text stringlengths 1 1.11k | source dict |
|---|---|
debian, cmake
The full error from ./build_isolated/rosmaster/CMakeFiles/CMakeError.log looks like:
$cat ./build_isolated/rosmaster/CMakeFiles/CMakeError.log
...
/usr/bin/gcc CMakeFiles/cmTryCompileExec1355085001.dir/CheckSymbolExists.c.o -o cmTryCompileExec1355085001 -rdynamic
CMakeFiles/cmTryCompileExec1355085001.dir/CheckSymbolExists.c.o: In function `main':
CheckSymbolExists.c:(.text+0x16): undefined reference to `pthread_create'
collect2: error: ld returned 1 exit status
make[1]: *** [cmTryCompileExec1355085001] Error 1
make[1]: Leaving directory `/home/nexix/Documents/source/ros_catkin_ws/build_isolated/rosmaster/CMakeFiles/CMakeTmp'
make: *** [cmTryCompileExec1355085001/fast] Error 2
...
/usr/bin/gcc -DCHECK_FUNCTION_EXISTS=pthread_create CMakeFiles/cmTryCompileExec2282391930.dir/CheckFunctionExists.c.o -o cmTryCompileExec2282391930 -rdynamic -lpthreads
/usr/bin/ld: cannot find -lpthreads
collect2: error: ld returned 1 exit status | {
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java, beginner, chess
in Java, it is customary to place curly brackets on the same line as the opening statements.
use curly brackets even for one line statements for improved readability and to avoid future bugs.
you should not have magic numbers in your code; extract these to static variables at class level. That way, it is clear which numbers actually belong together, it's easier to change, and the name will tell a reader what the number means (eg why 97?) | {
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javascript, beginner, jquery, html
It can be replaced by:
const regexDays = /Sun\. |Mon\. |Tues\. |Wed\. |Thurs\. |Fri\. |Sat\. /g;
// ...
var events = $($.parseHTML(data)).find('p').text().split(regexDays).splice(0, 1);
Factorization
The most obviously visible case where factorization can take place is the one of the showMonth() and showNextMonth() functions.
What's done in their body differ only by one using monthIndex and the other using monthIndex + 1, so they can be replaced by a unique function accepting monthIndex as an argument:
function showMonth(monthIndex) {
$.each(events, function(i, item) {
var eventsstr = JSON.stringify(item);
if (eventsstr.match(months[monthIndex] || months[0])) {
var concert = JSON.parse(eventsstr);
$('#concerts').append('<p>' + concert + '</p>');
}
});
}
And where they were called:
showMonth(); becomes showMonth(monthIndex);
showNextMonth(); becomes showMonth(monthIndex + 1); | {
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"url": null
} |
Yes. It is C. I had taken mean for parameter. Changed now :)
@csegate2 >> Reciprocal will be the mean.
Now, we use P{X>a} for our problem because our concerned variable Z is min of X and Y.
because in f(x) we solve for x>=0
I didnt get this line
P{X<a}=1−e−λaP{X<a}=1−e−λa
and
P{X>a}=e−λa
Now, we use P{X>a} for our problem because our concerned variable Z is min of X and Y.
What is the meaning and conclusion of this line.
Sir, what is the role of min(x, y) here ?
If it would have been max instead of min, then what would be the answer?
Arjun sir why are you taking this?
we use P{X>a} for our problem because our concerned variable Z is min of X and Y | {
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"lm_q2_score": 0.8221891305219504,
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"openwebmath_score": 0.9574569463729858,
"tags": null,
"url": "https://gateoverflow.in/3676/gate2004-it-33"
} |
orbitals, periodic-trends
You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen. This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern. The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619). | {
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"tags": "orbitals, periodic-trends",
"url": null
} |
javascript, portability
SMARTQUOTES.displayQuote = function () {
var quote = SMARTQUOTES.quotes[SMARTQUOTES.quoteIndex];
SMARTQUOTES.element.fadeOut('slow', function() {
SMARTQUOTES.element.html(quote);
});
SMARTQUOTES.element.fadeIn();
SMARTQUOTES.incrementQuote();
SMARTQUOTES.startTimer();
}
SMARTQUOTES.startTimer = function () {
var t = setTimeout('SMARTQUOTES.displayQuote()', SMARTQUOTES.duration);
}
SMARTQUOTES.start = function() {
SMARTQUOTES.displayQuote();
}
}
Take advantage of jQuery's .ready(). It's an abstraction that includes window.onload. Here's a shorthand version:
$(function(){
//DOM ready
}); | {
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"url": null
} |
c#, asp.net-core
Title: Removing a specific key of ModelState for a specific controller's method I have a controller for Post and Put request for the same entity: MyModelDTO with some ModelState's attributes.
The users can make a Post request without the Id attribute, because it is being generated on server side, but for Put request the Id attribute is needed (And has to be different than an empty guid).
I want to allow get a Post request without the "Id" property, and still use the ModelState functionality, the best solution I've came across is overriding the OnActionExecuting and removing the specific key:
public class ExceptPropertiesAttribute : ActionFilterAttribute
{
private IEnumerable<string> _propertiesKeys;
public ExceptPropertiesAttribute(string commaSeperatedPropertiesKeys)
{
if (!string.IsNullOrEmpty(commaSeperatedPropertiesKeys))
{
this._propertiesKeys = commaSeperatedPropertiesKeys.Split(',');
}
} | {
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} |
php, object-oriented, laravel
public function ajaxUpdate()
{
if (!Request::ajax())
{
Response::json(
array(
'message' => 'Invalid request',
'redirect' => 'some url'
)
);
}
$validator = $this->validateInput(Input::all());
//process resulst
}
public function postUpdate()
{
if (Request::ajax())
{
Response::json(
array(
'message' => 'Invalid request',
'redirect' => 'some url'
)
);
}
$validator = $this->validateInput(Input::all());
//process resulst
} | {
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"tags": "php, object-oriented, laravel",
"url": null
} |
The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 1/6, and so the probability of drawing a given card is 1/3. In particular, the probability of drawing the double-white card is 1/3, and the probability of drawing a different card is 2/3.
In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you.
### Solutions:
#### Intuition | {
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"openwebmath_score": 0.4683186709880829,
"tags": null,
"url": "https://math10blog.wordpress.com/tag/wiki/"
} |
reference-request, markov-chains, zero-knowledge
Title: Reference Request: soundness in a ZKP achieved by walking along a doubly-stochastic Markov chain? Consider the following variant of a zero-knowledge proof that two graphs, $G_1$ and $G_2$, given by adjacency matrices $M_1$ and $M_2$, respectively, are not isomorphic.
Here Peggy the prover wants to show Vicky the verifier that $G_1 \not\cong G_2$, where both $G_1$ and $G_2$ have $n$ vertices. | {
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"tags": "reference-request, markov-chains, zero-knowledge",
"url": null
} |
c++, recursion, template, c++20, constrained-templates
/* Override make_view to catch dangling references. A borrowed range is
* safe from dangling..
*/
template <std::ranges::input_range T>
requires (!std::ranges::borrowed_range<T>)
constexpr std::ranges::dangling make_view(T&&) noexcept
{
return std::ranges::dangling();
} | {
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"url": null
} |
osmosis
In the soil, water is transported predominantly by bulk flow.
However, when water comes in contact with the root surface,
the nature of water transport becomes more complex.
From the epidermis to the endodermis of the root, there are
three pathways through which water can flow:
the apoplast, transmembrane, and symplast pathways.
In the apoplast pathway, water moves exclusively
through the cell wall without crossing any membranes.
The apoplast is the continuous system of cell
walls and intercellular air spaces in plant tissues.
The transmembrane pathway is the route followed
by water that sequentially enters a cell on one side,
exits the cell on the other side, enters the next in the
series, and so on. In this pathway, water crosses at
least two membranes for each cell in its path (the
plasma membrane on entering and on exiting).
Transport across the tonoplast may also be involved.
In the symplast pathway, water travels from one cell | {
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python, algorithm, programming-challenge
Details:
O(n) Method using Hash Table.
For unsorted array, by using extra space
we can solve the problem in O(n) time. We maintain a hash table to
hash the scanned numbers. We scan each numbers from the head to tail,
when we scan the i-th number y, we do following steps:
Check if (x-y) is hashed.
If hashed, print out (j,i) where j is the elements of the hashed
value with index (x-y).
Add (key=x-y, value=i) into the hash table.
Therefore, the time complexity is O(n) and the space complexity is O(n).
Resources:
Find a pair of elements from an array whose sum equals a given number
Find two numbers in an array whose sum is x | {
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java, beginner, game, swing, awt
A good example is the way you unlock the buildings.
You introduce boolean variables to track the building states. A more OO-ish approach would be to hold the locked buildings in a List.
Then I'd remove the first element from that list and unlock it until the list is empty:
Building dummyBuildingToAvoidAnotherOddBallSolution = new Building("",0,0,0){ // anonymous inner class
@Override
public int hasReachedMinimumLevel() {
return 2 <= clicker;
}
}
List<Building> lockedBuildings =
new ArrayList<>(
Arrays.asList(dummyBuildingToAvoidAnotherOddBallSolution, bakery, robot, factory)); | {
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$1 = 1$
$3 = 2 + 1$
$5 = 4 + 1$
$7 = 4 + 2 + 1$
$9 = 8 + 1$
$11 = 8 + 2 + 1$,
and so on. On the right-hand side, I’ve written each number as the sum of powers of 2 (for the numbers at hand, that means 1, 2, 4, 8, 16, and 32). Notice that each expansion on the right hand side contains a 1. So, if the audience member tells me that her number is on the upper-left card, that tells me that there’s a 1 in the binary representation of her number.
Let’s now take a look at the first few number in the upper-middle card:
$2 = 2$
$3 = 2+1$
$6 = 4+2$
$7 = 4+2+1$
$10 = 8+2$
$11 = 8+2+1$,
and so on. Notice that each expansion on the right hand side contains a 2. So, if the audience member tells me that her number is on the upper-middle card, that tells me that there’s a 2 in the binary representation of her number. | {
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"url": "https://meangreenmath.com/tag/magic-trick/page/2/"
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quantum-mechanics, solid-state-physics
Title: Conservation of crystal momentum in time-dependent potential Consider a time-dependent problem. A particle is placed in a periodic potential with time-dependent amplitude
$$H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V_0(t)\cos^2(x)$$
Instantenous eigenstates are labelled by quasi-momentum $k$ nad band index $n$ - Bloch states $\phi_{n,k}(x,t)$. I wonder if there are some consequences of the discrete translational symmetry of lattice on the time-evolution?
Does the initial state $\phi_{n,k}(x,t=0)$ evolves into a superposition of different Bloch states $\phi_{m,k}(x,t)$ but with the same quasi-momentum $k$ as the inital state?
UPDATE
I checked some literature and found only one reference with time-dependent potential, but I think it does not answer my question completely.
Consider the translation operator $\hat{T}_{a_j}$ which has the property
$$\hat{T}_{a} \psi(x,t) = \psi(x - a,t).$$ | {
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homework-and-exercises, electric-circuits, capacitance
Title: Why isn't this a valid derivation of the formula for capacitors in series? I had to derive the formula for capacitors
(I decided to use m capacitors in my derivation) in series, and this is what I did.
The formula for a capacitor is
$$Q=CV,$$
which is the same as saying
$$\int \dfrac{dI}{dt} dt= C_n V_n.$$
Since they are all in series, $I_1=I_2=...=I_n=I$, thus $Q_1=Q_2....=Q$.
We also know that
$$V=IR \Longrightarrow \sum \frac{Q}{C_n} = V=\frac{Q}{C_{eq}}.$$
Since they are in series, we can apply kirchoffs voltage law, obtaining
$$ \frac{Q}{C_1}+\frac{Q}{C_2}+...\frac{Q}{C_n}= \frac{Q}{C_{eq}} \Longrightarrow Q \sum_{n=0}^{m} C_n^{-1} = \frac{Q}{C_{eq}}.$$
Thus we conclude that
$$ \sum_{n=0}^{m} C_n^{-1} = \frac{1}{C_{eq}}.$$
Is this a proper derivation of the result? You bring up an $R$, for no known reason, and immediately after that write down something equivalent to what you want to show that also comes out of nowhere, so it's very very confusing as written. | {
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 May 2014 Posts: 23 GPA: 3.55 Followers: 0 Kudos [?]: 2 [0], given: 13 Re: In a certain business, production index p is directly [#permalink] ### Show Tags 10 Jun 2015, 20:43 VeritasPrepKarishma wrote: Joint variation gives you the relation between 2 quantities keeping the third (or more) constant. p will vary inversely with i if and only if e is kept constant. Think of it this way, if p increases, e increases. But we need to keep e constant, we will have to decrease i to decrease e back to original value. So an increase in p leads to a decrease in i to keep e constant. But if we don't have to keep e constant, an increase in p will lead to an increase in e which will increase i. Here, we are not given that e needs to be kept constant. So we will not use the joint variation approach. Hi Karishma, Thanks for your reply. How will I know whether the question asks that a certain variable needs to be | {
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The pigeonhole principle allows an easy proof of the statement for $n \leq 24$ that doesn't work for $n \geq 25$. I tried to apply the following algorithm to create simple counter-examples: To prevent the first $n$ boxes from contributing to a collection, we need $n+1$ balls in the $n$th box. Then we do the same with the following boxes, starting at $n+1$ and so on.
The algorithm only works for $33 \leq n \leq 49$. For smaller $n$, we would need two boxes with $n+1$ balls, but since we have at most $125$ balls altogether and each box contains at least one ball, there are not enough spare balls. For $50 \leq n \leq 75$, there are not even enough balls to put $n+1$ balls in any box.
If the algorithm I mentioned above is optimal, then the statement is true for all $n$, except those between $33$ and $49$. But if that's true, how can I prove it for $25 \leq n \leq 32$ and $50 \leq n \leq 75$? | {
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ros
Title: unable to find install package
I have been attempting to install diamondback.
I enter "sudo apt-get install ros-diamondback-desktop-full"
and get
"unable to locate package ros-diamondback-desktop-full"
I am completely new to linux, so any help would be greatly appreciated
Originally posted by Dave Everett on ROS Answers with karma: 48 on 2011-04-14
Post score: 0
After adding the ROS repositories to your apt sources and adding the WillowGarage keys (steps 1.2 and 1.3 of the Ubuntu installation guide), make sure you also run sudo apt-get update (beginning of step 1.4).
sudo apt-get update will instruct apt to refresh it's list of available software to take into account the fact that you just added new repositories.
After this, you should be able to run sudo apt-get install ros-diamondback-desktop-full with no problems.
Originally posted by Eric Perko with karma: 8406 on 2011-04-14
This answer was ACCEPTED on the original site
Post score: 1 | {
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fourier-transform, complex
Title: FFTs of a complex signal - separating the real and imaginary parts I have a complex time varying signal at a single frequency x = a + jb where a represents the contribution from the cosine basis function and b represents the contribution from the sinusoid basis function.
I am trying to understand how the output differs if I was to take a complex FFT of x and inspect the real and imaginary components compared to taking two FFTs of the real and imaginary parts separately. I naively expected these to be equal when a=b. What am I missing?
% signal parameters
fs = 1e3;
f0 = 50;
t = (0:999)/fs;
wt = 0.5; %weighting between cosine and sin basis functions
NFFT = 1024;
freq = linspace(-fs/2,fs/2, NFFT);
freq2 = linspace(0,fs/2, NFFT/2);
wnd = hanning(length(t)).';
% Complex signal
x = wt*cos(2*pi*f0.*t) + (1-wt)*1j*sin(2*pi*f0.*t); | {
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nuclear-physics
In order to get these numbers, the nuclear shell model starts from an average potential with a shape something between the square well and the harmonic oscillator. To this potential a spin orbit term is added. Even so, the total perturbation does not coincide with experiment, and an empirical spin orbit coupling, named the Nilsson Term, must be added with at least two or three different values of its coupling constant, depending on the nuclei being studied.
Nevertheless, the magic numbers of nucleons, as well as other properties, can be arrived at by approximating the model with a three-dimensional harmonic oscillator plus a spin-orbit interaction.
Nuclei that are well described by this model, with the magic number of nucleons, might be characterized with the "harmonic" adjective (or quasi-harmonic etc). | {
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quantum-mechanics, homework-and-exercises, wavefunction, hilbert-space
$$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$
The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that
$$\begin{aligned}
\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= \langle \psi_1 - \psi_2 | \psi_1 \rangle + \langle \psi_1 - \psi_2 | \psi_2 \rangle \\
&= \langle \psi_1 | \psi_1 \rangle - \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle - \langle \psi_2 | \psi_2 \rangle
\end{aligned}
$$
Because $\psi_1$ and $\psi_2$ are orthogonal and normalized, you know $\langle \psi_i | \psi_j \rangle = \delta_{i j}$. Substituting, the above expression evaluates to $1 - 0 + 0 - 1 = 0$, demonstrating that the two vectors are indeed orthogonal. | {
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thermodynamics
Title: How to calculate overflow of liquid being expanded thermally? There is a question on OpenStax college physics that is phrased:
If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a
temperature of 5.00ºC, how much will overflow when its temperature
reaches 22.0ºC? | {
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newtonian-mechanics, work, everyday-life, estimation
The mouse does 0.001J work on me while the button is rising again, but I have not noticed any invigorating effects from this.
Note that I have ignored the work required to move my finger, i.e. I have assumed that I am 100% efficient (an approximation that my colleagues would question).
All suggestions for refinements to this calculation will be gratefully ignored. | {
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ros, gps-common, ros-indigo
Original comments
Comment by stevejp on 2018-04-09:
Is the version of gps_common that you have installed locally the same as the version of gps_common which was installed on whatever device was used to collect the bagged data?
Comment by haysgh on 2018-04-10:
I don't have an idea about the version of gps_common used to collect the data. Assuming the two versions are different, is there a solution to overcome this problem like using another tool to do the conversion or creating my own node.
Comment by stevejp on 2018-04-10:
There's some info here on fixing md5sums in bag files. You could also upload a piece of the bag file so others can test. | {
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measurements, power, statistics
Title: Averaging decibels Wikipedia:
The decibel (dB) is a logarithmic unit that indicates the ratio of a
physical quantity (usually power or intensity) relative to a specified
or implied reference level. | {
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python, roman-numerals
Title: Decimal to roman python I'm a beginner as you can see and I would like to know how I can improve my code. Studying for 6 months now. Thank you.
roman_dict = {1: 'I', 4: 'IV', 5: 'V', 9: 'IX', 10: 'X', 40: 'XL', 50: 'L', 90: 'XC', 100: 'C', 400: 'CD',
500: 'D', 900: 'CM', 1000: 'M'}
divide_list = [1000, 100, 10, 1]
def not_in_dict(fixed_decimal, divide_num):
sub_count = 0
sub_roman_multi = roman_dict[divide_num]
temp_decimal = fixed_decimal
while temp_decimal not in roman_dict:
temp_decimal -= divide_num
sub_count += 1
return roman_dict[temp_decimal]+(sub_count*sub_roman_multi) | {
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rust
// A mapping of "major.minor" to a struct representing "major.minor[.patch]"
let mut latest_versions: AHashMap<&str, Versioning> = AHashMap::new();
// For each Godot version fetched from tuxfamily
for godot_version in VERSIONS {
for stable_godot_version in &STABLES {
// We check if it is one of the versions marked as stable in our
// custom data set
if godot_version.starts_with(stable_godot_version) {
// If yes, we check if we already have a version set as
// "latest" in our hashmap.
let latest: Option<&Versioning> = latest_versions.get(stable_godot_version);
match latest {
Some(latest) => {
// If yes, we compare our currently-iterated-over version
// with the one in the hashmap, and if it is greater, we
// set it as the latest
let godot_version = Versioning::new(godot_version.as_str()).unwrap(); | {
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`sym(1333333333333333333)`
`ans = $1333333333333333248$`
`sym('1333333333333333333')`
`ans = $1333333333333333333$`
You can specify the technique used by `sym` to convert floating-point numbers using the optional second argument, which can be `'f'`, `'r'`, `'e'`, or `'d'`. The default flag is `'r'`, for rational form.
### Conversion to Rational Symbolic Form
Convert input to exact rational form by calling `sym` with the `'r'` flag. This is the default behavior when you call `sym` without flags.
```t = 0.1; sym(t,'r')```
```ans = $\frac{1}{10}$```
### Conversion by Using Floating-Point Expansion
If you call `sym` with the flag `'f'`, `sym` converts double-precision, floating-point numbers to their numeric value by using $\mathit{N}\cdot {2}^{\mathit{e}}$, where $\mathit{N}$ and $\mathit{e}$ are the exponent and mantissa respectively.
Convert `t` by using a floating-point expansion.
`sym(t,'f')`
```ans = $\frac{3602879701896397}{36028797018963968}$``` | {
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java, hibernate, authorization, jpa
// 2.
userRepository.findAllById(rolesByUserId.keySet())
// 3.
.forEach(user -> user.setRoles(rolesByUserId.get(user.getId())
.stream().map(roles::get).collect(Collectors.toSet())));
}
So basically the flow is the following:
fetch all the Role entities from the DB into a Map<Long, Role> (where the key is the Role ID and the value is the Role entity itself)
fetch the User entities from the DB that needs to be modified
for each User, iterate through the Role IDs to be assigned, fetch the corresponding Role entities from the Map, collect them into a Set and assign it to the User | {
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javascript, performance, algorithm, stack, complexity
function recurse(sum, value){
sum += value;
if(value > 10) { return sum } // Returns functions state.sum
return recurse(sum, value + 1); // Returns function call result
// the functions state is not required after the recurse call.
}
console.log(recurse(0,0));
Non tail call recursion
function recurse(sum, value){
if(value > 10) { return sum + value }
sum += recurse(value, value + 1); // Note that the function state
// that holds sum must be retained
// in order to complete the function
return sum; // Return function state variable
}
console.log(recurse(0,0)); | {
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"tags": "javascript, performance, algorithm, stack, complexity",
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python, programming-challenge, circular-list
After each round, the gap between the survivors doubles. So after one round, the survivors are every 2nd person; after two rounds, every 4th person, after three rounds, every 8th person, and so on. When there are an even number of people, the first person remains the first survivor in the next round. But when there are an odd number of people, the third person becomes the first survivor in the next round.
So to solve this problem, we don't need to remember all the survivors, we only need to remember the first survivor, the gap between survivors, and the total number of survivors. Like this:
def survivor3(n):
"""Return the survivor of a circular firing squad of n people."""
first = 1
gap = 1
while n > 1:
gap *= 2
n, odd = divmod(n, 2)
if odd:
first += gap
return first
Once again, we should check that this is correct:
>>> all(survivor(i) == survivor3(i) for i in range(1, 1000))
True | {
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c++, asynchronous, c++17, logging
instead of
int main()
{
return 0;
}
Message_type
Make a stand, either CamelCase or snake_case, not some weird offspring.
while (running)
{
if (true)
if (true) is redundant here, you can use { .. } only.
if (queue_log.empty()) { work_available.wait(lock); }
instead go:
work_available.wait(lock, [&]() { return !running || !queue_log.empty(); });
std::condition_variable::wait() might awake spuriously (without reason), in which case your queue_log will be empty and you called write_all for nothing. wait call with predicate takes care of this.
Message_type& message = queue_write.front();
use auto:
auto& message = queue_write.front();
this will make refactoring the code easier.
while (!queue_write.empty())
{
Message_type& message = queue_write.front();
std::cout << message << std::endl;
file << message << std::endl; | {
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human-genetics
Phenotypes such as eye colour are coded on the autosomes (i.e. not the sex chromosomes), and as such are determined by alleles inherited from both parents. As Koustav Pal mentions in his comment, many such traits are multi-allelic (i.e. determined by multiple regions of the genome), and are therefore more complex than first appear; eye-colour can change over time, and there are many more complexities that do not fit into the simple "blue", "brown", "green/hazel" categories [3]. This also applies to hair, so the child you mention could become more like their father as they grow older (or not!). | {
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ros, buildfarm
Title: buildfarm fails for noetic with KeyError: libcurlpp-dev
We are trying to release the packages in pf_lidar_ros_driver for melodic and noetic. We have been able to release for melodic successfully in the past and even the latest job was successful. However the job for noetic failed with the error:
Looking for the '.dsc' file of package 'ros-noetic-pf-driver' with version '1.2.0-1'
Traceback (most recent call last):
File "/usr/lib/python3/dist-packages/apt/cache.py", line 297, in __getitem__
rawpkg = self._cache[key]
KeyError: 'libcurlpp-dev'
During handling of the above exception, another exception occurred: | {
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def swap(array, a, b):
array[a], array[b] = array[b], array[a]
def getPermutationsHelper(res, curr_perm, array, pos):
if pos >= len(array):
if len(curr_perm) > 0:
res.append(curr_perm)
return
for i in range(pos, len(array)):
# add the number(array[i]) to the permutation
# place the element of interest at the first position (pos)
# Example: for getPermutationsHelper(res, [], [1,2,3], 0), while in this for loop
# when at value 1 (i=0), we want 1 to be at pos(0), so that we can iterate through [2,3,4] next without adding 1 again
# when at 2, we want 2 to be at pos(0), so that we can iterate through [1,3,4] next ([2,1,3,4])
# when at 3, we want it to be at pos(0), so that we can iterate through [2,1,4] next ([3,2,1,4])
# so we have to manually place it there (via swapping with the element at pos), then we return it just before the loop ends
# and move pos forward
new_perm = curr_perm + [array[i]] # add number to the permutation | {
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Hint Use the Limit Comparison Test instead:
$$\lim_n \frac{\sin^2(1/n)}{1/n^2} =\left( \lim_n \frac{\sin(1/n)}{1/n} \right)^2=\left( \lim_{x_n \to 0} \frac{\sin(x_n)}{x_n} \right)^2=1$$
From the geometric definition of $\sin x:$
For $0<x<\pi /2,$ take triangle $AOB$ with $OA=OB=1$ and $\angle AOB=x.$ Let D lie on $OA$ with $DB\perp OA.$ The arc of the circle centered at $O,$ of radius $1,$ from $A$ to $B,$ has length $x,$ which is greater than the straight-line distance $AB.$ So $$\sin x=DB<\sqrt {DB^2+DA^2}\;=AB<x.$$
Since $\sin (-x)=-\sin x,$ therefore $|\sin x|<|x|$ for $0<|x|<\pi /2.$
For $|x|\geq \pi /2$ we have $|x|>1\geq |\sin x|.$
And of course for $x=0$ we have $|\sin x|=0=|x|.$
If you are allowed to use the limit comparison test, I recommend it. Use your same $b_n$. Hint to get you started:
$$\lim_{n\to+\infty} \frac{a_n}{b_n} = \lim_{n\to+\infty} \frac{\sin^2(1/n^2)}{1/n^2}$$ | {
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"url": "https://math.stackexchange.com/questions/2401841/how-to-show-that-sin21-n-leq-1-n2-forall-n-in-mathbbn"
} |
1: v . is: Deriving the Euclidean distance between two data points involves computing the square root of the sum of the squares of the differences between corresponding values. ... Percentile. u, is v . Let’s assume OA, OB and OC are three vectors as illustrated in the figure 1. It can be computed as: A vector space where Euclidean distances can be measured, such as , , , is called a Euclidean vector space. General Wikidot.com documentation and help section. The squared Euclidean distance is therefore d(x SquaredEuclideanDistance is equivalent to the squared Norm of a difference: The square root of SquaredEuclideanDistance is EuclideanDistance : Variance as a SquaredEuclideanDistance from the Mean : Euclidean distance, Euclidean distance. . So there is a bias towards the integer element. The following formula is used to calculate the euclidean distance between points. View/set parent page (used for creating breadcrumbs and structured layout). Two squared, lost three square until as | {
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"url": "http://radiojuventudnerja.es/f3y8d/euclidean-distance-between-two-vectors-32de3b"
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python, proteins, sequence-analysis, motifs, multi-fasta
The second part of my question is how do I modify my code such that, the matches and frequencies of the reference sequence(Ref) are excluded. The idea would be to get an output such as the one below. I do not even know whether that is possible (A disclaimer this second part is a question on a tutorial that I don't know how to do. I am still learning)
Sequence3 73 79 PFLPGK
Sequence3 281 287 PLPPKL
Sequence3 288 294 PSSPAH
total sequences 5 #this would be the total number of sequences in file including the refrence
frequency Counter({'73_79_PFLPGK': 1, '281_287_PLPPKL': 1, '288_294_PSSPAH': 1}) #frequency of matches, excluding the reference
The first part appears to be a simple bug if there is no copy/paste error in the code:
Original code,
num_sequence=0
fasta_sequences = SeqIO.parse(open(input_file),'fasta')
for fasta in fasta_sequences:
num_sequence=+1 # < its here
print (num_sequence) | {
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Problem:
Given that $p, q, r, s$ are all positive real numbers and they satisfy the system
$p+q+r+s=12$
$pqrs=27+pq+pr+ps+qr+qs+rs$
Determine $p, q, r$ and $s$.
Attempt:
The AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:
1 $\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$ 2 $\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$ which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$ $(pqrs-81)(pqrs-81) \ge 0$ $pqrs \le 9$ or $pqrs \ge 81$
After that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below
$pqrs \le 81$ and $pqrs \ge 81$ | {
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"url": "https://mathhelpboards.com/threads/determine-p-q-r-and-s.7859/"
} |
quantum-mechanics, angular-momentum, quantum-states, spherical-harmonics
Except, here's the thing: the pure-square monomials $x^2,y^2,z^2$ in $(*)$ are not linearly independent! Why is this? well, because we're on the unit sphere, which means that these terms satisfy the identity
$$
x^2+y^2+z^2 = 1,
$$
and the constant term $1$ is already in our basis set. So, from the combinations $x^2+y^2$ and $z^2$ we can only form a single polynomial, and the correct choice turns out to be $x^2+y^2-2z^2$, to make the set not only linearly independent but also orthogonal (with respect to a straightforward inner product).
As for the rest of the spherical harmonics $-$ the ladder keeps climbing, but the steps are all basically the same as the ones I've outlined already. | {
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gazebo
[roscpp_internal] [2011-08-13 21:00:24,131] [thread 0x7f8ff3f95700]: [DEBUG] TCP socket [26] closed
[roscpp_internal] [2011-08-13 21:00:24,133] [thread 0x7f8ff3f95700]: [DEBUG] Accepted connection on socket [23], new socket [26]
[roscpp_internal] [2011-08-13 21:00:24,134] [thread 0x7f8ff3f95700]: [DEBUG] TCPROS received a connection from [127.0.0.1:40943]
[roscpp_internal] [2011-08-13 21:00:24,134] [thread 0x7f8ff3f95700]: [DEBUG] Connection: Creating ServiceClientLink for service [/gazebo/spawn_urdf_model] connected to [callerid=[/unnamed] address=[TCPROS connection to [127.0.0.1:40943 on socket 26]]]
[roscpp_internal] [2011-08-13 21:00:24,134] [thread 0x7f8ff3f95700]: [DEBUG] Service client [/unnamed] wants service [/gazebo/spawn_urdf_model] with md5sum [9ed9c82c96abe1a00c3e8cdaeee24413]
[roscpp_internal] [2011-08-13 21:00:24,276] [thread 0x7f8ff1f91700]: [DEBUG] TCP socket [26] closed | {
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quantum-mechanics, schroedinger-equation, perturbation-theory
Later the Schrodinger equation is written and multiplied out fully, and the terms with same $\lambda$ coefficients are equated.
What I don't understand is how can one randomly do this? If $\lambda$ is just a marker to keep track of the nth order correction, then why is the same $\lambda$ being used in the Hamiltonian as well? In this context, $\lambda$ is just a parameter that appears in the Hamiltonian. Writing
$$H(\lambda) = H_0 + \lambda H'$$
makes it clear that when $\lambda=0$, then $H$ is simply the unperturbed (and easy to handle) Hamiltonian; when $\lambda \neq 0$, then the Hamiltonian has an extra term which may make the problem much more difficult.
Because $\lambda$ appears in the Hamiltonian, it should be pretty obvious that the energy eigenstates $\psi_n$ and the corresponding eigenvalues $E_n$ depend on $\lambda$ as well. In that sense, we can write $\psi_n(\lambda)$ and $E_n(\lambda)$ as explicit functions of the parameter $\lambda$. | {
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python, beginner, web-scraping
Which brings me to...
SQL INJECTION!
You should escape. The simple way is just
sql_query = (
"INSERT INTO prices (hotel, city, adate, price)"
"VALUES('{hotel_id}'','{city_id}','{date}','{price}');"
).format(
hotel_id=pymysql.escape_string(self.hotel_id),
city_id=pymysql.escape_string(self.city_id),
date=date,
price=price
)
In reality, any executing query should use better mechanisms (like interpolating directly with cursor.execute) but you're writing to a file so this isn't as easy.
And finally
You pass hotel_id as a parameter to save_result by setting an instance variable. This is ugly; just pass a parameter. city_id should probably passed the same way for symmetry.
I've done a little more cleaning and came up with this.
__author__ = 'Mateusz Ostaszewski'
import datetime
import lxml.html as lh
import re
import sys
import pymysql
from selenium import webdriver | {
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homework-and-exercises, energy, momentum
Title: Is stopping something work? If somebody pushes against a mass moving with $3 \frac{m}{s}$ to slow it down to $2 \frac{m}{s}$, he will drain the moving system of kinetic energy. Does he do work then or does he consume work?
My homework problem is that somebody is on a train, splits the cars and pushes one away making all other ones slower. The total energy of the system is lower, but he did some work obviously since he pushed it.
http://wstaw.org/m/2011/11/19/m12.png
So do I add the two changes in kinetic energy (which yields a negative number) or do I add their absolute values? If you actually gave us your complete homework problem, then your homework problem is wrong; it describes an impossible occurrence. | {
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find similarities and differences notation to describe the end is. And it 's pretty easy rational function, set the numerator equal to solve... The behavior $\ ; =0$ ) ∞ ) turning points are where!: Rises to the left and Rises to the left and Falls to the left and Rises the! Rises to the left and Falls to the left and Rises to the left and to! 'S pretty easy hand side seems to decrease forever and has no asymptote x2 +4 &. Slant asymptotes ( slope $\ ; =0$ ) an oblique asymptoteand found... Function has a horizontal asymptote y = 1−3x2 x2 +4 function, as well as the of! Case of slant asymptotes ( slope $\ ; =0$ ) co-efficient of the co-efficient. 2 find the end behavior '', and it 's pretty easy decreases without bound is end! The rational function depends on the degrees of the numerator equal to 0and solve increases decreases! As x approaches negative infinity forever and has no asymptote function is x ⇔... As x approaches negative infinity x ) as x approaches negative infinity 'll | {
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"url": "http://animent.hu/szbo44cf/how-to-find-end-behavior-of-a-function-c43993"
} |
autocorrelation, cross-correlation, adaptive-algorithms, beamforming
Title: Auto/cross correlation of a cosine in the context of an adaptive beamformer I am currently reading chapter 13 of 'Adaptive Signal Processing' by Widrow and Stearns, so if anyone happens to have a copy of the book to hand it could be helpful.
I am reading the chapter on "Introduction to Adaptive Arrays and Adaptive Beamforming" and am struggling understanding the auto-correlation/cross correlation between the primary and reference signals at a delay of one (as shown in the diagram below) and in the more general case (where the signal is arriving at an angle), where the general cross correlation vector is described by:
$$
\mathbf{P} = \begin{bmatrix}
\phi_{dx}(0) \\ \phi_{dx}(1)
\end{bmatrix} = \begin{bmatrix}
E [C \cos k\omega_0 \cdot C \cos (k + \delta_0)\omega_0] \\ E [C \cos k\omega_0 \cdot C \sin (k + \delta_0)\omega_0]
\end{bmatrix}
$$ | {
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Is this BASIC? Wow, that's a blast from the past. – Chris Taylor Feb 7 '12 at 9:30 | {
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"url": "http://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar/103034"
} |
c#, parsing, reinventing-the-wheel
currentLine++;
body.Nodes.Add(ifBlock);
}
else if (Regex.IsMatch(trimmed, @"For\s.*$"))
{
var forLoopBlock = new CodeBlockNode(content[currentLine]);
currentLine++;
while (content[currentLine].Trim() != ReservedKeywords.Next)
{
forLoopBlock.Nodes.Add(new CodeBlockNode(content[currentLine]));
currentLine++;
}
currentLine++;
body.Nodes.Add(forLoopBlock);
} | {
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"tags": "c#, parsing, reinventing-the-wheel",
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} |
weather-forecasting
I don't have any background in meteorology, and in particular I don't have much of a sense of how PoP is actually computed in practice. I agree that the PoP = C x A that we see on a lot of websites leaves something to be desired. It gets across the loose idea that the definition involves the area affected as well as the probability of occurrence, which is fine for most casual readers but can be frustrating for more inquisitive minds. How much are the confidence and the area estimated from ensemble forecasts versus expert judgement? How was it done before ensemble NWP started in the 1990s? | {
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python, hash-map
"""
from itertools import product
def dict_product(d):
sorted_keys = sorted(d.keys())
for element in product(*list(d[k] for k in sorted_keys)):
yield dict(zip(sorted_keys, element))
if __name__ == "__main__":
import doctest
doctest.testmod(verbose=True) You can use the fact that d.keys() and d.values() are always aligned. This gets rid of both the awkward way of constructing the values as well as the sorting of the keys.
def dict_product(d):
keys = d.keys()
for element in product(*d.values()):
yield dict(zip(keys, element))
The assignment of keys is not strictly necessary, but it avoids the look-up every loop iteration.
Note that list(d[k] for k in sorted_keys) should have been written as a list comprehension directly, instead of calling list on a generator comprehension. So [d[k] for k in sorted_keys]. | {
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homework-and-exercises, newtonian-mechanics, rotational-dynamics, reference-frames, moment-of-inertia
Title: Query regarding parallel axis theorem Suppose there are two points at (4,0) and (-4,0) each of mass 1 kg and the origin is (0,0). The rotational inertia of the combined system is 32kg.m^2. But if we shift the axis of rotation to left by 2 units to (-2,0). The new moment of inertia is 40 kg.m^2. Which is $$M_{sys}d^2$$ more than the original value. But the moment of inertia of one particles decreased and other ones increased. then why the moment of inertia of entire system increased by $$M_{sys}d^2$$. Consider a general sketch of the situation and let's try to figure this out algebraically.
The separation $\ell = a + b$ is fixed, and the pivot is some distance $d$ from the combined center of mass in the middle. The distance from the pivot to each mass is assigned the letters $a$ and $b$.
Each mass is $m$ such that $M_{\rm sys} = 2m$
I will find the MMOI about the pivot, two different ways. | {
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javascript, canvas
const enemyCount = 60;
const enemyShoot = 10;
const enemy = {
shoot: 0,
shootProjectile(){ return Math.random() }
}
const enemies = $setOf(enemyCount,() => ({...enemy}));
var soak = 0;
var time = 0;
function testA() {
enemies.sort(() => 0.5 - Math.random()).slice(0, enemyShoot).forEach(e => {
soak += e.shootProjectile();
});
time ++;
}
function testB() {
const enemiesCopy = [...enemies];
const enemiesToShoot = [];
while (enemiesToShoot.length < enemyShoot && enemiesCopy.length) {
const [enemy] = enemiesCopy.splice(Math.floor(Math.random() * enemiesCopy.length), 1);
enemiesToShoot.push(enemy);
}
for (const enemy of enemiesToShoot) {
soak += enemy.shootProjectile();
}
time ++;
} | {
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"tags": "javascript, canvas",
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gravity, mass, weight
Title: Can mass be directly measured without measuring its weight? From Wikipedia, http://en.wikipedia.org/wiki/Mass
Inertial mass measures an object's resistance to being accelerated by
a force (represented by the relationship F=ma).
Active gravitational
mass measures the gravitational force exerted by an object.
Passive
gravitational mass measures the gravitational force experienced by an
object in a known gravitational field.
Mass-Energy measures the total
amount of energy contained within a body, using E=mc² | {
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project-ideas
Title: Developing DSP software, am I 10 years too late? I'm interested in developing software related to digital signal processing. The most obvious ideas are audio and video editing. There's an overwhelming amount of audio and video editing software out there. I feel like developing such software would be in vain.
What are the most lucrative frontiers in DSP software today?
Do you think audio/video software still have room for improvement? DSP is a wide field and goes beyond audio/video processing. And, since new digital devices are developed in any field, DSP algorithm implementation, simulation and design is still a very important area.
Some examples that come to my mind beyond audio/video: | {
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thermodynamics
Title: Heating power of a room affecting object of different temperature I have a object with a surface area of about $16 \, \mathrm{cm}^2 ,$ and I'm trying to calculate the rate at which it will be heating up if placed into a warmer room, so I can apply the same amount of cooling to make the temperature constantly lower than the room temperature.
For instance my room temperature is $20 \sideset{^{\circ}}{}{\mathrm{C}}$ and the object's temperature is $0 \sideset{^{\circ}}{}{\mathrm{C}} .$ If it helps, the object is made out of a ceramic material $\left(\text{Al}_2 \text{O}_3 \right) .$
I know it can be somehow calculated with Newton's law of cooling, but I can't seem to get past the heat transfer coefficient. Assume the room's temperature to be $T_0$ and constant, the object's temperature $T$ (considered uniform), its surface area $A$, its mass $m$, its heat capacity $c_p$ and $h$ the heat transfer coefficient.
With Newton's Cooling/Heating Law we get:
$$\frac{dQ}{dt}=hA(T_0-T)$$ | {
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haskell, monads, telegram
-- Public
newtype Controller = Controller
{ handlers :: [Handler]
}
processUpdate :: Controller -> TTypes.Update -> ConfigT ()
processUpdate controller update = do
updateInfo <- getUpdateInfo update
let sid = sessionId updateInfo
let context = HandlerContext { userId = r @UpdateInfo user_id updateInfo
, messageId = r @UpdateInfo message_id updateInfo
}
result <- withSession sid $ do
handlerAction <- findHandler updateInfo $ handlers controller
runHandler handlerAction context
either (handleSessionError context) return result
-- Private
data UpdateInfo = UpdateInfo
{ request :: Request
, message :: String
, user_id :: Int
, message_id :: Int
, sessionId :: Maybe String
}
data Request
= MessageRequest { message :: TTypes.GetMessage }
| ResponseRequest { message :: TTypes.GetMessage }
| QueryRequest { query :: TTypes.CallbackQuery
, message :: TTypes.GetMessage }
r :: (r -> a) -> r -> a
r = ($) | {
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.(c) $\mathop {\lim }\limits_{x \to 0^+} f'(x) = -\infty$
.(d) $\mathop {\lim }\limits_{x \to -\infty} f(x) = -\infty$
.(d) $\mathop {\lim }\limits_{x \to \infty} f(x) = \infty$
Interpret each piece of information.
.
My interpretations are as follows:
.(a) y-intersect at (0,0)
.(b) Concave up for all values of $x$, except at $x = 0$ (possibly an asymptote there?)
.(c) As you approach 0 from the left, the slope approaches infinity and so the function gets higher.
.(c) As you approach 0 from the right, the slope approaches negative infinity and so the function gets lower.
.(d) As you approach negative infinity (left), the function gets lower and lower,
.(d) As you approach positive infinity (right), the function gets higher and higher.
I'm not to sure about them though, and it seems like (b) and (c) are contradictory (if C is true, that would mean concave down to the right of 0)
I would really appreciate a second opinion.
If and only if the second condition at d is | {
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all over the place in the above proofs. Free functions inverse calculator - find functions inverse step-by-step This website uses cookies to ensure you get the best experience. Example 1 Show that the function $$f:\mathbb{Z} \to \mathbb{Z}$$ defined by $$f\left( x \right) = x + 5$$ is bijective and find its inverse. It is a good exercise to try to prove these on your own as well, and to compare your proofs with those given here. The same argument shows that any other left inverse b Ⲡb' b Ⲡmust equal c, c, c, and hence b. b. b. {eq}\eqalign{ & {\text{We have the function }}\,f\left( x \right) = {\left( {x + 6} \right)^2} - 3,{\text{ for }}x \geqslant - 6. f(x) = \begin{cases} \tan(x) & \text{if } \sin(x) \ne 0 \\ To prove A has a left inverse C and that B = C. Homework Equations Matrix multiplication is asociative (AB)C=A(BC). Information and translations of left inverse in the most comprehensive dictionary definitions resource on the web. Since gʹ is a right inverse of f, we | {
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"tags": null,
"url": "http://stereotypes.ingenium-elearning.com/qrbqo/jake-wood-mgfn/qg6ek.php?5e26de=left-inverse-is-right-inverse"
} |
complexity-theory, np-complete, p-vs-np
Title: Are there languages L1 ⊆ L2 ⊆ L3 when L1 and L3 are NP-Complete languages and L2 ∈ P? Are there languages L1 ⊆ L2 ⊆ L3 where L1 and L3 are NP-Complete languages and L2 ∈ P? Would this imply P=NP?
Thanks Consider the following problems:
$L_1$:
Input: two graphs $G_1$ and $G_2$
Question: does $G_1$ contain an hamiltonian path and is $G_2$ complete?
$L_2$:
Input: two graphs $G_1$ and $G_2$
Question: is $G_2$ complete?
$L_3$:
Input: two graphs $G_1$ and $G_2$
Question: does $G_2$ contain an hamiltonian path?
Then $L_1 \subseteq L_2 \subseteq L_3$. $L_1$ and $L_3$ are $\mathsf{NP}$-complete, but $L_2$ is in $\mathsf{P}$. | {
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electromagnetic-radiation, electricity, electric-current, radio-frequency
$P_{rad}=\eta \frac { J_0^2 \ }{8 \pi^2 k^2} \int_0^{2 \pi} d\phi \int_0^{\pi} d\theta \sin\theta \ (\cos^2\theta \cos^2\phi+\sin^2\phi)\frac{ \sin^2\bigr(k \frac {a}{2}(\sin\theta\cos\phi)\bigr)}{(\sin\theta\cos\phi)^2} \frac{ \sin^2\bigr(k \frac {b}{2}(\sin\theta\sin\phi)\bigr)}{(\sin\theta\sin\phi)^2}$
As you can probably see, the above integral is far too complicated to be calculated by hand. I suggest plugging in your numbers and solving the integral numerically with some software package (e.g. MATLAB).
I should probably also tell you that I have no idea on how to make the current uniform in such a sheet, as its current would also have a wave behavior $\mathbf J_s=J_0 e^{-ikz} \hat {\mathbf x}$; and that is also not necessarily how the current distribution will be in practice, because the distribution highly depends on where you exactly feed current through it. But as you can see even the simplest case of a uniform current is really hard to solve. | {
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-
Arturo is a canonical example of an enterprising soul! :) – Mariano Suárez-Alvarez Feb 11 '11 at 20:48
@Mariano: You are affirming the consequent! Your statement was basically "If $X$ is enterprising, then $X$ might post a proof"; I posted a proof, it does not follow I am enterprising...(-: (Besides, I was in the middle of it when your post came through...) – Arturo Magidin Feb 11 '11 at 20:53
I knew of your enterprisingness (?) long before this question came along! :) – Mariano Suárez-Alvarez Feb 11 '11 at 20:54
Thank you Arturo. A wonderful answer, much obliged. – user6927 Feb 11 '11 at 22:23
And yet, a downvote... Go figure. – Arturo Magidin Mar 25 '12 at 19:05
That «one can drop the parentheses» really means that «no matter how you put the parentheses in, the result will be the same». | {
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"openwebmath_score": 0.9600741267204285,
"tags": null,
"url": "http://math.stackexchange.com/questions/21581/how-does-one-actually-show-from-associativity-that-one-can-drop-parentheses"
} |
radioactivity
Title: Has the half-life of an isotope ever been accurately predicted before it could be measured? The half lives of many radioactive isotopes have been measured directly. But are their half-lives known to be a function of their composition?
In other words, can someone say "given an atom with this many protons, neutrons and electrons, and given the known strengths and ranges of the various forces at play inside an atom, we can quantify how unstable such an atom would be by saying that a collection of them would have a half-life of X"?
Has such a prediction ever been made about an isotope before people had encountered it, then later borne out by experiment? A quick perusal of the post-WWII literature (particularly Glenn Seaborg) pulls up papers like The New Element Californium (Atomic Number 98) where comparisons of the observed alpha decay half life are said to be in line with predictions. Such predictions follow along from papers such as Systematics of Alpha-Radioactivity. | {
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complexity-theory, polynomial-time, encoding-scheme
understanding is that two encodings are polynomially related if converting from one another requires a polynomial amount of time. Can anybody clarify things a bit? Garey and Johnson are referring to the fact that any encoding scheme for some instance $I$ of a problem $\Pi$ will only differ in length (i.e. number of bits) by a polynomial amount. For example, consider two possible ways to encode a graph: adjacency matrix, and adjacency list. | {
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"tags": "complexity-theory, polynomial-time, encoding-scheme",
"url": null
} |
quantum-mechanics, gravity, particle-physics, standard-model
Likewise, if Supersymmetry exists, it must be broken, as otherwise all super-partners would have the same mass as the known particles, selectrons would also have .511 MeV, and already long been observed. Therefore any supersymmetric theory must be broken to accomodate the SM as we know it (no susy particles upto order TeV).
Any theory that will include Gravity and Quantum Mechanics and GR must resemble either in their respective limits.
Theories don't just vanish, they'll stay relevant in the regime they explained well. | {
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entanglement, concurrence
Title: Can two states with the same entanglement be transformed into each other using local unitaries? Take two pure bi-partite states $\psi$ and $\phi$ that have the same amount of entanglement in them as quantified by concurrence (does the measure make a difference?). Can any such states be transformed into each other using local unitaries? Any two bipartite pure states $\psi$ and $\phi$ can be transformed into each other with local unitaries if and only if they have the same Schmidt coefficients. (To prove the 'only if' part, note that the reduced density matrix of either qubit has eigenvalues that are the squares of the Schmidt coefficients, and are unchanged by unitaries). | {
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a θ. Of two or more notes is the longest distance from center of the.!, Dsus4, Gmaj7, etc and it is called the radius and central angle: chords a... One is called chord of circle and one of its chords come across chords like Dom7, and... Sinθ x r 2 ) | Quantitative aptitude | basic Mathematics | Reasoning through the center the... To explain playing a circle is called radius of the circumference of circle. Also a circle chord formulas of the circle is the largest chord is a list of important formulas used in maths Physics... Line joining 2 points on the circumference of circle, PNG, clip art transparent... Segment within a circle 's diameter ): it is a segment a... A list of important formulas used in this free math video tutorial by Mario 's math Tutoring radius (! Then you can convert angle units the figure is a line perpendicular to center... They determine two central angles that are congruent, then click calculate connects any two of! To each other in terms of the circle given | {
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"url": "http://www.webdynamix.com/pb2lib5/9c2122-circle-chord-formulas"
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gazebo-4, gazebo-3
Title: Downgrade from Gazebo 4
I'm trying to downgrade from Gazebo 4 to Gazebo 3 and make it work with ROS Indigo. I'm using the following steps:
sudo sh -c 'echo "deb http://packages.osrfoundation.org/gazebo/ubuntu `lsb_release -cs` main" > /etc/apt/sources.list.d/gazebo-latest.list'
wget http://packages.osrfoundation.org/gazebo.key -O - | sudo apt-key add -
sudo apt-get update
sudo apt-get install ros-$distro-gazebo4-ros-pkgs
The problem is that I get the message "E: Unable to locate package ros-indigo-gazebo3-ros-pkgs". What am I doing wrong in the steps I'm using?
UPDATE: Does the package ros-indigo-gazebo3-ros-pkgs actually exist?
Originally posted by K. Zeng on Gazebo Answers with karma: 103 on 2014-10-29
Post score: 0 | {
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special-relativity, experimental-physics, time-dilation
Title: How does the Kennedy-Thorndike experiment test for time-dilation? From the Wikipedia page (as of 3/6/2023), | {
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electromagnetic-radiation, visible-light, reflection
The reason behind your observation is that the boundary condition at a smooth perfect conductor forces the tangential E field to be zero at the surface. How can the field on one side of the surface achieve that? Say the wall exist in a Cartesian 3D, filling the half space $x>0$, and imagine a plane wave incident onto such surface from the left (either right angle incidence or obliquely). Then in the region $x\leq 0$ we have both the incident and reflected electric fields $\vec{E}_{tot}=\vec{E}_{inc}+\vec{E}_{ref}$. Their components tangential to the surface, say along $\hat{y}$, will therefore be
$$ E_{y,tot}=E_{y,inc}+E_{y,ref},$$
which must be now forced to zero. This can only be achieved if $E_{y,inc}=-E_{y,ref}$, which gives the 180 deg shift that puzzled you. You can now extend this to the H field using similar steps. | {
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investor how much income they will receive each year, as a percentage, if he or she buys the stock today. To understand the uses of the function, let’s consider an example: We can use the function to find out the yield. Therefore, the current yield of the bond is 6.06%. We also provide a Current Yield Calculator with downloadable excel template. Find out the best practices for most financial modeling to price a bonds, calculate coupon payments, then learn how to calculate a bond's yield to maturity in Microsoft Excel. This function uses the following arguments: The settlement and maturity dates should be supplied to the YIELD function as either: As a worksheet function, YIELD can be entered as part of a formula in a cell of a worksheet. By taking the time to learn and master these functions, you’ll significantly speed up your financial modeling. To calculate the current yield of a bond in Microsoft Excel, enter the bond value, the coupon rate, and the bond price into adjacent cells | {
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python, machine-learning
Title: ML Retraining project Tear me to shreds.
The class RandomForestRetrainer will be used to retrain a machine learning algorithm. It has functionality for taking in a directory containing malware or benignware files and splitting them into training and testing sets, creating statistics from these files, creating concatenated mw/bw training/testing stat files, balancing the mw count with bw count via a reduction algorithm, and finally sending them to the ML classifier for training.
p.s. let me know if you want to see code from the other classes
import os
import datetime
from retraining.StatsFile import StatsFile
from retraining.Dataset import Dataset
from retraining.Partitioner import Partitioner
from retraining.GridBasedBalancingRandom import GridBasedBalancingRandom
from retraining.Classifier import Classifier
import config
class RandomForestRetrainer(object):
def __init__(self, previous_mw_dataset=None,
previous_bw_dataset=None): | {
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I have tried to do some calculations, but I don't think I am working it out properly.
- Is the following correct?
((0.0596/100)+(0.0312/100)+(0.0135/100)+(0.0098/100))*100 = 0.1141% (or 100% - 99.8859% = 0.1141%)
- Is the following correct?
((1/1678)+(1/3205)+(1/7407)+(1/10204))*100 = 0.1141%
- Is the following correct?
(((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100 = 0.0285%
If I take 1,400 coins from the magic bag, what is the chance/probability to receive either an iron, bronze, silver or gold coin?
(Receiving any of these four coins would be a success, and receiving any of the other 96 coins would be a failure).
I really have no idea how to calculate this; all I have managed to do is repeat one of the formulas above and multiply by 1,400.
- Is the following correct?
((((1/1678)+(1/3205)+(1/7407)+(1/10204))/4)*100)*1400 = 39.9339%
I understand that I am probably completely wrong about everything, so thank you very much to anyone willing to provide assistance. | {
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ros, svn
Title: Willow PR2 svn repo
How can I clone this SVN repo: http://code.ros.org/svn/wg-ros-pkg/all.
I understand that this repo isn't used for development anymore, but I'd still like to clone it in order to navigate through revision history.
$ svn co http://code.ros.org/svn/wg-ros-pkg/all
svn: Repository moved permanently to 'https://code.ros.org/svn/wg-ros-pkg/all'; please relocate
$ svn co https://code.ros.org/svn/wg-ros-pkg/all
Error validating server certificate for 'https://code.ros.org:443':
The certificate has expired.
Certificate information:
Hostname: *.ros.org
Valid: from Wed, 25 Apr 2012 00:00:00 GMT until Fri, 25 Apr 2014 23:59:59 GMT
Issuer: COMODO CA Limited, Salford, Greater Manchester, GB
Fingerprint: 4b:0b:52:6b:10:f4:d0:dc:42:03:22:f0:04:60:8f:df:e0:33:ff:c5
(R)eject, accept (t)emporarily or accept (p)ermanently? t
svn: URL 'https://code.ros.org/svn/wg-ros-pkg/all' doesn't exist | {
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say c and m function my_phase() IC = [0. East Facing Plot A north-east facing plot is best for all type of constructions, whether a house or a business establishment. For autonomous} ewline \textrm{systems, we plot the slope field and attempt to plot equilibria. The phase plane representation of the system, [x';y']=[1,-1;3,-1][x;y], would be concentric ellipses about the Eigen vectors. Graph functions, plot data, evaluate equations, explore transformations, and much more – for free! Start Graphing. See if the plot that it creates looks similar to what you want. Phase Plane Analysis is a graphical method for studying first and second-order systems by. The frequency at which the Nyquist plot is having the magnitude of one is known as the gain cross over frequency. To interpret this diagram, imagine that the antenna is at the origin. When all of the points are plotted, put a circle around the group of high probability values (probability >. Muslakar plot (long oval like plot with cut in | {
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spacetime, space
So "empty space" really is a kind of "stuff" that can have inhomogeneous properties: in General Relativity, one can quite definitely say that this piece of spacetime over here can have different geometrical properties from another piece over there.
In answer to your title: we, just like empty space, are made of quantum fields, so we couldn't exist without there at least being the possibility of quantum fields in their ground state, i.e. the existence of empty space is a necessary condition for our existence. | {
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genetics, biostatistics
Cc x Cc occurs 25% of the time, and 75% of offspring are affected
Cc x cc occurs 50% of the time, and 50% of offspring are affected
cc x cc occurs 25% of the time, and 0% of offspring are affected
Multiply these probabilities and you should have 43.75% expressing the phenotype; your prediction for the chi-square test is that you will see this same percentage regardless of sex.
From there, doing a chi-square test is simple and there are many many resources online for this (here is just one). | {
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c, homework, formatting
Title: Print character + the ASCII value, 10 pairs per line My exercise was:
Write a program that reads input as a stream of characters until encountering
EOF. Have the program print each input character and its ASCII decimal value.
Note that characters preceding the space character in the ASCII sequence are
nonprinting characters. Treat them specially. If the nonprinting character is a
newline or tab, print \n or \t, respectively.
Print 10 pairs per line, except start a fresh line each time a newline character is encountered.
This is my code (regarding the special characters, I defined only \n and \t):
#include <stdio.h>
int special_chars(int ch);
int main(void)
{
int x;
printf("please enter a some characters, and ctrl + d to quit\n");
special_chars(x);
return 0;
}
int special_chars(int ch)
{
int pairsNum = 0;
while ((ch = getchar()) != EOF)// testing charecters while not end of file.
{ | {
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signal-analysis, continuous-signals
I followed the following algorithm:
$$ y(t) =x(2t) $$
$$ y_1(t) = x_1(2t)$$
Let $$x_2(t) = x_1(t-t_0) ~~~\text{and}~~~ y_2(t) = x_2(2t) $$
On this following step (time sampling of the shifted argument) you make the usual mistake:
$$\implies y_2(t) = x_1(2(t - t_0))$$ which should instead be :
$$\implies y_2(t) = x_2(2t) = x_1(t - t_0)|_{t=2t} = x_1(2t - t_0)$$
The remaining parts follow as usual to show that the time scaler is a time-varying system... | {
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observational-astronomy, radio-astronomy, spectroscopy, t-tauri-stars
Title: Why would radio astronomers choose ¹³CO and C¹⁸O spectral lines instead of the most isotopically common combination? Wikipedia's T Tauri explains that this system is an atypical example of T Tauri stars.
It says:
As typical for the young stars, all three stars of T Tauri system are surrounded by a compact disks trimmed by star-star interaction. The disk around T Tauri N has a gap around 12 AU radius, indicating a presence of orbiting Saturn-mass planet within a gap.13
13ALMA Super-resolution Imaging of T Tau: r = 12 au Gap in the Compact Dust Disk around T Tau N
That article says: | {
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php, unit-testing
/**
* @param $syslogSeverity The severity of the syslog message to store
* @param $message The message to store in syslog
* @return true if the message was logged into syslog, false if it wasn't
* @see http://www.php.net/syslog
*/
protected function logToSyslog($syslogSeverity, $message) {
if ($this->syslogOpened) {
return \syslog($syslogSeverity, $message);
}
return false;
}
/**
* @param $message The message to log with error_log
* @return true if logged, false if not
* @see http://php.net/manual/en/function.error-log.php
*/
protected function logToErrorLog($message) {
$message = \date('M-d-Y H:i:s') . ' := ' . $message;
return \error_log($message);
}
/**
* @brief Ensures that the syslog is properly closed if it was opened.
*/
public function __destruct() {
if ($this->syslogOpened) {
\closelog();
}
}
} | {
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homework-and-exercises, klein-gordon-equation, qft-in-curved-spacetime, spherical-harmonics, black-hole-thermodynamics
Title: Klein Gordon equation in Schwarzschild spacetime (spherical harmonic mode expansion) My Question: In his GR text, Robert Wald claims that solutions, $\phi$, to the Klein-Gordon equation, $\nabla_a\nabla^a \phi = m^2 \phi$, in Schwarzschild spacetime can be expanded in modes of the form $\phi(r,t,\theta,\varphi) = r^{-1}f(r,t)Y_{lm}(\theta,\varphi)$ to obtain
\begin{equation}
\frac{\partial^2 f}{\partial t^2}-\frac{\partial^2f}{\partial r_*}+\left(1-\frac{2M}{r}\right)\left[\frac{l(l+1)}{r^2}+\frac{2M}{r^3}+m^2 \right]f=0, \label{eq 1} \tag{Wald 14.3.1}
\end{equation}
where we have the Regge-Wheeler tortoise coordinate $r_{*}\equiv r+2M\ln(r/2M-1)$, with $r$ the usual Schwarzschild "radial" coordinate, and $Y_{lm}(\theta,\varphi)$ the spherical harmonics.
I simply want to verify \eqref{eq 1}, using the definition of the tortoise coordinate and spherical harmonics.
My Attempt: I first note that the Schwarzschild metric in its usual "radial" coordinate, $r$, is | {
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general-relativity, differential-geometry, metric-tensor, tensor-calculus, curvature
In actuality, the only such solution is to have space be flat everywhere! However, when solving the equations people usually discount the origin, allowing for more interesting solutions.
Note that if $R_{\mu \nu \rho \sigma} = 0$ your space must be Minkowski everwhere. However, you can have $R_{\mu \nu \rho \sigma} \neq 0$ somewhere and still have $R_{\mu \nu} = 0$. This is the case in the Kerr solution. Kerr space time is not flat. It is curved (even outside the singularity) even though $R_{\mu \nu} = 0$ there. There is no contradiction. Ricci flat is much weaker than being completely flat. | {
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java, object-oriented, design-patterns, gui, physics
direction / position
You might want to wrap direction and position into vectors. The main benefit would be, that you have a lot less parameters to pass and set, and it's usually quite clear, what a Vector is. Also not really sure if the calculation of the positions should be within the Projectile, since those are very common calculations - you might want to calculate those in a separate "MyMaths" type, or even in a Vector itself, maybe. The "maybe 3d in the future" problem should then be easier to implement (I have to admit, I suck at maths, so not quite sure if that would really help)
Other | {
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homework-and-exercises, kinematics, projectile
for the two balls from the starting height is the same, the difference in time will be the air time of the upward thrown ball, which is $2v/g$. The initial height $H$does not matter because the ball's path starts and ends at the same height when measuring air time. | {
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homework-and-exercises, newtonian-mechanics, energy, rotational-dynamics, reference-frames
To see how to break this up into center of mass quantities, we will need to define the center of mass quantites. Let us define the total mass $m$ of the object to be $m=\int \rho(\r)d\r$, and let us define the center of mass position $\r_{cm}$ by $\r_{cm} = \dfrac{1}{m} \int \r \rho(\r) d \r$. Let us define the center of mass velocity $\v_{cm}$ to be the derivate of the center of mass position, which is given by the expression $\v_{cm} = \dfrac{1}{m} \int \v(\r) \rho(\r) d \r$.
First we will prove an intuitive result, which is that $\v(\r_{cm}) = \v_{cm}$. This is intuitive because the body is rigid so the center of mass has to move with the object. To prove this just see that
$\begin{equation}
\begin{aligned}
\v_{cm} &= \dfrac{1}{m} \int \v(\r) \rho(\r) d \r \\
&= \dfrac{1}{m} \int (\mathbf{v}_0 + \w \times \r) \rho(\r) d \r \\
&= \dfrac{1}{m} \int \mathbf{v}_0 \rho(\r) d \r + \dfrac{1}{m} \int \w \times \r\rho(\r) d \r \\ | {
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c#, wpf, windows, xaml
public static void LoadVideoView(AddVideoFolder videoView)
{
VideoView = videoView;
ReadVideoFolderSettings();
}
private static async void CreateFolders()
{
StorageFolder localFolder = ApplicationData.Current.LocalFolder;
VideosFolder = await localFolder.CreateFolderAsync("\\Videos\\", CreationCollisionOption.OpenIfExists);
StorageFolder settingsFolder = await localFolder.CreateFolderAsync("Settings\\", CreationCollisionOption.OpenIfExists);
SettingsFile = await settingsFolder.CreateFileAsync("settings.dat", CreationCollisionOption.OpenIfExists);
} | {
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python, game, gui
count += 1
def do_move(self):
"""computer picks a move to make"""
#mix it up a little by starting with the center for first move sometimes
if (not 1 in self.bState) and (not -1 in self.bState): #i.e. first move
if random.random() < 0.20: #20% of the time start in center
self.do_button(5)
return
#handle the case where player as made first move to a corner
if (1 in self.bState) and (not -1 in self.bState):
if self.bState.index(1) in [1,3,7,9]:
self.do_button(5)
self.specialDefense = TRUE
return | {
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school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100. . 200 xx (1+200)/2 = 20100 The sum of a finite arithmetic sequence is equal to the count of the number of terms multiplied by the average of the first and last terms. Put them in pairs: 1 & 500 2 & 499 3 & 498 etc There are 250 pairs, each pair = 501 250*501 = 125250-----How about only the odds: 1+3+5+7+...493+495+497+499? Therefore, 62500 is the sum of first 250 odd numbers. . Let there be n terms in this A.P. The sum of the odd numbers (from 1) up to to 500 is 62500. What is the sum of first 500 odd numbers? This is an A.P. The sum of 1+2+3+4+...496+497+498+499+500? Find the number and sum of all integer between 100 and 200, divisible by 9: ----- Numbers between 100 and 200, divisible by 9: 108 117 126 135 144 153 162 171 180 189 198 The sum : 1683 Flowchart: C++ Code Editor: Contribute your code and comments through Disqus. step 2 apply the input parameter values in the formulaSum = n/2 x (a + | {
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"url": "http://uffol.com/men-in-kbnu/7a4c40-sum-of-numbers-1-to-250"
} |
The angle between the minute hand and the hour hand of a clock when the time is 4.15 is
A) 0 B) 37.5 C) 27 D) 15
Answer & Explanation Answer: B) 37.5
Explanation:
Angle between hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H 'o clock
=> $\fn_jvn \small 30\left ( H -\frac{M}{5} \right )+\frac{M}{2}$ degrees
Here H = 4, M = 15 and the minute hand is behind the hour hand.
Hence the angle
$\fn_jvn \small 30\left ( H -\frac{M}{5} \right )+\frac{M}{2}$ = 30[4-(15/5)]+15/2 = 30(1)+7.5 = 37.5 degrees
Report Error
10 1594
Q:
What is the angle made by the hour hand and the minute hand, if the clock shows 9:15 pm ?
A) 165 degrees B) 172.5 degrees C) 112.5 degrees D) 125.5 degrees
Answer & Explanation Answer: B) 172.5 degrees
Explanation: | {
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} |
a mechanism to calculate and represent time and space complexity for any algorithm. This is an example of abuse of notation, the practice of redefining some standard bit of notation (in this case, equations) to make calculation easier. Denote G n ( ) = @gn( ) @ , 2[0;^], G^ n n(^), G n G n ( ), G = EG n ( 0), = E g (z; 0)g (z; 0) 0. The goal of computational complexity is to classify algorithms according to their performances. yx(,ε) depends on a small parameter ε, and the solution of the. (2007),Wu (2009),Meterelliyoz et al. So I can't figure out how to add text to my image but this is my question. An example is a. The ω notation makes the table nice and symmetric, but is almost never used in practice. For a simple real zero the piecewise linear asymptotic Bode plot for magnitude is at 0 dB until the break frequency and then rises at +20 dB per decade (i. Now having the asymptotic results for, we can turn to the question of testing hypothesis and construction of the classical tests | {
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"openwebmath_score": 0.7279811501502991,
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"url": "http://xfat.kab-dinkelsbuehl.de/asymptotic-functions-examples.html"
} |
I agree; this would be the easiest fix.
Also change the help for this command to state that it calculates the sample standard deviation. In the help, just change the word "population" to sample.
I'm afraid stddevs is also confusing, stddevpfroms would be better.
05-29-2016, 04:56 PM
Post: #19
Mike Elzinga Junior Member Posts: 16 Joined: May 2016
RE: stdevp( ) appears to be mislabeled
(05-29-2016 05:57 AM)parisse Wrote:
(05-28-2016 08:12 PM)Mike Elzinga Wrote: I agree; this would be the easiest fix.
Also change the help for this command to state that it calculates the sample standard deviation. In the help, just change the word "population" to sample.
I'm afraid stddevs is also confusing, stddevpfroms would be better. | {
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"lm_q2_score": 0.8947894696095782,
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"openwebmath_score": 0.5120582580566406,
"tags": null,
"url": "https://www.hpmuseum.org/forum/thread-6313.html"
} |
data-structures, probability-theory, hash, hash-tables
$$
b = v_1 - Ax_1, \\
A(x_2-x_1) = v_2 - v_1.
$$
Regarding 3-wise independence: I could be missing something, but it seems that the family is 3-wise independent.
Hint for refuting 4-wise independence: Take $x_3 = x_1 + x_2$ and $x_4 = 0$. | {
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quantum-operation
meaning the action of any finite extension of $\Phi$ on unit-rank projections returns the partial trace of a unit-rank projection, which is always a positive semidefinite operator. | {
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} |
soft-question, history, mathematics, epistemology
Throughout most of its history mathematics developed as a tool of science and technology, from the time of Archimedes to the era of Euler, Lagrange, Gauss and Legendre. So it should not be surprising that it "works" in physics. It was not until about 1850 that Pure Mathematics became recognised as a separate subject.
As Paul T points out, the issue was addressed by Eugene Wigner in a famous essay, "The Unreasonable Effectiveness of Mathematics in the Natural Sciences" (
http://www.maths.ed.ac.uk/~aar/papers/wigner.pdf.) However, I think this description of "unreasonable effectiveness" clashes with the reality of mathematical physics. | {
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newtonian-mechanics, angular-momentum, reference-frames, momentum
Is the linear momentum added to the system really the same regardless of where you push? It seems to me that a push at the center of mass should add more linear momentum due to not adding any angular momentum, but that's apparently not the case. You are not doing the same amount of work in both cases. You can see this in two ways. 1) Total work is the change in kinetic energy, and the kinetic energy differs in the two cases (the one that rotates has an extra rotating kinetic energy term), 2) The displacement of the point where you apply the force is different, it will be larger in the rotational case, so the work will be different too. | {
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} |
$$z = (((w_1 \oplus w_2) \oplus (w_3 \oplus w_4)) \oplus ((w_5 \oplus w_6) \oplus (w_7 \oplus w_8))).$$
The benefit of the heap structure is that there are $\mathcal{O}(\log n)$ intermediate quantities that depend on any input, whereas the linear structure has $\mathcal{O}(n)$. The intermediate quantities correspond to the values of each of the parenthesized expressions.
Since fewer intermediate quantities depend on a given input, fewer intermediates need to be adjusted upon a change to the input. Therefore, we get faster algorithms for maintaining the output quantity $z$ as the inputs change. | {
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"openwebmath_score": 0.6590133905410767,
"tags": null,
"url": "http://timvieira.github.io/blog/"
} |
c, interpreter, brainfuck, c99
buffer = calloc(1, lSize+1);
if(!buffer) {
fclose(fp);
fputs("memory alloc fails",stderr);
exit(IO_ERR);
}
if(fread( buffer , lSize, 1 , fp) != 1) {
fclose(fp);
free(buffer);
fputs("entire read fails",stderr);
exit(IO_ERR);
}
fclose(fp);
SourceFile sourceFile = {
buffer,
lSize
};
return sourceFile;
}
// check validity of a brainfuck program
bool checkSourceFileValidity(const SourceFile* sourceFile){
/*
Keep track of how many lines we have scanned and the amount of chars in those lines
*/
size_t newlines = 0;
size_t lineCharsScanned = 0; | {
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"tags": "c, interpreter, brainfuck, c99",
"url": null
} |
python, sql, database, sqlite
# if cost data is proposed, update the cost data of the paper
if days_cost is not None:
for day_id, cost in enumerate(days_cost):
connection.execute(
"UPDATE papers_days_cost SET cost = ? WHERE paper_id = ? AND day_id = ?;",
(cost, paper_id, day_id)
)
connection.commit()
return True, f"Paper {paper_id} edited."
return False, "Something went wrong."
## delete an existing paper
# do not allow if the paper does not exist
def delete_existing_paper(paper_id: int) -> tuple[bool, str]:
with connect(DATABASE_PATH) as connection:
# get the IDs of all papers that already exist
paper = connection.execute(
generate_sql_query('papers', columns=['paper_id'], conditions={'paper_id': paper_id})
).fetchone() | {
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"tags": "python, sql, database, sqlite",
"url": null
} |
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