text stringlengths 1 1.11k | source dict |
|---|---|
# Are there any visual proofs for $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$?
I was flipping through Proofs Without Words (PWW) and saw many visual proofs for sequences and series. However, I saw none for $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
Are there any visual proofs for the above series? | {
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"lm_q1q2_score": 0.8234861571096462,
"lm_q2_score": 0.8479677545357568,
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"openwebmath_score": 0.47237011790275574,
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"url": "https://math.stackexchange.com/questions/1044022/are-there-any-visual-proofs-for-sum-n-1-infty-frac1n2-frac-pi2"
} |
following (quadratic) equation logπ 1 − 1 2 log(σ2)− (x−µ 1)2 2σ2 1 = logπ 2 − 1 2 log(σ2)− (x−µ 2)2 2σ2 2 To simplify the math, we assume that the two components have equal variance(i. Discriminant analysis¶. Quadratic discriminant analysis falls somewhere between the linear approaches of linear discriminant analysis and logistic regression and the non-parametric approach of K-nearest neighbors. QDA¶ class sklearn. Trending AI Articles: 1. Least squares linear regression in which linear decision boundary is assumed is the most rigid one and stable to fit. 6 Linear decision. Though not as flexible as KNN, QDA can perform better in the presence of a limited number of training observations because it does make some assumptions abou the form of the decision boundary. The shading indicates the QDA decision rule. I The common covariance matrix Σ = 1. Linear discriminant analysis and logistic regression will perform well when the true decision boundary is linear. The boundary is a affine | {
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"lm_q1q2_score": 0.8386081185009281,
"lm_q2_score": 0.8519528076067262,
"openwebmath_perplexity": 1224.9299710586267,
"openwebmath_score": 0.6168559193611145,
"tags": null,
"url": "http://tuningfly.it/saiv/qda-decision-boundary.html"
} |
beginner, haskell, interpreter, brainfuck
usage :: String
usage = "Usage: bf-interpreter-ast-exe [-sp] [-sm] [-f file] [-p program | program]"
validateOptions :: Options -> Either String Options
validateOptions opts@(Options _ aOpts)
| Program `Map.member` aOpts && File `Map.member` aOpts = Left "Error: Only one of the options File and Program can be present"
| Program `Map.member` aOpts || File `Map.member` aOpts = Right opts
| otherwise = Left "Error: One of the options File and Program must be present"
Tape.hs
module Tape
( Tape(..)
, makeTape
, forwardTape
, reverseTape
, tapeValue
, onTapeValue
) where
data Tape a = Tape [a] !a [a] deriving (Eq)
instance Show a => Show (Tape a) where
show (Tape ls v rs) = show (reverse ls) ++ " " ++ show v ++ " " ++ show rs
makeTape :: a -> Tape a
makeTape def = Tape [] def [] | {
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"tags": "beginner, haskell, interpreter, brainfuck",
"url": null
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c++, ai, visitor-pattern
class Visitor {
public:
virtual void visit (Human*) = 0;
virtual void visit (AI*) = 0;
};
class Human : public Inputter {
using Inputter::Inputter;
void accept (Visitor& visitor) override { visitor.visit(this); }
};
class AI : public Inputter {
using Inputter::Inputter;
void accept (Visitor& visitor) override { visitor.visit(this); }
};
// Use separate Visitor classes for each query to follow the open-closed principle.
class MealVisitor : public Visitor {
const std::vector<std::string> mealOptions;
std::string mealChosen;
public:
MealVisitor (const std::vector<std::string>& m) : mealOptions(m) { }
std::string getMealChosen() const { return mealChosen; }
private:
void visit (Human*) override;
void visit (AI*) override;
};
class TipVisitor : public Visitor {
int tipGiven;
public:
int getTipGiven() const { return tipGiven; }
private:
void visit (Human*) override;
void visit (AI*) override;
}; | {
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"tags": "c++, ai, visitor-pattern",
"url": null
} |
condensed-matter, hamiltonian, parity
\mathcal U\boldsymbol=\frac{\sqrt{2}}{2}
\begin{bmatrix}
\hphantom{\boldsymbol-}\mathrm I & \hphantom{\boldsymbol-}\mathrm I\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\hphantom{-}\\
\hphantom{\boldsymbol-}\mathcal T & \boldsymbol-\mathcal T\vphantom{\tfrac{\dfrac{a}{b}}{\tfrac{a}{b}}}\hphantom{\boldsymbol-}
\end{bmatrix}\,,\quad
\mathcal U^{\boldsymbol-1}\boldsymbol=\frac{\sqrt{2}}{2}
\begin{bmatrix}
\hphantom{\boldsymbol-}\mathrm I & \hphantom{\boldsymbol-}\mathcal T\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\hphantom{-}\\
\hphantom{\boldsymbol-}\mathrm I & \boldsymbol-\mathcal T\vphantom{\tfrac{\dfrac{a}{b}}{\tfrac{a}{b}}}\hphantom{\boldsymbol-}
\end{bmatrix}\,,\quad
\mathcal T f\left(z\right)\boldsymbol=f\left(\boldsymbol-z\right)
\tag{04}\label{04}
\end{equation}
where $\:\mathcal T\:$ the parity transformation, transforms the Hamiltonian \eqref{03} onto diagonal form.
We have | {
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"tags": "condensed-matter, hamiltonian, parity",
"url": null
} |
regular-languages, regular-expressions, pumping-lemma, discrete-mathematics
Title: Pumping Lemma,regular languages Lets say that we have the language L = { $a^n$$b^m$$c^{m+n}$ $|$ $m$,$n$ $>=0$ }
What is the way that i should follow to prove
that the language is not regular?
Assume that the language is regular.
So,there is a string $w\in L$ such that $w = xyz$ , $|xy|\leq n, y\neqε $, n is the pumping length | {
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"tags": "regular-languages, regular-expressions, pumping-lemma, discrete-mathematics",
"url": null
} |
Search
## Search Loci: Convergence:
Keyword
Random Quotation
The most painful thing about mathematics is how far away you are from being able to use it after you have learned it.
In J. R. Newman (ed.) The World of Mathematics, New York: Simon and Schuster, 1956.
See more quotations
# Sums of Powers of Positive Integers
## Solutions to Exercises 15-17
Exercise 15. Beginning with our simplification of Fermat’s formula, we have
$${{n(n + 1)(2n + 1)(3n^2 + 3n - 1)} \over {30}} = {{n(n + 1)(2n + 1)} \over 6} \cdot {{(3n^2 + 3n - 1)} \over 5}.$$
Beginning with Fermat’s formula, we have | {
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"lm_q2_score": 0.8152324983301568,
"openwebmath_perplexity": 1590.3728346885275,
"openwebmath_score": 0.9998376369476318,
"tags": null,
"url": "http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3284&bodyId=3591"
} |
python, beginner, tic-tac-toe
print "The Player Will Then Take Another Turn And the Game Will Resume Until Either Player Wins "
print '\n'
Player_Board = [['|',' ','|',' ','|',' ','|'],['|',' ','|',' ','|',' ','|'], ['|',' ','|',' ','|',' ','|']] | {
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"url": null
} |
image-processing, matlab, computer-vision, object-recognition, 3d
Create a fake volume with noise of intensity comprised between .1 and .5, add two solid 3D balls to it, and rescale all the values to range from 0 to 1:
data = Rescale[ RandomReal[{.1, .5}, {200, 100, 100}] +
Join[DiskMatrix[10, {100, 100, 100}], DiskMatrix[20, {100, 100, 100}]]];
Construct a 3D image (which can be manipulated, whose color transfer function can be interactively edited, etc.), and segment it by looking for voxel values between .5 and 1:
image3d = Image3D[data];
segmented = Binarize[image3d, {.5, 1}];
Compute the morphological connected components from the binary 3D image:
components = MorphologicalComponents[segmented];
Compute the centroid of each shape by averaging the voxel positions for each shape:
Table[
p = N@Position[components, i];
Mean[p],
{i, 1, Max[components]}
] | {
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"url": null
} |
javascript, jquery, dom
$('.CurrentAverage').text('Current Average: ' + weightedAvg.toFixed(2)); // Finally, set the 'Current Average' text to the newly-calculated average
});
$(document).on('grade-change', '.script-grade', function handleGradeChange() { // Listen for when grades change
var avg = $(this).closest('tbody').find('.script-avg'); // Find closest average
var gradeArr = $(this).closest('tbody').children().children('.script-grade').contents(); // Get array of all grades in current section
gradeArr = Array.from(gradeArr); // Same process as with the weightArr above
gradeArr = gradeArr.map(function parseGrades(gradeStr) {
return parseInt($(gradeStr).text()); // Parse the text into a float
}); | {
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You can get at least one correct element overall by trying random permutations. I think the expected number of tries for just a simple strategy of "pick some half of remaining elements, randomly permute them" before you see a change in the number of matches would have a finite upper bound over all n. That'd allow you to bisect to the location of one correct element in O(log(n)).
uau 2021-06-05 18:42
It occurred to me that you can also get O(n*log(n)) by kind of the opposite strategy: instead of proving for one element at a time what its correct value must be, you can also collect permutations with zero matches that prove where every element can not be, and get enough of those in O(n*log(n)) to rule out any wrong value. | {
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"openwebmath_perplexity": 591.0678449678438,
"openwebmath_score": 0.6592031717300415,
"tags": null,
"url": "https://mersenneforum.org/printthread.php?s=043379c55fe2686d9b3306fafd8500da&t=26868&pp=97"
} |
venus
But the moon's orbit "in its entirety is between the Earth and Sun", and is "fully lit from the perspective of the Earth" once a month, so how can we conclude that Venus revolves around the Sun? The Moon orbits the Earth in such a way that it sometimes can be on the opposite side of the Earth as seen from the Sun. In other words, it’s possible to have Sun-Earth-Moon along an imaginary straight line, in that order.
However, by Ptolemy’s time, it had already been noticed that Venus is always seen close to the Sun (maximum angular distance about 47°). So it can never get opposite to the Sun as seen from Earth.
Ptolemy didn’t know why (the reason is that Venus orbits the Sun, not the Earth, which was inconceivable to Ptolemy), but his system made it so that Venus (and Mercury, for that matter) were orbiting the Earth closer to it than the Sun, and in such a way as to always be near it. | {
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electrostatics, charge, work, potential, capacitance
$$
\mathscr{E}_p = \frac{q^2}{8\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)
$$
which is negative (the conductor attracts the inner charged particle). When the inner charged particle is removed and put far away, the electrostatic potential energy of the system is zero. So the minimum work needed to take the inner charge out and far away, in ideal case, is minus the above value.
Of course, in reality, the work can be much higher, due to losses into heat or radiation - but this is meant to be ignored in the original question. | {
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"url": null
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What do i use to get started? All i know is the distance formula $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$.
Kindly Help
-
Try to find the global minima of your distance formula when plugging in the line for $x_2,y_2$ – Listing Nov 26 '11 at 12:43
I have no idea how to find the global minima. – alok Nov 26 '11 at 12:55
A geometric way of looking at it: your problem is equivalent to finding the radius of a circle that is tangent to a given line and centered at a given point. Remember that tangency of a circle and a line corresponds to the resulting quadratic equation having a double root. – Guess who it is. Nov 26 '11 at 13:49
@J.M. Assuming the OP knows the tangency condition (or in the worst case to get tangent replace $x^2$ by $xx'$ etc), this is the best way, IMHO. – Tapu Nov 26 '11 at 14:05 | {
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"openwebmath_score": 0.8813772797584534,
"tags": null,
"url": "http://math.stackexchange.com/questions/85761/distance-between-a-point-and-a-line"
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python, python-3.x, physics
I think that’s easier to read, and makes it clear where they’re coming from.
I see no reason to define a c2 variable, especially when you use it inconsistently (sometimes you have c2, sometimes c * c.
The scipy.constants module contains lots of these constants built-in to high levels of precision, but that may be inappropriate for your work.
At the point when E is defined, the variable mass hasn’t been defined. In fact, I can’t find any mention of the variable in the script. What’s it doing here?
Something only happens if the user types exactly yes or no when asked “Show work?” You might be better off defining a list of “acceptable” responses which means the user wants their work shown.
For example:
yes_list = ['yes', 'y', 'true', 't', 'yep']
if input('Show work?').lower() in yes_list:
show_input = True
else:
show_input = False | {
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quantum-field-theory, electromagnetic-radiation, photons, polarization, helicity
$$\hbar \left<\psi|a_+^\dagger a_+ - a_-^\dagger a_- | \psi\right> = (2\,\alpha^2-1)\,\hbar$$
One photon emissions are always the right pure quantum superpositions that make sure angular momentum is conserved. For example, if a fluorophore absorbs linearly polarized light, and if it does not exert torque on its surroundings before spontaneous emission, the emission must be linearly polarized. Likewise for any other general polarization state absorbed by fluorophores. See the answer https://physics.stackexchange.com/a/73439/26076 for more details, and you also might like to look at the work of Gregorio Weber in the 1950s on the polarization of fluorescence. See see G. Weber, “Rotational Brownian motion and polarization of the fluorescence of solutions,” Adv. Protein Chem. 8, 415–459 (1953) and other works.
My own work in this field is summarised in J. Opt. Soc. Am. B, Vol. 24, No. 6/June 2007 p1369. | {
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"url": null
} |
gravity, expansion, gravitational-waves, dark-energy, fate-of-universe
In principle gravitational waves might allow us to fractions of a second after the big bang. Electromagnetic waves can see back to where the cosmic background radiation formed, about 400,000 years after the big bang.
You are right, the universe has expanded. At the present epoch it is estimated that the observable universe, containing objects that have emitted light or GWs that may reach us now, is around 46 billion light years.
However, it seems quite likely that the universe continues well beyond this horizon, and sources beyond this horizon can never have emitted light or GWs that will reach us. | {
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"url": null
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• The difference of two uniform distributions is a triangular distribution, so if you ask if it is possible to get uniform of a difference of i.i.d. uniforms, then the answer is not. – Tim May 15 '18 at 6:48
• Same Q asked here: math.stackexchange.com/questions/2048939/… so far without answers! – kjetil b halvorsen May 15 '18 at 7:48
• It would indeed seem difficult to avoid realizations outside $[-1,1]$ when both $j$ and $k$ have probability mass close to these endpoints. – Christoph Hanck May 15 '18 at 9:54
• It is not possible. To my recollection this is (in slightly different form) already answered somewhere on site. I'll see if I can locate it – Glen_b -Reinstate Monica Jun 19 '18 at 10:07 | {
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"openwebmath_score": 0.9975154995918274,
"tags": null,
"url": "https://stats.stackexchange.com/questions/346269/uniform-pdf-of-the-difference-of-two-r-v"
} |
c++, template-meta-programming, variadic
StunRFC(const std::string& res) : m_res{res}{}
bool valid() const {
if (!strcmp("stun", m_res.c_str()))
return true;
return false;
}
};
struct NonValid
{
};
template<>
struct is_validator<Null>
{
static const bool value = true;
};
template<>
struct is_validator<RtpRFC>
{
static const bool value = true;
};
template<>
struct is_validator<StunRFC>
{
static const bool value = true;
};
template<>
struct is_validator<RtspRFC>
{
static const bool value = true;
};
/*terminator*/
bool VParse(...) {
return false;
}
template <class T,
//typename std::enable_if<is_validator<T>::value>::type,
typename...ARgs>
bool VParse(T type, ARgs&&... FArgs)
{
if constexpr (is_validator<T>::value) {
if (type.valid()) {
return true;
}
else {
return VParse(std::forward<ARgs>(FArgs)...);
}
}
return VParse(std::forward<ARgs>(FArgs)...);
} | {
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"tags": "c++, template-meta-programming, variadic",
"url": null
} |
python, assembly
The posted code has violates PEP8 on many counts. I suggest to review the guidelines and apply carefully
The indentation is odd at a few places. For example the return statement in tobin
snake_case is recommended for variable and function names. For example, symbol_table would be better than SymbolTable. Also to_bin better than tobin. Just to name a few.
A more idiomatic way of writing line[0]=='(' is line.startswith('(')
Prefer Python 3
Judging by the print statement without parentheses,
it seems you're on Python 2.
Consider switching to Python 3. | {
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java, beginner, io
fos.write(r); // Writes encoded character in fileout
}
} catch (IOException e) {
System.err.println("Caught IOException: " + e.getMessage());
}
finally {
if (fis != null) {
try {
fis.close();
} catch (IOException e) {
e.printStackTrace();
} // Closes FileInputStream resources
}
if (fos != null) {
try {
fos.close();
} catch (IOException e) {
e.printStackTrace();
} // Closes FileOutputStream resources
}
}
}
} | {
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"tags": "java, beginner, io",
"url": null
} |
organic-chemistry, hydrocarbons, halogenation
When the compound is chlorinated on the extreme carbons, there is a chiral carbon in both the resulting compounds, which gives a total of 4 stereo isomers. When compound is chlorinated on 2nd or 3rd carbon, it still has a chiral carbon in both resulting compounds. So the total stereo isomers add up to 8 according to me, but it doesn't match the options given. First of all, welcome to ChemSE! I presume that you are a student, which I am not, and that this problem arises from a preparatory examination. Too many of these types of questions are ambiguous at best and this one is no exception. The sec-butylbenzene as drawn in question 33 must be assumed to be racemic because there is no indication that it is a single enantiomer. The products are presumed to be monochlorinated and that the fractionating column is achiral and infinite in plates. In the diagram below, the Products (P) are presumed to be constitutional and stereoisomeric while the Fractions are designated as (F). | {
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newtonian-mechanics, forces, free-body-diagram, string, elasticity
Also can ideal strings be bent since they will still have the same length which complies with the assumption of constant length with time?
Yes. There is no problem in changing the shape of an ideal string in any way as long as you don't stretch(or shrink) it. If you weren't able to bend an ideal string, then how would you get it around a pulley?!(you will encounter many questions about the Atwood machine if you haven't yet, and these all won't have existed if an ideal string could not be bent).
Will $T'=T$? | {
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python, validation, calculator, tkinter
clear_button = Button(root, bg="#302e2e", fg="white", text="AC", font=("Helvetica", 14, 'bold'), pady=10,command=self.clear_screen)
clear_button.grid(row=1, columnspan=2, sticky="WE")
backspace_button = Button(root, bg="#302e2e", fg="white", text="Backspace", font=("Helvetica", 14, 'bold'), pady=10, command=self.backspace)
backspace_button.grid(row=1, column=3, sticky="NSEW")
author_button = Button(root, bg="#302e2e", fg="white", font=("Helvetica", 14, 'bold'), text="Info", pady=10, command=self.show_info_popup)
author_button.grid(row=1, column=2, sticky="NSEW") | {
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3. Feb 20, 2013
### kwal0203
Wow thanks, this is one question in a group of six and seems to be significantly harder than the others for some reason.
Can you provide me with any insight as to how you came up with the solution? Is it just a lot of practice and seeing similar problems in the past?
I want to develop my problem solving skills but I would never have come up with this method!
4. Feb 20, 2013
### Mathitalian
It is a classic exercise on Taylor series :) It is very important to know the macLaurin series of common functions and all the properties of "elementary functions". Practice will help you! :)
[Sorry, my English is awful :|] | {
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"url": "https://www.physicsforums.com/threads/messy-taylor-polynomial-question.673152/"
} |
c++, c++20
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstddef>
#include <iostream>
#include <iomanip>
#include <iterator>
#include <map>
#include <ranges>
template<typename T>
using counter = std::map<T, std::size_t>;
static bool isprint(char c)
{
return std::isprint(static_cast<unsigned char>(c));
}
int main()
{
auto hist = counter<char>{};
std::ranges::for_each(std::istreambuf_iterator<char>{std::cin},
std::istreambuf_iterator<char>{},
[&hist](char c){ ++hist[c]; });
auto const keys = hist | std::views::keys;
auto const values = hist | std::views::values;
auto const max_val = std::ranges::max(values);
// width of the scale column
unsigned count_width = static_cast<unsigned>(std::ceil(std::log10(max_val)));
// scale to fit a 24-line display (but just assume it's wide enough)
auto constexpr rows = 23u;
auto const scale = std::log(static_cast<double>(max_val - 1)) / rows; | {
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zoology, eyes, anatomy
As you are particularly interested in mongooses, the paper "Characteristics of the eye of the Indian mongoose (Herpestes auropunctatus)" reports the following concerning pupils dilation induced by topical application of pilocarpine: "The dilated pupil is approximately 3.80 mm in diameter and slightly oval, the long axis being horizontal. [...] The constricted pupil is teardrop-shaped, with horizontal and vertical dimensions of 2.5 and 1.1 mm, respectively, the narrow end of the aperture being the nasal end." However that shape seems specific to Indian mongooses because if you search for images of mongooses you'll see that most of them have oblong pupils when contracted. But all of them seem to have almost circular pupils when dilated. Here you can compare the pupil of a mongoose during the day (strongly contracted) and at night (mostly dilated, the color comes from the night photograph): | {
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structural-engineering, construction-management
Title: Sheathing carrying vertical load more than studs When wooden walls are created and sheathed in place before erecting (sheathing does go all the way to sill and top plates) some studs don’t have a great connection to top or sill plates because of imperfections in the end cut of the studs or less than perfect construction. They’re locked in place by sheathing before a vertical load (the roof) can settle those gaps. Even after the roof has been added, I’ve still seen gaps between studs and plates. Will strong sheathing actually carry the vertical load for years, or even the lifetime of the structure? Or will settling effectively work on the sheathing/stud connections, until those weaken and move until the studs create full contact with the plates? It depends on the primary function of the stud that is floating. Two results are expected: | {
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php, object-oriented, mvc, spl
Title: Using `exec()` in an autoloader After reviewing autoloader functions on the site, I didn't see anything like this, and I was wonder what may be wrong with this?
function findNloader($classFile){
$file = exec("find ./ -name ".$classFile.".php");
if(file_exists($file)){
return require $file;
} else {
return false;
}
} | {
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3. Apr 12, 2013
### raopeng
Thanks for the reply. That's my idea exactly. But that problem appears in a chapter well before the author introduces the L'Hospital's Theorem so there must be another way to do it... I am torturing myself...
4. Apr 13, 2013
### Staff: Mentor
I would check to see if the author has discussed this limit in the chapter you're in: $\lim_{x \to 0} x ln(x)$. Since he doesn't present L'Hopital's Rule until later, it's possible he has said something about this limit, which is the crux of the problem you're trying to solve.
5. Apr 13, 2013
### Infrared
You don't really need l'Hopital for $\lim_{x\rightarrow 0} {xlnx}$. I would let $x = e^{-n}$ and take the limit as n→∞.
$\lim_{n\rightarrow \infty} {e^{-n}ln(e^{-n})} = \lim_{n\rightarrow \infty} {\frac {-n}{e^n}}$. I guess this limit is technically a l'Hopital problem, but it isn't hard to see that it goes to zero.
6. Apr 16, 2013
### raopeng | {
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algorithm, c, comparative-review, generics, c-preprocessor
static inline void FIFO_AdvanceIndex(size_t *const pIndex, const size_t size, const size_t n)
{
*pIndex = (*pIndex < (size - n)) ? *pIndex + n : *pIndex - (size - n);
}
static inline void FIFO_AdvanceHead(FIFO_Control_s *const pControl, const size_t size, const size_t n)
{
FIFO_AdvanceIndex(&pControl->Head, size, n);
pControl->NofAdds += n;
}
static inline void FIFO_AdvanceTail(FIFO_Control_s *const pControl, const size_t size, const size_t n)
{
FIFO_AdvanceIndex(&pControl->Tail, size, n);
pControl->NofTakes += n;
}
static inline size_t FIFO_Head(const FIFO_Control_s *const pControl)
{
return pControl->Head;
}
static inline size_t FIFO_Tail(const FIFO_Control_s *const pControl)
{
return pControl->Tail;
}
static inline size_t FIFO_NofOccupied(const FIFO_Control_s *const pControl)
{
return (pControl->NofAdds - pControl->NofTakes);
} | {
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statistical-mechanics, quantum-statistics, chemical-potential, fermi-energy
Title: General formula for the variation of the chemical potential with temperature For small temperatures $T$, such that $k_\mathrm{B}T\ll \mu(T=0)\equiv \mu(0)$, the variation of chemical potential with temperature is given by $$\mu(T)=\mu(0)\left[1-\frac{\pi^2}{12}\left(\frac{k_\mathrm{B}T}{\mu(0)}\right)^2\right].\label{eqn:1}\tag{1}$$ Is there an exact analytical solution for the chemical potential at any temperature $T$ in a way that $\eqref{eqn:1}$ and its high temperature limit $(k_\mathrm{B}T\gg\mu(0))$ can be deduced from that? I am interested in such a situation because one often has to deal with Fermi gases at very high temperatures. The short answer is that there is no explicit formula for $\mu(T)$ valid at any $T$. In Peter Young's notes you can see his attempt to connect the low-$T$ and high-$T$ limits, and the basic equations can be found in most books on condensed matter physics. I'll just give the starting point here. One can equate the standard expression for $N$ at | {
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swift, unit-conversion
private func convertCoordinate(string: String) -> Double {
var separatedCoordinate = string.characters.split(separator: " ").map(String.init)
let direction = separatedCoordinate[0].components(separatedBy: CharacterSet.letters.inverted).first
let degrees = Double(separatedCoordinate[0].components(separatedBy: CharacterSet.decimalDigits.inverted)[1])
let minutes = Double(separatedCoordinate[1].components(separatedBy: CharacterSet.decimalDigits.inverted)[0])
let seconds = Double(separatedCoordinate[2].components(separatedBy: CharacterSet.decimalDigits.inverted)[0])
return convert(degrees: degrees!, minutes: minutes!, seconds: seconds!, direction: direction!)
}
private func convert(degrees: Double, minutes: Double, seconds: Double, direction: String) -> Double {
let sign = (direction == "W" || direction == "S") ? -1.0 : 1.0
return (degrees + (minutes + seconds/60.0)/60.0) * sign
}
} | {
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c#, wpf, easing
private FillBehavior _opacityBehavior = FillBehavior.HoldEnd;
/// <summary>
/// Gets or sets the fill behavior of the opacity sub-animation. Default is HoldEnd.
/// </summary>
public FillBehavior OpacityBehavior
{
get { return _opacityBehavior; }
set { _opacityBehavior = value; }
}
public FadeAnimation(UIElement target)
{
Target = target;
}
public FadeAnimation(UIElement target, AnimationType type, Direction direction)
: this(target)
{
Type = type;
Direction = direction;
}
protected override void BeginAnimationDetail()
{
Transform tempTrans = Target.RenderTransform; | {
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quadcopter, battery, brushless-motor, flight-controller
This is just an example to help understand the concept behind. You can adopt this approach to choose correct parts for the specifications provided by you.
Coming to the cost part which you want to be <=30k rupees is something no one can assure because the cost of the parts will differ from manufacturer to manufacturer and there might be other factors involved into it like customs. So, you will have to compare the prices on your own or you can check with various forums to get used parts for a cheaper price for overall cost reduction.
Hope this helped ;)
Happy flying!! | {
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formal-languages, reductions, undecidability
Title: Reducing a non-RE language to its complement Is there a language $L$ such that both $L$ and $L$'s complement are non turing recognizable languages, but there is a reduction between them?
I couldn't find one... Such a undecidable language $L$ can not be in $RE$, since if there was a computable reduction from $L$ to $\overline{L}$, then $L$ would be in $R$.
Let $x'$ be the value to which $x$ reduces. $x' \in L$ if and only if $x\not \in L$. We could then decide $x\in L$ by enumerating $L$, and stopping to report that $x\in L$ as soon as we encounter $x$, and report that $x\not \in L$ when we encounter $x'$. That would make $L$ decidable, which is a contradiction. So there can be no reduction from a language in $RE$ to its complement.
By a similar argument, such a language can not be in $co-RE$ either. | {
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Your steps are all clear (that's why I didn't downvote). But I was raised to think that it is simply wrong not to add the equivalence arrows. I'm not the only one thinking that way either! But years of experience teaching mathematically talented youth fresh out of high school have convinced me that they are taught differently these days :-). See the comments under the accepted answer (and +1 to you). – Jyrki Lahtonen Jul 19 '13 at 12:45
@JyrkiLahtonen Yeah you are right about to add implication sign in each step In India we also used to learn this thing but here I didn't write cause I take new step to write new expression.And I just want to show the method.this is not my exam so I should worry about it.I am updating it in my answer thanks – iostream007 Jul 19 '13 at 22:43 | {
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java, performance, multithreading, finance
It could be argued that an ArrayList would perform better than a LinkedList despite this really being the LinkedList's turf. This is because we're talking about small amounts of data and a predictable amount of it (4); arraycopy-ing has good cache locality in this case, and won't take extra allocations. (*)
Code duplication? You branch on the size of priceQueue in getDirection, but the code is very similar.
Fewer than 4 elements: It is still possible to calculate an average and have an oldest pricing when the window is not full. The window can not be empty, because the relevant methods are called with a new price as parameter.
Whether this is desirable depends on the expectations of who will be evaluating the solution, but I find it helpful to limit the number of null-pointers in circulation. It also allows for some code merging.
Putting it together: Squeaking these changes into place cut wall time to about 25% in my local measurements.
@Override | {
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quantum-field-theory, symmetry, quantum-electrodynamics, gauge-invariance, cpt-symmetry
Title: Is it possible to have renormalisable C-, P-, or T-violating terms in QED? In a recent paper (1710.01791), Witten claims that there are no renormalisable $\mathrm C,\mathrm P,\mathrm T$-violating terms that can be included in the QED Lagrangian: | {
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turing-machines
You can actually construct an equivalent TM that uses at most one more state and requires at most the same number of steps.
The sketch is as follows:
consider a transition of the original TM that keeps the head still. This transition writes $\alpha$ to the current tape cell and changes the state to $q$. Then you can find the next transition that will be executed by the machine, i.e., the transition performed when the current tape symbol is $\alpha$ and the current state is $q$.
Suppose that the latter transition writes $\beta$ and moves to $q'$.
If the transition also moves the head, then you can "shortcut" the first transition by directly writing $\beta$ and moving to $q'$. | {
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Let $K$ be any field and let $t$ be an indeterminate. For each choice of integer $n\geq 1$, we get a different subfield $K(t^n)\subseteq K(t)$.
In case you're not familiar with it, here is the definition of $K(X)$: it is the field of fractions of $K[X]$, the ring of polynomials in the indeterminates $X$ over $K$. That is, the elements of $K[X]$ are polynomials where the coefficients are elements of $K$, and the "variables" or "indeterminates" are the elements of $X$. Thus for example, if $K=\mathbb{Q}$ and $X=\{s,t\}$, then $K[X]=\mathbb{Q}[s,t]$ consists of polynomials such as $\frac{1}{2}+5s+t^2+3st$, and $K(X)=\mathbb{Q}(s,t)$ consists of rational functions such as $$\frac{\frac{1}{2}+5s+t^2+3st}{7-3s^3+4st^2}.$$ | {
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ros, cmake, catkin-make-isolated
isolated/controller_manager_msgs;/wg/stor2a/ahendrix/hydro/devel_isolated/control_toolbox;/wg/stor2a/ahendrix/hydro/devel_isolated/audio_play;/wg/stor2a/ahendrix/hydro/devel_isolated/audio_capture;/wg/stor2a/ahendrix/hydro/devel_isolated/audio_common_msgs;/wg/stor2a/ahendrix/hydro/devel_isolated/audio_common;/opt/ros/hydro | {
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"tags": "ros, cmake, catkin-make-isolated",
"url": null
} |
human-biology, immunology
Title: Are there any viruses or bacteria which have evolved to withstand higher temperatures due to fever My question was raised after receiving this information:
The primary reason the body raises its temperature (via the Hypothalamus in this case) is that bacteria and viruses tend to optimally thrive at 98.6F, which is also your body's optimal operating temperature. | {
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data, computer
http://en.wikipedia.org/wiki/Flash_memory
which evolved from EEPROM memory chips. Flash memory is composed of flash memory cells which primary contain the floating-gate transistors; such cells have a control gate (CG) and de facto isolated floating gate (FG). FG can keep electric charge for years and the electric field from that affects the threshold voltage of the cell (voltage at which the cell becomes conductive). Some extra operations are needed to erase and rewrite the chips which can't be done indefinitely; after less than a million of erasures, the cell breaks down.
There are many other physical mechanisms in which the information is being stored. All the "relatively easily" rewritable technologies use some kind of variable electric or magnetic fields that may be produced by matter. Compact disks (CD/DVD) use optical information and so on. | {
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. . $(a)\;\tfrac{8}{33} \qquad(b)\;\tfrac{62}{165} \qquad (c)\;\tfrac{17}{33}\qquad (d)\;\tfrac{103}{165} \qquad (e)\;\tfrac{25}{33}$
We will find the probability that there are no matching pairs among the four cards.
The first card can be any card (it doesn't matter): $\tfrac{12}{12}$
The second card can be any of the other 10 non-matching cards: $\tfrac{10}{11}$
The third card can be any of the other 8 non-matching cards: $\tfrac{8}{10}$
The fourth card can be any of the other 6 non-matching cards: $\ftrac{6}{9}$
Hence: . $P(\text{no match}) \;=\;\frac{12}{12}\cdot\frac{10}{11}\cdot\frac{8}{ 10}\cdot\frac{6}{9} \:=\:\frac{16}{33}$
Therefore: . $P(\text{at least one match}) \;=\;1-\frac{16}{33} \;=\;\frac{17}{33}$ | {
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"url": "http://mathhelpforum.com/statistics/208657-combinations-probability-question-please-help-print.html"
} |
operating-systems, virtual-memory
1. How many pages can be stored in physical memory at once?
2^(24-10) = 2^14
2. How many entries are ther in the page table?
2^22
3. How many bits are necessary per entry in the page table? (Assume each entry has PPN, residentbit, dirty bit)
16
4. How many pages does the table require?
2^23 bytes = 2^13 pages
5. What is the largest fraction of VM that might be resident?
1/(2^8)
6. A portion of the page table is given to the left. What is the physical address for virtual address 0x1804?
VPN = 6 --> PPN = 2 --> PA = 0x804
$$\begin{array}{c|c|c|}
\text{VPN} & \text{D} & \text{R} & \text{PPN} \\ \hline
\text{0} & 0 & 0 & 7 \\ \hline
\text{1} & 1 & 1 & 9 \\ \hline
\text{2} & 1 & 0 & 0 \\ \hline
\text{3} & 0 & 0 & 5 \\ \hline
\text{4} & 1 & 0 & 5 \\ \hline
\text{5} & 0 & 0 & 3 \\ \hline
\text{6} & 1 & 1 & 2 \\ \hline
\text{7} & 1 & 0 & 4 \\ \hline
\text{8} & 1 & 0 & 1 \\ \hline
\text{...} & ... & ... & ... \\ \hline
\end{array}$$
Could you help me with this two exercises? | {
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to power. The function f is differentiable on the interval b. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. This is an advantage in physical applications where one is dealing with very small numbers or a small difference between two functions. 9 Representation of Functions by Power Series 671 Operations with Power Series The versatility of geometric power series will be shown later in this section, following a discussion of power series operations. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. A power series [centered at x = 0] is a series of the form X1 n=0 c nx n = c 0 + c 1x+ c 2x2 + c 3x3 + c 4x4 + ::: (Informally, we can think of a power series as an \in nite polynomial. Josh Taylor vs Regis Prograis RESULT: Taylor wins World Boxing Super Series to become unified champion. 812) that the series of Example 11. In another video, I will find a Taylor series expansion, so look for | {
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statistical-mechanics, fluid-statics
If the cereal clumps touching he sides of the bowl, the buoyant pieces at the sides are climbing the meniscus, caused by milk's attraction to glass or porcelain. Other pieces of cereal then gravitate toward the slight depression away from the meniscus caused by each piece floating at the side. If the milk is repelled by the sides of the bowl, cereal gravitates toward the inverse meniscus which curves downward at the sides.
The mechanism is different for small objects that are heavier than water but are supported by surface tension.
Vella and Mahadevan think that buoyancy, surface tension and the meniscus explain the cheerios effect.
The hexagonal pattern is the best arrangement to fit the most pieces of cereal into a given surface area. The hexagon is one of only three regular polygons which can tile the Euclidean plane. The other two are the triangle and the square, but they do not provide the densest arrangement of cereal. | {
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quantum-mechanics, electrons, schroedinger-equation, discrete, orbitals
If I draw a comparison with a similar orbital system, Total energy of a satellite revolving earth is given by
$E=-\frac{Gm_1m_2}{2r}$
And it is continuous.
I understand that Schrodinger equation is a wave equation and from linked question I take that confined strings can have quantized number of hops. However an electron in an isolated atom is of course excited at one end by potential supplied by nucleus but what confines electron at other end?
I'd like to draw your attention to another similar analogy. Please consider wave equation of EM wave
$\nabla^2E=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2} $
Also has a solution with continuous angular momentum and energy.
We know quantized energy levels are proportional to $\frac{1}{n^2}$ and summation of this series is finite. Is this a reason for energy levels being discrete? (Atom cannot supply infinite energy) If this is so why this doesn't apply to gravitational system? | {
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botany, plant-physiology, development
Note that it says nothing about the tree's age, simply stating that the wood must be healthy. That's because old wood/young wood produce the same result in the new trees. I was using a 40 year old razor russet to graft onto all the Malling V rootstocks I had for that variety. And of course they grew as expected, as a young sapling should. I've heard of a nearby nursery taking budwood from a 100 year old tree that still looked good. | {
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scala, exception-handling
What we see here is a monad.map(f).flatten which is the same as monad.flatMap(f) which is the same as for (m <- monad) yield f(m). Applying this fact to your code snippet results in:
def processTmpFile(as: List[A], bs: List[B]): Try[Unit] =
if (values.isEmpty)
Success(())
else for (file <- createTmpFile(fields, values))
yield runProcessWithFile(file)
def run(): Try[Unit] = {
for {
fields <- getFields()
values <- getValues()
} yield processTmpFile(fields, values)
}
Operating on a deeper level of the monad (what you did with your condition) means that you can either nest the monad and unpack it later at the end or that you can unpack it immediately after the nesting happened (that is what a for comprehension does). | {
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python, python-3.x, game
player.lvl_up()
continue
elif choice > len(self.locations) or choice <= 0:
cls()
print('--- Invalid choice')
continue
cls()
self.locations[choice - 1].visit(player)
class Shop(Location): | {
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discrete-signals, periodic
Title: Very simple question about signal periodicity $$x[n] = u[n]+u[-n]$$
Is it periodic or not? My answer is
$$u[n] = {1 , n\geqslant0}$$
$$u[-n] = {1 , n\leqslant0}$$
which means that the signal $x[n]$ is always equal to $1$ from $-\infty$ to $+\infty$, but it equals $2$ at $n=0$. Am I right? For the step function $u[n]$ defined as $$u[n]=\begin{cases}1\text{ $n\geq0$,}\\ 0\text{ otherwise,}\end{cases}$$ the function $x[n]=u[n]+u[-n]$ is given by $$x[n]=\begin{cases}2\text{ $n=0$,}\\1\text{ otherwise.}\end{cases}$$ Clearly the signal $x[n]$ is not periodic. | {
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What is the remainder of $(14^{2010}+1) \div 6$?
What is the remainder of $(14^{2010}+1) \div 6$?
Someone showed me a way to do this by finding a pattern, i.e.:
$14^1\div6$ has remainder 2
$14^2\div6$ has remainder 4
$14^3\div6$ has remainder 2
$14^4\div6$ has remainder 4
And it seems that when the power is odd, the answer is 2, and when it's even, the answer is 4.
2010 is even, so the remainder is 4, but we have that +1, so the final remainder is 5. Which is correct.
But this method doesn't seem very concrete to me, and I have a feeling the pattern may not be easy to find (or exist?) for every question. What theorem or algorithm can I use to solve this instead?
- | {
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homework-and-exercises, statistical-mechanics, partition-function
Title: Help in calculating Partition function and choosing the ensemble I've been given this situation
"A surface contains $N$ identical atoms in a fixed position. Every atom can occupy one of two states with energies $E_1$ or $E_2$ and the temperature is $T$." | {
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ros-kinetic, turtlebot3
Also, when building packages from source, It is recommended to use the rosdep command. This command installs all the packages that the packages in your catkin workspace depend upon but are missing on your computer.
rosdep install --from-paths /path/to/your/catkin_ws/src --ignore-src -r -y
Originally posted by Pyo with karma: 358 on 2018-04-16
This answer was ACCEPTED on the original site
Post score: 0 | {
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biochemistry, amines, drugs, medicinal-chemistry, pharmacology
* The table on Wikipedia does actually show one example where the methyl group is modified to an ethyl group. However, I guess that would still be considered a very small change, compared to the piperidine ring in Ritalin. | {
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powershell
Second is that if your .psm1 file has but one purpose, and that is a singular function, why not just make it a script with parameters? Then you don't have to first import it then execute it.
Contents of remote script on server c:\CreateNewObject.ps1:
Param([Parameter(Mandatory=$true)]$firstStanza,[Parameter(Mandatory=$true)]$secondStanza)
return ($firstStanza + $secondStanza).GetEnumerator() | sort -Property name
Putting both of those into use, you end up with something like:
$firstStanza=@{
a = "Mary had"
b = "a little"
c = "lamb"
}
$secondStanza=@{
d = "its fleece"
e = "was white"
f = "as snow"
}
$scriptBlock = {
. C:\CreateNewObject.ps1 -firstStanza $using:firstStanza -secondStanza $using:secondStanza
}
Invoke-Command -ComputerName serverName.internalDomain.com -ScriptBlock $scriptBlock | {
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# Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel to (a) X axis (b) parallel to Y axis.
Find the points on the curve $x^2+xy+y^2=7$ where tangent is parallel to (a) X axis (b) parallel to Y axis.
For part (a) if we differentiate w.r.t $x$ we get $y'=\large-\frac{2x+y}{x+2y}$, this when equated to zero (parallel to X axis means slope is zero) gives $x=-y/2$, this when substituted in the equation of the curve gives $\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right),\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)$
For part (b) we find $dx/dy$, equate it to zero and find that the points are $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right),\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$
$\underline{\text{My question}}$ is if we consider two points on the curve $(x_1, y_1),(x_2, y_2)$ such that $x_1 = x_2$, because for the line parallel to Y axis coordinates of x remain the same. Substituting these two points in the equation of the curve we get | {
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discrete-signals, frequency-spectrum, frequency, frequency-domain
Why do we lose correlation versus frequency offset? Consider the correlation process sample by sample, and frequency as a rotating phasor on the complex plane. When the two signals are at the same frequency, the complex conjugate multiplication in the correlation formula causes the resulting phasor from each product to be aligned as in the graphic below, and the resulting summation grows to a very large number (such as the scaling by $N$ in the DFT!):
However if there is a slight frequency offset, the resulting phasors from each product will have an increasing rotation from sample to sample. When we are off in frequency by $1/T$, the total summation over that time will sum to 0!. | {
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ros, slam, ros-melodic, hector-localization, 3dslam
In comparison to the backpack tutorial (with 2 velodyne scanners), I have a simple horizontal lidar.
Accordingly, I have changed the parameters:
num_laser_scans = 1,
num_point_clouds = 0
To create the tf, we call the robot urdf file with the robot state publisher in the launchfile
<launch>
<param name="robot_description"
textfile="$(find robot_navigation)/urdf/robot_3d.urdf" />
<node name="robot_state_publisher" pkg="robot_state_publisher"
type="robot_state_publisher" />
<node name="cartographer_node" pkg="cartographer_ros"
type="cartographer_node" args="
-configuration_directory $(find robot_navigation)/configuration_files
-configuration_basename robot_3d.lua"
output="screen">
<remap from="imu" to="imu/data_raw" />
<remap from="scan" to="sick_s300_laser/scan" />
</node>
<node name="cartographer_occupancy_grid_node" pkg="cartographer_ros"
type="cartographer_occupancy_grid_node" args="-resolution 0.05" />
</launch> | {
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java, datetime
public boolean within(Time fromIncluded, Time toIncluded) {
return (!fromIncluded.after(this)) && (!this.after(toIncluded));
}
} I'd name the class TimeOfDay to be absolutely clear about its purpose.
Consider adding validation to the constructor (0 ≤ hour ≤ 23 and 0 ≤ minutes ≤ 59).
Because before() and after() are predicates, I'd rename them to isBefore(TimeOfDay t) and isAfter(TimeOfDay t).
I prefer getHours() over hours(). | {
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python, svm, libsvm
I was wondering is there a way I can change implement this in sklearn, its for a paper that im working.
Kind regards. This is already implemented, with the sample_weights parameter of the fit method. They play the role of your $\nu_i$. | {
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slam, navigation, ros2, opencv, gpu
Is there a different git for rtabmap with GPU support (I used branch foxy-devel)?
Shoud I do a clean install with CUDA 10.x, OpenCV3 and try a new build of rtabmap?
Am I completely missing out on some other issue, that might block the GPU-Features on my system? (Standard programs using OpenCL or T-API are working just fine on the recent setup, so the problem seems to me to be centered around rtabmap itself, not the availability of GPU for computational purposes ...)
Any helpful suggestion will be greatly appreciated. | {
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python, beginner, object-oriented, unit-testing
Title: Bowling scoring kata I've been writing basic Python scripts for a while now to help process data or automate some task but I've decided I should start picking up unit testing and objective orientated programming (the vast majority of my scripts so far have been procedural).
As a starter I decided to follow along with Uncle Bob's bowling scoring kata to try and get my mind around TDD and the idea of writing the absolute minimal code at every step to either make the test go red or green (plus any refactoring steps).
As it's a bare bones example of following TDD the main program doesn't actually have an entry point other than via the tests.
Things that stand out to my beginner's eye: | {
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machine-learning, deep-learning, nlp
Then after you created, email addresses with all possible (almost) patterns you can get help from Machine Learning techniques. So the model might give you a probability with the relevant name and surname. (Also, you can configure the output so that it would give you top n name and surnames which have higher probability)
Another thing that is needed to be considered is the possibility of two different persons having the same name and surname. Lastly, independently from using which approach your solution cannot be perfect because, for example, it is not possible to understand whether the character 'j' stands for John or Jake in the email address. Thus, if you can integrate the email content into your solution, that will increase the performance drastically.
Update Accordingly: Check this answer which does not exactly answer your problem, but the context is the same. | {
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P(m|M)=C^M_mp^m(1-p)^M,\quad 0\leq m\leq M \end{align} and zero otherwise. We assume that total number of bookings has uniform probability distribution: \begin{align} P(M)=\frac{1}{N'+1},\quad 0\leq M\leq N' \end{align} The probability thatmpeople show up irrespective of how many have booked is: \begin{align} P(m)=\sum_{M=m}^{N'}P(M)P(m|M)=\frac{1}{N'+1}\sum_{M=m}^{N'}C^M_mp^m(1-p)^M,\quad 0\leq m\leq N' \end{align} and zero otherwise. The summation begins fromM=m$because for$M<m,P(m|M)=0. The expected loss is therefore: \begin{align} L & =\sum_{m=0}^{N-1}P(m)~(N-m)S \\ & =\frac{S}{N'+1}\sum_{m=0}^{N-1}(N-m)\sum_{M=m}^{N'}C^M_mp^m(1-p)^M \end{align} in which the summation onm$ends at$N-1$because for$m>N-1\$ the airline suffers zero loss of money. | {
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python, performance, python-2.x, iterator, iteration
quiz[ str(activitiesDict[actClass][classSec]["requirements"][requirement][requirementInfo]) ]["variable"] = "False" #Not changing
selectedClasses[ str(activitiesDict[actClass][classSec]["title"]) + " Quiz " + str(activitiesDict[actClass][classSec]["requirements"][requirement][requirementInfo])[-1] ] = quiz # Add freshman quiz | {
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machine-learning, deep-learning, neural-network, computer-vision, convolutional-neural-network
Overfitting doesn't seem to be a problem with either model: training and validation sets tend to have similar performance, give or take 5%
I'm using initial = tf.truncated_normal(shape, stddev=0.1) to initialize all weights
I'm using initial = tf.constant(0.1, shape=shape) to initialize all biases
It could be specifically the convolution that's the problem, since a shallow single-hidden-layer convnet did poorly but a two-hidden-layer fully-connected neural net with no convolution gets around 65% accuracy | {
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"url": null
} |
quantum-chromodynamics, symmetry-breaking, chirality
$$ F_0\neq 0 \color{red}\Leftrightarrow \text{ChSB}\quad\text{and}\quad \langle \bar qq\rangle\neq 0 \color{red}\Rightarrow \text{ChSB} $$ Yes, you understand correctly,
$$ F_0\neq 0 \qquad \color{red}\Leftrightarrow \qquad\chi SB \tag{↯} \\
\langle \bar qq\rangle\neq 0 \qquad\color{red}\Rightarrow \qquad\chi SB . $$
In the latter case, of course, a charge annihilating the vacuum in (3.18) dictates a vanishing r.h.s.
However, conversely, imagine a (complete, fantasy) contrapositive hypothetical, in which
$$
\langle \bar qq\rangle = 0
$$
but, for whatever unrealistic reason, you also have
$$
\langle \bar uu + \bar dd -2\bar ss \rangle =c\neq 0, \qquad \Longrightarrow \\
\langle \bar uu + \bar dd \rangle = \langle 2\bar ss \rangle +c,
$$
which is to say $Q^A_8|0\rangle \neq 0$; so, for $c=-3 \langle \bar ss \rangle $ you may still have $\chi SB$, cf. (3.17), a=8. | {
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co.combinatorics, notation
I would recommend that you treat multisets as first-class entities, which have characteristic functions, but are not the same as characteristic functions.
That is, that you should keep the distinction between the collection and the function, unless you feel that identifying the two is very useful or important. | {
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Imaginary Numbers are not "imaginary", they really exist and have many uses. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The square root of minus one â(â1) is the "unit" Imaginary Number, the equivalent of 1 for Real Numbers. Based on the mathematical number Step 1 number translation, English definition... Beginning to end 0, 10 j ( because i '' already means current, but they ’! Purely imaginary if it has no real part is equal to 0. generally ' i '.! The # 1 tool for creating Demonstrations and anything technical this tutorial shows you steps... Please make sure that the real number, you get a complex number whose real part, such 5i... Form a + bi music is playing, you get a complex number a number. result! Is expressed as any real number, i, has an interesting property more than one of these description apply! ' is n't a variable, it 's an imaginary number. ( because i '' already current. And a real part, such | {
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"url": "https://s147164.gridserver.com/innovative-industrial-lgdzje/393838-pure-imaginary-numbers-examples"
} |
electromagnetism, electricity, electric-circuits, inductance
Title: Inductor with ac supply When an ac source is connected with an inductor we have a continuous change in voltage and current across the inductor.
According to farady and lenz law we will get an equal and opposite voltage against the source voltage for each change of voltage or current across the inductor.
Now since we always get an equal and opposite voltage across the inductor for each change in source voltage so it means that the current will never be able to flow in the inductor but yet current is always flowing in each text book. I am confused about it. Please guide me. I shall be grateful to you. Note that it is an ideal case and we have no resistance in the circuit.
Now since we always get an equal and opposite voltage across the inductor for each change in source voltage so it means that the current will never be able to flow in the inductor . . . . | {
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quantum-mechanics, perturbation-theory, approximations, fermis-golden-rule
Title: Why is it intuitively unreasonable for this transition probability to grow quadratically in $t$? In Sakurai's "Modern Quantum Mechanics" section 5.6, there is a seemingly simple statement made that I do not understand the logic of. The author is considering a physical situation in which we "turn-on" a time-independent potential at $t=0$, and ask what the relevant transition probabilities are.
$$\hat{V}(t)=\begin{cases}0 &,t<0 \\ \hat{V} &, t\geq 0\end{cases}$$
If the system under consideration begins in state $|i\rangle$, then to first-order (in the Dyson series), the transition amplitude between that initial state and a final state $|n\rangle$ as a function of time is:
$$c^{(1)}_f(t)=-\frac{i}{\hbar}\int_0^tdt'\,e^{i\omega_{ni}t}V_{fi}=\frac{2V_{ni}}{\hbar\omega_{ni}}e^{\frac{1}{2}i\omega_{ni}t}\sin\left(\frac{\omega_{ni}t}{2}\right)$$
or, more usefully, the transition probability as a function of the energy difference $\Delta E\equiv \hbar \omega_{ni}$: | {
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and the second is
$4 \displaystyle \int_{\substack{\xi, \eta>0 \\ \xi \eta \leq 1}} \frac{ d \xi \, d \eta}{(1+\xi^2)(1+\eta^2)}$;
They are equal by a change of variables. For the first quantity, expand $(1-xy)^{-1}$ as a geometric series and integrate term-wise to get $3 \cdot \zeta(2)$. The second can be computed to $\pi^2/2$. This can be found in detail in Kontsevich and Zagier's Periods (bottom of page 8), where they attribute the idea to Calabi. (You should easily be able to hunt down an identity of $\zeta (n)$ being equal to an integral over the unit square in $\mathbb{R}^n$, like the first above. It might be a fun exercise to see if this idea is can be adapted for even $n$.)
One geometric/probabilistic proof is by Eugenio Calabi. | {
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## Use Calculator to Calculate Angle Bewteen two Vectors in Spherical Coordinates
1 - Enter the spherical coordinates $\rho_1$ , $\theta_1$, $\phi_1$ of point $P_1$, and the spherical coordinates $\rho_2$ , $\theta_2$, $\phi_2$ of point $P_2$, selecting the desired units for the angles, and press the button "Calculate". You may also change the number of decimal places as needed; it has to be a positive integer.
$\rho_1 =$ 1 $\theta1 =$ 45 degrees radians $\phi_1 =$ 45 degrees radians $\rho_2 =$ 2 $\theta2 =$ 90 degrees radians $\phi_2 =$ 30 degrees radians
Number of Decimal Places =
$\alpha =$ (degrees)
$\alpha =$ (radians) | {
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ros, navigation, odometry, base-link, robot-localization
The way it computes those things is via sensor fusion. If you have wheel encoders that say you are driving at 1.0 m/s forward, then the EKF integrates those velocities over time, so after 10 seconds, it says you went 10 meters straight forward. If you also have an IMU that says you are turning with some velocity, it integrates those at the same time, so if your wheel odometry forward velocity and IMU angular velocity are constant, the EKF will output your path as an arc, because it's integrating the velocity measurements from both sensors.
It also can fuse pose data directly, i.e., rather than integrating velocities, we directly measure poses and the EKF uses those to directly update its state estimate.
Originally posted by Tom Moore with karma: 13689 on 2018-01-26
This answer was ACCEPTED on the original site
Post score: 7 | {
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lagrangian-formalism, conservation-laws, symmetry, conformal-field-theory, noethers-theorem
\sqrt{g}\,\omega(x)T_{\mu\nu}g^{\mu\nu}.
\end{cases}$$
Since conformal symmetry implies that the trace of $T$ vanishes, we see that
$$J^{\mu}=\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}$$
is conserved. This becomes the statement you want to prove when we take the flat limit. | {
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kinematics, velocity, distance
2. solution:
We can calculate the distance between two points using pythagorean theorem: $d=\sqrt{(v_1t)^2+(v_2t+\frac{a_2t^2}{2})^2}$
$\textbf{It's derivative is relative velocity.}$
If we now graph those two functions, we don't get the same graph:
Why is there a diference between those two methods?
The distance between two points... it’s derivative is relative velocity. | {
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navigation, ekf, robot-localization
Original comments
Comment by shashank on 2016-12-07:
according to my understanding your are saying that my frame ids should be ( base_link-->laser_odom , base_link--> camera_odom, base_link-->wheel_odom) . The ekf_localization should publish me odom-->base_link. Am i correct ???
Comment by paulbovbel on 2016-12-07:
that's correct
Comment by Tom Moore on 2016-12-19:
Note that the odometry messages/inputs can have the same frame_id, but unless the measurements are in agreement with each other (e.g., odom0 and odom1 both provide measurements that don't diverge), it's going to produce ugly results. I try to stick to one pose and N velocity inputs.
Comment by shashank on 2016-12-20:
So u are saying i should choose the pose input from the most reliable source of odometry(say laser odometry) and for rest odometry (camera,wheel) source i should only choose twist(velocity) messages and omit out pose. Am i correct??
Comment by Tom Moore on 2017-01-04:
Yes, correct. | {
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CHAIN RULE . This is just a review, how does this relate to u-substitution? Integration by substitution allows changing the basic variable of an integrand (usually x at the start) to another variable (usually u or v). When it is possible to perform an apparently difficult piece of integration by first making a substitution, it has the effect of changing the variable & integrand. If you're seeing this message, it means we're having trouble loading external resources on our website. Times, actually, I'll do this in a, let me do this in a different color. Which one of these concepts is not part of logistical integration objectives? would be to put the squared right over here, but I'm This exercise uses u-substitution in a more intensive way to find integrals of functions. This calculus video tutorial provides a basic introduction into u-substitution. what's the derivative of that? And if you want to see it in the other notation, I guess you So when we talk about bit closer, g of f of x, g | {
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newtonian-mechanics, reference-frames, inertial-frames
Title: How to prove that a particle or a body moves in a circular trajectory with respect to centre of mass? In some situations in mechanics, observing motion of bodies with respect to centre of mass often gives useful insights to visualize the situation and obtain many results.In some cases the motion of particles/rigid bodies with respect to centre of mass (COM) is circular. I would like to illustrate with the help of an example: | {
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microbiology, homework
The mec gene complex is composed of mecA, its regulatory genes, and
associated insertion sequences. The class A mec gene complex (class A
mec) is the prototype complex, which contains mecA, the complete mecR1
and mecI regulatory genes upstream of mecA, and the hypervariable
region (HVR) and insertion sequence IS431 downstream of mecA. The
class B mec gene complex is composed of mecA, a truncated mecR1
resulting from the insertion of IS1272 upstream of mecA, and HVR and
IS431 downstream of mecA. The class C mec gene complex contains mecA
and truncated mecR1 by the insertion of IS431 upstream of mecA and HVR
and IS431 downstream of mecA. There are two distinct class C mec gene
complexes; in the class C1 mec gene complex, the IS431 upstream of
mecA has the same orientation as the IS431 downstream of mecA (next to
HVR), while in the class C2 mec gene complex, the orientation of IS431
upstream of mecA is reversed. C1 and C2 are regarded as different mec | {
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c#, unit-testing, bdd
public class Chestnut : Product
{
public override bool Check(Customer customer)
{
if (customer._expenditure > 100 && customer._gender =="M")
{
return true;
}
else
{
return false;
}
}
}
public abstract class Product
{
public abstract bool Check(Customer customer);
}
public interface IOfferCalculator
{
IEnumerable<Product> Calculate(Customer customer, IList<Product> products);
}
public class Customer
{
public string _gender { get; protected set; }
public decimal _expenditure { get; protected set; }
public IList<Product> _assignedProducts = new List<Product>();
public Customer(string gender, decimal expenditure)
{
_gender = gender;
_expenditure = expenditure;
} | {
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java
testCollection.add(test01);
testCollection.add(test02);
testCollection.add(test03);
testCollection.add(test04);
testCollection.add(test05);
testCollection.add(test06);
The integerList has indexes which are present on testCollection. I want to filter out the testCollection which will have object consisting the index 1, 3, 5. Index 6 is not present on object collection.
I wrote the code as below example. Is there any better as like Java 8 way?
List<Test> testList = new ArrayList<>(testCollection);
List<Test> newTestList = new ArrayList<>();
for (Integer integer : integerList) {
for (int j = 0; j < testList.size(); j++) {
if (integer == j) {
newTestList.add(testList.get(j));
}
}
}
System.out.println(newTestList);
It will have the following output as result:
[Test{id=1, name='B'}, Test{id=3, name='D'}, Test{id=5, name='F'}] | {
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ros-indigo, raspberrypi, camera
Originally posted by Amrac on ROS Answers with karma: 11 on 2015-07-10
Post score: 0
Original comments
Comment by allenh1 on 2015-07-11:
Could you please be a little more specific with your commands? It would greatly help us figure out the best way to assist if we knew more. What does the rosinstall command output? Have you tried rosdep install image-transport?
The problem is that pesky underscore.
allenh1@tuxbox $ rosdep what-needs image_transport
compressed_depth_image_transport
image_rotate
image_proc
rviz
image_view
compressed_image_transport
camera_info_manager
rqt_image_view
theora_image_transport
polled_camera
stereo_image_proc
depth_image_proc
You should be able to run rosdep install image_transport and get these resolved.
Originally posted by allenh1 with karma: 3055 on 2015-07-12
This answer was ACCEPTED on the original site
Post score: 1 | {
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## Input Arguments
collapse all
Condition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.
Example: x > 2
Value when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Value if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If otherwiseVal is not specified, its value is NaN.
## Output Arguments
collapse all
Piecewise expression or function, returned as a symbolic expression or function. The value of pw is the value val of the first condition cond that is true. To find the value of pw, use subs to substitute for variables in pw.
## Tips | {
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"openwebmath_score": 0.8195955753326416,
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"url": "https://fr.mathworks.com/help/symbolic/piecewise.html"
} |
homework-and-exercises, symmetry, tensor-calculus, elasticity, stress-strain
Title: Why are the components of elasticity tensor 21? I know there's a duplicate but I didn't understand the whole answer which was:
A 6×6 matrix has 36 different components. When you reduce it to a symmetric case it has 1+2+3+4+5+6=21, where we are summing the number of entries without double counting. | {
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us insight into the nature of the relationship between the dependent and the independent variables. So, clearly, the units don’t automa And the law for centripetal force is: F = m v 2 r. P none The relationship between an object’s mass and its terminal velocity Report by Victor Jeung, Terry Tong, Cathy Liu, Jason Feng relationship between the speed and the mass is a square root relation. Solve for the required rate of change of the square. Last ride. 93 1. For example, take 2 otherwise identical planes flying at the same speed, but increase the chord on one of them by 2x. Lift and drag increase exponentially with speed—if speed is doubled, drag or lift will be quadrupled. Continue the takeoff on the remaining engines and compute the additional distance for the vehicle to clear a 50 ft obstacle, determining the takeoff distance 3. The relationship for the kinetic energy per unit volume of water is thus proportional to its velocity and can be expressed as:. Since power is speed times | {
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"openwebmath_score": 0.7215793132781982,
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"url": "https://smartclinic2.diapindia.org/wp-content/uploads/2022/01/rhl9jwj/the-increase-of-speed-is-a-squared-relationship.html"
} |
quantum-mechanics, quantum-spin, quantum-entanglement
Title: Which is the difference between entangled spins and classical coins? I'm trying to understand why entanglement has no classical analog.
I imagine the usual well known example: Charlie prepares two spins in the entangled states. Alice and Bob measure them using their own detectors.
Why the results cannot be explained as for a classical couple of coins (head/tail)?
My understanding so far is that - if Alice and Bob measured only the z components, the two situations (quantum one and classical analog) would be really indistinguishable: when Alice measures -1 then Bob gets +1 and viceversa. | {
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string-theory, ads-cft, discrete, supergravity, duality
The approximation of the moduli space by "one point" is basically equivalent to assuming that the charges of all the brane-like objects are continuous, not quantized, and ignoring to the Dirac quantization rules and all similar quantum effects. This approximation is good enough at "low energies". When we consider the low-energy limit, the only objects that carry such charges are "large branes" that look like macroscopically large extended black holes. And those carry so huge charges, relatively to the minimum units, that these charges may be considered continuous. | {
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java, primes
if (num % 2 == 0 || num % 3 == 0 || num % 5 == 0) {
return false;
}
long limit = (long)Math.sqrt(num);
for (long i = 7; i <= limit; i += 30) {
if (
num % (i+0) == 0 ||
num % (i+4) == 0 ||
num % (i+6) == 0 ||
num % (i+10) == 0 ||
num % (i+12) == 0 ||
num % (i+16) == 0 ||
num % (i+22) == 0 ||
num % (i+24) == 0
) {
return false;
}
}
return true;
} | {
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• Ohhhh got what you mean by "brushed". When I formally did it, I prove it, but for sake of this forum post, I "brushed" it to make it more direct! In a test or assignment, I would do it with a little more rigor! Thanks anyway for the hints to improve it mate – Bruno Reis May 22 at 5:18
• @BrunoReis Great! Then ignore points 3, 4. – Theo Bendit May 22 at 5:19 | {
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A square matrix $$Q$$ is called an intensity matrix if $$Q$$ has zero row sums and $$Q(x, y) \geq 0$$ whenever $$x \not= y$$.
Theorem 5.1
If $$Q$$ is a matrix on $$S$$ and $$P_t := e^{tQ}$$, then the following statements are equivalent:
1. $$P_t$$ is a Markov matrix for all $$t$$.
2. $$Q$$ is an intensity matrix.
The proof is related to that of Lemma 4.2 and is found as a solved exercise below.
Corollary 5.1
If $$Q$$ is an intensity matrix on finite $$S$$ and $$P_t = e^{tQ}$$ for all $$t \geq 0$$, then $$(P_t)$$ is a Markov semigroup.
We call $$(P_t)$$ the Markov semigroup generated by $$Q$$.
Later we will see that this result extends to the case $$|S| = \infty$$ under some mild restrictions on $$Q$$.
## 5.4. Jump Chains¶
Let’s return to the chain $$(X_t)$$ created from jump chain pair $$(\lambda, K)$$ in Algorithm 4.1.
We found that the semigroup is given by
$P_t = e^{tQ} \quad \text{where} \quad Q(x, y) := \lambda(x) (K(x, y) - I(x, y))$ | {
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python, python-3.x, mergesort
'shopping', 'sideboards', 'slate', 'sleeps', 'soaping', 'southwesters', 'stubbly', 'subscribers', 'sulfides', 'taxies', 'tillable', 'toastiest', 'tombstone', 'train', 'truculent', 'underlie', 'unsatisfying', 'uptight', 'wannabe', 'waugh', 'workbooks' | {
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"tags": "python, python-3.x, mergesort",
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computability, context-free, decidability
instance $\Pi$
question $I(\Pi)=\{0,\dots,n-1\}$? | {
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predictive-modeling, recommender-system
Title: POC - Get an idea to create a Predictive Model I'm trying to look for an idea to create a predictive model having the following data:
Customer_ID - Integer
Catalog_ID - Integer
Country_Code - Integet
Year - Integet
Month - Integer
Day - Integer
Quantity_Purchased - Integer
Product_Purchased - Double | {
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genetics, evolution, population-genetics, book-recommendation
Intuitively introduces key concepts.
Emphasis on modelling with examples of problems and their solutions.
Relatively short (I'm planning to read from cover to cover). TL;DR
I'd recommend Population Genetics: A Concise Guide (Gillespie) for an introduction to population/evolutionary genetics (thanks AGS for highlighting this big miss of mine on the first version of my answer).
I'd recommend A Biologist's guide to mathematical modelling in evolution and ecology (Otto and Day) if you want to ensure your knowledge in mathematics by learning their application to evolutionary biology.
General Entry Books to Population Genetics | {
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"tags": "genetics, evolution, population-genetics, book-recommendation",
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string-theory, experimental-physics, quantum-gravity, theory-of-everything, loop-quantum-gravity
Certainly What I find particularly odd is that string theory is not listed under "fully compatible with observation" but rather "disputed" -- which would be at odds with the second, weaker interpretation (since general relativity arises as an effective theory of string theory at the low-energy classical limit).
To add to the confusion, Loop Quantum gravity is listed under "Experimentally constrained", although it contradicts Lorentz symmetry 1. I suppose similar comments apply to "BEC vacuum theory".
Can anyone make any sense out of what these terms mean here? "Fully compatible with observations" is a rather vague statement. Actually, two aspects of adequacy to reality have to be distinguished when a new theory reaches a degree of explicitation. These are | {
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# Definition:Submatrix
## Definition
Let $\mathbf A$ be a matrix with $m$ rows and $n$ columns.
A submatrix of $\mathbf A$ is a matrix formed by selecting from $\mathbf A$:
a subset of the rows
and:
a subset of the columns
and forming a new matrix by using those entries, in the same relative positions, that appear in both the rows and columns of those selected.
## Notation
A submatrix of $\mathbf A$ is denoted as follows.
Let:
$\left\{ {a_1, a_2, \ldots, a_r}\right\}$ be the indices of the $r$ selected rows
$\left\{ {b_1, b_2, \ldots, b_s}\right\}$ be the indices of the $s$ selected columns
where all of $a_1, \ldots, a_r$ are between $1$ and $m$, and all of $b_1, \ldots, b_s$ are between $1$ and $n$.
Then the submatrix formed from rows $\left\{ {a_1, a_2, \ldots, a_r}\right\}$ and columns $\left\{ {b_1, b_2, \ldots, b_s}\right\}$ is denoted as:
$\mathbf A \left[{a_1, a_2, \ldots, a_r; b_1, b_2, \ldots, b_s}\right]$ | {
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