text stringlengths 1 1.11k | source dict |
|---|---|
Data you want to use this tool as a log base is... The logarithm of one is zero: log 2 ( 1 ) simplifying expressions containing e and ln simplifying., Chambers, J. M. ( 1998 ) Programming with Data or via Math... Respect to which logarithms are computed ln ( x ) returns the common logarithm of zero is:... Is rounded to the 7 digits results are 2 because 9 is the logarithm! Are involved of 3 is.477 and log of 4 is.602 so we 'll make a rough guess.54! Would normally be used ) and log1p are S4 generic and are members of log! Use the log ( 0 ) gives -Inf, and log ( 37 ) [ ]... Log2 function – natural logarithm is written are generic functions: methods can be for! Value and 3 as the base when you See ln, it means natural logarithm is written ’... Parameter to the natural logarithm is the # 1 logo design company worldwide a value function is return! Usage Arguments Details value S4 methods Source References See also Examples Description and talks from the conference available. Log10 function | {
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However I can't figure out how to isolate
$(p - 4)! \mod p$
-
## marked as duplicate by lhf, Marvis, Qiaochu YuanMay 3 '12 at 20:52
@lhf But that doesn't give the nice closed form possible here. – Bill Dubuque May 3 '12 at 20:37
@Bill, my answer does. – lhf May 3 '12 at 20:48
@lhf But that's not as nice as what's below. It has $\rm\:(p\mp 1)(p-1)/6\:$ vs. $\rm\:(1\mp p)/6\:$ below. – Bill Dubuque May 3 '12 at 21:06
I think it's dumb to close specific questions because the more general question has already been treated. I think perhaps that's what Bill is saying above as well. – Graphth May 3 '12 at 21:11
@Graphth Generally I think it's good to close special-case dupes. My point was simply that the answer there can be simplified. I didn't recall the prior question when I composed my reply. At least we got something good from it (a simpler result). – Bill Dubuque May 3 '12 at 21:14
Hint $\rm\ mod\ p\!:\: -1 \equiv (p-1)! = (p-1)(p-2)(p-3)(p-4)! \equiv (-1)(-2)(-3)(p-4)!$ | {
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"url": "http://math.stackexchange.com/questions/140563/help-manipulating-wilsons-theorem"
} |
image-recognition, classification, convolutional-neural-networks
If it's by design, maybe there is some better version of CNN that can perform better? Wolfram's image id system is specifically meant to figure out what the image is depicting, not the medium.
To get what you want you'd simply have to create your own system where the training data is labeled by the medium rather than the content, and probably fiddle with it to pay more attention to texture and things as such as that. The neural net doesn't care which we want - it has no inherent bias. It just knows what it's been trained for.
That's really all there is to it. It's all to do with the training labels and the focus of the system (e.g. a system that looks for edge patterns that form shapes, compared to a system that checks if the lines in the image are perfectly computer-generated straight and clean vs imperfect brush strokes vs spraypaint).
Now, if you want me to tell you how to build that system, I'm not the right person to ask haha | {
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"tags": "image-recognition, classification, convolutional-neural-networks",
"url": null
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quantum-mechanics, operators, hilbert-space, hamiltonian, observables
Title: What is $\langle \phi | H | \psi \rangle$ in QM? I know that $\langle \phi | \psi \rangle$ is the probability of going from the $\psi$-state to the $\phi$-state, and that $\langle \phi | H | \phi \rangle$ is the expectation value of the energy for the $\phi$-state.
But how should I interpret $\langle \phi | H | \psi \rangle$? This is a supplement to freude's correct answer: | {
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electrical-engineering
sin(t) = n1/n2
Where n1 and n2 are the refractive indices of the air and rod. Then, you have a geometry problem to work out how that fits in with the curve of the rod. Here's a diagram; the edges of the rod are black, the ray of light is red, the blue and green are constructions to help solve it. | {
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# singularities and residues
#### pantboio
##### Member
Consider $$f(z)=\frac{z+5}{e^\frac{1}{z}-3}$$ Find and classify its singularities and compute residues.
I think singularities are: $0,\infty$ and zeroes of denominators. We have $e^{\frac{1}{z}}=3$ for $z=(log(3)+2k\pi i)^{-1}$. I think $0$ is essential, $\infty$ a simple pole and zeroes of denominator are simple poles.
How can i prove this facts, and how to compute residues?
#### chisigma
##### Well-known member
Consider $$f(z)=\frac{z+5}{e^\frac{1}{z}-3}$$ Find and classify its singularities and compute residues.
I think singularities are: $0,\infty$ and zeroes of denominators. We have $e^{\frac{1}{z}}=3$ for $z=(log(3)+2k\pi i)^{-1}$. I think $0$ is essential, $\infty$ a simple pole and zeroes of denominator are simple poles.
How can i prove this facts, and how to compute residues? | {
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"url": "https://mathhelpboards.com/threads/singularities-and-residues.2588/"
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Reference
1. Alster, K., Pol, R., On function spaces of compact subspaces of $\Sigma$-products of the real line, Fund. Math., 107, 35-46, 1980.
2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
3. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
4. Gruenhage, G., Covering properties on $X^2 \backslash \Delta$, W-sets, and compact subsets of $\Sigma$-products, Topology Appl., 17, 287-304, 1984.
5. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
6. Michael, E., Rudin, M. E., A note on Eberlein compacts, Pacific J. Math., 128, 149-156 1987.
7. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984. | {
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"openwebmath_score": 0.9472842812538147,
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"url": "https://dantopology.wordpress.com/category/compact-space/"
} |
gazebo
One thing I would try is to make the
robot's collision objects into actual
geometries, rather than using CAD
files. Also try to add a small amount
of friction in the joints and see what
happens, especially the joint above
the casters. | {
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complexity-theory, terminology, runtime-analysis, efficiency
This conflicting terminology is obviously confusing. Of course, any (good) formal discussion of algorithmic complexity will precisely define its terminology, but in general, are either of the two usages above clearly more conventional than the other? If someone uses the term "efficient" in an informal discussion of computational complexity without defining it, is it more likely that they mean "polynomial time" or "polylogarithmic time"? There is no one true answer. It depends on context. The most common context is one where polynomial-time is taken as more or less synonymous with efficient, so if you had no further context, I would certainly guess "polynomial time". Polylogarithmic time is used only in very narrow contexts.
In general, if you think your audience might not be sure about the meaning, then I recommend you to define your terms. | {
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newtonian-mechanics, thermodynamics, statistical-mechanics, computational-physics
This was the equilibrium stat mech view. In kinetic theory, the situation is a bit more complex. You may have heard the term "molecular chaos". This is an assumption, firstly made by Boltzmann that assumes that velocities before a collision are uncorrelated. That is, when two particles collide, we can assume that they forgot everything about their past and their previous collisions. This assumption holds true in a lot of cases (except maybe in very very dense system, with the so called ring collisions or in non-equilibrium matter such as granular gases) and is well verified numerically. By assuming that you can recover the Maxwell-Boltzmann distribution from the Boltzmann equation. Note however that, assuming that two velocities are always uncorrelated and that colliding particles are uncorrelated (in term of velocities) is not the same thing. | {
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c++, c++11, tic-tac-toe
//reading player move
void read_move()
{
int *p_moveX = new int;
int *p_moveY = new int;
std::cout << "\nChoose a field (Player " << which_player << "): ";
std::cin >> *p_moveX >> *p_moveY;
write_move(*p_moveX, *p_moveY, which_player);
delete p_moveX; p_moveX = NULL;
delete p_moveY; p_moveY = NULL;
}
//saving player's move to table
void write_move( int X, int Y, int plr_number)
{
if(table_board[X-1][Y-1]!=0 || X<=0 || Y<=0 || X>3 || Y>3)
{
std::cout << "You can't do this!";
read_move();
}
else
{
table_board[X-1][Y-1] = plr_number;
}
} | {
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fft, frequency-spectrum, analog-to-digital
23 dB is SIGNIFICANT and I suspect your sine wave is not really a sine wave and you are seeing a harmonic (or an aliased harmonic).
Here are some possible culprits, some of which you may have ruled out already:
You did not specify what frequency the next highest peak was, but harmonics are usually excluded from SFDR. (400KHz, 600KHz, 800KHz, etc)
Did you sufficiently filter the analog signal prior to sampling? If not any higher aliases will fold into band. (Changing the sampling rate and retesting, not the FFT size, would give clues if this is occuring with such a strong spur).
Did you observe your test signal in the analog domain to see if it is cleaner than the SFDR you are trying to measure? If you did not look at this (with a spectrum analyzer for example) is possible your test signal itself has such a spur. | {
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c++, beginner, arduino
switches();
}
String cancel = Serial2.readString();
if (cancel.toUpperCase() == "CANCEL") {
break;
}
}
}
void letterA() {
for (int x = 0; x <= 100000; x++) {
dw(0, HIGH);
dw(1, HIGH);
dw(6, HIGH);
dw(7, HIGH);
dw(21, HIGH);
dw(3, LOW);
dw(4, LOW);
delay(0.5);
dw(21, LOW);
dw(3, HIGH);
dw(4, HIGH);
dw(2, LOW);
dw(5, LOW);
dw(20, HIGH);
delay(0.5);
dw(20, LOW);
dw(2, HIGH);
dw(5, HIGH);
dw(1, LOW);
dw(6, LOW);
dw(19, HIGH);
dw(18, HIGH);
dw(17, HIGH);
dw(16, HIGH);
dw(15, HIGH);
dw(14, HIGH);
delay(0.5);
dw(14, LOW);
dw(15, LOW);
dw(16, LOW);
dw(17, LOW);
dw(19, LOW);
dw(1, HIGH);
dw(6, HIGH);
delay(0.5);
}
delay(200);
} | {
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coll->data[coll->count++] = entry;
return coll;
}
int main(int argc, char **argv)
{
int mx = 1;
if(argc>1){
int mxcmd = atoi(argv[1]);
if(1 <= mxcmd && mxcmd <= MXDST){
mx = mxcmd;
}
else{
fprintf(stderr, "invalid maxdist value, "
"got %d\n", mxcmd);
exit(-1);
}
}
int nmx = 1 << mx;
coll_ptr table[nmx+1][mx+1];
int n, dst;
for(n=0; n <= nmx; n++){
for(dst = 0; dst <= mx; dst++){
table[n][dst] = coll_new();
}
}
int grand[mx+1];
for(dst = 0; dst <= mx; dst++){
grand[dst] = 0;
}
tree_inst base;
base.distsub[0] = NULL;
base.size = 0;
coll_record(table[0][0], &base);
for(n=1; n <= nmx; n++){
int m; time_t time_begin, time_end;
time_begin = time(NULL);
for(m=0; m <= n-1; m++){
int dst1, dst2;
for(dst1 = 0; dst1 < mx; dst1++){
for(dst2 = 0; dst2 < mx; dst2++){
coll_ptr
ca = table[m][dst1],
cb = table[n-1-m][dst2];
int c1, c2;
for(c1 = 0; c1 < ca->count; c1++){
for(c2 = 0; c2 < cb->count; c2++){
tree_ptr
t1 = ca->data[c1],
t2 = cb->data[c2]; | {
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"url": "https://math.stackexchange.com/questions/1132147/egf-of-rooted-minimal-directed-acylic-graph"
} |
ros-melodic
pcl::removeNaNFromPointCloud(*cloud,*outputCloud, indices);
sor.setInputCloud(outputCloud);
sor.setLeafSize (0.01f, 0.01f, 0.01f);
sor.filter (*outputCloudFilt);
}
All the objects used here were pre-initialised before main(). The code runs just as slowly if I remove the filtering from the callback.
I should also add that the speed is slow independently of how many points are in the cloud.
EDIT: Using fromROSMsg() doesn't help with my problem.
Originally posted by christophecricket on ROS Answers with karma: 21 on 2020-03-24
Post score: 0 | {
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python, iteration
Title: Python line condenser function This is a function I just wrote that tries to condense a set of strings into grouped lines. It actually works, but looks ugly. Is there a better way to achieve the same thing?
Take 4 filtering empty strings first and dropping the if line on the final yield
def mergeToLines(strings, length=40, sep=" "):
strs = (st for st in sorted(strings, key=len, reverse=True) if st)
line = strs.next()
for s in strs:
if (len(line) + len(s) + len(sep)) >= length:
yield line
line = s
else:
line += sep + s
yield line | {
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plus wolfram.com 's definition before asking comulative negative binomial distribution e.g... P ) y = nbincdf ( x, R, p, 'upper ' Description..., e.g, cumulative distribution function fixed and the number of trials varies discrete and unimodal,,! Validity despite possibly degrading runtime performance variance for given parameters way to tweek code! Probability textbook, plus wolfram.com 's definition before asking negative binomial distribution cdf new math.. You observe the dispersion ( variance ) higher than expected by Poisson algebraic way to tweek the code get. Alternative to Poisson when you observe the dispersion ( variance ) higher than expected Poisson... To Poisson when you observe the dispersion ( variance ) higher than expected by Poisson a of! An alternative to Poisson when you observe the dispersion ( variance ) higher than expected by Poisson the function the... Checked for validity despite possibly degrading runtime performance or one of its parameters, of the numbers. | {
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"url": "http://vivifive.com/baby-desire-wnbgatc/article.php?903884=negative-binomial-distribution-cdf"
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quantum-mechanics, operators
Title: What is the definition of variance for a non-Hermitian operator? I am trying to understand what is the correct way to compute the variance of a non-Hermitian operator.
I was thinking that it was simply something that:
$$
\langle (\Delta a)^2 \rangle = |\langle \psi| \hat{a}^2 |\psi\rangle| -| \langle \psi |\hat{a}|\psi\rangle|^2
$$
But now I have read on Wikipedia that the variance for a random complex variable can be written as:
$$ Var[Z] = \mathbb{E}[|Z|^2] - |\mathbb{E}[Z]|^2 $$
In the first term, the absolute value is computed before the expectation value, so I think the formula I have written before may be wrong.
Now I am thinking that it may be
$$
\langle (\Delta a)^2 \rangle = |\langle \psi| \hat{a}^\dagger \hat{a} |\psi\rangle| -| \langle \psi |\hat{a}|\psi\rangle|^2
$$ | {
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Now all $$d+k\cdot \lambda(n)$$ are valid private key where $$k \in [0,g]$$, listing;
1. 835
2. 1957
3. 3079
4. 4201
5. 5323
6. 6445
SageMath code to find the above example;
p = random_prime(200, 400) #upper and lower range
q = random_prime(200, 400)
n = p*q
e = 43
print("n = ",n)
print("factors %s = " % n, factor(n))
phi = (p-1)*(q-1) # or call euler_phi(n)
print("phi = ",phi)
if gcd(e,phi) != 1:
print( gcd(e,phi))
lmd = lcm(p-1,q-1) #or call carmichael_lambda(n)
print("lambda = ",lmd)
print("gcd(%s,%s) = " % (p-1,q-1), gcd(p-1,q-1))
print("inverse of %s by phi " %e, inverse_mod(43,phi))
print("inverse of %s by lambda" %e, inverse_mod(43,lmd))
d = inverse_mod(43,lmd)
for k in range(gcd(p-1,q-1)):
print(d+k*lmd) | {
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Originally Posted by AllanCuz
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html
If you draw the xy domain you can see we want the area under the line but inside the circle,
If we convert this to polar co-ordinates this means
$0 \le \theta \le \frac{ \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis.
For r, we are going from the origin to the radius of the circle. So the interval of r must be
$0 \le r \le \sqrt{3}$
We can now compute this integral,
$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$
$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$
$= \iint_Q rcos^2 \theta dr d \theta$
$= \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$
Which can be easily computed using double angle formulas
Thanks a lot for the reply, really helped. | {
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} |
bash, installer
func_ptexists() {
if [ "$1" == "-update" ] ; then
func_clean -save
else
[ -e "$installdir/Popcorn-Time" ] &&
read -p "
WARNING: Popcorn-Time is already installed in '$installdir' and will be erased. Do you want to keep the configuration files (bookmarks, watched list, settings, ...) [y/n] ? "
if [ "$REPLY" == "y" ] ; then
func_clean -save
else
sudo rm -rf $HOME/.config/Popcorn-Time/
fi
sudo rm -rf /usr/share/applications/popcorn-time.desktop
fi
func_clean -all
} | {
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"tags": "bash, installer",
"url": null
} |
javascript, php, html
document.getElementById(right).click();
else if(S.includes(event.which) || S.includes(event.keyCode)) //s //d
document.getElementById(down).click();
else if(A.includes(event.which) || A.includes(event.keyCode)) //a //d
document.getElementById(left).click();
else if(W.includes(event.which) || W.includes(event.keyCode)) //d
document.getElementById(up).click();
}
</script>
<div class="container">
<div class="topleft">
<?php
if(isset($_SESSION["name"])) {
echo "Your name: " . $_SESSION['name'] . "<br>";
echo "Hitpoints: " . $_SESSION['hp'] . "<br>";
echo "Gold: " . $_SESSION['gold'] . "<br>";
echo "XP: " . $levelxp[1] . "<br>";
echo "Level: " . $levelxp[0] . "<br>";
echo "Location: " . $_SESSION['xy'][0] . ", " . $_SESSION['xy'][1] . "<br>";
echo "<a href='reset.php'>RESET ALL DATA</a><br>"; | {
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"tags": "javascript, php, html",
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ros-melodic
<transmission name="tran1">
<type>transmission_interface/SimpleTransmission</type>
<joint name="base_to_lb_wheel">
<hardwareInterface>hardware_interface/VelocityJointInterface</hardwareInterface>
</joint>
<actuator name="motor1">
<hardwareInterface>hardware_interface/VelocityJointInterface</hardwareInterface>
<mechanicalReduction>1</mechanicalReduction>
</actuator>
</transmission>
<link name="lf_wheel">
<visual>
<origin xyz="0 0 0" rpy="0 0 0"/>
<geometry>
<cylinder radius="0.0525" length="0.05"/>
</geometry>
<material name="black">
<color rgba="0 0 0 1"/>
</material>
</visual>
<inertial>
<origin xyz="0 0 0" rpy="0 0 0"/>
<mass value="0.104"/>
<inertia ixx="9.332916666666665e-05" ixy="0" ixz="0"
iyy="9.332916666666665e-05" iyz="0" izz="0.00014332499999999997"/>
</inertial> | {
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c#, algorithm, linked-list
while (left != null && right != null)
{
(carriedValue, onesValue) = AdditionValues(left.Data, right.Data, carriedValue);
ll.AddNode(new Node(onesValue));
left = left.Next;
right = right.Next;
}
if (left == null)
{
while (right != null)
{
(carriedValue, onesValue) = AdditionValues(0, right.Data, carriedValue);
ll.AddNode(onesValue);
right = right.Next;
}
if (carriedValue != 0)
{
ll.AddNode(carriedValue);
}
}
if (right == null)
{
while (left != null)
{
(carriedValue, onesValue) = AdditionValues(left.Data, 0, carriedValue);
ll.AddNode(onesValue); | {
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# Counting how many ways 30 green balls…
How many ways are there to distribute 30 green balls to 4 persons if Alice and Eve together get no more than 20 and Lucky gets at least 7?
The answer given to me was $2464 = C(26, 3) − 66 − 46-24$ but I got $C(26, 3) - 6$.
Here is what I did:
Using "stars and bars" (or whatever the actual name for it is), I found the total number of possible combinations. Since 7 balls must immediately go to Lucky, there are 23 balls left to distribute. There are 3 'bars', so the total number of possible combinations is $\binom{26}{3}$.
Because this number accounts for the situations where Alice and Eve have $\geq 21$ balls, I must subtract those out. (Here is where I think I messed something up).
When Alice and Eve have 21: This leaves 2 balls left to distribute, so $\binom{3}{2}$
When Alice and Eve have 22: This leaves 1 ball left to distribute, so $\binom{2}{1}$
When Alice and Eve have 23: This leaves no balls left to distribute, so it equals 1. | {
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"url": "https://math.stackexchange.com/questions/2722673/counting-how-many-ways-30-green-balls"
} |
electromagnetism
However, it seems to me that the question has arisen as a consequence of studying electric induction as it is presented in high school or university E&M courses. These courses are dealing with the electrodynamics of continuous media, that is all the scales - time scales, spatial scales, strengths of fields, etc. are assumed to be such, that we can neglect the microscopic structure of the materials. In particular, the electrostatic induction is discussed under the assumption of an ideal conductor, i.e. material with infinitely many mobile electrons (for all practical purposes).
Theoretical concepts and limits of applicability of theories are just (or even more) important in physics as math. | {
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reaction-mechanism, catalysis
Also there is no such thing as an irreversible reaction. If the backward reaction is very unfavourable energy-wise, the catalyst won't change that, but you cannot have two highly exothermic reactions in your funny circle anyway. Where should the energy come from? | {
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"tags": "reaction-mechanism, catalysis",
"url": null
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angular-momentum, rotational-dynamics, torque
Title: Doubt related to rotational motion (rolling down an inclined plane) I have doubt regarding when an object ( sphere , ring , cylinder etc ) rolls down a FRICTIONLESS inclined plane.
Let us suppose the angle of inclination of inclined plane is θ.
Now the forces acting on it will be weight force (mg) and normal reaction. A component of mg perpendicular to incline mgcosθ will be balanced by the normal reaction. There will be another component of mg along the incline mgsinθ which will give the object acceleration gsinθ. There is no friction.
Now,
acceleration, a = radius ( r ) x angular acceleration (α)
Now as there is, a = gsinθ so there must be nonzero value of α.
But force mg and normal reaction are forces whose line of action passes through the centre of the object ( ring etc ) ( passes through centre of mass). So net torque about centre should be zero and there should be no angular acceleration. | {
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algorithms, graphs, traveling-salesman, hamiltonian-circuit
Let $P_G$ denote the set of Hamiltonian pairs of $G$.
For $e, e'\in E$, let $P_G(e, e')$ be the set of Hamiltonian pairs of $G$ for which $e$ and $e'$ lie in the same cycle.
Note that if $x$ is a vertex of $G$ and $G$ is $4$-regular, then $x$ is the endpoint of exactly four edges, say $e_1$, $e_2$, $e_3$, $e_4$, and $P_G = \bigsqcup_{i=2}^4 P_G(e_1,e_i)$, where the union is disjoint.
It follows that $|P_G| = \sum_{i=2}^4 |P_G(e_1,e_i)|$.
THEOREM (Thomason): If $G$ is a $4$-regular multigraph with at least three vertices and $e$ and $e'$ are two edges of $G$, then $|P_G(e, e')|$ is even.
By our earlier observation, which is made in Thomason's paper, if one can prove that $|P_G(e_1, e_i)|$ is even for all $i = 2, \dots, 4$, it will follow that $|P_G|$, the number of Hamiltonian pairs, is even.
Now, let $G$ be the union of two edge-disjoint Hamiltonian cycles on $n$ vertices, viewed as subgraphs of $K_n$ (the complete graph on $n$ vertices).
Then $G$ is a $4$-regular graph. | {
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"tags": "algorithms, graphs, traveling-salesman, hamiltonian-circuit",
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c#, recursion, hash-map
Title: Recursive conversion from ExpandoObject to Dictionary For my blazor library which is a modification of this awesome library I have to convert an ExpandoObject into a Dictionary<string, object> since ExpandoObjects aren't serialized properly in the newest preview versions of dotnet-core. See my question related to this for more details.
My current approach goes as follows:
Dictionary<string, object> ConvertDynamicToDictonary(IDictionary<string, object> value)
{
return value.ToDictionary(
p => p.Key,
p =>
{
// if it's another IDict (might be a ExpandoObject or could also be an actual Dict containing ExpandoObjects) just go through it recursively
if (p.Value is IDictionary<string, object> dict)
{
return ConvertDynamicToDictonary(dict);
} | {
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2. ### Some help me- Math
Chris tells Adam that the decimal value of −1/13 is not a repeating decimal. Is Chris correct? A) No, because the fraction has a decimal period of 3 B) No, because the fraction has a decimal period of 6 C) No, because the
3. ### math
Name 3 fractions whose decimal equivalent is 0.25. Explain how you know each fraction is equivalent to 0.25.
4. ### Math
What is 5/9 written as a decimal? 0.45 0.4 repeating 0.5 0.5 repeating What is 0.26 written as a fraction in simplest form? 26 over 100 13 over 50 26 over 10 13 over 25 In a fraction with a denominator of 15,which value could the
1. ### Math
How do you write 24% as a decimal? 2.4 0.24*** 0.024 How do you write 135% as a decimal? 1.35*** 13.5 0.135 How do you write 4% as a decimal? 0.4*** 0.40 0.04 Which fraction is equivalent to 72%? (No answer, not sure on this one.)
2. ### One Last Help | {
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"url": "https://www.jiskha.com/questions/55003/every-fraction-has-a-decimal-equivalent-that-either-terminates-for-example-1-4-0-25-or"
} |
evolution, reproduction, psychology
Since I have a little time today, I'm adding a couple of tangential references: Mice sperm have responded to sperm competition so much by evolving hooks on their heads and chain up together and swim together which enhances the chances of paternity. Ducks are quite different. they have a lot of forced copulation, but female choice of mate is still quite influential. Duck vaginas are very complicated, with corkscrews and dead ends to make it difficult to copulate without 'consent'. | {
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php, object-oriented, file-system
return $return;
}
}
The class has one method and that method has a couple of tasks:
get files
parse only PHP files
instantiate non abstracts | {
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can just draw a line is very important for many topics! Gradually increasing or … linear function Known Ys many more transformations for this one than constant equations represents. The steepness of a function functions, f ( x ) from the.... | {
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"openwebmath_score": 0.4739937484264374,
"tags": null,
"url": "http://ruthsutton.co.uk/kch34en/linear-parent-function-slope-28edd5"
} |
amateur-observing, data-analysis, spectra, image-processing
where e.g. $T_{G,H{\beta}}$ would be the product of the G-pixel QE curve and the narrowband-filter transmission at the wavelength of H$\beta$ (something like $0.6 \times 0.9$), and so forth.
Since we now know $I_{H\alpha}$, we can subtract $T_{G,H{\alpha}} I_{H\alpha}$ from Equation (2) to remove the contribution to the G image from H$\alpha$ photons (call the result $F^{\prime}_{G}$); similarly, we can subtract $T_{B,H{\alpha}} I_{H\alpha}$ from Equation (3) to remove the contribution to the B image from H$\alpha$ photons (to get $F^{\prime}_{B}$).
Now we have:
(4) $F^{\prime}_{G} = T_{G,H{\beta}} I_{H\beta} + T_{G,O3} I_{O3}$
and
(5) $F^{\prime}_{B} = T_{B,H{\beta}} I_{H\beta} + T_{B,O3} I_{O3}$
This is two equations in two unknowns ($I_{H\beta}$ and $I_{O3}$), and so is solvable; some manipulation produces (assuming I didn't make any silly algebra mistakes):
(II) $I_{H\beta} = (F^{\prime}_{B} - T_{B,O3} F^{\prime}_{G} / T_{G,O3}) | {
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"tags": "amateur-observing, data-analysis, spectra, image-processing",
"url": null
} |
kinematics, orbital-motion
Title: Why are the Sun's outer layers(photosphere) moving slower than the inner layers? Why are the Sun's outer layers(photosphere) moving slower than the inner layers? Veritasium made a video on this but couldn't make any sense of it. Please explain it to me at the level which a high-school-er can understand. Brief summary of what was said in the referenced video:
Photons being emitted from the sun collide with particles of dust in the solar system, slowing down the particles of dust through the transfer of momentum in the collision.
Equally, photons collide with material inside the sun (and on the surface), slowing down the particles of the sun. | {
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$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$
• Very nice link to expansion for $e^x$. (+1) – Hypergeometricx Jan 7 '17 at 15:51
• Weird. I don't see the underline in the definition of $\binom nk$. Am I the only one with this problem? I'm on Chromium 55.0.2883.87 (64-bit). – rubik Jan 7 '17 at 21:55
• @rubik I had the same problem -- edited the post to make it displaystyle so it would show up. I think we should bring up this on meta as a bug with mathjax, would you like to do it or should I? – 6005 Jan 7 '17 at 22:54
• The frac displays fine in my mobile Chrome (both with dfrac and frac). – Martin Argerami Jan 8 '17 at 6:35
• @6005 You're right, done. Here's the discussion. – rubik Jan 8 '17 at 9:05
If you're not seeing the utility of the binomial theorem, then I think you're missing an important observation: | {
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star, positional-astronomy, astropy, skyfield
Title: Calculating the position of stars at a given point in time Is there any particular methods that help us calculate the position of a star at a certain point in time? This question is what I am trying to achieve, but for me, I do not want to use the data because the data that we can access to calculate the position is limited to only a few years. I want a method which uses the current position, speed, velocity of stars and calculate its position for many years back (like in 11000 BCE to 13000 BCE). I am trying to use astropy and I am really new to this, so I just need a push in the right direction; helpful resources will serve a great deal as well, thank you.
Edited:
There are a few stars' positions that are recorded and mentioned in one of our history books. SO what I am trying to do is collect all the recorded stars' positions and try to map them in real-time at a particular year and see if the recorded data exists or not. | {
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thermodynamics, energy, applied-physics
What's probably going on:
If you've poured $200mL$ out of the bottle, the remaining liquid will cool faster, because the rate of heat transfer is roughly proportional to the surface area, while heat capacity is proportional to the volume. Surface area varies as the $2/3$ root of volume, so if you multiply the volume by a factor of $0.8$, you multiply the volume by $0.8^{2/3}$, and so the cooling time is multiplied by $0.8/0.8^{2/3} \approx 0.93$. | {
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} |
.swiper-slide .content p,.full-width-content .vc_span12 .swiper-slide .content p{font-size:13.2px!important;line-height:24px!important;}}@media only screen and (max-width :690px){.nectar-slider-wrap[data-full-width="true"][data-fullscreen="false"] .swiper-slide .content h2,.nectar-slider-wrap[data-full-width="boxed-full-width"][data-fullscreen="false"] .swiper-slide .content h2,.full-width-content .vc_span12 .nectar-slider-wrap[data-fullscreen="false"] .swiper-slide .content h2{font-size:15px!important;line-height:21px!important;}.nectar-slider-wrap[data-full-width="true"][data-fullscreen="false"] .swiper-slide .content p,.nectar-slider-wrap[data-full-width="boxed-full-width"][data-fullscreen="false"] .swiper-slide .content p,.full-width-content .vc_span12 .nectar-slider-wrap[data-fullscreen="false"] .swiper-slide .content p{font-size:10px!important;line-height:17.52px!important;}}@media only screen and (min-width:1000px){.container,.woocommerce-tabs .full-width-content | {
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"tags": null,
"url": "http://www.wolfpacknyc.com/review/738508-inverse-laplace-transform"
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electromagnetism, electric-fields, vectors, vector-fields, linear-systems
Title: Does electric field obey the triangle law of vector addition and subtraction? I know that electric field strength is force per unit charge but what I have not yet understood properly is that how electric field can obey the laws of vector addition and subtraction excluding the linear situation?
Does electric field completely obey the triangle law of vectors? Does it produce a resultant like other vectors when two electric fields meet at an angle?
If yes, then is it used practically anywhere in our world. Electric Field is force per unit charge. Since force is a vector, a vector divided by scalar also gives a vector. Just think in this way; at a point the net force is obtained by laws of vector addition, so electric field is also effectively obtained by same way. That makes sense | {
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analytical-chemistry, mass-spectrometry
Title: Defining sensitivity of TOF Mass Spec - Does the detector play a role? Looking at TOF MS systems, my current understanding is that temporal resolution results in mass resolution - so by controlling the flight time (with distance) and sampling rate, you can start to distinguish smaller and smaller changes in mass. Hopefully, this is correct.
Where I'm confused is when people start talking about sensitivity.
I heard a talk where they mentioned femtogram and attogram sensitivity.
Does this refer to the amount of the sample they are using?
OR is this referring to the minimum amount of the analyte they can detect?
Does that translate to needing a certain number of ions to hit the detector in a single scan to have a signal significantly above the noise?
Finally, what types of detectors are capable of measuring so low and are there any issues with them? | {
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c++, performance, c++11, reinventing-the-wheel, classes
namespace DigitalImageProcessing
{
template<class SizeT = unsigned int>
class MonoImage
{
public:
MonoImage()
{
}
constexpr auto GetimageSizeX()
{
return image_size.xsize;
}
constexpr auto GetimageSizeY()
{
return this->image_size.ysize;
}
constexpr auto GetimageSize()
{
return this->image_size;
} | {
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"tags": "c++, performance, c++11, reinventing-the-wheel, classes",
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rcservo, walking-robot
Title: Servo motor considerations for a quadruped I'm building a quadruped and I'm not sure of the features I should be looking for in a servo motor. e.g. digital vs analog, signal vs dual bearings. Some of the ones I'm considering are here If you are not on the budget, go for the digital with metal gears. Basically, higher the torque better your bot will be. For the legged bots speed of the servo is not an issue but torque is.
To be sure that servo will have enough torque, you will need to roughly calculate the mass of the bot, divide it by number of legs - 2. As during the walk two legs will be in the air and other legs need to be able to hold the mass of the bot. | {
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Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.
• ah, I was trying to deal with the case where, say, we have $x= 0.99$ and $y = 1.01$. Considering $f$ on $[0,2]$ instead of $[0,1]$ is a smart way to deal with that!
– MCT
Mar 4, 2015 at 22:06
• @MCT I always stuck with this (why) in many proofs could you please explain this little bit? why $f$ is bounded all over $\mathbb{R}$? Jul 8, 2021 at 4:33
• @Priya because for every $x \in \mathbb{R}$, there is a point $y \in [0,2]$ with $|y-x|$ being a multiple of $1$, so $f(x) = f(y)$. So if $M$ is larger than all $f(y)$ for $y \in [0,2]$, it must also be larger than all points $f(x)$ for $x \in \mathbb{R}$
– MCT
Jul 11, 2021 at 14:44
A periodic continuous function is simply a function defined on a circle, $$f:S^1 \longrightarrow R.$$ Circle is a compact space. A continuous function on a compact space is both bounded and uniformly continuous. | {
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java, algorithm, tree, pathfinding, breadth-first-search
/**
* This class implements the IDDFS algorithm (iterative deepening depth-first
* search) described in
* <a href="http://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search">Wikipedia</a>.
*
* @author Rodion Efremov
* @version 1.6
* @param <N> the actual type of a tree node.
*/
public class IDDFSNodeFinder<N extends AbstractBinaryTreeNode<?, N>>
extends AbstractTreeNodeFinder<N> {
private int previousReachedNodes;
private int reachedNodes;
@Override
public List<N> search(final N source,
final Predicate<N> goalNodePredicate) {
if (source == null) {
// 'path' is empty, thus denoting a non-existent path.
return new ArrayList<>(0);
}
final Map<N, N> parentMap = new HashMap<>();
parentMap.put(source, null);
previousReachedNodes = 0;
reachedNodes = 0; | {
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} |
functional-programming, event-handling, react.js, typescript
/*
React hooks
*/
// this is an optional error message
const [errorMessage, setErrorMessage] = React.useState<string | undefined>(undefined);
function assert(assertion: boolean, message: string, extra?: () => object): void {
if (!assertion) {
if (extra) {
const o: object = extra();
const json = JSON.stringify(o, null, 2);
message = `${message} -- ${json}`;
}
// write to errorMessage state means it's displayed by the `<ErrorMessage errorMessage={errorMessage} />` element
setTimeout(() => {
// do it after a timeout because otherwise if we do this during a render then React will complain with:
// "Too many re-renders. React limits the number of renders to prevent an infinite loop."
setErrorMessage(message);
}, 0);
console.error(message);
}
} | {
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} |
black-holes, black-hole-firewall
As a last note, I think your confusion concerning the extracts above, is that what happens in extract 1, is when a person reaches dangerously close to the singularity - however this depends on the size of the black hole and could happen a long time after passing the event horizon (although Alice would be fried so wouldn't get past the event horizon.), or with a tiny black hole you would be stretched before the event horizon.
So it would depend on the size of the black hole as to whether Bob sees her stretched out or burnt up. Hope that helps :-) | {
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decision making function in.... Optimal separating hyper-plane as the classification decision making function in SVM assumption that the underlying distributions... About it requires the mean and variance-covariance matrix of the data derived above using the Mahalanobis distance that... D, as explained here average of the metric learning existing techniques is proposed [... Improve this Question | follow | mahalanobis distance classifier May 31 '13 at 5:57. mostar mostar be decided upon additive. That captures most of the distance between a point p and a distribution D, as explained here takes consideration! Use Mahalanobis distance: a direction-sensitive distance classifier that uses statistics for each class the squared Mahalanobis.! Each dataset into 70 % for the binary classification algorithms to handle data... Between the pixels and requires the mean and variance-covariance matrix [ 4 ] generalized Euclidean is! Is the most promising binary classification for estimating | {
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"openwebmath_score": 0.7011394500732422,
"tags": null,
"url": "http://portcetate.ro/grille-or-lktg/mahalanobis-distance-classifier-c2ddb4"
} |
c#, .net, validation, error-handling, crud
/// <summary>
/// To hold the results, be it the list of errors or just the success message.
/// </summary>
public sealed class ResultModel
{
private object _success; // to hold boolean and int value.
public object Success { get { return _success; } }
private List<string> _errors; // to hold multiple errors, could be validation errors.
public List<string> Errors { get { return _errors; } }
public void AddSuccess(object any)
{
this._success = any;
}
public void AddError(string text)
{
if (string.IsNullOrEmpty(text))
{
return;
}
if (this._errors == null)
{
this._errors = new List<string>();
}
this._errors.Add(text);
}
} | {
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"tags": "c#, .net, validation, error-handling, crud",
"url": null
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Theorem: If $Y$ is a finite set, then any injection $X \to Y$ is an isomorphism (of sets) if and only if it is surjective.
• $f : \mathbb{Z} \to \mathbb{Q}$ is not surjective
However, most of your experience is probably about finite sets, and so your experience has mislead you to intuit "$X$ and $Y$ have the same cardinality" and "all injections $X \to Y$ are bijections" the same way.
Another possibility of where your intuition has gone astray is
Theorem: If $Y$ is a finite set, then if any one morphism $X \to Y$ is an isomorphism, then all injections $X \to Y$ are isomorphisms.
and your experience with finite sets has led you to intuit "same cardinality" with "all injections are bijections" the same. | {
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} |
quantum-mechanics, hilbert-space, operators, mathematical-physics, observables
which suggests to me that $E_\lambda$ must be differentiable and hence continuous. Am I right to think that this, probably among other things, means that in the range of the integration in $(1)$, there cannot be an eigenvalue embedded in the continuum$?^\dagger$ I think I could make sense of $(1)$ if the spectrum is split like e.g. in the case of the hydrogen atom: For all energies below $0$, we have bound states and hence eigenvalues and eigenvectors and then the continuum part follows, not 'interrupted' by bound states again, such that we could integrate from $0$ to $\infty$.
But what happens if there is an eigenvalue embedded in the continuum? Are the corresponding projectors already in the sum in $(1)$? Then how can one appropriately split the integration domain?
My concern is due to the fact that at an eigenvalue $\lambda_k$, $E_{\lambda_k}$ is not left continuous (although right continuous by definition), so I guess there could arise some problems. | {
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homework-and-exercises, kinematics, units, differentiation
Title: Why do these equations result an incorrect unit for acceleration?
Hello everyone.
Imagine an object moving around a certain point on a circular orbit. Magnitude of the velocity is constant during the motion ($|v|$). The orbit radius is $r$. (I'd better notice that we're just talking about kinematic view of this motion.)
According to the image I've uploaded, we'll have:
$\large v_x(\theta)=|v|\cdot \cos\theta$
$\large v_y(\theta)=|v|\cdot \sin\theta$
Since perimeter of the circular path is $2\pi r$, and magnitude of the velocity is constant, we'll have:
$\large\theta (t)=\frac{|v|\cdot t}{2\pi r} \times 2\pi =\frac{|v|\cdot t}{ r}$
Now we can combine these equations:
$\large v_x(\theta)=|v|\cdot \cos(\frac{|v|\cdot t}{ r})$
$\large v_y(\theta)=|v|\cdot \sin(\frac{|v|\cdot t}{ r})$
By this point, everything is okay. But the problem occurs here, where we try to get derivative of $v_x(t)$ and $v_y(t)$ in order to find $a_x(t)$ and $a_y(t)$. As we know by differentiation we have: | {
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It's the same result, if you change the parentheses for modulus:
$$-\ln \left| \cos \frac x2 - \sin \frac x2 \right|+ \ln \left| \sin \frac x2 + \cos \frac x2 \right| = \ln \frac{\left| \sin \frac x2 + \cos \frac x2 \right|}{\left| \cos \frac x2 - \sin \frac x2 \right|} =$$
$$\ln \left| \frac{\left( \sin \frac x2 + \cos \frac x2 \right)^2}{\left( \cos \frac x2 - \sin \frac x2 \right)\left( \sin \frac x2 + \cos \frac x2 \right)} \right | = \ln \left| \frac{1 + \sin x}{\cos x} \right | = \ln |\sec x + \tan x|$$
And
$$-\ln {|\csc x + \cot x|} = \ln \left|\frac{\sin x}{1+\cos x} \right| = \ln \left|\frac{2\sin \frac x2\cos \frac x2}{2\cos^2 \frac x2} \right| = \ln \left |\tan \frac x2 \right| = \ln \left | \sin \frac x2 \right| - \ln \left |\cos \frac x2 \right|$$
• Please rectify the last part – lab bhattacharjee Feb 12 '17 at 6:59
• @labbhattacharjee thanks! – Rafael Deiga Feb 12 '17 at 16:34 | {
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php, unit-testing
Is this a good work flow for TDD? If not, what could I improve?
You understand the basic principle, however I think it would have been possible to write all the tests before writing the real code. This is also often done in real life: The tests are known before the code is written. The writing of the code is driven by the tests. Or, to put it in another way: If you don't know what you're going to test, then how on earth could you write any code?
Did I run enough tests? Do I need to write more?
As I illustrated before, by writing the inverse function as well, you can do thousands of tests, without having to be very selective about it.
Is it okay that, during intVal() re-writes, previous tests broke?
Yes, that's fine. That's the whole idea behind test-driver-development: The tests tell you what's wrong.
Remember that there are frameworks for writing tests. It is clearly possible to work without them, but why reinvent the wheel again? See, for instance: PHPUnit | {
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algorithm, matrix, r
anyx_r <- function(x,y){
for(x_ in x) if(x_ == y) return(TRUE)
FALSE
}
vec <- 1:1e7
x <- 5e6
microbenchmark::microbenchmark(
rloop = anyx_r(vec,x),
cpp = anyx_cpp(vec,x),
native = any(vec==x)
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# rloop 166.5758 171.34355 203.15277 179.9776 198.8560 990.1650 100
# cpp 39.5462 40.60585 57.84617 41.4594 46.1232 690.1746 100
# native 36.9900 37.86090 51.80317 38.9640 43.6510 888.3059 100
Almost but not quite ;).
So bottom line, in general you can trust vectorized R functions, even if it might seem they're working too much at first sight. | {
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"tags": "algorithm, matrix, r",
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} |
# When there are consecutive integers and if their range is equal to the
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Math Revolution GMAT Instructor
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When there are consecutive integers and if their range is equal to the [#permalink]
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29 Feb 2016, 20:39
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When there are consecutive integers and if their range is equal to their median, what is the number of them?
1) The smallest number of them is 5
2) The largest number of them is 15
* A solution will be posted in two days.
_________________ | {
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appropriate boundary conditions. 18 describes conservation of energy. This technique is known as the method of descent. di erential equation to a collection of ordinary di erential equations along each of its ow lines is called the method of characteristics. Your diffusive equation leads always to the conservation of energy in your spatial domain if Neumann BC are imposed. DEPARTMENT OF PHYSICS AND ASTRONOMY 4. Let be a kinematically admissible variation of the deflection, satisfying at. We start by changing the Laplacian operator in the 2-D heat equation from rectangular to cylindrical coordinates by the following definition::= (,) × (,). The catalyst particles are supposed spherical and surrounded by the uniform concentration and temperature of the fluid at that same point. The basic equation of radiant heat transfer which governs the radiation field in a media that absorbs, emits, and scatters thermal radiation was derived. Now, general heat conduction equation for sphere is given | {
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• How do you get this: $n$ good, then $2n + 179$ good? Thanks! – Darya Schedrina Dec 1 '19 at 18:36
• This comes from the paper of Graham. Under our notation here: $1/3 + 1/7 + 1/78 + 1/91 + a_1/2 + \dotsc + a_m/2 = 1$, and hence $3 + 7 + 78 + 91 + 2n = 2n + 179$ is good. Note that Graham needs all $a_i$ to be different, so for us it's enough to have e.g. $1/3 + 1/6 + a_1/2 + \dotsc + a_m/2 = 1$ and hence $2n + 9$ is good. – WhatsUp Dec 1 '19 at 18:41
• OEIS A028229 implies that all $n>23$ are good. – Max Alekseyev Dec 1 '19 at 18:53
• Strictly speaking, OEIS A028229 is not the result of Graham. Also, you gave a link to the same paper of him. – Max Alekseyev Dec 1 '19 at 22:12
• @MaxAlekseyev Ah... thanks for pointing out. You see, I simply copied that link from the OEIS page without looking into the details. The OEIS page is kind of misleading, by putting the link to Graham's paper under "The table of n, a(n)". – WhatsUp Dec 1 '19 at 22:36 | {
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"url": "https://mathoverflow.net/questions/347377/harmonic-sums-and-elementary-number-theory/347388"
} |
$$bt^2-2at-b=b\left[t^2-2\frac abt-1\right]=b\left[\left(t-\frac ab\right)^2-\left(1+\frac{a^2}{b^2}\right)\right]$$ now use substitution to get it in the form: $$\int\frac{dx}{x^2-1}=\frac 12\ln\left(\frac{x-1}{x+1}\right)+C$$
If I were doing this myself, I'd prefer José Carlos Santos' method---which among other unadvertised benefits (1) doesn't introduce a problem at odd integer multiples of $$\pi$$ (see below) and (2) doesn't break the symmetry of the roles of $$a$$ and $$b$$.
Here's a conventional way to finish using your approach: Completing the square in the denominator (temporarily assuming $$b \neq 0$$) gives $$-b t^2 + 2 a t + b = - b \left[\left(t - \frac{a}{b}\right)^2 - \left(\frac{a^2}{b^2} + 1\right)\right] .$$ So, applying the translation substitution $$u = t - \frac{a}{b}$$ leaves the standard integral $$-\frac{2}{b} \int \frac{du}{u^2 - \lambda^2}, \qquad \lambda := \frac{\sqrt{a^2 + b^2}}{b} .$$ | {
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electromagnetic-radiation, electric-fields, reflection, refraction
The trick is that not all of the energy is concentrated at one frequency. You basically have a step function which has an infinitely wide spectrum, including a DC component.
Those low frequencies behave differently. At the extreme end, they move electrons around to create the static field you refer to. If you wanted to study it more, you could use Fourier transforms, or use Laplace transforms to study the transients. | {
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## Loujoelou 2 years ago Explain, in complete sentences, how you would expand (3x + 7y)^4 using Pascal’s Triangle. I got 3x^4+84x^3y+126x^2y^2+84xy^3+7y^4. Can anyone verify to see if this is correct? :) Thank you very much.
1. ajprincess
It is nt correct.
2. Loujoelou
Oh okay, can you show me what I did wrong?
3. TuringTest
what is the 5th row of pascal's triangle?
4. Loujoelou
1,4,6,4,1
5. TuringTest
yes, and those will be your coefficients now what are your terms?
6. Loujoelou
x^4 + x^3y + x^2y^2 + xy^3 + y^4
7. TuringTest
no, you have ignored the coefficients in the terms themselves they must remain...
8. Loujoelou
1x^4+4x^3y+6x^2y^2+4xy^3+1y^4 ?
9. TuringTest
the terms are 3x and 7y so you have 1 4 6 4 1 1(3x)^4(7y)^0+4(3x)^3(7y)^1+6(3x)^2(7y)^2+4(3x)^1(7y)^3+1(3x)^0(7y)^4
10. Loujoelou | {
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javascript, jquery, html
Event Handler -
You've attached the event listener to the document which is fine, but you've used an Id selector to filter it to. There can be only one element with in the document referenced by that Id.1 Unless the #updateBtn element is expected to be destroyed and recreated during the documents lifetime, it would be better to specify that element explicitly.
Unused Variable -
The locations variable is never referenced after it is instantiated, remove it.
Improper Scope -
The rows variable is pushed to, but never read from. If it is to be useful to other code it will need to be moved out of the event handler's scope.
The content variable is never used at this level of the scope, move its declaration to the narrowest scope required.
Exactly One - | {
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ros
Title: Reg: Subscribing to Depth and rgb image at the same time
Hi,
I am new in ROS. I wanted to get both the depth and rgb image of an object at the same time and use it in the callback function. if I use it as mentioned below. It works for me.
void imageCallback(const sensor_msgs::ImageConstPtr& kinect_image)
image_transport::Subscriber sub = it.subscribe("camera/rgb/image_color", 1, imageCallback);
or
image_transport::Subscriber sub = it.subscribe("camera/depth/image", 1, imageCallback);
I dont know how can I get both the rgb and depth image in the callback funtion. The below way doesn't work.
void imageCallback(const sensor_msgs::ImageConstPtr& kinect_image,const sensor_msgs::ImageConstPtr& depth_image)
image_transport::Subscriber sub = it.subscribe("camera/rgb/image_color","camera/depth/image", 1, imageCallback);
I know, may be the question sounds very naive. But, I would really be grateful if someone can guide me. | {
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experimental-physics, gravitational-waves, interferometry, ligo, earthquake
Title: Could LIGO discovery be due to e.g. earthquakes or have a terrestrial source? I mean, could it have been earthquakes or anything else? I want to add one very compelling argument which clearly shows that Ligo could not have been because of earthquakes or terrestrial phenomena, or at least their probabilities would be outrageously low.
What was announced yesterday was that two different detectors picked up almost entirely identical signals with a spacing of a few milliseconds difference. The distance between the two detectors in Livingston, Lousiana and Hanford, Washington as the crow flies on the surface of the Earth is about 3042 km.
But gravitational waves would not care about the ATCF distance along the surface of the earth but the actual geodesic distance between the two points which is much smaller (about 3000km). | {
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flipped a coin 450 times. If you flip three coins, it's eight - two for the first times two for the second times two for the third. The game is played in stages. The order does not matter as long as there are two head and two tails in the flip. e head or tail. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. Flipping more coins¶ If we want to flip more coins, it's going to be a pain in the neck to make that table over and over. Then, it displays the results, as well as the theoretical and observed probabilities of each event happening. Given this information, what is the probability that it is a. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ | {
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ros
Have you done the ROS tutorials? If not, I recommend you do so. ROS is a complicated system, and you need to really grasp the concepts before you attempt coding. It is very important you understand what a Message is, and how to use them.
However, to answer your question, it depends on how you are modifying it. For the float message, you will need to add to the top of your code the following line.
#include <sensor_msgs/ChannelFloat32>
Now, you can change the function to read
void floatCallback(const sensor_msgs::ChannelFloat32::ConstPtr& float_msg)
and that should work.
Originally posted by allenh1 with karma: 3055 on 2013-06-30
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by Martin Günther on 2013-07-15:
Hmm... sensor_msgs/ChannelFloat32.msg isn't "the float message", std_msgs/Float32 (or Float64) is. And for velocities, geometry_msgs/Twist is the standard. | {
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quantum-mechanics, dirac-delta-distributions, fermis-golden-rule
$$ \lim_{t\to\infty} \frac{F\left(t,\frac{E-E_i}{\hbar}\right)}{t}~=~\pi ~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar ~\delta(E-E_i) , \tag{C-32'} $$
cf. this Phys.SE post. This in turn implies that for large enough times $t$ the probability $P(t)$ grows proportionally with $t$, and that the transition rate
$$ \frac{dP(t)}{dt}~=~\frac{P(t)}{t} \tag{*}$$
is given by the proportionality factor.
For more details of the proof of Fermi's golden rule, see my Phys.SE answer here. | {
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As the arguments of both functions in the integral are not single variables, we can start to consider a change of variable in order to just have one variable in the secant and tangent function. This can be made using the change of variable $$\label{cambio1} \left\lbrace\begin{array}{l} y = 2x-1 \\ dy = 2 \,dx \end{array}\right.$$ Notice that the derivative $dy$ is easy to obtain from the expression of the integral: $dx$ can be expressed in terms of $dy$ as $$\label{eq1} dx = \frac{dy}{2}.$$ Substituting the change of variable \eqref{cambio1} in the integral, we obtain \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \int 3 \sec(y)\tan(y) \,\frac{dy}{2} \tag{\footnotesize by \eqref{cambio1} and \eqref{eq1}}\\ &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \label{da1} \end{align} We have now reduced our problem to a more simple integral. We want to integrate the expression $\sec(y)\tan(y)$, which is the derivative of\footnote{This comes from the following calculation: $(\sec(y))'= | {
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neural-network, classification
I have tried with up to 500k cases. I separate 10% for generalization checks after training, and train on the remaining 90% with a 50/50 validation split.
I've tried with the test data weighted 75% toward rule A, 50/50, and 75% toward ruleB.
I've tried 0-10 hidden layers, and neuron counts from 2 to 256 (each hidden layer gets the same number of neurons).
I change the number epochs as time allows, but generally it's 10-100. My longest runs have been several hours (with giant case numbers, and dropouts to prevent overfitting).
I've used batch sizes of 1-50.
I've tried learning rates of 0.0001 - 0.1.
I'm currently using ReLU activation, initializing bias to const(0.1) and kernel w/ heNormal. I have tried several other approaches for all three.
I standardize the inputs to center on zero w/ variance of one.
The loss function is categoricalCrossentropy.
The optimizer is Adam. Yes, they absolutely can 'learn' conditional rules. | {
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ruby, coffeescript, haml
- if (((@text.subtitle.nil? || @text.subtitle.empty?) && (current_page?(edit_text_url(@text)) || current_page?(new_text_url))) || !@text.subtitle.empty?)
%h2#subtitle.col-xs-12.col-sm-11.col-md-11.col-lg-11.placeholdify{class: ('editable' if current_page?(edit_text_url(@text)) || current_page?(new_text_url)), contenteditable: (true if current_page?(edit_text_url(@text)) || current_page?(new_text_url)), data: {placeholder: t(:sub_title)}}= raw(@text.subtitle) unless @text.subtitle.nil?
%section#text-commands-bar.new-text-container
-if @text.is_draft
%span.pull-left= t(:text_still_draft)
%ul.pull-right.options
- if (current_page?(new_text_url) || current_page?(edit_text_url(@text)))
%li.drop_down_menu#menu_categoria.text-center
= select("text", "category_id", Category.all.collect { |c| [t(c.locale_key.to_sym), c.id]}, {include_blank: t(:select_cat)})
%li.drop_down_menu#menu_lingua.text-center | {
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} |
filters, z-transform, transfer-function, impulse-response, causality
$$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{z-2} \Longrightarrow Y(z)z-2Y(z) = X(z) \Longrightarrow
y_{n+1} - 2y_n = x_n,$$ where the last step is due to the property $$Z[x_{n+k}](z) = z^kX(z).$$ The difference equation implies the system is causal and we infer the ROC is $$2 < |z| < \infty.$$ Where is my mistake? Where did I make an implicit assumption? $y_{n+1} - 2y_n = x_n$ still doesn't tell you. You could take it to mean a causal, unstable system: $y_{n+1} = x_n + 2y_n$, or a stable, non-causal system $y_n = \frac 1 2 y_{n+1} - \frac 1 2 x_n$.
You just have to know in advance which one is being talked about*. As a corollary, people are usually going to assume that any pole $|a| > 1$ is an unstable pole of a causal system: it's up to you to tell people when you're deep enough into theory-land that you're taking noncausal systems seriously. | {
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machine-learning, neural-network, deep-learning, regression, linear-regression
Title: One Neural network with multiple outputs or multiple neural networks with a single output? I an building a feed forward deep learning model using tabular data. The inputs are numeric features or categorical features (represented with embeddings). The outputs are the same number of numeric input features.
Is there any known research or models out there which verifies that using a single model with multiple outputs would be better/worse than multiple models, each with a single output?
In essence, with N observations and M outputs, a single model minimizes:
$$
\frac{1}{N}\sum_n^N\sum_m^M \left(y_m^{(n)} - \hat{y}_m^{(n)} \right)^2
$$
while multiple models with single output, each minimize:
$$
\frac{1}{N}\sum_n^N \left(y_m^{(n)} - \hat{y}_m^{(n)} \right)^2
$$
For a single value of $m$. | {
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general-relativity, differential-geometry, tensor-calculus
Then in the submanifold $N$ if we fix an orientation we can likewise set $\mu(h)=\pm\sqrt{\mathfrak h}dy^1\wedge\dots\wedge dy^{n}$ with the $+$ sign chosen for a positive coordinate system $y$ and the negative sign for a negative coordinate system $y$.
Note that $\mu(h)$ is an $n$-form on $N$ and not on $M$, so eg. we cannot write in index notation in a coordinate system of $M$. Not by default. But because of the metric tensor, we have a reasonably unique extension of $\mu(h)$ into $M$ (by that I mean that $\mu(h)$ is extended into an $M$-tensor but is still only defined at points of $N$).
Namely, by the assumptions on the metric and the causal type of $N$ for any $p\in N$ we have $$ T_pM=(T_pN)^\perp\oplus T_pN, $$i.e. each tangent spaces decomposes into a direct sum. Let $\Pi:TM|_N\rightarrow TN$ denote the corresponding projection operator which projects tangentially. | {
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c#, performance, linq, set
object result2 = null;
var timer2 = Stopwatch.StartNew();
for (var i = 0; i < Iterations; i++)
result2 = collections.IntersectAll2().ToList();
timer2.Stop();
result2.Dump(String.Format("result2: {0}", timer2.Elapsed));
object controlResult = null;
var controlTimer = Stopwatch.StartNew();
for (var i = 0; i < Iterations; i++)
controlResult = collections.IntersectAllControl().ToList();
controlTimer.Stop();
controlResult.Dump(String.Format("controlResult: {0}", controlTimer.Elapsed));
}
Note: The ToList() calls were needed on all invocations to ensure that the intersections are actually generated.
Run on my machine with multiple collections:
result1: 00:00:01.0210291
result2: 00:00:01.0285069
controlResult: 00:00:03.1512838
And with a single collection:
result1: 00:00:02.9254441
result2: 00:00:04.8505489
controlResult: 00:00:00.2102433 | {
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quantum-gate, quantum-algorithms, qiskit, programming, circuit-construction
circ.h(control)
# create and append oracle
oracle = create_oracle(train_register, control) # returns an Instruction
circ.append(oracle, [train_register, control])
If my train_register and control are both QuantumRegisters with length 1, this works perfectly fine. But I want to make this work for variable length of these registers, I get the error qiskit.circuit.exceptions.CircuitError: 'The amount of qubit arguments does not match the instruction expectation.'.
This is logical, since this error occurs because qiskit checks the length of my provided qargs in QuantumCircuit.append(), and I provided two registers in a list with both one qubit, so the length of the list happens to be the same as the number of qubits.
However, I want this to work for variable length. If I know what the length of my QuantumRegisters are, say len(train_register) is 3 and len(control) is 1 I can use
circ.append(oracle, [0, 1, 2, 3]) | {
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mechanical-engineering, modeling, vibration, finite-element-method
Title: Modelling splashing water in FEA I'm running an FEA on a vessel that has a "splashing water" effect with vibrations. The problem I'm really coming up against is what kind of Young's Modulus and Poisson's ratio to use for the fluids? - air and water. I have been successful in using the bulk Modulus, and treat it as uncompressible - i.e. Poisson's ratio of 0.5, but I never used it for vibrational analysis, just static analysis. Are these still good assumptions in a dynamic analysis? This may be only a partial answer since I don't have any idea what to do for the Young's modulus of the fluid.
Consider the paper Dynamic Pressures on Accelerated Fluid Containers by G. W. Housner. This is referenced by ASCE 4-98 Seismic Analysis of Safety-Related Nuclear Structures, Commentary Section C3.1, for analysis of hydrodynamic loads on tanks. In the paper, Housner states that: | {
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javascript, jquery
// has to get implmented
// normal value
} else {
o[this.name] = this.value;
}
});
return o;
}; Here's what I came up with:
var addNestedPropToObj = function (obj, name, value) {
var path = name.split('.'),
current = obj,
len = path.length - 1,
i = 0;
for (; i < len; i++) {
current[path[i]] = current[path[i]] || {};
current = current[path[i]];
}
if ( 0 < path[i].indexOf( "[]" ) ) {
name = path[i].replace('[]', '');
current[name] = current[name] || [];
current[name].push(value);
} else {
current[path[i]] = value;
}
return obj;
};
jQuery.fn.serializeObject = function () {
var o = {},
a = this.serializeArray(),
i = 0,
len = a.length;
for (; i < len; i++) {
o = addNestedPropToObj(o, a[i].name, a[i].value);
}
return o;
};
Demo and test cases
Tips: | {
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data-structures
If you're worried about high-volume times where you need to respond quickly, you might even be able to skip the hit/total counters and do the Bayesian analysis on the individual cache hits and cache misses. | {
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quantum-mechanics, double-slit-experiment, interference, quantum-eraser
"Interact" is definitely the wrong word here. Light doesn't interact with other light; if you cross two beams of light and examine them after the crossing point, you will not be able to detect that the two beams had crossed. What's happening here is, specifically, interference, which is not an interaction between two light beams; instead, it just derives from the simultaneous presence of two electromagnetic fields at the same point in space. These two electromagnetic fields just add their intensities, because electromagnetism is linear. | {
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catkin
the force-cmake flag is neccessary, I just tried it out ;)
Originally posted by kalectro with karma: 1554 on 2013-01-28
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by BigBlueDart on 2013-01-29:
It appears that I only get this issue when running catkin_make_isolated. Running catkin_make does not appear to have the same problem recognizing flags. | {
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observational-astronomy, black-hole, mass, gaia, mass-gap
Title: How do OGLE-III and GAIA measure the mass of free microlensing black holes? What is the "hypothesized lower mass gap" between 2.5 and 5 solar masses? eventually links to Constraining the masses of microlensing black holes and the mass gap with Gaia DR2.
The angle of deflection of light passing a massive object is given by:
$$\theta = \frac{4GM}{r c^2}$$
where $r$ is the minimum distance from the mass that the light passes.
If two black holes pass by a line-of-sight to a distant object and their speeds and distances of closest approach $r$ both scale linearly with their mass, they produce the identical deflection magnitude and time dependence. | {
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autocorrelation standard error
Autocorrelation Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Autocorrelation Variance a li li a href Autocorrelation Standard Error Underestimate a li li a href Robust Standard Errors Autocorrelation a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss relatedl the workings and policies of this site About Us Learn multicollinearity standard error more about Stack Overflow the company Business Learn more about hiring developers or posting autocorrelation confidence interval ads
average standard error | {
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"openwebmath_score": 0.3121330142021179,
"tags": null,
"url": "http://winbytes.org/help/standard-error/proportion-standard-error.html"
} |
java, object-oriented, tree, search, generics
public interface MctsDomainAgent
<A extends MctsDomainAgent<A, M, S>,
M,
S extends MctsDomainState<S, M, A>> { ..... }
Note that the first generic is a self-referential one, and has the pattern X extends MySelf<X, ....>. This is how self referencing generics are done. The most important part about this declaration, is that it declares ourselves! Now, with that declaration, we can reuse the generic token in other parts of the declaration. For example, here:
public interface MctsDomainState
<S extends MctsDomainState<S, M, A>,
M,
A extends MctsDomainAgent<A, M, S>> { ..... }
we declare S to be self-referential, and then we use S to declare the type of A which is a MctsDomainAgent.
Using these classes is surprisingly easy:
public class TicTacToePlayer
implements MctsDomainAgent<TicTacToePlayer, String, TicTacToeState> { .... | {
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"tags": "java, object-oriented, tree, search, generics",
"url": null
} |
control, pid
And you want to move on it with a velocity profile like this:
Then follow this example:
clc; clear all; close all
%% Path setup
path_step = transpose(0:.01:500);
x_path = path_step/17 + 2*sin(path_step/50);
y_path = sqrt(path_step) + 5*sin(path_step/10);
position_matrix = [x_path,y_path];
distance_between_points = diff(position_matrix,1);
dist_from_vertex_to_vertex = hypot(distance_between_points(:,1), distance_between_points(:,2));
cumulative_dist_along_path = [0; cumsum(dist_from_vertex_to_vertex,1)];
%% Cart setup
saturation_limit = 4;
motor_coef = .5;
motor_model = @(input_command) min(saturation_limit, max(-saturation_limit, input_command * motor_coef)) + .3 * randn(1);
k_p = 3;
k_i = 1;
k_d = .1;
%% Simulation and initial conditions setup and array initialization
time_step = .1;
sim_time = 100;
t = transpose(0:time_step:sim_time);
distance_actual = zeros(size(t));
err = zeros(size(t));
err_integrated = 0;
err_derivative = 0;
distance_actual(1) = 20; % actual starting point | {
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quantum-mechanics, hilbert-space, mathematical-physics
But I cannot feel it completely unless I see an example of such a sequence. What is an example of a Cauchy sequence of vectors $(\psi_1,\psi_2,...)$ that we encounter in quantum mechanics? Here's a concrete example. For a particle in an infinite potential well of width $a$, the normalized energy eigenvectors are of the form
$$\psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right)$$
Most wavefunctions - such as the $\Psi(x) = \frac{1}{\sqrt{a}}$, corresponding to a uniform spatial probability density throughout the well - cannot be written as a finite linear combination of energy eigenvectors. It can, however, be expressed as the sum of the convergent series
$$ \sum_{n=1}^\infty\frac{2\sqrt{2}}{(2n-1)\pi} \psi_{2n-1} \rightarrow \Psi(x)$$
as illustrated with the following plot of the first $n$ partial sums:
The sequence of partial sums $\Psi^{(n)}:= \sum_{k=1}^n\frac{2\sqrt{2}}{(2n-1)\pi}\psi_{2n-1}$ is indeed Cauchy (which can be verified as a nice exercise), as requested. | {
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experiment is referred to as This calculator is featured to generate the complete work with steps for any corresponding input values to solve Poisson distribution worksheet or homework problems. The Poisson Probability Calculator can calculate the probability of an event occurring in a given time interval. For help in using the calculator, read the Now need to convert these averages into probability. This has been a guide to Poisson Distribution. See the above The properties of the Poisson distribution have relation to those of the binomial distribution:. that the Poisson random variable (X) falls within a certain range. View all posts by Zach Post navigation. We might, for example, ask how many customers visit a The average rate of success 6. Poisson Distribution Calculator. Here, n would be a Poisson Online help is just a mouse click away. Poisson probability. A Poisson distribution is a probability distribution of a Poisson random variable. Poisson Distribution in Excel. question, | {
"domain": "marqueindia.org",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9759464485047916,
"lm_q1q2_score": 0.8174722323678324,
"lm_q2_score": 0.8376199673867852,
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"openwebmath_score": 0.7082796096801758,
"tags": null,
"url": "https://marqueindia.org/bo-ness-nyvu/15d57f-poisson-distribution-calculator"
} |
computer-vision, projection, geometry
bring into camera coordinates: Xc = R^-1 * X - R^1 * translation
transform into image coordinates: x = Xc * focal length / Xc.z + principal point
I have read the theory from various sources and it seems I am following the rules.
This is just artificial data, so not sure if focal length and principal point can have any effect on it? What am I doing wrong? Why is the cube not centered?
EDIT:
I construct the cube like this (duplicate vertices are there because I need them to draw lines):
cv::Matx31d center(0,0,1);
double delta = 0.2;
std::vector<cv::Matx31d> volumeCorners = {
// back
(center + cv::Matx31d(-delta, delta, delta)),
(center + cv::Matx31d(delta, delta, delta)),
(center + cv::Matx31d(delta, delta, delta)),
(center + cv::Matx31d(delta, -delta, delta)),
(center + cv::Matx31d(delta, -delta, delta)),
(center + cv::Matx31d(-delta, -delta, delta)), | {
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scikit-learn, mse, ridge-regression
Finally, when such an exact relationship is the truth, it makes sense not to penalize the regression. Your coefficients here are a little bit larger, and the errors drop away to nearly nothing, especially considering the scale of the target. | {
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The numerical value of this answer is 214.853, as expected.
• +1 clear, concise and provides the analytic expression :) – ubpdqn Jul 22 '15 at 3:12
• Such a result is what was really expected. – LCFactorization Jul 22 '15 at 4:55
• Sorry, just to understand your sayings: How comes my answer isn't "a solution symbolic at every step" ? – Dr. belisarius Jul 23 '15 at 21:50
• @Belisarius Certainly, I meant no criticism. Your solution is creative and produces a quite satisfactory result. I merely meant to convey that the approach I used did not require using Root functions at an intermediate point. Would you like me to revise my first sentence? Best wishes. – bbgodfrey Jul 24 '15 at 1:35
• Ah,ok. Kind of a language barrier. Agree! – Dr. belisarius Jul 24 '15 at 2:16
The plan is first get the "external" contour and then use Green's theorem to find its area.
r[t_] := {-9 Sin[2 t] - 5 Sin[3 t], 9 Cos[2 t] - 5 Cos[3 t], 0} | {
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"lm_q1q2_score": 0.841915183867935,
"lm_q2_score": 0.8596637451167995,
"openwebmath_perplexity": 3799.1356710434197,
"openwebmath_score": 0.3483494520187378,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/88623/symbolic-area-calculation-for-a-parametric-self-intersecting-closed-curve/88637"
} |
to the origin in 1024 points. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of a triangular pulse is a sinc 2 function. Compare the result with pan (b). Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T 0 where it has a height of one:. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. Fourier Transform of the Gaussian Konstantinos G. The reason why Fourier analysis | {
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"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9875683520739038,
"lm_q1q2_score": 0.8807243687639972,
"lm_q2_score": 0.8918110497511051,
"openwebmath_perplexity": 790.5836348459956,
"openwebmath_score": 0.8125916719436646,
"tags": null,
"url": "http://xdhq.visitmarzamemi.it/fourier-transform-of-triangular-pulse.html"
} |
Why Surgery? | {
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226287518853,
"lm_q1q2_score": 0.8162863257019526,
"lm_q2_score": 0.8354835411997897,
"openwebmath_perplexity": 233.14372944331421,
"openwebmath_score": 0.8773239254951477,
"tags": null,
"url": "https://mathoverflow.net/questions/261018/homeomorphisms-of-sn-times-s1/261023"
} |
wavelet
Title: what is Shannon entropy? I was reading a paper about finding optimal morlet wavelet function for CWT, and they were minimizing Shannon entropy.
what is Shannon entropy and it uses?
(specially on continuous wavelet transform) Shannon entropy is a way of measuring the degree of unexpectedness or unpredictability of a random variable. For example rolling a die has higher entropy than flipping a coin because the die has more possible outcomes making it harder to predict. Same goes for a biased coin versus a fair coin. | {
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quantum-field-theory, cosmology, vacuum, dark-energy, cosmological-constant
term "cosmological constant" synonymously with the total vacuum energy density. In the continuum, each mode of a free scalar field contributes an energy $\frac12\hbar\omega_k$, where $\omega_k$ is the frequency of the oscillator, and the bare (unregularised) value for the vacuum energy density is given by an integral over all the zero modes: | {
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} |
quantum-mechanics, homework-and-exercises, operators, wavefunction
Not able to understand how to determine expectation value when ground state and first excited states are given. I'm not sure why the state is called $\hat A$ since $\hat A$ is also an operator, but if you call the state $|\psi\rangle$ such that $|\psi\rangle=(3|\psi_1\rangle+4|\psi_2\rangle)/5$ then the expectation value of $\hat A$ is given by
$$\langle\hat A\rangle=\langle\psi|\hat A|\psi\rangle$$
Since $\hat A$ is linear ( most quantum operators are linear) this can be expanded as follows:
$$\tfrac{1}{5}\left(3\langle\psi_1|+4\langle\psi_2|\right)\hat A\tfrac{1}{5}\left(3|\psi_1\rangle+4|\psi_2\rangle\right)=
\\\tfrac{1}{25}\left(9\langle\psi_1|\hat A|\psi_1\rangle+12\langle\psi_1|\hat A|\psi_2\rangle+12\langle\psi_2|\hat A|\psi_1\rangle+16\langle\psi_2|\hat A|\psi_2\rangle\right)$$
You can now replace $\hat A|\psi_1\rangle$ by $|\psi_2\rangle$ and $\hat A|\psi_2\rangle$ by $|\psi_1\rangle$.
What do you get as an expectation value now? | {
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"tags": "quantum-mechanics, homework-and-exercises, operators, wavefunction",
"url": null
} |
complexity-theory, space-complexity, nondeterminism, kleene-star
$$
\exists \text{$s$-$t$ path in $G$} \longleftrightarrow (s,t) \in G \lor \exists x,y \text{ s.t. } (s,x)|\langle G \rangle^n|(y,t) \in A^*.
$$
If we assume that $\mathsf{L}$ is closed under Kleene star then $A^* \in \mathsf{L}$, and so we can evaluate the right-hand side formula in logspace. Since $s$-$t$ reachability is $\mathsf{NL}$-complete, we deduce that $\mathsf{L}=\mathsf{NL}$. | {
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"tags": "complexity-theory, space-complexity, nondeterminism, kleene-star",
"url": null
} |
c#, programming-challenge, strings
StringBuilder res = new StringBuilder();
int curr = 0;//current s pointer
foreach (var item in index2strings)
{
var index = item.Key;
var source = item.Value.Item1;
var target = item.Value.Item2;
//check each index if source appears in s
for (int k = curr; k < index; k++)
{
res.Append(S[k]);
curr++;
}
//check the entire prefix is found
bool isFound = true;
for (int sIndx = index, j = 0; sIndx < index + source.Length; sIndx++, j++)
{
if (S[sIndx] != source[j])
{
isFound = false;
break;
}
}
if (!isFound)
{
continue;
}
curr = index + source.Length;
//append new string
foreach (var t in target)
{ | {
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"tags": "c#, programming-challenge, strings",
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} |
spacetime, astrophysics
Title: Whats with the 8 in Einstein's field equation in mathematical form? So, I've been trying to learn a lot of physics and teach myself it. But recently I came across a problem. When I read through Einstein's field equation in mathematical form, I noticed there's an $8\pi$. What does this mean? Why is it there
$$\large {} R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}$$
Thanks so much! All helps appreciated!
By the way, please explain the reason for the downvotes so I can ask a better question later.
Nevermind, thanks so much guys! I guess my thought was mostly right. Thank you so much for that comment The $8\pi$ appears when you take the classical limit of the field equations.
Let's suppose we have derived the field equations through the variational principle
\begin{equation}
\delta S = \delta \int_\Omega \mathcal{L} \sqrt{-g} \ d^4x = 0,
\end{equation} | {
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objective-c, library
+ (NSArray *)allProfiles;
+ (instancetype)profile;
+ (instancetype)profileNamed:(NSString *)profileName;
// etc, other tools you need for instantiating profiles, plus tools for deleting
@property NSString *profileName;
@property NSMutableDictionary *profileInfo;
@end
And a subclass might look like this:
@interface DTProfile: MPProfile
@property NSInteger lastLevel;
@property BOOL showHelpTips;
@end
static NSString * const kDTProfileInfoLastLevel = @"com.example.dt.lastLevel";
static NSString * const kDTProfileInfoShowHelpTips = @"com.example.dt.showHelpTips";
@implementation DTProfile
- (void)setLastLevel:(NSInteger)lastLevel {
self.profileInfo[kDTProfileInfoLastLevel] = @(lastLevel);
}
- (NSInteger)lastLevel {
return [self.profileInfo[kDTProfileInfoLastLevel] integerValue];
}
// etc.
@end | {
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"tags": "objective-c, library",
"url": null
} |
-
So from the above it follows that $$\log(1+|z|)=kz\,\,,\,k\in\Bbb C\,\,\,\text{a constant}?$$ – DonAntonio Jul 16 '12 at 18:55 @DonAntonio I don't see how that follows from the above. – marlu Jul 16 '12 at 19:07 Perhaps I misunderstood something @Marlu, but isn't this what you wrote in the last two lines of your answer? – DonAntonio Jul 16 '12 at 19:11 Do you mean that the same reasoning as above would show that $\log(1+|z|)/z$ is constant? It doesn't because $g$ needs to be entire to apply Liouville's theorem. – marlu Jul 16 '12 at 19:29 thanks. "Entire" precisely was what I missed. +1 – DonAntonio Jul 16 '12 at 23:57
We will prove a slightly more genaral fact. The proof is based on this answer
Theorem. Let $f\in\mathcal{O}(\mathbb{C})$ and for all $z\in\mathbb{C}$ we have $|f(z)|\leq\varphi(|z|)$. Assume that $$\lim\limits_{R\to+\infty}\frac{\varphi(R)}{R^{p+1}}=0$$ then $f$ is a polynimial with $\deg (f)\leq p$. | {
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"url": "http://math.stackexchange.com/questions/171610/an-entire-function-is-identically-zero/171618"
} |
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