text stringlengths 1 1.11k | source dict |
|---|---|
bahwa f(x) = 0 memiliki satu akar antara x = a dan x = b ; maka f(a) dan f(b) memiliki tanda berlawanan (diasumsikan bahwa grafik f(x) adalah menerus antara a dan b ) sekarang kita lihat bahwa c adalah pertengahan antara a dan b. Similarities with Bisection Method: Same Assumptions: This method also assumes that function is continuous in [a, b] and given two numbers ‘a’ and ‘b’ are such that f(a) * f(b) < 0. Then α is 0. The Bisection Method will cut the interval into 2 halves and check which. The Newton-Raphson method assumes the analytical expressions of all partial derivatives can be made available based on the functions , so that the Jacobian matrix can be computed. 3 Least Squares Approximations It often happens that Ax Db has no solution. Implementing the bisection method in Excel 8:59. Previous Slide Section 4. See full list on ece. In these cases, deflection (due to the component’s own weight and due to applied loads and forces) can affect the running properties of the bearings | {
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• Thanks for making it easier to understand. But wouldn't you think that because person A probably didn't choose it, B and C would have a harder time. Giving the most advantage to A. – Brandon Pickert Sep 8 '13 at 19:44
• As you pointed out yourself, the probability that A doesn't get the number is about $66.66\%$, so the probability she gets the number is about $33.34\%$, or more exactly, $\frac{1}{3}$. That is exactly what my answer says. So you in fact agree with your friends. – André Nicolas Sep 8 '13 at 19:53 | {
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navigation, ros-hydro, tf2, local-planner, global-planner
Comment by robotiqsguy on 2013-12-17:
To debug this problem I was thinking of pushing values on to the error_string. That way it would only get printed when an error happens.
Comment by tfoote on 2013-12-17:
There are debugging tutorials here: http://wiki.ros.org/tf/Tutorials/Debugging%20tf%20problems and feel free to put debugging code in whereever you need. | {
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"url": null
} |
newtonian-mechanics, newtonian-gravity, time
In all treatments I haven't seen this phenomena being discussed, so, how would one justify they can ignore this effect? | {
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The solution appears to be correct. Just for the sake of sanity, here's a different argument, based on the idea that knowing derivatives is knowing many limits.
First, find the limit of the logarithm of the beast, which is best treated also with the substitution $$x=1/t$$, which makes us trying to find $$\lim_{t\to0^+}\frac{1}{t}\log\left(\frac{\sqrt{1+2t+3t^2}-\sqrt{1+3t^2}}{t}\right) = \lim_{t\to0^+}\frac{1}{t}\log\left(\frac{2}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\right)$$ This can be rewritten as $$\lim_{t\to0^+}-\frac{\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)-\log2}{t}$$ which is the negative of the derivative at $$0$$ of $$f(t)=\log\bigl(\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}\,\bigr)$$ Since $$f'(t)=\frac{1}{\sqrt{1+2t+3t^2}+\sqrt{1+3t^2}}\left(\frac{1+3t}{\sqrt{1+2t+3t^2}}+\frac{3t}{\sqrt{1+3t^2}}\right)$$ we have $$f'(0)=1/2$$ and therefore the limit is $$-1/2$$, so your given limit $$e^{-1/2}$$ | {
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} |
standard-model, scattering, quantum-chromodynamics
Title: How to compute scattering amplitude $\gamma\pi^+\to\pi^+\pi^0$ I wish to find the amplitude for process $\gamma\pi^+\to\pi^+\pi^0$ at low energies. I am familiar with the basic concepts and techniques of QFT but have never dealt with the scattering processes including hadrons. In this case, I do not have a clue where to start. I am basically asking for a reference where this process is computed and/or what are the key points one needs to have in mind thinking about the reaction: appropriate effective Lagrangian, certain selection rules etc. The standard chiral lagrangian has a $\pi\to-\pi$ symmetry that forbids processes with an odd number of Goldstone bosons. This is not a symmetry of QCD, however, and these reactions are described by the Wess-Zumino term. The Wess-Zumino term is not a local lagrangian in 4-d, but it can be written as a local lagrangian on a 5-d manifold which has 4-d Minkowski space as a boundary. This is discussed in many text books, see, for example, Section | {
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"tags": "standard-model, scattering, quantum-chromodynamics",
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ros, spinonce, dynamic-reconfigure
// Instantiate the controller
force_controller::controller myControl(node);
// Set up the Dynamic Reconfigure Server to update controller gains: force, moment both proportional and derivative.
if(myControl.dynamic_reconfigure_flag)
{
// (i) Set up the dynamic reconfigure server
dynamic_reconfigure::Server<force_error_constants::force_error_constantsConfig> srv;
// (ii) Create a callback object of type force_error_constantsConfig
dynamic_reconfigure::Server<force_error_constants::force_error_constantsConfig>::CallbackType f;
// (iii) Bind that object to the actual callback function
//f=boost::bind(&force_controller::callback, _1, _2); // Used to pass two params to a callback. | {
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"url": null
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python, python-3.x
218 5 1000 - HIDSL latest XX.Y.Z.79 - ✓ ✗ Musterstraße 42, 123456 Irgendwo -
219 6 1000 - HIDSL latest XX.Y.Z.80 - ✓ ✗ Musterstraße 42, 123456 Irgendwo -
330 7 1000 399 HIDSL latest XX.Y.Z.182 - ✗ ✗ Musterstraße 42, 123456 Irgendwo -
508 8 1000 - HIDSL latest XX.Y.Z.189 - ✗ ✗ N/A - Because the code snippet is so small and missing any other context, it's hard to identify any problems in scaling / that may arise elsewhere in the code. Thus if you want a more tailored answer, you should add more code. | {
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java, mvc, caesar-cipher
public String askForClearText() {
System.out.println("Please enter clear text: ");
final String clearText = new Scanner(System.in).nextLine();
return clearText;
}
public int askForKey() {
System.out.println("Please enter a positive integer number between 1 - 25");
final int key = new Scanner(System.in).nextInt();
return key;
}
public void warnUser() {
System.out.println("Please enter a valid key value!");
}
public void presentEncrpytedText(String encryptText) {
System.out.println("Here is your encrypted text: ");
System.out.println(encryptText);
}
}
The controller:
package biz.tugay;
import biz.tugay.caesarcipher.CaesarCipher;
public class CaesarCipherController {
private final CaesarCipherView caesarCipherView;
private final CaesarCipher caesarCipher; | {
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ros, opencv
{-------------------------------------------------------------------------------
[ 0%] Built target rospack_genmsg_libexe
[ 0%] Built target rosbuild_precompile
[100%] Building CXX object CMakeFiles/cv_bridge.dir/src/cv_bridge.o
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp: In function ‘int cv_bridge::getCvType(const std::string&)’:
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:51: error: ‘BGR16’ is not a member of ‘enc’
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:52: error: ‘RGB16’ is not a member of ‘enc’
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:55: error: ‘BGRA16’ is not a member of ‘enc’
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:56: error: ‘RGBA16’ is not a member of ‘enc’
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:63: error: ‘BAYER_RGGB16’ is not a member of ‘enc’
/home/metallo/ros/trunk/cv_bridge/src/cv_bridge.cpp:64: error: ‘BAYER_BGGR16’ is not a member of ‘enc’ | {
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same entry.. A square matrix of data called a, writing code in mathematics, matrix! Or lower triangle part of this and can be observed for the and... Matrices of the original matrix to columns in the transposed matrix new in... Ensure you get the best experience on our website, will take the form =∗∗∗∗∗∗, will., to properly illustrate that this is indeed the case with all skew-symmetric that... A matrix, then the transpose, which is validated in the transpose. N'T need to transpose the dataframe and change the column index are,. 'Re back where you started matrix can be thought of as another matrix with and... Next, we consider the following theorem to specify the index and the column index are switched the! Row 2, etc of a two Dimensional array ) Syntax nagwa is an educational technology aiming. Way of viewing this operation is that the matrix transpose would switch the rows in the first of. =−422−7 transpose of a column matrix is answer =44−1−7, does ( − ) =− MxN matrix the | {
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} |
human-biology, genetics, dna
By the time we get to third cousins, "we are getting down near the baseline probability that a particular gene possessed by A will be shared by any random individual taken from the population".
Dawkins was ball-parking that a bit, although he was really talking only about a specific gene. The inbreeding coefficient is properly defined as:
The probability that an individual carries two identical-by-descent alleles at a locus. | {
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$\hspace{1.3cm}$
Using the principle of superposition, $$I = I_{\text{centre}} + 6I_{\text{edge}},$$ where $$I_{\text{centre}} =\gamma \frac{M}{7}\left(\frac{l}{\sqrt{7}}\right)^2 = \gamma \frac{Ml^2}{49} = \frac{I}{49}.$$
Now, by the parallel axis theorem $\displaystyle I_{\text{edge}} = I_{\text{COM}} + Md^2$ where $$\displaystyle I_{\text{COM}} = \frac{I}{49}$$ and $\displaystyle d= \frac{\sqrt{3} l}{2}$ (this was one source of error), so $\displaystyle I_{\text{edge}} = \frac{I}{49} + \frac{3Ml^2}{4},$ and \begin{align*} I &= \frac{I}{49} + 6\left(\frac{I}{49}+ \frac{3Ml^2}{4}\right), \\ I & = \frac{I}{7} + \frac{9Ml^2}{2}, \\ \frac{6I}{7} & = \frac{9Ml^2}{2}, \\ I & = \frac{21Ml^2}{4}. \end{align*}
This seems incorrect? It feels wrong, comparing to a disk of radius $l/2$ which has moment of inertia $Ml^2/4$ it seems far too large.
It would also be nice if we could verify our answer numerically or otherwise. Any references are also appreciated. | {
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} |
python, python-2.x, file-system
Python has a number of functions like os.path.relpath and os.path.join which would probably be a better choice then string manipulation. They'll handle the funky corner cases.
def transfer_file(self, src_path):
dest_path = self.get_remote_path(src_path)
logger.info('Copying\n\t%s\nto\n\t%s:%s' % (src_path, self._ssh_prefix, dest_path))
try:
# Make sure the intermediate destination path to this file actually exists on the remote machine
self._connection.execute('mkdir -p "' + os.path.split(dest_path)[0] + '"')
self._connection.put(src_path, dest_path)
except Exception as e:
logger.error('Caught exception while copying:')
logger.exception(e) | {
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image-processing, signal-analysis, noise, image-compression, jpeg
NO-REFERENCE PERCEPTUAL QUALITY ASSESSMENT OF JPEG COMPRESSED
IMAGES
https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1038064
No-Reference JPEG Image Quality Assessment
Based on Blockiness and Luminance Change
https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=7888493 | {
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operating-systems, memory-management, virtual-memory, paging
Here is a nice answer. | {
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particle-physics, scattering, neutrinos, antimatter, scattering-cross-section
We'll see the consequences few lines below.
Squaring the amplitudes, averaging over the spin of the initial quark and summing over the spins of the outgoing quark and lepton yields:
$$|\overline{\mathcal{M}}_1|^2 = 16 G_F^2 s^2$$
$$|\overline{\mathcal{M}}_2|^2 = 16 G_F^2 u^2$$
where $s$ and $u$ are the 2 Mandelstam variables. (Denoting by $p_1,p_2$ the 4 momenta of the 2 particles in the initial state and $p_3,p_4$ the ones of the 2 particles in the final states, we have $s=(p_1+p_2)^2$ and $u=(p_1-p_4)^2$). Using a well known formula for the differential cross-section $\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s} |\overline{\mathcal{M}}|^2$ (valid in our massless approximation, $\Omega$ being the solid angle in the center of mass frame) gives:
$$\frac{d\sigma_1}{d\Omega}= \frac{G^2_F}{4\pi^2}s$$ | {
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physical-chemistry, thermodynamics, electrochemistry
My another doubt is why even bother defining a reference electrode if we can find the electrode potentials of a half cell with its change in gibbs energy.
Or is it the other way around, that we calculate the value of ΔG based on the measured value of E. If we use the tabelated reduction potential $E°$ of a half-reaction in the equation $ΔG° = -nFE°$ the we implicitly assume the reaction $\ce{Ox(aq) + $\frac n2$ H2(g) -> Red(aq) + n H+(aq)}$.
If other than SHE half reaction had been chosen as the conventional potential reference, the respective implicit reaction would have been different, with different $ΔG°$ and different $E°$ for both the reaction and the zinc half reaction.
If we consider two general half-reactions then $Δ_\text{r}G{^\circ} = -nF(E{^\circ}_\text{cathode} - E{^\circ}_\text{anode} )$ | {
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It's not possible access element of the array in SQL the same way as in PL/SQL - i.e. a(i), thus function f was declared in with clause for that purpose. Otherwise solution would have been much shorter.
Other limitations
• throws an exception for arrays shorter than 3 elements (instead of returning 1)
• there is an assumption that powermultiset_by_cardinality preserves order even though it's not explicitly stated in the documentation
sqlplus listing | {
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algorithm, datetime, vba
End If
Next innerItem
Next outerItem
Implementing that in a clean way would be easier said than done though, because you can't overwrite a value in a collection. That means you will need to keep single collection of objects that represent your data structure. For simplicity's sake, you could create a class module as simple as this.
Option Explicit
Public DateValue As Date
Public Count As Long
Alternatively, assuming your original data that you want to merge has only one instance of each date, then a Scripting.Dictionary could be a great data structure to use. See my example below.
Public Sub test()
Dim dates1 As New Scripting.Dictionary
dates1.Add #1/14/2013#, 1
dates1.Add #1/15/2014#, 3
dates1.Add #1/17/2016#, 4
Dim dates2 As New Scripting.Dictionary
dates2.Add #12/1/2014#, 4
dates2.Add #1/15/2014#, 7
dates2.Add #1/17/2018#, 12
' first copy the first dict to a new one to return
Dim results As New Dictionary | {
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java, object-oriented, clustering
/**
* Encode document using Term Frequency - Inverse Document Frequency
*/
private void encode(Document document) {
// Normalize word histogram by maximum word frequency
Vector vector = document.getHistogram().clone();
// Allow histogram to be deallocated as it is no longer needed
document.setHistogram(null);
vector.divide(vector.max());
// Normalize by inverseDocumentFrequency
vector.multiply(inverseDocumentFrequency);
// Store feature vecotr in document
document.setVector(vector);
// Precalculate norm for use in distance calculations
document.setNorm(vector.norm());
}
/**
* Hash word into integer between 0 and numFeatures-1. Used to form document
* feature vector
*
* @param word
* String to be hashed
* @return hashed integer
*/
private int hashWord(String word) {
return Math.abs(word.hashCode()) % numFeatures;
} | {
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organic-chemistry, reaction-mechanism, alcohols, hydrocarbons, rearrangements
My questions are:
1- Is tert-butyl (don't know the correct term) migration possible in pathway- 2?
2- What are the major and minor products? Which factor will decide the major product: stability of final compound or the most stable intermediate? Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on labeling to elaborate the mechanism. | {
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javascript, jquery
Title: Created a DialogBuilder object, but having issues with subscribing/unsubscribing to dialog events var WorkflowDialogBuilder = _.once(function () {
'use strict';
var workflowDialog = $('#WorkflowDialog');
var workflowDialogContent = $('#WorkflowDialogContent');
var events = {
onApplyChangesSuccess: 'onApplyChangesSuccess',
onValidationFailed: 'onValidationFailed',
onDialogOpen: 'onDialogOpen'
};
var dialogConfig = {
autoOpen: false,
buttons: {
OK: function () {
var form = $(this).find('form'); | {
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ros
Original comments
Comment by phivu123 on 2017-06-12:
Hi Tommi, thanks a lot for your recommendation. How was you pedestrian detection using dlib? Is it good? Can you share something about the result? I made a brief survey on c++ libraries for pedestrian detection but so far dlib seems to be a feasible one that is better than OpenCV...
Comment by phivu123 on 2017-06-12:
... but very few users share the result on this.
Comment by Tommi on 2017-06-13:
I'm really happy with dlib, it's a good quality library. Nowadays you'd get a better performance with Deep Neural Networks but with a bigger computational cost in general.
Comment by phivu123 on 2017-06-13:
That sounds great, do you know any workable c++ library that supports deep neural network training for pedestrian detection, or can dlib do that? Thanks!
Comment by Tommi on 2017-06-13:
I don't know but you might want to look into Python libraries instead. | {
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to the party that made the content available or to third parties such Here we go. Here that symmetric matrix has lambda as 2 and 4. If I want the length of x, I have to take-- I would usually take x transpose x, right? The entries in the diagonal matrix â are the square roots of the eigenvalues. It's important. Made for sharing. which specific portion of the question â an image, a link, the text, etc â your complaint refers to; Please be advised that you will be liable for damages (including costs and attorneysâ fees) if you materially In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ If A is an n x n symmetric matrix, then any two eigenvectors that come from distinct eigenvalues are orthogonal.. So I have a complex matrix. Proof: We have uTAv = (uTv). And x would be 1 and minus 1 for 2. Complex numbers. And | {
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"url": "https://www.talknativ.com/mncwueub/why-are-eigenvectors-of-symmetric-matrices-orthogonal-824e67"
} |
(n-1) + (n-2) + … + 2 + 1 = n(n-1)/2
So X = n(n-1)/2.
Therefore, the time complexity of the worst case scenario is O(n2).
Time Complexity Model of the Worst Case:
Average Case:
Random Variables:
A random variable is called discrete if its range is finite or countably infinite.
Bernoulli Random Variable – type of discrete random variable with a probability mass function p(x) = P{X = x}
Bernoulli Random Variable:
X is a Bernoulli Random Variable with probability p if
px(x) = px(1-p)1-x, x = {0,1}.
Other notes: E[X] = p
Performace:
Although popular, bubble sort is nearly universally derided for its poor performance on random data. This derision is justified as shown in Figure 1 where bubble sort is nearly three times as slow as insertion sort.
posted Apr 22, 2014
## Related Articles | {
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"openwebmath_score": 0.728144645690918,
"tags": null,
"url": "https://www.queryhome.com/tech/41373/analysis-of-bubble-sort"
} |
bash
# Create a temporary working directory
# ====================================
temp_backup=$(mktemp -d -p "${backup_directory}")
# Manage previous backups
# =======================
# List every previous backup and put it into an array
backups=()
while read -r -d ''; do
backups+=("${REPLY}")
done < <( find "${backup_directory}" -mindepth 1 -maxdepth 1 -name "$(scheme \*)" -printf "%A@:%p\0" | \
sort -z -t: -n -r | \
tr '\n\0' '\0\n' | cut -d: -f2 - | tr '\n\0' '\0\n' \
)
# If it exists, select the latest backup as a reference for rsync --link-dest
if (( ${#backups[@]} > 0 )); then
latest_backup="${backups[0]}"
else
latest_backup=""
fi
# Compute the backups to remove
# We add one backup before cleaning up
# Thus, we keep $backup_count - 1 from the ${backups[@]}
old_backups=("${backups[@]:${backup_count} - 1}")
# Cleanup function
# ================ | {
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python, programming-challenge, time-limit-exceeded
class Solution:
def knightProbability(self, N: int, K: int, r: int, c: int) -> float:
i = 0
pos_moves = deque([[r, c, 1]]) # (r, c, prob)
self.moves_memo = {}
while i < K and len(pos_moves):
increase_i_after = len(pos_moves)
for j in range(increase_i_after):
move = pos_moves.popleft()
for next_move in self.get_pos_moves(N, move[0], move[1], move[2]):
pos_moves.append((next_move[0], next_move[1], next_move[2]))
i += 1
ans = 0
for m in pos_moves:
ans += m[2]
return ans/len(pos_moves) if len(pos_moves) > 0 else 0
def get_pos_moves(self, n, r, c, prev_p):
# Returns a list of possible moves
if (r, c) in self.moves_memo:
pos_moves = self.moves_memo[(r, c)]
else:
pos_moves = deque([])
if r+2 < n: | {
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} |
javascript, html, tic-tac-toe, socket.io
*
* We have these numbers in the Player.wins array and for the current
* player, we've stored this information in playsArr.
*/
Game.prototype.checkWinner = function(){
var currentPlayerPositions = player.getPlaysArr();
Player.wins.forEach(function(winningPosition){
if(winningPosition & currentPlayerPositions == winningPosition){
game.announceWinner();
}
}); | {
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} |
newtonian-mechanics, rotation
I'm not understanding part b. Why is the rotation about O? Shouldn't it be about C? If you do it about C you get a different answer because you'd get the same angular velocity but a different radius to find a_y. Why did they even add point c to the diagram if it's rotating about O? If I drew the diagram for the rotation about O, I'd just have the alpha about O, what are all those other vectors? On my rolling motion diagram you'd just have an a_y and an a_x. Why separate a_o from those two? Consider the very first image, where the object is standing on end. Points G and the point marked later as the "instantaneous center" C are at height $r$ above the table. As the object rolls, they both drop down until C is on the table. So from the kinematical view, neither G nor C can be the center because they are rotating around another point, namely the center of translation O. | {
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c#, error-handling, csv
{
importModel.Error = "Error2";
return importModel;
}
else
{
var products = _productService.GetProducts(productSkuList).ToList();
if (!importModel.InvalidSkuList.Any())
{
bool isImported = _productService.Import(mappings);
if (!isImported)
{
importModel.Error = "Error3";
}
}
else
{
return importModel;
}
}
}
else
{
importModel.Error = "Error4";
return importModel;
}
} Generally speaking whenever you are facing the problem like the above one what you can do is to perform an assessment against your current flow control and/or try to replace some part of your logic to reduce code complexity. The former one tries to logically reduce the complexity while the latter one mechanically.
Assessment | {
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linear-regression
Title: Calculate coefficient w* I'm learning ML from Bishop's book. But I don't know that How should I calculate w* in the below picture. Just looked into the book. It was an example of the data that is presented in Fig 1.4. Numpy is a good package to derive the sum of squares fit to the data. This is an optimization problem. Look at the notes section of numpy.polynomial.polynomial.polyfit. I think the data is not given (you can digitize it!). The book is showing some qualitative behavior to give some intuition. | {
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machine-learning, deep-learning, keras, tensorflow, time-series
Full code
import numpy as np
import pandas as pd
train_df = pd.DataFrame(np.random.randint(0,100,size=(100, 16)))
train_df.columns=['var1(t-3)','var2(t-3)','var3(t-3)','var4(t-3)','var1(t-2)','var2(t-2)','var3(t-2)','var4(t-2)','var1(t-1)','var2(t-1)','var3(t-1)','var4(t-1)','var1(t)','var2(t)','var3(t)','var4(t)']
train_X=train_df.drop(['var1(t)'],axis=1)
train_y=train_df[['var1(t)']]
#subset the 3 previous timesteps of the 4 variables for the time sries part
train_X_LSTM=train_X[train_X.columns[:12]].reset_index(drop=True).values
train_X_LSTM=train_X_LSTM.reshape(-1, 3, 4)
#target is always var1(t)
train_y_LSTM=train_y.values
#take the current timestep fatures which are var2,var3,var4 which are realised at t=t
train_X_MLP=train_X[train_X.columns[-3:]].reset_index(drop=True).values
#target is always var1(t)
train_y_MLP=train_y.values | {
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python, performance, programming-challenge
Python provides the function enumerate for simultaneously iterating over items and their indexes:
for i, count in enumerate(counts):
# code using i and count
For collating items based on their keys, it is handy to use collections.defaultdict, like this:
groups = defaultdict(list)
for i, count in enumerate(counts):
groups[count].append(i)
To split a sequence into groups of length \$n\$, there's a well-known trick that's described in the documentation for zip, where it says:
This makes possible an idiom for clustering a data series into n-length groups using zip(*[iter(s)]*n). This repeats the same iterator n times so that each output tuple has the result of n calls to the iterator. This has the effect of dividing the input into n-length chunks.
There is no need to pass a key function to sort. Sequences are compared lexicographically by their items, and because we know that all the items are different, the comparison will never have to look past the first item. | {
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visible-light, units, estimation, si-units
$2.8\times10^{-9}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{sr}^{-1}$, or
$3.6\times10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}$ for the full sky. Note though that the intensity in this region is rather uncertain, since COB observations are challenging due to the Milky Way's dust emission. | {
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"url": null
} |
ros, kinect, rviz, wireless, turtlebot
core service [/rosout] found
process[kinect_breaker_enabler-1]: started with pid [2877]
process[openni_manager-2]: started with pid [2878]
process[openni_camera-3]: started with pid [2882]
process[pointcloud_throttle-4]: started with pid [2883]
process[kinect_laser-5]: started with pid [2884]
process[kinect_laser_narrow-6]: started with pid [2885]
[kinect_breaker_enabler-1] process has finished cleanly.
log file: /home/turtlebot/.ros/log/f79eacf2-5bd6-11e1-9247-485d607d9b81/kinect_breaker_enabler-1*.log
[ INFO] [1329751680.368102392]: [/openni_camera] Number devices connected: 1
[ INFO] [1329751680.368697517]: [/openni_camera] 1. device on bus 001:09 is a Xbox NUI Camera (2ae) from Microsoft (45e) with serial id 'A00363914970053A'
[ WARN] [1329751680.372477514]: [/openni_camera] device_id is not set! Using first device.
[ INFO] [1329751680.445963003]: [/openni_camera] Opened 'Xbox NUI Camera' on bus 1:9 with serial number 'A00363914970053A' | {
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"tags": "ros, kinect, rviz, wireless, turtlebot",
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adaptive-optics
Title: How difficult is adaptive optics computationally? What kind of computer do you need to do adaptive optics of big telescopes? Are we talking supercomputer clusters or can my desktop do it, or my smartphone? I suppose it depends on the size of the telescope and the wavelength that is targeted. I'm interested in both the kind of computers that are used and the actual number of FLOPs per second that is required. If you're only trying to correct a small field, using a single guide star (or a single laser guide star), then the requirements aren't very arduous. The MACAO adaptive-optics module used on the Very Large Telescope at the European Southern Observatory dates from the early 2000s and is based around two 400 MHz PowerPC 604 chips (one for the "supervisory computer", the other for the "real time computer"). You can read more about it here (PDF file). | {
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python
So far I've attempted to use numpy.fromfile and reading the data in using:
with open(filename, "rb") as file:
...
and attempting to decode it with a mixture of different codecs. The numpy method gave floating point values back, unfortunately they were nonsensical. The decoding didn't work for any of the codecs I tried. I found this Python package from the MIT Laboratory for Computational Physiology that seems to be specifically meant to process that kind of data.
This notebook demonstrates the functionalities of the library. | {
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newtonian-mechanics, gravity
Title: What are some ways to prove that centripetal force is a center seeking force? What are some examples of cases that illustrate the center-seeking behavior of centripetal force? I cannot find any examples of the center-seeking behavior of object in a circular path without any other source of force such as tension or another external force. In these cases, how are we able to say that the force is center-seeking, it seems to me that it can also be an outward force.
In the case of object following semi-circular paths such as parabolic motion and in a bowl, the "center" seeking force is clearly demonstrated, but I cannot find such examples in the case of complete revolutions.
I cannot find any examples of the center-seeking behavior of object in a circular path without any other source of force such as tension or another external force. | {
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newtonian-mechanics, classical-mechanics, particle-physics, rotational-dynamics, rotation
$$\ \epsilon = \frac {L (I_1 - I_2)} {I_1 I_2} = L \frac {(\frac {-dI}{dt})}{I_1I_2} \tag 6 $$
We count the change in moment of inertia over time:
$$\ \frac {dI}{dt} = - \epsilon \frac { (I_1 I_2)} {L} \tag 7 $$
Now it is easier to record (6):
$$\ \frac {d \omega} {dt} = - \frac {dI} {dt} \frac {\omega_1} {I_2} = = - \frac {dI} {dt} \frac {\omega_2} {I_1} \tag 8 $$
We can now complete the formula (2):
$$\ \frac {d \omega} {dt} I_2= - \frac {dI} {dt} \omega_1 \tag 9 $$
or
$$\ \frac {d \omega} {dt} I_1= - \frac {dI} {dt} \omega_2 $$
We know that the angular acceleration times the moment of inertia is the moment of force, $\ \epsilon I =M $
So we can save the formula (2) as follows
$$\ \frac {dL} {dt} = \frac {dI} {dt} \omega_1 + I_2 \frac {d\omega} {dt}= M_I + M_\omega = 0 \tag {10} $$
So we have here two opposite inside moments of forces (their source is inside the object) which are zeroing and nonzero angular acceleration which shows the pattern (8). | {
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} |
biochemistry, periodic-table, isotope, astrochemistry
Title: Biological Consequences of Asteroid Mining—Death by Isotope? It's been documented that NASA hope to capture an asteroid in 2025, and have subsequent aims to mine that asteroid. If if this is successful, we would expect other asteroids to be mined in the future.
A consequence of this is that relative atomic masses of elements mined—those with two or more stable isotopes—will no longer be faithful to our current periodic table. Ruthenium alone, a high-value rare earth, has five stable isotopes alone, ranging from $^{98}\ce{Ru}$ to $^{102}\ce{Ru}$. The relative abundances of these isotopes are bound to differ in other places other than Earth, and so would the weighted average. | {
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$$f(x):= x^1+x^2+x^3+x^4+x^5+x^6$$
pack the sequence of number of sides of a fair die (note that the power of each monomial represent one of the sides of a dice).
Now: multiplication of generating functions have the effect that the new sequence, after multiplication, is a sum of products of the old ones, where the indices of every product in each sum add up to the exponent of the monomial that will accompany.
Its easy to check that, as we are throwing three dice, then the generating function
$$g(x):=f(x)^3=(x^1+x^2+x^3+x^4+x^5+x^6)^3$$
pack as coefficients the total amounts of different ways to add up to the exponent of each monomial.
Now: the polynomial $$f$$ can be seen as the partial sum of a geometric series, i.e.
\begin{align*} f(x)&=x^1+x^2+x^3+x^4+x^5+x^6\\ &=x(x^0+x^1+x^2+x^3+x^4+x^5)\\ &=x\sum_{k=0}^{5}x^k\\ &=x\frac{1-x^6}{1-x} \end{align*}
Then $$g(x)=x^3\left(\frac{1-x^6}{1-x}\right)^3=x^3\color{red}{(1-x^6)^3}\color{green}{(1-x)^{-3}}$$ | {
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c++, recursion, template, c++20, constrained-templates
recursive_reduce_string function test with initial value, specified operation, std::vector<std::string>:
_ foo bar baz quux
recursive_reduce_string function test with execution policy, initial value, specified operation, std::vector<std::string>:
_ foo bar baz quux
recursive_reduce_string function test with std::deque<std::string>:
1234
recursive_reduce_string function test with std::deque<std::deque<std::string>>:
123412341234
recursive_reduce_string function test with std::vector<std::wstring>:
0123
recursive_reduce_string function test with std::vector<std::vector<std::wstring>>:
0123012301230123
recursive_reduce_string function test with std::array<std::wstring, 4>:
1234
recursive_reduce_string function test with std::array<std::array<std::wstring, 4>, 4>:
1234123412341234
recursive_reduce_string function test with std::deque<std::wstring>:
0123
recursive_reduce_string function test with std::deque<std::deque<std::wstring>>:
0123012301230123 | {
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python, beginner, k-sum
Use an Exception for exceptions
It is generally bad practice to put prints into your computation routines. Instead consider using exceptions.
if not list1:
raise ValueError("list is empty")
Then catch the exception like:
try:
for j in find_pairs_sum_n(l, 10):
print(j)
except ValueError as exc:
print(exc) | {
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"tags": "python, beginner, k-sum",
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$Ze$= Zed is a knight
• possible duplicate of Knights and knaves: Who are B and C? (task 26 from "What Is the Name of This Book?") – DonAntonio Dec 4 '12 at 3:12
• @Don: Could you please describe the isomorphism you see between this problem and the duplicate you suggested? Superficially they look quite different. – joriki Dec 4 '12 at 3:20
• @amWhy: The same question to you; your suggested duplicate also looks quite different superficially. – joriki Dec 4 '12 at 3:22
• To all, I mistakenly voted to close, prematurely. The "possible duplicate" that was generated as a comment was not accurate. I have deleted that comment. Apologies. @joriki - You needn't be so argumentative, joriki. – amWhy Dec 4 '12 at 3:24
• @amWhy: I'm sorry if I came across as argumentative; that wasn't my intention; I was only asking for an explanation and expressing a different view. Please explain why my comment struck you as argumentative to help me avoid that in the future. – joriki Dec 4 '12 at 3:33 | {
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c++, c++11, template-meta-programming, macros, sfinae
They are global symbols, causing surprise replacements in the most unlikely of places
It is easy to write a macro that looks like it does what you want, when it actually compiles into something else. | {
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navigation, ekf, ros-kinetic, robot-localization, ekf-localization-node
ekf_se_map:
frequency: 30
sensor_timeout: 0.1
two_d_mode: false
transform_time_offset: 0.0
transform_timeout: 0.0
print_diagnostics: true
debug: false
map_frame: map
odom_frame: odom
base_link_frame: m100/base_link
world_frame: map
odom0: dji_odom
odom0_config: [true, true, true,
true, true, true,
true, true, true,
true, true, true,
true, true, true]
odom0_queue_size: 10
odom0_nodelay: true
odom0_differential: false
odom0_relative: false
odom1: odometry/gps
odom1_config: [true, true, true,
false, false, false,
false, false, false,
false, false, false,
false, false, false]
odom1_queue_size: 10
odom1_nodelay: true
odom1_differential: false
odom1_relative: false | {
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Now we will use your reformulations, with $P(x,y) \equiv x\geq y , Q(x,y) \equiv x=y$, and just translate:
$$Q(x,y) \implies P(x,y) \implies P(x+z,y+z)$$ $$Q(x,y) \implies Q(y,x) \implies P(y,x) \implies P(y+z,x+z)$$ $$P(x+z,y+z)\wedge P(y+z,x+z)\implies Q(x+z,y+z)$$
Is that fine? | {
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"url": "https://math.stackexchange.com/questions/1921144/proving-some-properties-of-real-numbers-using-predicate-logic"
} |
particle-physics, wavefunction, wave-particle-duality, wavefunction-collapse
$$\left<x\right> = \left<\psi\right|x\left|\psi\right> = \int_{-\infty}^\infty \psi^*(x) \cdot x \cdot \psi(x) \cdot d x.$$
There is no "incomplete information" about the particle: if we have the full quantum state $\left|\psi\right>$, we have all the information we could ever (mathematically) obtain about the particle. There is actually a very famous result about this called Bell's theorem, which (paraphrased, simplified, and applied to your specific question) says that quantum mechanics cannot be created with a "local hidden variable" theory. That is, there cannot simply exist a deterministic, but arbitrarily complex, mechanism for selecting the point in space the wavefunction collapses to - it is "truly" random.
Similarly fundamental is the uncertainty between two non-commuting observables (more on this in a second), such as position ($x$) and momentum ($p$). You've probably seen the famous Heisenberg uncertainty principle, summarized as:
$$\Delta x \Delta p \ge \frac{\hbar}{2}.$$ | {
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The only reason to restrict this to subgroups of finite index is that the definition of the symmetric group for an infinite set is not a given. For some, the symmetric group on an infinite set, $$S_X$$, just means all bijections $$X\to X$$; others require the bijections to have finite support (that is, $$\mathrm{supp}(\sigma) = \{x\in X\mid\sigma(x)\neq x\}$$ is finite for each bijection $$\sigma$$), so that the elements of $$S_X$$ can still be described as consisting of a product of disjoint cycles, etc. (Also, its most common application is that a subgroup of finite index contains a normal subgroup of finite index, so the more general statement does not provide wider applications).
If you simply define $$S_X$$ to be the group of bijections $$X\to X$$, a group under composition, then this theorem holds: | {
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} |
quantum-mechanics, mathematics
The dimension of the vector space corresponds to the size of the phase space, so to speak. Spin of an electron can be either up or down and these are all the possibilities there are, therefore the dimension is 2. If you have $k$ electrons then each of them can be up or down and consequently the phase space is $2^k$-dimensional (this relates to the fact that the space of the total system is obtained as a tensor product of the subsystems). If one is instead dealing with particle with position that can be any $x \in \mathbb R^3$ then the vector space must be infinite-dimensional to encode all the independent possibilities. | {
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organic-chemistry, heat
&\mathrm{E1}{:} & \ce{RCH2-CR2-X + B- &-> RCH2-CR2+ + X- + B- -> RCH=CR2 + X- + HB}\\
&\mathrm{E2}{:} & \ce{RCH2-CH2-X + B- &-> RCH=CH2 + X- + HB}\end{align}$$
Clearly, each substitution keeps the number of freely diffusing particles the same while each elimination increases it by one. Therefore, the entropic term of the eliminations is likely to be higher than that of the substitutions. The rest is given by the Gibbs free energy equation:
$$\Delta G = \Delta H - T\Delta S$$
The entropic factor scales with temperature as Bryce already mentioned so this outweighs any other effects at high temperatures. | {
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performance, image, swift
if (pixelShouldBeInOriginalColor) {
newImagePointer[currentPixelRedIndex] = redComponent
newImagePointer[currentPixelGreenIndex] = greenComponent
newImagePointer[currentPixelBlueIndex] = blueComponent
newImagePointer[currentPixelAlphaIndex] = data[currentPixelAlphaIndex]
} else {
let greyScale = greyScaleFromRed(redComponent, green: greenComponent, blue: blueComponent)
newImagePointer[currentPixelRedIndex] = greyScale
newImagePointer[currentPixelGreenIndex] = greyScale
newImagePointer[currentPixelBlueIndex] = greyScale
newImagePointer[currentPixelAlphaIndex] = 255
}
}
return newImagePointer
} | {
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"tags": "performance, image, swift",
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units, experimental-technique, si-units, metrology
Title: How is a standard unit divided into equally smaller or fractional units physically/experimentally? Consider the standard unit of length: meter. How was it divided into decimeter, centimeter, millimeter, etc. when there were no shorter lengths than the standard? What is the physical/experimental process involved? Well historically people were using all sorts of accurate and inaccurate units, remember there were times when measurements were done in arm length, foot length etc. So absolutely there were both smaller and larger units before the establishment of any system. | {
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php5, networking
You have three places where you exit with the "unauthorised" error message. Ideally, there should be one place where you do that.
Furthermore, instead of just dying, your script should output a "403 Forbidden" HTTP header, and exit successfully.
<?php
function isGoogleBotRequest()
{
// Check User-Agent and verify using reverse-DNS lookup as recommended
// in https://support.google.com/webmasters/answer/80553
if (strpos($_SERVER['HTTP_USER_AGENT'], 'Google') !== false) {
$hostname = gethostbyaddr($_SERVER['REMOTE_ADDR']);
return preg_match('/\.googlebot\.com$/i', $hostname);
}
return false;
} | {
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"tags": "php5, networking",
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c#, programming-challenge, depth-first-search
str.Append(boggle[i, j]);
if (IsWord(str.ToString(), dictionary))
{
_list.Add(str.ToString());
}
for (int row = i - 1; row <= i + 1 && row < boggle.GetLength(0); row++)
{
for (int col = j - 1; col <= j + 1 && col < boggle.GetLength(1); col++)
{
if (col >= 0 && row >= 0 && !visited[row, col])
{
DFS(row, col, boggle, dictionary, str, visited);
}
}
}
visited[i, j] = false;
str.Remove(str.Length - 1, 1);
} | {
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mobile-robot, kinematics, dynamics
Any help would be really appreciated.
After some search, I am only able to find the geometric solution which makes sense to me. As $\dot \phi$ is the instantaneous angular velocity about $Z_v$ axis, then it has two components on $Z_b$ and $X_b$ axes. Note that $\dot \phi sin \psi$ component is on the negative $X_b$ axis. Finally I get:
$\Omega_g= [-\dot \phi sin\psi \ \dot \psi \ \dot\phi cos\psi]$
as given in the paper.
P.S: If you have a mathematical solution please contribute here. Also, if my solution is wrong let me know. | {
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ros, ros-kinetic, roscpp, callbacks
Originally posted by PeteBlackerThe3rd with karma: 9529 on 2019-05-28
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by gvdhoorn on 2019-05-28:
If this is just about the topic name I would suggest to use a MessageEvent.
Comment by CharlieMAC on 2019-05-29:
@PeteBlackerThe3rd, how could I do this if my callback is class method?
Comment by gvdhoorn on 2019-05-29:
@CharlieMAC: this topic has come up multiple times on ROS Answers. If you search for something like subscriber extra argument site:answers.ros.org (with Google) it should return quite a few posts that answer your question. | {
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electromagnetism, second-quantization
$$\begin{align*}
I &\equiv \frac{1}{T} \int_{t=0}^T u(\mathbf{x},t) \,\mathrm{d}t \\
&= \frac{1}{T} \int_{t=0}^T \epsilon_0 \left[ E(\mathbf{x},t) \right]^2 \,\mathrm{d}t \,,
\end{align*}$$
where I've written $u$ for the energy density of the electromagnetic fields and $T$ for the period of the wave. I've also taken $\mathbf{x}$ as constant and left out a uninteresting factor of $A/A$. (Of course you can do the integral in space if you prefer.)
To recover the expression you exhibited, just write
$$ E(\mathbf{x},t) = E \sin \left(\mathbf{k}\cdot\mathbf{x} - \omega t \right) \,,$$
and evaluate the integral.
Anyway, the kernel of the integral is your guide for how to interpret the $E^2$ in performing the second quantization. You want the average of the square (which is not zero) rather than the square of the average (which is zero). | {
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# Give an example of a linear transformation whose kernel is the line spanned by vector: $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$
I know that a linear transformation could be a projection onto the plane with normal vector $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$, but finding the projection would be too difficult.
I could easily think up a matrix where multiplied by $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T = 0$, but I'm not sure on how to choose a matrix where $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$ is the only element of the kernel.
Also, can you please explain this Hint: "to describe a subset as a kernel means to describe it as an intersection of planes?" | {
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Systems [Book]. Casper and F. Choosing a Backup Generator Plus 3 LEGAL House Connection Options - Transfer Switch and More - Duration: 12:39. PPT – Second order circuits (i). RLC Simulation: Impulse Response Input voltage is pulse => Capacitor stores energy And then releases the energy Dr. Frequency Response of a Circuit The cutoff frequencies in terms of βand ω 0 A Serial RLC Circuit 2 2 c1022 ββ ωω =− + + 2 2 c2022 ββ ωω =+ + The cutoff frequencies in terms of Q and ω 0 2 10 11 1 c 22QQ ωω =−++ 2 10 11 1 c 22QQ ωω =++ ECE 307-5 8 Frequency Response of a Circuit Example Using serial RLC circuit, design band. • To measure the step response of second-order circuits and. All math explained Program3: See how the impulse response is derived. The RC circuit is formed by connecting a resistance in series with the capacitor and a battery source is provided to charge the capacitor. More generally, an impulse response is the reaction of any dynamic system in response to some external change. | {
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"url": "http://brjf.socialimmobiliare.it/impulse-response-of-rlc-circuit.html"
} |
thermodynamics, water, phase-transition, evaporation, gas
The difference between these two scenarios is the fact that for the first case the system was in a sealed container and the pressure increased as the temperature increased whereas in the second case the pressure was kept constant. | {
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ros, rviz, markerarray, tf2, transform
Title: Displaying TF Data
I'm looking to display tf data in a similar fashion to what rviz does (and how it displays an arrow from each child frame origin to parent frame origin). Right now, I subscribe to the /tf topic. Every time I receive a new tf message, I use tf.TransformListener to compute the positions for each arrow marker (see code below). When I try to display the newly created markers in real time, I find that the markers are considerably behind the tf tree. The transformPose function takes up a significant amount of time. Thus, I am wondering how rviz displays tf data using markers in real team. They must have a more efficient method?
def callback_tf(self, msg):
self.pub_rm.publish(self.getMarkers(msg))
def getMarkers(self, msg):
mks = MarkerArray()
counter = 0
for transform in msg.transforms:
child_frame = transform.child_frame_id
parent_frame = transform.header.frame_id | {
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$1: p_2^2+p_2p_3+p_3^2$
$2 :p_1^2+p_3^2$
$3: p_1^2 + p_2^2.$
But generalizing to any $n$ is proving more difficult than expected.
EDIT: Corrected mistakes pointed our by @elias.
• I think to calculate $K_1$ you have to add up the probabilities calculated before instead of multiplying them. Also you should care about $\left\{2,3\right\}$ containing basically two different situations, based on which of the other two players chose $2$. Jan 29 '17 at 10:20
• Also, the main idea of a mixed strategy is to play according the $K$ probability distribution instead of playing the number with the highest probability every time. Jan 29 '17 at 10:24
• You are still missing a term from $K_1$. In fact $K_1=p_2^2+2p_2p_3+p_3^2=(p_2+p_3)^2=(1-p_1)^2$, which is not a big surprise: you win exactly if no one else chooses 1. You got $K_2$ right, but $K_3=p_1^2+p_2^2$. Jan 31 '17 at 10:15 | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/2118796/how-to-choose-the-smallest-number-not-chosen"
} |
homework-and-exercises, newtonian-mechanics, rocket-science
Title: Max altitude of rocket
A rocket is launched perpendicularly with an initial acceleration of 10 m/s2, if the fuel runs out after 1 minute, what will be the maximum height reached by the rocket? I assuming that the acceleration remains constant. So using Newton's equations-
$$v=at$$ So after the fuel runs out, $$v=(10)(60)=600m/s$$ The distance it travelled in this time can be calculated by $$s=1/2at^2=(0.5)(10)(60)^2=18000m$$
Now, after the fuel runs out, the rocket will still go up a little bit, that can be calculated by- $$v^2-u^2=2gs$$ The $u$ here will be $v$ from the previous case and gravity is acting downwards and hence $g$ will be approximately $-10m/s^2$. So, $$s=-600^2/-20=18000m$$
Thus, Maximum Height $=18000+18000=36000m$ | {
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php
return gmstrftime($format, $this->getTimestamp());
} There is still some redundant code in there(' - %H:%M' and '%A %e %B %Y'). You could put these in variabels, but i like it so more. Sory for my bad English.
public function formatToFullDate($countryCode, $hideTime = false)
{
$format = '';
$time_format = '';
switch($countryCode)
{
case 'us':
case 'ca':
$time_format = ' - %I:%M %P';
$date_format = '%A %e %B %Y';
break;
case 'de':
$time_format = ' - %H:%M';
$date_format = '%A, den %e. %B %Y';
break;
default:
$date_format = '%A %e %B %Y';
$time_format = ' - %H:%M';
}
if($hideTime)$time_format = '';
return gmstrftime($date_format.$time_format, $this->getTimestamp());
} | {
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Moment of Inertia A vertical differential element of area is Cho. The second moment of area, also known as area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. 5-1 We needed to find the average angular acceleration in order to attempt to cancel out the torque due to friction. + A3 (4) Moment of inertia of the composite section about AB is given by: IAB = I1 + I2 + I3 = Ig1 + A1 y12. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. 01 18-Jun-2003 1. 12 cm 1y = 12,380 cm 10 cm 8 cm. Solution 12. Ibrahim Moment of Inertia of a beam Extended objects can be treated as a sum of small masses. 1 kg m 2 Initial moment of inertia of the system, I i = 7. Properties of the Centroid. Find the 2nd moment of area for the shape shown the about the axis s – s. Solution: There is no reference origin | {
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1: Estimating the Sum of an Infinite Geometric. If and are convergent series, then and are convergent. Neelon (San Marcos, CA) To Professor J ozef Siciak on his 80th birthday Abstract. Conditional Convergence. We can't apply the integral test here, because the terms of this series are not decreasing. The limit. We encounter here the main di erence between convergent and divergent series. Innovation is serendipitous but manageable; mysterious, but solvable; from divergent to convergent thinking, creativity can emerge from chaos to the order and innovation becomes the light organizations can reach out. A series is an infinite sum, written. While the introductory story about Achilles and the Tortoise introduces an apparent paradox which we were able to resolve using a convergent (geometric) series, this story uses the properties of a divergent (harmonic) series to shed light on an unbelievable but true situation. Age of the Islands. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n | {
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c#, razor
Then replace it in your other view with:
@{ Html.RenderPartial("~/path/ReceiptDetails.cshtml", Model.Receipt); }
Then you've got a reusable chunk of HTML and you now don't need to worry about your original concern.
As a side note, there may be a more semantically correct choice than using <label> here. For example, a definition list could be appropriate.
Also, don't use <br/>, style the elements with CSS to be block. | {
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You seem to know where the binomial formula for counting these tuples comes from, but for others reading, this formula is determined as explained here.
We basically reduced the problem "distributing $n$ balls into $r$ urns with restriction $m_i$ on each urn" to "distributing $n-M$ balls into $r$ urns" by translating solutions from one to the other. You can think of it physically as starting out by putting all of the $m_i$ balls into each urn as required -- you have $n-M$ left over to distribute freely, and you can use what you already know to count how to do that.
A simpler version of this problem might be "how many ways can you give your two children \$5 allowance (in whole dollar amounts) if each child needs at least \$1?" Rather than count some complicated configuration, you can really just think of this \$1 each as already handed out to them and then solve the easier problem "how many ways can you give your two children \$3 allowance?" (In this case, the answer is 4 ways.) | {
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kinect, opencv2, nodes, cv-bridge, callbacks
The callback function doesn't get called at all. It just kind of gets stuck and doesn't do anything.
I am attempting to execute this node above that I have created at specific intervals and every interval it should take the necessary images and move forward to the next command (without having to ctrl+c myself), but it gets stuck without entering the callback function.
Update I didn't mention properly that I use rosrun to call this node. But I also use rosrun right before this to call another node. Does that make any difference? I don't have an issue with the one before this (but that's not my own code/node).
I noticed similar posts here which all mentioned spinOnce() which I included based on their suggestion right at the beginning. | {
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What is your first identity right now is called vandermonde's identity.Another way to see it is that you want the subsets of $k$ elements from disjoint sets $A$ and $B$ where the quantity of elements in $A$ goes from $0$ to $k$. Hence you want all of the $k$-subsets of $A\cup B$
For calculating the $\sum\limits_{i=0}^n \binom{n}{i}^2 (-1)^i$ Alex R's comment is good. Although it only works for odd $n$.
The point is that $\binom{n}{i}=\binom{n}{n-i}$ and $i$ and $n-i$ have different parity when $n$ is odd. So they cancel out. so if $n=2k$ you obtain $\sum_{i=0}^{2k+1}\binom{2k+1}{i}^2(-1)^i=0$ | {
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quantum-field-theory, gauge-invariance, bosonization
I'm particularly confused since I've seen use of bosonization of a fermion coupled to a gauge field to explain things like the chiral anomaly and lack of dependence on the vacuum angle. (see e.g. Witten Nucl Phys B149, 285)
For instance suppose we have a gauge invariant Lagrangian like
$$\bar{\psi}i(\displaystyle{\not}\partial+i\displaystyle{\not}A)\psi -\frac{1}{4}F^2$$
and after naive bosonization we get something like (forgive me for ignoring the constant factors)
$$\frac{1}{2}(\partial{\phi})^2+\partial_\mu\phi A^\mu -\frac{1}{4}F^2$$
but this is missing the $\frac{1}{2}A^2$ term that would be needed for gauge invariance! You are completely correct that the gauge symmetry "mysteriously" disappears from the boson. The shift symmetry does not become gauged (this would be a non-linear realization of the gauge theory). Bosonization is not a simple change of variables! | {
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java, object-oriented
public void deleteAllBooks() {
books.clear();
}
public int getBookCount() {
return books.size();
}
public void addAuthor(String name) {
authors.add(new Author(name));
}
public void clearBookDirectory() {
books.clear();
authors.clear();
}
public void addBookTitle(String title, Author author, int id) {
bookTitles.add(new BookTitle(title, author, id));
}
public int getBookTitlesCount() {
return bookTitles.size();
}
public String toString() {
return books.toString();
}
public List<Book> getAllBooks() {
return books;
}
public HashSet<BookTitle> getAllBookTitles() {
return bookTitles;
}
}
package com.library.models;
public class BookTitle {
private String title;
private Author author;
private int id; | {
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machine-learning
Title: Can I use the test dataset to select a model? I'm trying to understand if the test dataset can be used to select a final trained model. Let's assume this scenario:
I first split the whole dataset: 70% training, 30% test. Then I fit several models (let's say NN, RandomForest, AdaBoost,..) on the training dataset with cross-validation and tune the hyperparameters to get the best performance on the train data. I know that these scores are biased, since I was tuning the hyperparameters on this data.
Then I use the test dataset to get the true performance on the unbiased data and select which model performs the best.
Is this a correct way to use the test dataset? Some confusion comes from the internet definition of the test dataset:
The sample of data used to provide an unbiased evaluation of a final model fit on the training dataset. | {
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voice, mfcc
Title: Acoustic features distinguishing the male and female voice? I would like to use MFCC analysis to distinguish between a male voice and a female voice.
Are there any particular quantifiable characteristics consistent with, and exclusive to, either gender? (E.g., energy, fundamental frequency, excitation, the presence of certain waveforms, etc.) Main difference is the frequency of the fundamental, which is about an octave higher for women with the split point around 160 Hz or so. A fundamental lower than 160 Hz is most likely a male and a fundamental higher is most likely a female. A good overview over a number of studies on the topic can be found at http://www2.ling.su.se/staff/hartmut/f0_m%26f.pdf | {
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beginner, c
} Overall
Aside from the atrocious format, very nice code for a learner.
Use an auto-formatter!
Avoid manual formatting, yet always run an auto-formatter before code review.
Ding, Ding, Ding!!!
No warnings, even with many enabled. Yeah!
Bug: direction never assigned before use.
char direction;
...
control(direction,&snake[0][0],&snake[0][1]);
Avoid naked magic numbers
//int board[25][64];
//for (int i = 0; i < 25; i++) {
// for (int a = 0; a < 64; a++) {
#define BOARD_ROWS 25
#define BOARD_COLS 64
int board[BOARD_ROWS][BOARD_COLS];
for (int i = 0; i < BOARD_ROWS; i++) {
for (int a = 0; a < BOARD_COLS; a++) {
// if(i == 0 || i == 24){
if(i == 0 || i == BOARD_ROWS - 1){ | {
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python, nlp, gensim
Then search for the antonyms from the result. If you are getting a big list of antonyms, use only the top n results for your searching.
Although, W2V gives most commonly used words in the context of the keyword, it's hard to guess which is the antonym of the keyword. | {
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python, python-3.x, numpy
b1 = round((y + 1) / ratio1 + (x + 1) * ratio2)
else:
a = round(x / ratio1 + y * ratio2)
a1 = round((x + 1) / ratio1 + (y + 1) * ratio2)
rhomb[b:b1, a:a1] = arr[y, x]
if reverse:
rhomb = np.flipud(rhomb) if dimension == 'vertical' else np.fliplr(rhomb)
return rhomb | {
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c, programming-challenge, sorting, linked-list, factors
Title: Deficient Numbers I have found the following interesting challenge on the web:
Deficient Numbers
A number is considered deficient if the sum of its factors is less than twice that number.
For example: 10 is a
deficient number because its factors are 1, 2, 5, 10 and their sum is
1 + 2 + 5 + 10 = 18 which is less than 10 * 2 = 20.
Challenges
Easy level: write a program to verify whether a given number is deficient or not.
Medium level: write a program to find all the deficient numbers in a range.
Hard level: given a number, write a program to display its factors, their sum and then verify whether it's deficient or not.
I implemented it in C:
#include <stdlib.h>
#include <stdio.h>
#define DEBUG 1
/*
XXX EASY-OPTION IMPLEMENTATION
A list data structure is used to store the factors of a given number.
A simple boolean-like data type is then returned by the main isDeficient function.
*/
typedef enum bool bool;
enum bool {
false, true
}; | {
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c++, beginner, image, steganography
Crypt.cpp
//Crypt.cpp
#include "Crypt.h"
#include <tclap/CmdLine.h>
int main(int argc, char *argv[]) {
auto args = parseCMD(argc, argv);
CImg<PTYPE> secret(args["secret0"].c_str());
CImg<PTYPE> apparent(args["apparent"].c_str());
if (args["resize"] == "true") {
enum class Interpolation { NoneRawMem = -1, NoneBoundaryCondition, NearestNeighbour, MovingAverage, Linear, Grid, Cubic, Lanczos };
secret.resize(apparent, (int)Interpolation::NearestNeighbour);
}
encrypt<PTYPE>(apparent, secret, args);
if (args["signature"] == "true") {
sign(apparent, std::stoi(args["secretbitdepth"]) / secret.spectrum());
}
apparent.save(args["output"].c_str());
return 0;
} | {
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cr.crypto-security, privacy
Question. How can both parties compute the dot product $x\cdot y$ in time $\mathcal O(n)$ without leaking their respective secrets ? I will assume you are in the honest-but-curious model. You can't represent real numbers in finite space, so I will assume all values are represented in fixed-point arithmetic, to $d$ bits of precision; thus $x$ is represented as $x = x'/2^d$ where $x'$ is an integer. Then $x \cdot y = x' \cdot y' / 2^{2d}$, so the problem is equivalent to computing $x \cdot y$ where $x,y \in \mathbb{Z}^n$. Here is a scheme for that. | {
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solar-system, jupiter, mercury, neptune, orbital-resonance
Any ideas how to explain the two-fold nature of the outcomes of planetary resonances ? Any references to some literature / mathematical proof would be highly appreciated.
Thank you in advance, Mercury and Spin-orbital resonance is pretty straight forward. Planets and Moons are gravitationally lumpy and large bodies are somewhat fluid, even rocky bodies. Both aspects are prone to tidal forces and that can lead to spin-orbital resonance if the tidal forces are strong enough. Mercury's somewhat high eccentric orbit balances out with a 3:2 spin-orbital resonance.
Resonance in 3 body orbital system is much more complicated and the math is above my pay-grade. A proper answer to this question would have lots of math where orbital stability can be measured over long time periods and thousands of orbits. Something precision is probably lost in an intuitive explanation. | {
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beginner, c, multithreading, pthreads, dining-philosophers
consistency_check[KG_2] += student->other_students[i].current_weight[KG_2];
consistency_check[KG_3] += student->other_students[i].current_weight[KG_3];
consistency_check[KG_5] += student->other_students[i].current_weight[KG_5];
}
if( consistency_check[KG_2] > MAX_2KG_3KG
|| consistency_check[KG_3] > MAX_2KG_3KG
|| consistency_check[KG_5] > MAX_5KG ) {
printf(RED "Inconsistent State\n" );
printf("[%d, %d, %d]\n" RESET,consistency_check[KG_2],
consistency_check[KG_3],
consistency_check[KG_5]);
}else {
printf("Supply: [%d, %d, %d]\n", student->weight_rack[KG_2],
student->weight_rack[KG_3],
student->weight_rack[KG_5]);
}
fflush(stdout);
} | {
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astronomy, planets, solar-system
From The Formation of Uranus and Neptune among Jupiter and Saturn over on The Astronomical Journal: | {
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magnitude, gaia
Title: How can I convert a GAIA-magnitude "G" to Johnson "V"? Question: How can I convert the Gaia-magnitude G (and G_ESTIMATE) into a V-magnitude? The solution probably can be found in the Gaia Data Release Documentation for Gaia's EDR3 ("Early Data Release 3"), specifically section 5.5.1, Relationships with other photometric systems, but I have to admit, that I can't make heads or tails of it.
Related Question: The question Relationships between G magnitude, Johnson V magnitude, and spectral type of stars? is related in that it wants to explore the relationship between G and Johnson-V. However, my separate question is not a "duplicate", because the other question has the relation between the magnitudes and spectral type as its focus (perhaps erroneously, shouldn't the relation be between a colour index and spectral type?). | {
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ros
<inertial>
<mass value="5.0"/>
<inertia ixx="1.0" ixy="0.0" ixz="0.0" iyy="1.0" iyz="0.0" izz="1.0"/>
<origin/>
</inertial>
</link>
<joint name="right_rotator_to_tricep" type="revolute">
<parent link="right_rotator_link"/>
<child link="right_tricep_link"/>
<origin xyz="-0.045 0.0075 0" rpy="0 0 0" />
<axis xyz="0 -1 0" />
<limit effort="1000.0" lower="-1.57" upper="1.57" velocity="0.5"/>
</joint>
<link name="left_upper_bicep_link">
<visual>
<geometry>
<cylinder length="0.145" radius="0.0325"/>
</geometry>
<material name="grey"/>
<origin rpy="0 0 0" xyz="0 -0.0325 0"/>
</visual>
<collision>
<geometry>
<cylinder length="0.145" radius="0.0325"/>
</geometry>
<contact_coefficients mu="0" kp="1000.0" kd="1.0"/>
<origin rpy="0 0 0" xyz="0 -0.0325 0"/>
</collision> | {
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Note: based on the form of the solution of $a_n$, I suspect there's a direct combinatorial argument (but I don't know it).
• I am very impressed. I should have a look at generating functions. Thanks a lot, I'll study this and try to understand what it all means. – coredump Feb 13 '15 at 21:04
• @coredump: If you want to take a look at generating functions, one excellent place to start is Herb Wilf's free book: math.upenn.edu/~wilf/gfologyLinked2.pdf. After reading his book, I was hooked. – Rus May Feb 13 '15 at 23:41
Let $m=|S|$ be the number of distinct characters, and let $a_n$ be the number of strings in $S^n$ having an even number of each character in $S$. Let
$$g(x)=\sum_{n\ge 0}\frac{a_n}{n!}x^n$$
be the exponential generating function (egf) for $\langle a_n:n\in\Bbb N\rangle$. (The egf is wanted because the order of the characters matters.) The egf for the sequence given by $$a_n=\begin{cases}0,&\text{if }n\text{ is odd}\\1,&\text{if }n\text{ is even}\end{cases}$$ is | {
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transportation, motors, compressed-air
Pneumatic Cylinders:
Provide more force and speed per unit size than any other actuator except hydraulic.
Force and speed are easily adjustable and are independent of each other.
Are most economical when the scale of deployment matches the capacity of the compressor.
Small compressors are efficient and economical when used for a small number of devices.
Low component costs.
High operating costs (install, replacement cylinders, electricity for compressor (76%)) | {
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java, object-oriented, android, homework
and call it like this:
if (playerWithColorMeetsWinningCondition(R.drawable.white)) {
//white wins.
}
your conditions there are horribly long because you've hardcoded the sizes and checking each element manually. I'm going to stop here even though there's a lot more to say.
Pick up a Java book before attempting to do anything I mentioned at least and don't let it go until you finish all of the exercises in it, and learn everything it teaches. It will pay off, since you seem to lack a basic understanding of programming here.
Be sure to come back with a revised version of your game after you're done learning so that we can have another round at it and praise you for your progress :) | {
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navigation, move-base
Title: Using move_base to turn wheels?
Hello,
I am currently using move_base(with default local and global planners) to get the robot in the desired position. My problem is that when my robot(2 wheeled, built using real wheelchair) is told to move forward right or forward left, it is making a path that goes around too much to get there. In other words, if the robot is located at (0,0) and is commanded to move to (1,1), it is not turning its wheels enough to get there more directly and rather makes a curvy path with large radius than necessary. Is there a parameter that I can tweak to make the radius of path smaller?
I would appreciate any help, Thank you.
Originally posted by bkim on ROS Answers with karma: 86 on 2012-08-12
Post score: 2 | {
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tensor-calculus
$R_{ij}=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$
I asked at math.stackexchange, but got no answer. Maybe this is more Phys Math Met as in Boas, which I was reading when this question came up. I think @mikestone's comment is correct. The MathWorld site is talking about tensors in 3 dimensions, not 2D. In 3 dimensions (actually, in all dimensions $\geq3$), it can be shown that any isotropic rank-2 tensor is proportional to the identity ($\delta_{ij}$), see Richard Fitzpatrick's notes here, for example.
In two dimensions, there are two rank-2 isotropic tensors, $\delta_{ij}$, and what you have called $\epsilon_{ij}$. Here's a quick way to list all the isotropic tensors in 2D that I have drawn heavily from the fantastic analyses here and here. There are two equivalent ways to define an isotropic tensor: one is the way you have defined it, saying that $A$ is an isotropic tensor iff $$A = R\cdot A\cdot R^T,$$ | {
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autocorrelation, random-process, wiener-filter
$$S_x(\omega)=F(\omega)F^*(\omega)\tag{4}$$
where $F(\omega)$ corresponds to a real-valued causal and stable system with all its poles and zeros in the left half-plane (i.e., $F(\omega)$ is a minimum-phase system). Note that $|F(\omega)|=|F^*(\omega)|=\sqrt{S_x(\omega)}$.
Using $(4)$, we can write the integrand of $(3)$ as follows:
$$|H(\omega)|^2S_x(\omega)-2\text{Re}\left\{H(\omega)S_x(\omega)e^{-j\omega\lambda}\right\}+S_x(\omega)=\left|H(\omega)F(\omega)-F(\omega)e^{j\omega\lambda}\right|^2\tag{5}$$ | {
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I have thought about this problem some time ago. I don't know any theorems or proofs, but my method for an $n\times n$ square was the following:
• Make chips marked with the numbers from $0$ to $n-1$, $2n$ chips for each number. Of these $2n$ chips, make $n$ different from the other $n$ chips (for example, $n$ white chips and $n$ red chips).
• On an empty $n\times n$ board, you must put two chips in each square: a red one and a white one. If a square has a red $j$ and a white $k$, this represents that the number $nj+k$ is in this square (that is: $j$ and $k$ are the digits of the number in base $n$).
• In each row and column there must be the $n$ different numbers for each color, in a similar way to a Sudoku.
• You must also watch for the diagonals.
This way, I was able to make a 5x5 square and a 6x6 square. By the way, I think (but I don't have any proof either) that if $n$ is odd, the central square must contain the arithmetic mean of the row. | {
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homework-and-exercises, newtonian-mechanics, fluid-statics, surface-tension
Title: In what case does the parameter of surface tension vary?
I was trying to solve a question regarding force of surface tension,it happens that I was unable to solve it using F = Tl,But it happens that the intended solution was to be done such that F =T2l,but I'm not sure if it should be this was this,there are no two surfaces to the needle or to the water.
How is it that the force varies without change number of surfaces? How is that the shape of the object can make the number of surfaces vary with ways? Surface tension acts tangentially to the surface of water, So when you immerse a nail in the water, water from both sides are being in effect
Hence 2*L is used in the formula | {
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inorganic-chemistry, electrochemistry, electrons, water
Title: Extraction of hydrogen and oxygen from water The apparatus that I am using is a glass container, with two electrodes,two glass cylinders, water, and a battery. In the glass container will be filled with water and two electrodes kept inside the glass cylinder and then dipped in water.
The two electrodes are then connected with the positive and negative terminals of the battery and the container is sealed. As the current is started the water will be separated into hydrogen and oxygen. After extraction of the gas, the two gases are passed separately. | {
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gauge-invariance, observables, correlation-functions, s-matrix-theory
Title: Gauge transformations at infinity Consider the following paragraph taken from page 15 of Thomas Hartman's lecture notes on Quantum Gravity:
In an ordinary quantum field theory without gravity, in flat spacetime, there two types of physical observables that we most often talk about are correlation functions of gauge-invariant operators $\langle O_{1}(x_{1}) \dots O_{n}(x_{n})\rangle$, and S-matrix elements. The correlators are obviously gauge-independent. S-matrix elements are also physical, even though electrons are not gauge invariant. The reason is that the states used to define the S-matrix have particles at infinity, and gauge transformations acting at infinity are true symmetries. They take one physical state to a different physical state - unlike
local gauge transformations, which map a physical state to a different description of the same physical state. | {
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# Math Help - Integral proofs
1. ## Integral proofs
Actually I have 2 questions?
1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
I tried using the antiderivative proof but I dont think that is the right direction on for this one.
2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3.
Bit clueless on this one.
2. Originally Posted by Caity
Actually I have 2 questions?
1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
I tried using the antiderivative proof but I dont think that is the right direction on for this one.
where are you stuck on this one? you've learned about partitions and all that, right?
2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3. | {
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zoology, species-identification, ornithology
Can someone here either confirm my identification or provide a more informed identification? 1. Carrion crow (Corvus corone)
The plumage of carrion crow is black with a green or purple sheen, much greener than the gloss of the rook. The bill, legs and feet are also black. It can be distinguished from the common raven by its size (48–52 cm or 19 to 20 inches in length as compared to an average of 63 centimetres (25 inches) for ravens) and from the hooded crow by its black plumage
3. Hooded crow (Corvus cornix)
ashy grey bird with black head, throat, wings, tail, and thigh feathers, as well as a black bill, eyes, and feet.
2. This one is a bit more difficult, but I think it is also a hooded crow or possibly a hybrid Carrion crow × Hooded crow (Corvus corone × Corvus cornix).
Hybrid
At first glance, #2 looked somewhat like a Western jackdaw (Corvus monedula) because the grey feathers travel fairly high up on its neck and because the bird appears to be somewhat smaller. | {
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An alternative way to carry out the same procedure is not to wait until all $n$ tickets are drawn from the $X$ box. Instead, after drawing each ticket, immediately read its value. If it says $X=0$, do nothing more. If it says $X=1$, though, immediately draw a ticket from the $Y$ box and read its value.
This alternative procedure can be described by drawing a single ticket from a new box. Up to two numbers are written on each ticket, called "$X$" and "$Y$", to record a single sequence of up to two draws. According to the foregoing description, which has three outcomes, there must be three kinds of corresponding tickets:
1. $X=0$. These tickets model drawing a value of $0$ from the $X$ box. Their proportion within the new box, in order to emulate the properties of the first step, must equal $1-p$. Don't bother to write any value for $Y$, because $Y$ will not be observed when such a ticket is drawn. | {
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