (see e.g. this question on Math SE). From the condition on the determinant, we thus have
$$C_{11}C_{22}-C_{12}^2\ge 0,$$
that is,
$$|C_{12}|\le\sqrt{C_{11}C_{22}}.$$
For further details and derivations for the general case, see:
R. M. Fano, L. J. Chu, and R. B. Adler, Electromagnetic fields, energy, and forces, MIT Press, 1968. | {
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"tags": "energy, capacitance, inductance",
"url": null
} |
newtonian-mechanics, forces, everyday-life
I think this topic may be confusing for students because the ground is doing no work, yet still exerts a force. The ground doesn't have any energy to use to push you up, so it may seem counterintuitive that the ground is the source of the force. The very word "exert" in the phrase "exert a force" makes us think of actively pushing or pulling like a person does, not sitting there letting things happen to you as the ground does. This is just something you need to get used to. Although in everyday language, we say "I forced the door open" or "I was forced against my will" and the word "force" refers to something that is moving or expending energy or exhibiting some sort of agency, that is not so in physics. The energy to jump comes from your legs (and other parts of your body), but the force comes from the ground. | {
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"tags": "newtonian-mechanics, forces, everyday-life",
"url": null
} |
python, python-3.x, numpy, coordinate-system, geospatial
def sigma_iteration(sigma_0: np.ndarray, sigma1, alpha, beta, distance) -> tuple[np.ndarray, np.ndarray]:
# cos 2σm = cos(2σ1 + σ)
cos_2sigma_m = np.cos(2 * sigma1 + sigma_0)
# Δσ = B · sin σ · {cos 2σm + B/4 · [cos σ · (−1 + 2 · cos² 2σm)
# − B/6 · cos 2σm · (−3 + 4 · sin² σ) · (−3 + 4 · cos² 2σm)]}
delta_sigma = (
# B · sin σ ·
beta * np.sin(sigma_0) *
# cos 2σm + B/4 ·
(
cos_2sigma_m + beta/4 *
# [cos σ ·
np.cos(sigma_0) *
(
# (−1 + 2 · cos² 2σm) -
-1 + 2*cos_2sigma_m**2 -
# − B/6 · cos 2σm ·
beta/6 * cos_2sigma_m *
# (−3 + 4 · sin² σ) ·
(-3 + 4*np.sin(sigma_0)**2) *
# (−3 + 4 · cos² 2σm)]
(-3 + 4*cos_2sigma_m**2)
)
)
)
# σʹ = s / b·A + Δσ
sigma = distance/B/alpha + delta_sigma
return sigma, cos_2sigma_m | {
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"id": 42999,
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"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "python, python-3.x, numpy, coordinate-system, geospatial",
"url": null
} |
ros2, colcon
--- stderr: examples_rclcpp_minimal_publisher
Traceback (most recent call last):
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_core\executor\__init__.py", line 91, in __call__
rc = await self.task(*args, **kwargs)
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_core\task\__init__.py", line 93, in __call__
return await task_method(*args, **kwargs)
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_ros\task\ament_cmake\build.py", line 59, in build
rc = await extension.build(
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_cmake\task\cmake\build.py", line 87, in build
rc = await self._reconfigure(args, env)
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_cmake\task\cmake\build.py", line 174, in _reconfigure
completed = await run(
File "c:\opt\ros\foxy\x64\lib\site-packages\colcon_core\task\__init__.py", line 177, in run
completed = await colcon_core_subprocess_run( | {
"domain": "robotics.stackexchange",
"id": 38740,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "ros2, colcon",
"url": null
} |
python, python-3.x, image
# check if data and dicts exist else download and generate
mnist_dir = './data'
mnist_images_fn = "./mnist_images"
mnist_images_fn_loc = os.path.join(mnist_dir, mnist_images_fn + ".npy")
mnist_idx_dict_fn = "./mnist_idx_dict"
mnist_idx_dict_fn_loc = os.path.join(mnist_dir, mnist_idx_dict_fn + ".pickle")
if os.path.isfile(mnist_images_fn_loc) and os.path.isfile(mnist_idx_dict_fn_loc):
print("Found image data, loading...", end="")
images = np.load(mnist_images_fn_loc)
with open(mnist_idx_dict_fn_loc, 'rb') as handle:
idx_dict = pickle.load(handle)
print("DONE")
else:
print("Not all image data found, preparing...")
import gzip
import shutil
import urllib.request
mnist_urls = ["http://yann.lecun.com/exdb/mnist/train-images-idx3-ubyte.gz",
"http://yann.lecun.com/exdb/mnist/train-labels-idx1-ubyte.gz"] | {
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"id": 37071,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "python, python-3.x, image",
"url": null
} |
length formula - Example 1 Discuss the formula for arc length to the... Arc of a circle when the angle for finding arc length to find is in degrees, we n't. Arc length 2πR/360 length for one degree: 1 degree corresponds to an arc is a segment of circle. In degrees, we ca n't use the formula for either degrees or radians for finding arc length the length! The circumference of the angle arc along the circumference of the central angle of a circle in radians an. The circumference 0.52 = 1.56 m 6 length and use it in a couple of examples and it! Finding arc length given radius and central angles radians ) of examples degree corresponds to an arc length to the. On 'Find the length of the angle is in radians. length Problems... Off by the angle angle of a circle in radians and r is the radius of the angle in! If the measure of the circle = 1.56 m 6 each arc length step-by-step... For either degrees or radians for finding arc length for one degree: 1 degree corresponds to arc! The central | {
"domain": "calcigarro.cat",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534398277177,
"lm_q1q2_score": 0.8091709915327061,
"lm_q2_score": 0.8244619242200081,
"openwebmath_perplexity": 497.4108081609029,
"openwebmath_score": 0.9313715100288391,
"tags": null,
"url": "http://booking.calcigarro.cat/restaurants-in-decuvlq/1e8d7d-how-to-find-arc-length-with-radians"
} |
machine-learning, datasets, generative-adversarial-networks, generative-model, social
Title: What are the possible social consequences of training neural networks with artificially generated data? Machine learning models and, in particular, neural networks are trained with data often collected from the real world, such as images of real people.
Meanwhile, neural networks (such as GANs) are also used for data generation. Each year, they become better at this task to the point that even humans are not able to distinguish real-world data from the artificially generated one.
So it is possible that neural networks will start to learn with data that was generated by other neural networks, because it will look as real even for a human, but naturally will be not related to the real world.
Will it lead to some machine learning collapse?
Might it lead to some changes in human's perception of the world, because people get a very big part of their knowledge using computers, connected to the Internet? | {
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"id": 397,
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"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "machine-learning, datasets, generative-adversarial-networks, generative-model, social",
"url": null
} |
newtonian-mechanics, torque, rotation, special-functions
To put short: basically, the way it wants to fall naturally
I do understand that there will be always a tendency for the chain to keep tilting but is there any way to make the tilting negligible?
Alternate methods:
I saw a video about how rotating chains are related to Bessel functions, so I am wondering if we can use that here refer 12:55 of this video. Segments of chain in the loop are subject to tension forces from each side. The resultant of these must have a radial component which provides the centripetal acceleration, and a vertical component which supports the weight (except at the point of support where the vertical component of the cord supports all of the weight). With a higher angular velocity, the tensions increase, and the angle of tilt gets smaller (but never zero). | {
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"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "newtonian-mechanics, torque, rotation, special-functions",
"url": null
} |
javascript, functional-programming, ecmascript-6
q.cleanUp = () => question;
return q;
}
}
let validationOptions = {
validateString
};
function validateString() {
// validate
return this;
}
That way the initial object is unchanged, and you don't have to concern yourself with deleting the validation functions from the question when your done.
Wrapper Class
I think you also ought to consider making the validator a wrapper class for your question. That way your validation functions stay out of your data, you can still chain your methods, and you don't have to assign the validation functions for each object:
export class Validator {
constructor(question) {
this.data = question;
}
validateString() {
return this;
}
cleanUp() {
return this.data;
}
}
If you want the validation functions to be free from an object, classes have prototypes that you can assign the methods to:
function validateString() {
return this;
} | {
"domain": "codereview.stackexchange",
"id": 20266,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "javascript, functional-programming, ecmascript-6",
"url": null
} |
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