text stringlengths 1 1.11k | source dict |
|---|---|
@Fedor: perhaps submodularity is hiding somewhere? – Suvrit Nov 17 '10 at 22:03
@Fedor: It is possible to view (an equivalent form of) Tutte's Linking Theorem as a special case of the Matroid Intersection Theorem, which is a min-max formula. I believe this is the approach that the encyclopedia (Schrijver) takes. See my answer to this question for a statement and application of matroid intersection. – Tony Huynh Nov 18 '10 at 11:57
@Tony: I do not see how to does it follow from the matroid intersection, or matroid union, or Rado transversal theorem or anything in such flavour. Would you please give a hint? – Fedor Petrov Nov 19 '10 at 19:08
@Fedor: If C is a maximum size common independent set of M / A \B and M \A /B, then C in fact gives equality in the min-max formula that you wrote – Tony Huynh Nov 19 '10 at 19:39 | {
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"openwebmath_score": 0.8949562907218933,
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"url": "http://mathoverflow.net/questions/46361/menger-theorem-via-matroids?sort=newest"
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A bench if either the team captain permutations Definition } is there a name for this type of are. 26 Red Lion Square London WC1R 4HQ, 2 ) when additional restrictions are that we include... Such permutation that fits is: { 3,1,1,1,2,2,3 } is there an algorithm count., will be 24 – 12 or 12 certain restrictions imposed, the of... Answers this page is on permutation and Combination 5 – 1 ) ( )! Surprisingly difficult Each condition: a. without restrictions ( solutions ) Date RHHS... 5 ) its content is subject to our Terms and Conditions allowed to be created using digits! Girls Boys = 5 of problem ABC and CBA are not same as order. On permutation and Combination '' as the order of arrangement is different 4 types the teacher! | {
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"openwebmath_score": 0.45561349391937256,
"tags": null,
"url": "http://justwilliet.com/andy-fordham-fwt/06i4428.php?page=d5d024-saris-sentinel-3-bike-bike-rack-220355"
} |
c#, performance, multithreading
Title: Sum of 2 arrays using multi threading I'm trying to implement sum of 2 array using multi threading.
I expected that when I use more threads it will be faster than when I use one thread, but that isn't the case.
public static int[] a = new int[100];
public static int[] b = new int[100];
public static int[] c = new int[a.Length];
public static int numberofthreads;
static void Main(string[] args)
{
Random r = new Random();
for (int i = 0; i < 100; i++)
{
a[i] = r.Next(5, 20);
b[i] = r.Next(10, 30);
}
Console.WriteLine("Enter the Number of threads to use: ");
numberofthreads = int.Parse(Console.ReadLine());
var watch = new System.Diagnostics.Stopwatch();
watch.Start();
Thread[] threadsarray = new Thread[numberofthreads];
List<Thread> threads = new List<Thread>(); | {
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"url": null
} |
volcanic-hazard, reference-request, seismic-hazards, natural-disasters
Let's say that the pollution level increased in USA and resulted in a catastrophe. Could we say that higher level of pollution and catastrophe occurred in USA can cause another catastrophe in somewhere else? I'll add to Ben MS's answer by mentioning the 2010 eruption of Eyjafjallajökull in Iceland, that shut down air traffic for quite some time.
Another example that comes to mind are tsunamis. The 2004 tsunami in the Indian Ocean mostly affect South East Asia: Indonesia, Thailand, Sri Lanka, and the eastern part of India. But other areas were affected as well, for example in Africa: Somalia, Kenya and the neighbouring Yemen. Here's a map: | {
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"tags": "volcanic-hazard, reference-request, seismic-hazards, natural-disasters",
"url": null
} |
velocity, rocket-science, propulsion
Title: Does an ion thrust engine consume more energy as it speeds up? This question goes to a very basic non-understanding of mine that I have had in the back of my mind for ages - I just read the following here:
ion thrusters are capable of propelling a spacecraft up to 90,000
meters per second (over 200,000 miles per hour (mph). To put that into
perspective, the space shuttle is capable of a top speed of around
18,000 mph. The tradeoff for this high top speed is low thrust (or low
acceleration). Thrust is the force that the thruster applies to the
spacecraft. Modern ion thrusters can deliver up to 0.5 Newtons (0.1
pounds) of thrust, which is equivalent to the force you would feel by
holding nine U.S. quarters in your hand. | {
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"tags": "velocity, rocket-science, propulsion",
"url": null
} |
algorithms, algorithm-analysis, divide-and-conquer, multiplication
So my question is, what's so special about 2x2? Why not break the original matrix multiplication to 3x3 matrix multiplication. Surely, if we can make the multiplication in 21 multiplications or less, then we would have a better result than dividing it to 2x2. And then why stop at 3x3, when you can divide the problem to 4x4, or in general axa? There is nothing special about 2×2 matrices. In fact you can do much better using larger matrices. The reason that you are only being explained the 2×2 algorithm is that it is simple to describe. The special thing about 2×2 matrices are that they are the smallest square matrices which are larger than 1×1.
You can read more in this recent survey by Dumas and Pan on fast matrix multiplication, which concentrate on practical algorithms.
From a pure mathematical point of view, the following theorem is known: | {
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"tags": "algorithms, algorithm-analysis, divide-and-conquer, multiplication",
"url": null
} |
c#, asp.net
/// <summary>
/// PopulateControls
/// </summary>
/// <param name="cust"></param>
private void PopulateControls(Customer cust)
{
ddl_title.SelectedValue = cust.Title;
txtb_firstName.Text = cust.FirstName;
txtb_lastName.Text = cust.LastName;
txtb_postion.Text = cust.Postion;
ddl_gender.SelectedValue = cust.Gender.ToString();
ddl_company.SelectedValue = cust.CompanyID.ToString();
btn_add.Visible = false;
} | {
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"url": null
} |
integrate to 0 over a full period, ie: Z. The Fourier series of the full wave rectified sine wave is (from here): The DC component has magnitude 2A/π, while the first AC component has magnitude 4A/3π. Square Wave. , the 0 th Fourier Series Coefficients) is a 0 =0. 33 percent Fourier series 50 harmonics. For half-wave rectifier, it is about 1. There is always an inherent phase difference between a sinusoidal input and output (response) for a linear passive causal system. Hence, Fourier series is the sum of these special trigonometric functions [5, 10, 12]. 1 Phase Half Wave Rectifier ( R-E and RLE Load) 10:04 mins. Fitting a single sine wave to a time series 4. Find its Fourier Series coefficients. The output of a full-wave input is the absolute value of its input, shown in Figure 15. of Fourier Coefficients you want to calculate //f=function which is to be approximated by Fourier Series // //*Some necessary. A sawtooth wave represented by a successively larger sum of trigonometric | {
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"openwebmath_perplexity": 733.3904003333771,
"openwebmath_score": 0.8957338929176331,
"tags": null,
"url": "http://chiavette-usb-personalizzate.it/opdw/rectified-wave-fourier-series.html"
} |
The analysis equation is designed to expose the damping value as a new addition to the 1st-order filter (which lacks it.) All 2nd-order low pass filters can be analyzed using it, keeping in mind that $$\\omega_{_0}=1\$$ at the time. So that's why you see the value $$\d=1.414\$$ for a Butterworth. They are presuming that you understand that this is in the context of $$\\omega_{_0}=1\$$ and that you know how to shift the values to get the cutoff frequency you want while keeping the damping factor the same. You can set these details independently. Which is an important facet of this process. | {
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"lm_q2_score": 0.8397339676722393,
"openwebmath_perplexity": 1890.6398661354,
"openwebmath_score": 0.8112914562225342,
"tags": null,
"url": "https://electronics.stackexchange.com/questions/367767/sallen-key-butterworth-response"
} |
thermodynamics, heat-transfer, cooling, heat-exchanger
Each radiator will have a temperature gradient across it. (Here I'm talking about how much the temperature of each fluid changes as it passes through the heat exchanger.) If you hook them in parallel, each one will be getting 1/N of the flow, but they will all have the same temperature gradient from input to output.
If you hook them in series, all of the flow will be going through all of them, but each one will have only roughly 1/N of the overall temperature difference across it — with the hottest one also having the highest differential, because it's transferring more heat to the other fluid.
Note that you can make this "series or parallel" decision independently for each of the two fluids. There is a total of four different ways you can configure them. | {
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"tags": "thermodynamics, heat-transfer, cooling, heat-exchanger",
"url": null
} |
#### Orodruin
Staff Emeritus
Homework Helper
Gold Member
2018 Award
My lecturer said that coordinatesystem is orthonormal if it is normalized and orthogonal, but I do not understand, which conditions a coordinatesystem must satisfy to be normalized.
This is not an appropriate way to describe a coordinate system. A set of coordinates being orthogonal means that its coordinate lines intersect at right angles. A set of vecors can be orthonormal and this is the case for the normalised basis vectors of orthogonal coordinates.
If you require the coordinate lines to have orthonormal tangent vectors, then you are restricting yourself to coordinate systems that preserve the form of the metric to be the Kronecker delta. This restricts coordinate transformations to translations and rotations and we already have a name for such coordinates: Cartesian coordinates.
#### olgerm | {
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"openwebmath_score": 0.9271495938301086,
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"url": "https://www.physicsforums.com/threads/how-are-basis-vectors-defined.956270/"
} |
java, object-oriented, interview-questions, csv
public FailedToFindPersonException() {
}
}
Looking at my code the areas I don't like / I'm not sure about are:
PeopleList doing too much: should I have just kept that class simple and added the methods for working with the list in the main method in ManipulateDataApplication?
Multiple constructors in Person
I wanted to add a generic counter method instead of the specific one I added but I ran out of time
I like catching exceptions and converting them to custom exceptions but maybe the way I have done it is overkill and I don't like my check for an empty peopleList then throwing a custom exception but I'm just not sure what a better alternative is. Design Princples and Patterns
Other than learning on the job I've read Clean Code
Uncle Bob would wag his finger at you, then open up that book to SRP. :D | {
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contribute more than others. Organization Weighted Regressions (Medians) As noted in the Methodology section, the Mercer regression methodology is "regressed market results. It is useful in many situations e. Click this link to view mean and median prices for the last 12 months. Weighted averages are frequently used when calculating a final grade for a specific course, as the final comprehensive exam usually counts more toward the course grade than each of the chapter tests. Inverse distance weighted interpolation is commonly used. Weighted blankets are the trendiest way to help poor sleepers get a good night's rest. w (sum of squares of weights). Example of using the Calculator to calculate the weighted mean and the standard deviation Learn more about Minitab 18 If you know the formula for a statistic, you can use the Calculator to calculate it. Sales-weighted fuel-economy rating (window sticker) of purchased new vehicles for October 2007 through December 2017. In the same manner as | {
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"openwebmath_score": 0.7859612703323364,
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"url": "http://agenziafunebretassoni.it/rxix/weighted-mean.html"
} |
taxonomy, ornithology, species
pp. 59–60. Several genetic studies of geese, including recent work with mitochondrial DNA (van Wagner and Baker 1986, Shields and Wilson 1987, Quinn et al. 1991, Paxinos et al. 2002, Scribner et al. 2003) have verified previous suggestions based on differences in voice, nesting habits, habitat, and timing of migration, as well as in color and size (e.g. Brooks 1914, Aldrich 1946, Hellmayr and Conover 1948), that the forms treated as the single species Branta canadensis by all previous AOU Check-lists and most other works actually constitute at least two species, and further that each of the two species may be more closely related to another member of the genus than to each other. Thus, we divide B. canadensis by recognizing a set of smaller-bodied forms as the species B. hutchinsii, and rearrange our representatives of the genus in the sequence bernicla, leucopsis, hutchinsii, canadensis, sandvicensis. Additional analysis may result in further splitting. | {
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electromagnetic-radiation, atmospheric-science
Light at longer wavelength infrared radiation interacts with various vibrational modes of multi-atomic molecules. These interactions result in the multiple peaks and valleys in your graphic between 1 and 30 µm. The atmosphere is a very good absorber of even longer wavelengths, 30 µm to 30 mm. Another series of vibrational interactions take place between 30 mm and 30 cm, resulting in another suite of peaks and valleys in that graphic.
The atmosphere is more or less transparent to radiation with a wavelength between 30 cm and 10 meters. These longer wavelengths represent another window to the universe, the radio window. This window closes at very long wavelengths. The ionosphere reflects that very long wavelength radiation back into space. | {
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natural-satellites, neptune
b) Neptune exerts tides onto the moons. Typically no moons can exist within the Roche limit. How close a moon can exist depends on the rigidity of the body, though. Big moons (mass $M_m$) cannot exist as far inward (distance $d$ to Neptune) as a small moon (Neptune's radius $R_N$ and mass $M_N$):
$$ M_m = \left(\frac{R_N}{d}\right)^3\cdot2M_N$$ or easier expressed as ratio of the moon's radius $r_{moon}$ and the moon's distance to Neptune $d$ as well as its density $ \varrho_{moon}$: $$ \left(\frac{r_{moon}}{d}\right)^3\cdot\varrho_{moon} = constant $$
Thus the further out, the larger the moon can be without falling apart due to tidal forces. Wikipedia has a page on Neptune's moons and references which indicates that the inner moons are somewhere at the tidal limit for bodies which are of somewhat granular nature as is expected for at least the smaller moons or maybe even all with the exception of Triton. | {
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python, formatting
Insights regarding options to this issue, or more general comments would be appreciated. The line in question is flagged with a comment. Apologies, if I am not supposed to included doc information with posted scripts.
"""
Script: empty_tests.py
Modified: 2015-05-25
Purpose:
checks on objects that I use, that have a __len__ property
Notes:
- collections.OrderedDictionary and other classes in collections behave
in a similar fashion
- for NumPy arrays, use size property rather than __len__
"""
import numpy as np
import collections
c0 = collections.Counter()
c1 = collections.Counter([0])
objs = [ [],[1],(),(1),{},{1:"one"},"","1",None,True,1,False,0,c0,c1 ]
is_empty = [ True if not i else False for i in objs ]
t = [ type(i).__name__ for i in objs ] # correct based on comment
#t = [str(type(i)).split("\'")[1] for i in objs ] # line in question
print("\n{:<15} {:<6} {:<10}".format("Object","Empty","Type"))
for i in range(len(objs)): | {
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## Running the Simulation
The simulation can be run from the command line; within the directory examples/tutorial_1/ execute python3 -m opencmp config.
Several messages will print out to warn that default values are being used for some of the commonly specified configuration file parameters. A message will also print out at the end of the simulation when saved results are converted to .vtu files.
To visualize the simulation results go to the newly created “output” subdirectory and open “transient.pvd” in ParaView. | {
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"url": "https://opencmp.io/tutorials/tutorial_1.html"
} |
biochemistry, proteins, synthetic-biology, prion
Title: What's the difference between prions and prion-like proteins? If I added a prion domain to a protein, does that make the protein a prion-like protein or would it be considered a prion at that point?
I'm trying to understand what prions are, how they aggregate and what macroscopic shapes they conform to (i.e. a ring vs smaller micelles).
Thank you very much! My understanding is that the principal quality of a prion is that it behaves like a prion in cells- the logic is a little circular.
A few of these criteria for acting like a prion (in yeast, where prions are most common): | {
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} |
algorithms, time-complexity, binary-trees
$
f(v) = \begin{cases}
v & \text{if}\quad \texttt{height}(v.left) = \texttt{height}(v.right) \\
f(v.left) & \text{if}\quad \texttt{height}(v.left) > \texttt{height}(v.right) \\
f(v.right) & \text{if}\quad \texttt{height}(v.left) < \texttt{height}(v.right)
\end{cases}
$
If the heights of the children of a node $v$ are the same, then clearly $v$ is the LCA of the deepest nodes of the subtree rooted at $v$. If the left subtree is taller, then we want the LCA of the deepest nodes of the subtree rooted at $v.left$, since they are deeper than the deepest nodes in the subtree rooted at $v.right$. The same logic follows for $v.right$ when it is taller.
The values for $\texttt{height}$ and $f(v)$ can be computed in a post-order traversal of $T$ in linear time.
Calling $f(root)$ should return the LCA of the deepest nodes in the tree. | {
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the-sun
base_cmd = """
MAKE_EPHEM = 'YES'
OBJ_DATA = 'NO'
COMMAND = '301'
EPHEM_TYPE = 'ELEMENTS'
CAL_TYPE = 'GREGORIAN'
CSV_FORMAT = 'YES'
"""
def fetch_data(center, plane, start, stop, step, verbose=False):
cmd = f"""
!$$SOF
{base_cmd}
CENTER = '{center}'
REF_PLANE= '{plane}'
START_TIME = '{start}'
STOP_TIME = '{stop}'
STEP_SIZE = '{step}'
!$$EOF
"""
#print(cmd)
req = requests.post(url, data={'format': 'text'}, files={'input': ('cmd', cmd)})
version = re.search(r"API VERSION:\s*(\S*)", req.text).group(1)
if version != api_version:
print(f"Warning: API version is {version}, but this script is for {api_version}")
m = re.search(r"(?s)\\\$\\\$SOE(.*)\\\$\\\$EOE", req.text)
if m is None:
print("NO EPHEMERIS")
print(req.text)
return None
if verbose:
print(req.text)
else:
lines = req.text.splitlines()
print("\n".join(lines[5:16]))
return m.group(1)[1:] | {
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digital-communications
The only caveat that needs to be stated is that baseband data
communication will always have a zero imaginary component, while for
bandpass communication the imaginary component of the complex envelope
might be nonzero. | {
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"tags": "digital-communications",
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} |
cosmology, stars, error-analysis
Title: Age of the universe and age of stars The age of the universe is 13.798±0.037 billion years, yet the age of HD 140283 is 14.46±0.8 billion years, how this can be the case? It's down the fact that different properties are used to calculate the ages and if you look at the margin of error in the calculations they're not incompatible.
The lower limit for the age of HD140283 is:
14.46 - 0.8 = 13.68 billion years
which is within the range for the age of the universe.
Once better measurements of HD140283 are made it's age will probably be revised to a more "sensible" figure that matches more closely the age of the universe. | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "cosmology, stars, error-analysis",
"url": null
} |
python, python-3.x
d != "rien"
):
print("!!!DESOLE la session est deja occupee!!! ")
elif (
x == '2' and y == '2' and z == '2'
):
if (
e == "rien"
):
e=E2
elif (
e != "rien"
):
print("!!!DESOLE la session est deja occupee!!! ")
elif (
x == '2' and y == '2' and z == '3'
):
if (
f == "rien"
):
f=E2
elif (
f != "rien"
):
print("!!!DESOLE la session est deja occupee!!! ")
elif (
x == '2' and y == '3' and z == '1'
):
if (
g == "rien"
):
g=E2
elif (
g != "rien"
): | {
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"url": null
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along its arc Lc Length between any two points on a circular curve R Radius of a circular curve ∆ Total intersection (or central) angle between back and forward tangents. They relate areas between. 1 Area of a Region Between Two Curves 447. Area Between Two Curves Calculator. be the vertical distance between the graphs of. A Diophantine equation is a polynomial equation in two or more unknowns for which only the integer solutions are sought (an integer solution is a solution such that all the unknowns take integer values). 1, the graphs of both and lie above the axis, and the graph of. solution curves for. Here is a set of practice problems to accompany the Area Between Curves section of the Applications of Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. Suppose the region is bounded above and below by the two curves and , and on the sides by and. As a measure for the resemblance of curves in arbitrary dimensions we consider the so-called | {
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"tags": null,
"url": "http://vbsu.carlosfassanelli.de/area-between-two-curves-problems-and-solutions-pdf.html"
} |
general-relativity, black-holes, event-horizon, hawking-radiation, black-hole-thermodynamics
$$\dfrac{dr}{d\nu}<-\dfrac{1}{2}\bigg(1-\dfrac{2M(\nu)}{r}\bigg)\tag{1}$$
Now, suppose that the trajectory of a particle (which, of course, is supposed to be either a lightlike or a timelike trajectory) connects two events--one outside the horizon and one inside the horizon. Also, let's take these events to be very close to the horizon. Then, the coordinates of these events can be taken as $$\Big(\nu, 2M(\nu)+ \delta \xi_{o}\Big)$$ and $$\Big(\nu+d\nu, 2M(\nu+d\nu)-\delta\xi_{i}\Big)$$ where $\delta\xi_{i}$ and $\delta\xi_{o}$ are infinitesimally small parameters that we can change to make the chosen events as close to the horizon as we wish.
Now, for the pair of these two events
$$\dfrac{dr}{d\nu}=\dfrac{2M(\nu+d\nu)-\delta\xi_{i}-2M(\nu)-\delta\xi_{o}}{d\nu}=2\dot{M}(\nu)-\bigg(\dfrac{\delta\xi_{i}+\delta\xi_{o}}{d\nu}\bigg)\tag{2}$$
Combining $(1)$ and $(2)$, we get that our stipulation that a particle connects the considered two events can be true only if | {
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> int(x+y+x^2*dy, x); lprint(%);
(1/2)*x^2-(1/4)*(2-2*x)*(2*x-x^2)^(1/2)-(1/2)*arcsin(-1+x)+(1/3)*x^2*(2*x-x^2)^(1/2)+(1/3)*x*(2*x-x^2)^(1/2)+(2*x-x^2)^(1/2)
This is copied and pasted directly from the Maple 11 worksheet
Thank you both for your responses. I see now that I was dropping the x^2 and keeping the y'(X) term. *sigh*
Simon Bridge | {
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"url": "https://www.physicsforums.com/threads/integral-over-a-curve.836174/"
} |
• Is it possible to solve this modular equation (assuming there is a solution) in general given these restrictions?
• Suppose that for a given $p$ there is a solution $k$ such that $k^2+k+1\equiv 0\pmod{2^p-1}$; then $k\cdot(k+1)\equiv -1\pmod{2^p-1}$. Then (as noted in a comment) $k^3\equiv -(k+1)^3\equiv 1\pmod{2^p-1}$. Is it possible to conclusively prove whether $k$ or $k+1$ is a quadratic non-residue from this information?
-
For any $p$, $(2^p-1,2^n)=1$, in particular for $n=1,2$ so you can write $$m^2=(i+2^{-1})^2=-3\cdot 4^{-1}\mod 2^p-1$$ and it becomes a problem w.r.t to residues $\mod 2^p-1$. – Pedro Tamaroff Feb 6 at 16:54
I don't know if you used 'suppose further' to imply that it was an extra hypothesis needed, but it's easy to show that you must have $k^3\equiv 1$; $k^3-1=(k^2+k+1)(k-1)\equiv 0$. – Steven Stadnicki Feb 6 at 16:54 | {
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"url": "http://math.stackexchange.com/questions/666269/is-it-possible-to-solve-i2i1-equiv-0-pmod2p-1-in-general"
} |
c#, wpf, active-directory
else if (data is DirectReports)
{
var directReports = data as DirectReports;
if (directReports.Reports == null) continue;
userPrincipal = directReports.User;
foreach (var directReport in directReports.Reports)
{
dynamic result = new ExpandoObject();
AddAttributesToResult(
null, null, directReport, userPrincipal, result);
results.Add(result);
}
directReports.Dispose();
}
else
{
var computerPrincipal = data as ComputerPrincipal;
groupPrincipal = data as GroupPrincipal;
userPrincipal = data as UserPrincipal;
dynamic result = new ExpandoObject();
AddAttributesToResult( | {
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But I claim that also player II cannot have a winning strategy. If there were a winning strategy for player II, then we could use that strategy to generate a countable subset of $X$, since it provides a systematic way to extend any particular finite sequence of distinct elements of $X$. Just have player I play that finite sequence, and then $a$, and player II will respond with a fresh element of $X$. Since $X$ has no countably infinite subsets, there can be no such strategy.
So the game is not determined. QED
This game is related to the classical fact that clopen determinacy implies the axiom of choice, since in the comparatively trivial game, where player I plays a nonempty set from a fixed collection, and player II playing an element of that set (and winning upon doing so), it is clear that the first player cannot have a winning strategy, but a winning strategy for player II provides a choice function. | {
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"tags": null,
"url": "https://mathoverflow.net/questions/269168/is-determinacy-on-an-infinite-dedekind-finite-set-consistent"
} |
java, object-oriented, game, mvc, swing
Title: MVC Layout - Which way to add listeners is better? So I'm doing a basic MVC layout for a pretty basic game that I am making. The game requires the user to move up/down/left/right via buttons on the GUI. Since I'm using an MVC layout and my buttons are in a different class than the ActionListeners, I was wondering what the best way to add the action listeners are?
Method 1:
View Class ActionListener method:
public void addMovementListeners(ActionListener u, ActionListener d, ActionListener l, ActionListener r){
moveUp.addActionListener(u);
moveDown.addActionListener(d);
moveLeft.addActionListener(l);
moveRight.addActionListener(r);
} | {
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neuroscience, biophysics
To counteract this discharge the Sodium/Potassium pumps expends energy in the form of ATP in order to recharge the batteries. The amount of current that the pumps moves is very small relative to the larger currents of sodium rushing in and the potassium rushing out during an Action Potential. Thus when a neuron fires there is very little overall current contribution from the pumps.
When the neuron is at rest (not firing) then the Sodium/Potassium pump is able to recharge the membrane and the current becomes important. Still this is a very slow process, relative to action potentials.
Another thing to note, even though the cell is at rest the batteries are discharging slightly. Their is a leakiness to the membrane, sodium leaks in and potassium leaks out. Thus the pump's current will counteract these leaky currents at rest, keeping the batteries charged and thus the overall net current across the membrane at rest will be zero. | {
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there's no need to shout that
ok :)
@Silent yes
6:02 AM
they span K_1 K_2 because they include the generators $\alpha_i$, $\beta_j$
so the ring they generate contains $F[\alpha_i, \beta_j]$
which is $K_1 K_2$
@Silent Yeah
Well, $\sum a_{ij}\alpha_i\beta_j$
sorry
It's a field, it contains $\alpha$ and $\beta$, and anything that contains $\alpha$ and $\beta$ contains $\sum a_{ij}\alpha_i\beta_j$
Therefore it's the field generated by $\alpha$ and $\beta$
Are we implicitly assuming that $K_1,K_2$ are contained in some larger field $L$? Because otherwise we need to check closure under inversion as well, perhaps
6:06 AM
leaky, does tensor product help in field theory?
@AkivaWeinberger oh yes
If $A=\sum a\alpha\beta$ then $x\mapsto Ax$ is an $F$-linear map
and we have a generating set with $mn$ things
which means there's a basis with at most $mn$ things
so it's a finite-dimensional space
so that's a surjective map
so $1=Ax$ has a solution
so $1/A$ is in our set
thank you!! | {
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"tags": null,
"url": "https://chat.stackexchange.com/transcript/36/2019/3/27"
} |
quantum-field-theory, particle-physics
In QFT, you have to abandon the notion that there is a real value for the field at every point - that would be the classical picture of the field just being the solution of the classical equation of motion. All you can have are expectation values, and the idea that there is some kind of field actually being "excited" is...ill-defined. It isn't clear at all what one would want to say with that. It's just a cute picture that many like, not a mathematically rigorous statement.
Of course, the time evolution will act differently on many-particle states than it will on one-particle states. But these states are states in the Hilbert space of the theory, and only by analogy (since they are created by Fourier components of free fields) identified with "excitations", an analogy that breaks down if we would not consider the asymptotic Fock space but instead the full (but sadly, mostly unknown) Hilbert space of the interacting QFT. | {
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IBVP for a heat equation on ${x>0,t>0. The result of these inverse transforms is the solution of the IBVP. The direct and ISPs for a time fractional diffusion equation with nonlocal boundary condition involving a parameter have been investigated. Viewed 254 times 1. In this section, we discuss the initial boundary value problems (IBVPs for short) for wave equation. Thus the heat equation takes the form: = + (,) where k is our diffusivity constant and h(x,t) is the representation of internal heat sources. The temper-ature distribution in the bar is u. In Chapter 2 we review the well known solutions of IBVP's for the standard Heat equation on the semi-infinite line. IBVP: Heat Equation. 8 BVPs in Polar Coordinates 396 Chapter 13 Solution of the Heat IBVP in General 407 13. Nonlinear problems for the one-dimensional heat equation in a bounded and homogeneous medium with temperature data on the boundaries x = 0 and x = 1, and a uniform spatial heat source depending on the heat flux (or the | {
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"openwebmath_score": 0.7227175831794739,
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"url": "http://wasb.aquatique-freiburg.de/ibvp-heat-equation.html"
} |
quantum-mechanics, homework-and-exercises, potential, perturbation-theory, atoms
e^x\int_{\lambda x/v}^{\infty} dt e^{i\omega_n t}e^{-v t/\lambda}
\right)\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\left(\frac{\lambda}{v+ i\omega_n\lambda}+
\frac{\lambda}{v- i\omega_n\lambda}
\right)\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\frac{2\lambda v}{v^2+\omega_n^2\lambda^2}\right|^2\\
&\vdots
\end{aligned} This should get you on your way! | {
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$\displaystyle \frac{450}{a}=-150$
$\displaystyle a=-\frac{450}{150}$
$\displaystyle a=-3.0\:km/s^2$
Therefore, the magnitude of the acceleration should be $\displaystyle a=3.0\:km/s^2$
## Calculating the unit and value of k given a velocity function| University Physics
### PART A. Determine the units of k in terms of m and s.
ANSWER: $m/s^3$
We know that the unit for velocity is m/s and the unit for time is s. Therefore,
$v=kt^2$
$m/s=k\left(s\right)^2$
$k=\frac{m/s}{s^2}$
$k=m/s^3$
### PART B. Determine the value of the constant k.
ANSWER: $k=49.8\:m/s^3$
$v_x=\frac{dx}{dt}$
$dx=v_xdt$
$\int \:dx=\int \:v_xdt$
$x=\int \:kt^2dt$
$x=\frac{kt^3}{3}+C$
Solve for C using the pairs $x=-7.90\:m,\:\:t=0\:s$
$C=-7.9\:$
The position function therefore is
$x\left(t\right)=\frac{kt^3}{3}-7.90$
Solve for k using the pair $x=8.70\:m,\:t=1.00\:s$
$8.70=\frac{k\left(1\right)^3}{3}-7.90$
$\frac{k}{3}=8.70+7.90$
$k=3\left(16.6\right)$
$k=49.8\:m/s^3$ | {
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"url": "https://engineering-math.org/tag/randall/"
} |
c#, generics, inheritance
public int Get(TEnum type)
{
return this[type];
}
public void Set(TEnum type, int value)
{
this[type] = value;
}
public int Compute(ModificationType modificationType, TEnum type, int value)
{
switch (modificationType)
{
case ModificationType.Increase:
return (int)(value * (1.0f + (Get(type) / 100f)));
case ModificationType.Decrease:
return (int)(value - ((value * (Get(type) / 100f))));
default:
throw new ArgumentOutOfRangeException(nameof(modificationType), modificationType, null);
}
}
private int Clamp(int value)
{
return (value < MinimumModification) ? MinimumModification : (value > MaximumModification) ? MaximumModification : value;
} | {
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cosmology, metric-tensor, space-expansion, distance, redshift
Does this mean that the distance in the Hubble law is actually not a physical distance that one would measure with a ruler? The physical distance would be obtained by using the relations above which I have worked out!?
Hubble's law, in the sense $v = H d$, is only valid when the distances are small enough that $z \ll 1$, in which case all the distance metrics are approximately the same. However you are right that to measure the acceleration of the Universe (which goes beyond Hubble's law), you have to be more precise The supernova measurements which were the first evidence of the acceleration of the Universe used the luminosity distance.
Finally, as a friendly tip for the future: in general it's a good idea to ask focused questions with 1 or at most 2 questions per post. For little niggling doubts about definitions, often doing a few problems is a good way to build confidence and clean up confusion, and filter for larger conceptual issues that are harder to answer on your own. | {
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} |
quantum-mechanics, momentum, wavefunction, fourier-transform
Title: Momentum Wave Function gives strange expectation values Suppose there's a particle with the wave function $\psi(x)=\frac{1}{\sqrt{L}}$ for $0<x <L$ and 0 everywhere else.
One way to get the associated Momentum Wave function is direct integration on $1/\sqrt{L}$.
$$\phi(k)=\frac{2e^{ikL/2}\sin{kL/2}}{k\sqrt{2\pi L}}$$
The first x derivative of the space wave function is zero so I'm thinking that makes the momentum expectation value zero. That's consistent with the expectation value calculated using the momentum wave function. It becomes necessary to integrate an anti-symmetric function over a symmetric interval.
Things break down for the expectation value for the expectation value of the square of the momentum. The second derivative of the spatial wave function is zero.
The integral you need to find the expectation value of the square of the momentum in momentum space is
$$ \int_{-\infty}^\infty \frac{4\sin^2({kL/2})}{2\pi L}dk$$
This integral does not converge. | {
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Now, for each $x\in[0,1]$, we have \begin{align} |f(x)-p_n(x)| &= \Big|\mathbb{E}\big[f(x)-f(\overline{X}_n(x))\big]\Big| \\ &\leq \mathbb{E}\big|f(x)-f(\overline{X}_n(x))\big| \\ &< \underbrace{\mathbb{P}\big(|\overline{X}_n(x)-x|<\delta\big)}_{\leq 1} \frac{\varepsilon}{2} + \underbrace{\mathbb{P}\big(|\overline{X}_n(x)-x|\geq\delta\big)}_{ \text{via Chebyshev's} }\,c \;. \end{align} By Chebyshev's inequality, we have $$\mathbb{P}\big(|\overline{X}_n(x)-x|\geq\delta\big) \leq \frac{\mathrm{Var}[\overline{X}_n(x)]}{\delta^2} = \frac{x(1-x)}{n\delta^2} \leq \frac{1}{4n\delta^2} \;,$$ which is smaller than $\frac{\varepsilon}{2c}$ for $n\geq\frac{c}{2\varepsilon\delta^2}$. It follows that $|f(x)-p_n(x)|<\varepsilon$ for all $x\in[0,1]$, provided $n\geq \frac{c}{2\varepsilon\delta^2}$, and this concludes the proof. | {
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} |
It is convenient to write $c=ab$. A simple argument shows that we can assume $m\leq n\leq k$ and $d=k+2$. [The equation $b*a=bcb^{-1}$ shows we can swap $m$ and $n$ as $|bcb^{-1}|=k$. The equation $a^{-1}c=b$ shows we can swap $n$ and $k$ as $|a^{-1}|=m$. Thus we may assume $m\leq n\leq k$.]
It is easy to prove the result for small cases such as $m=n=2$. With these simple ideas Stefan's table of data can be simplified, and extended. It seems that there may be results already in the literature. Can an expert help? What about the special case when $m,n,k$ are each powers of the same prime? | {
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"url": "https://mathoverflow.net/questions/118092/order-of-products-of-elements-in-symmetric-groups/397646"
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java, regex, performance
Pattern pattern = Pattern.compile(regex);
int[] stack = new int[options.length];
boolean done = false;
List<List<SymbolChoice>> listOfSymbolChoices = new ArrayList<List<SymbolChoice>>();
StringBuilder sb = new StringBuilder(stack.length);
while (!done) {
sb.setLength(0);
// check the stack as a match.
for (int i = 0; i < stack.length; i++) {
sb.append(options[i][stack[i]]);
}
if (pattern.matcher(sb.toString()).matches()) {
List<SymbolChoice> match = new ArrayList<>(stack.length);
for (int i = 0; i < stack.length; i++) {
match.add(options[i][stack[i]]);
}
listOfSymbolChoices.add(match);
}
// increment the stack
int depth = stack.length;
done = true;
while (--depth >= 0) {
stack[depth]++;
if (stack[depth] > options[depth].length) {
stack[depth] = 0;
} else { | {
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#### Rule 2: An = in the definition of a symbol should be read as “is defined as.”
In other words, in ‘real’ math, ☼ is meaningless. So the GMAT testmaker has to give that symbol a definition. Paraphrasing the question above into English, we get the following: “When two numbers are connected by the ☼ sign, that operation is defined as four times the product of those two numbers, minus the first over the second.” Or perhaps more simply: “When two numbers are on either side of the ☼, we plug the first number into every x in the definition in the question stem and we plug the second number into every y in that definition.”
So if we see them complete version of the problem:
If x☼y = 4xy – x/y, for all and for all y0, then 6☼3 equals
Then the problem, despite looking intimidating at first, turns out to be quite simple.
#### Rule 3: The vast majority of symbolism problems test no math more complex than simple substitution. | {
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c#, sliding-tile-puzzle
public SliderState()
{
Grid = new List<List<char>>();
Previous = null;
}
public SliderState(List<List<char>> tGrid, SliderState tPrev = null)
{
Grid = tGrid.ConvertAll(x => new List<char>(x)); // Create a deep copy of the list
Previous = tPrev;
}
public bool Equals(SliderState other) // Two states are considered equal if their grids are identical (the ancestor state is disregarded)
{
for (int i = 0; i < Grid.Count; ++i)
{
for (int j = 0; j < Grid[i].Count; ++j)
{
if (Grid[i][j] != other.Grid[i][j])
{
return false;
}
}
}
return true;
} | {
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optimization, algorithm, haskell
type AlgoArray = IOArray Int Int -- allows for experimentation with IOUArray or others
lenarr :: AlgoArray -> IO Int -- simplified type
lenarr arr = getBounds arr >>= return . snd
fillArray :: [Int] -> IO (AlgoArray, AlgoArray)
fillArray ns = do -- could be Applicative style: (,) <$> newListArray <*> newArray
x <- newListArray (0, lcount) (2:ns)
y <- newArray (0, lcount) 0
return (x, y)
where
lcount = length ns
visit :: AlgoArray -> Int -> IO ()
visit x i = lenarr x >>= visit' x i >> putStrLn ""
where
visit' x i c
| i > c = return ()
| otherwise = readArray x i >>= putStr . show >> visit' x (i + 1) c
maj :: AlgoArray -> AlgoArray -> Int -> IO (AlgoArray, Int)
maj m a j = do
valaj <- readArray a j
valmj <- readArray m j
maj' valaj valmj
where
maj' valaj valmj
| valaj == (valmj - 1) = writeArray a j 0 >> maj m a (j - 1)
| otherwise = return (a, j) | {
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python, clustering, scikit-learn
This is equal to the Kullback-Leibler divergence of the joint distribution with the product distribution of the marginals.
KL divergence (and any other such measure) expects the input data to have a sum of 1. Otherwise, they are not proper probability distributions. If your data does not have a sum of 1, most likely it is usually not proper to use KL divergence! (In some cases, it may be admissible to have a sum of less than 1, e.g. in the case of missing data.)
Also note that it is common to use base 2 logarithms. This only yields a constant scaling factor in difference, but base 2 logarithms are easier to interpret and have a more intuitive scale (0 to 1 instead of 0 to log2=0.69314..., measuring the information in bits instead of nats).
> sklearn.metrics.mutual_info_score([0,1],[1,0])
0.69314718055994529 | {
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quantum-field-theory, quantum-spin, group-theory, group-representations, spinors
I'm being a bit fussy about the fact that $\mathfrak{su}(2)$ is a real vector space, because I want to make the following point: If someone gives you generators $J_i$ ($i=1,2,3$) for a representation of $\mathfrak{su}(2)$, you can construct a representation of the compact group $SU(2)$ by taking real linear combinations and exponentiating. But if they give you two sets of generators $A_i$ and $B_i$, then you by taking certain linear combinations with complex coefficients and exponentiating, you get a representation of $Spin(3,1)$, aka, a projective representation of $L$. If memory serves, the 6 generators are $A_i + B_i$ (rotations) and $-i(A_i - B_i)$ (boosts). See Weinberg Volume I, Ch 5.6 for details.
The upshot of all this is that complex projective irreps of $L$ are labelled by pairs of half-integers $(a,b) \in \frac{1}{2}\mathbb{Z} \times \frac{1}{2}\mathbb{Z}$. The compex dimension of the representation labelled by $a$,$b$ is $(2a + 1)(2b+1)$. | {
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first paper calculus. By-Sa 2.5 Theorem 1.2 of c that satisfy the Theorem introduction into 's... The Value ( s ) guaranteed by the Theorem on Brilliant, the plan arrives at its.! Calculus, but later changed his mind and proving this very important Theorem Theorems are some of the MVT when. And applying the Theorem on Local Extrema MATH 123 at State University of.! A f b ' 0 then there is a matter of examining cases and applying the.! On Brilliant, the largest community of MATH and science problem solvers up in finding the Value ( )! ) guaranteed by the Theorem of examining cases and applying the Theorem on Local Extrema the... Rolles Theorem with the intermediate Value Theorem in calculus and they are into... Determine whether the MVT can be applied to f on the closed interval various types the largest community MATH. The foundational Theorems in differential calculus the special case of the most important theoretical tools in calculus Theorem was proven... In this post we give a | {
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newtonian-mechanics, spacetime, reference-frames, coordinate-systems, relative-motion
These are my thoughts and I am a bit confused... So, how should we interpret this sentence? What is the meaning of relativity of space here? Yes, you can take a random point as your origin, but how do you know which random point you have picked? Imagine you are floating alone in empty space with nothing for a thousand km in any direction. If you pick a point somewhere around you at random to be your origin, how do you know where it is? The only way you can do it is to specify it in relation to something physical. You could place a pin at the spot you believe you have picked, but you would be faced with the challenge of placing it in such a way that you left it there motionless with respect to the point it was intended to mark. But how would you know you had? How would you know that the pin was not drifting away from your intended point? You couldn't know, of course, because a point in empty space is an utterly abstract notion. The only way in practice in which you can define an origin | {
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general-relativity, tensor-calculus
Proof. We will prove this at a point $p\in\mathcal{M}$ using Riemann normal coordinates. The result can be extended pointwise to all of spacetime.
We set $q_a:=A_{ab}u^b$ and by antisymmetry of $A_{ab}$ it is clear that $q_au^a=0$. Next, define $$B_{ab}:=A_{ab}-u_aq_b+q_au_b.$$ Note that $B_{ab}u^b=0$. Let $u^a,X^a,Y^a,Z^a$ be an orthonormal basis on $T_p\mathcal{M}$. Let us define at $p$ the three numbers
$$b^1:=B_{ab}Y^aZ^b,b^2:=B_{ab}Z^aX^b,b^3:=B_{ab}X^aY^b$$
and introduce the vector $$b^a:=b^1X^a+b^2Y^a+b^3Z^a$$ | {
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quantum-mechanics, quantum-information, quantum-states
Title: Why does the space of pure qudit states have dimension $2(D-1)$, rather than $D^2-2$? It is well known that two-dimensional states, that is, qubits, can be represented using the Bloch sphere: all pure states lie on the surface of a three-dimensional sphere, while all the mixed states are in its interior.
This is consistent with a simple parameter counting argument:
the space of all qubits is the space of all $2\times 2$ positive Hermitian operators with unit trace, which has dimension $2^2-1=3$, while for the pure states there are only $2\times2 - 2=2$ real degrees of freedom (two complex coefficients minus normalization and global phase).
The same conclusion can be reached by considering that the pure states are a subset of all states satisfying the single additional constraint that $\operatorname{Tr}(\rho^2)=1$.
What happens for higher dimensional qudits? Let's consider the states living in a Hilbert space of dimension $D$. | {
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velodyne
<max_angle>0</max_angle>
</vertical>
</scan>
<range>
<min>0.05</min>
<max>50.0</max>
</range>
</ray>
<plugin name="sonar_plugin" filename="libgazebo_ros_block_laser.so">
<gaussianNoise>0.00</gaussianNoise>
<alwaysOn>true</alwaysOn>
<updateRate>100.0</updateRate>
<topicName>/output_point_cloud</topicName>
<frameName>sonar_ring</frameName>
</plugin>
<always_on>true</always_on>
<update_rate>100.0</update_rate>
</sensor>
</gazebo> | {
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venus, rotation
Title: What is the current accepted theory as to why Venus has a slow retrograde rotation? According to this NASA overview, the planet Venus is unique (amongst the major planets), Venus has a slow retrograde axial rotation, taking 243 Earth days to make one rotation (which is longer than its orbital revolution).
What is the current accepted theory as to why (and how) Venus developed this anomalous slow retrograde axial rotation? There seem to be a few, and none are accepted by the whole scientific community. The main ones:
Venus was struck by a large body during its early formation
The spin axis flipped, as can happen with a gyroscope
The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle | {
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cc.complexity-theory, reference-request, lo.logic, computability
Title: Proof and computational complexity I couldn't find documents elaborating on this: if the Curry Howard correspondence is to be interpreted as establishing a strong relation between proofs and programs, should there not be a strong relation between proof and computational complexity?
I'm asking this as a reference request. Maybe the keyword you are looking for is "Implicit Complexity". It is more general than Curry-Howard correspondence, but several lines of research investigate along the axis you are interested in. You can check for instance the publications of Patrick Baillot for many references and pointers.
For a little self-promotion, here are for instance two recent papers [KPP1,KPP2] characterizing via the Curry-Howard correspondence on certain cyclic proofs the following complexity classes, depending on the logical rules incorporated or not in the cyclic proof system: | {
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quantum-mechanics, hilbert-space, operators, atomic-physics, hydrogen
$${\cal N}=\sum_{n=1}^{\infty}\sum_{l=0}^{n-1}\sum_{m=-l}^{l}n|n,l,m\rangle\langle n,l,m|,$$
for which the eigenvalues are the principal quantum numbers for the bound states and zero otherwise.
In the position space representation of this operator, each term in the sum involves the explicit Schrödinger wave function. For example, when the $n=1$ term acts on a general wave function $\Psi(\vec{r})$, it takes the form
$$\psi_{1,0,0}(\vec{r})\int d^{3}r'\,\psi^{\dagger}_{1,0,0}\left(\vec{r}\,'\right)\Psi\left(\vec{r}\,'\right)
=\frac{1}{\pi a_{0}^{3}}\exp\left(-r/a_{0}\right)\int d^{3}r'\,
\exp\left(-r'/a_{0}\right)\Psi\left(\vec{r}\,'\right).$$
You could also write this operator in a different basis of energy eigenstates (such as that obtained by solving the Coulomb problem in parabolic coordinates). | {
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general-relativity, special-relativity, acceleration, time, velocity
Title: How does the clock postulate apply in non-inertial frames? I've read countless answers and other sources on the question of whether time dilation is caused both by velocity and acceleration or only by velocity, but they all look at things only from an inertial frame and/or use the popular version of the clock postulate, so they don't seem to answer the questions I pose here. Many of them also talk about it in terms of relative lengths of different paths through spacetime, but I think that sidesteps the question of whether acceleration itself can cause those lengths to change in accelerating frames (e.g., by causing a different metric to apply). | {
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filters, noise, kalman-filters, digital-filters, stochastic
s & 0 \\
0 & s \\ \end{bmatrix} - \begin{bmatrix}
0 & 1 \\
-b & -a \\ \end{bmatrix}\right)^{-1}\right\}\\
&= \mathscr{L}^{-1}\left\{\left(\begin{bmatrix}
s & -1 \\
b & s+a \\ \end{bmatrix}\right)^{-1}\right\}
\end{align}
... (at this point I get stuck as I am using Sympy in python to calculate the inverse Laplace transform and it fails for this matrix).
But I want to know how I should treat this Stochastic effect in a Kalman filter. Uh, this is the whole point of a Kálmán filter. It has a stochastic measurement model (noisy measurements) and a stochastic process state update model and uses the measurements for estimating the current state of the stochastic process.
If the process were deterministic, you would approach perfect information over time.
So you just plug the variation of your acceleration into the covariance matrix for the state update.
The exact way to do that depends on the conventions your course material employs (lots to choose from). | {
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The function FindComplexRoots described below does exactly that. The function and some more documentation can be found in this GitHub repository.
The calling syntax is of the form
FindComplexRoots[e1==e2, {z, zmin, zmax}]
and it will attempt to find roots of the equation in the complex rectangle whose corners are zmin and zmax; its output is identical to that of NSolve and FindRoot. To do multiple equations, you then do
FindComplexRoots[{e1==e2, e3==e4, …}, {z1, z1min, z1max}, {z2, z2min, z2max}, …]
It comes with a Seeds option to control how many random seeds are used (50 by default), a SeedGenerator option to control how the seeds are created, and a few other niceties.
Let me start with a bunch of options and documentation. This includes syntax information so you'll get pretty colouring and a usage message so autocomplete templates will work in versions that have them. | {
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ros, archlinux
Title: Installing ros-indigo-desktop-full on ArchLinux: pyqt4-10-common and pyqt4-common are in conflict (pyqt-common). Remove pyqt4-common?
I can't overcome the above problem. If I start to remove packages to install pyqt4-10, too much other apps would need to be removed (some of them I really need). Does anyone have a solution for this? Manual installation on an other machine didn't really help, the dependency problem strikes me there too.
Cheers
Paul | {
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python, numpy, pandas
they are being called in a React d3 application [...] frequently
Cutting a couple of characters in your field names is premature and mis-directed optimisation. There are plenty of opportunities for actual optimisation elsewhere.
Put 273.15 into a constant rather than leaving it as a magic number. When you subtract this number, it's probably a good idea to round(); I was conservative and rounded to 12 decimals to cut the error. This is only possible because your real data stop at two decimals.
Most of your difficulty comes from the fact that your data frame is effectively rotated, and needs to be rotated again to be sane. In Pandas terminology this rotation is called stacking. You need to unstack your property names to columns, and you need to stack your time columns to indices. Once this is done, the data are much, much easier to manipulate with (nearly) stock Pandas functions.
Suggested
from pprint import pprint
import pandas as pd
ABSOLUTE_ZERO = -273.15 | {
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ros, cameras, usb
EDIT2: @kaliatech If I run the front camera at 1280x720, I can get up to 34fps with 1 CPU core maxing out at 100% at a time and the rest at more or less 25%. If I run just the down-facing camera at 640x480, I can get up to 60fps with CPU usage around 55% for 1 core while the rest are less than 20%. If I have both the down-facing cam and the whycon node running, I still have 60fps from usb_cam but CPU usage maxes out at 100% and 70% for 2 cores. Finally, if I run all three, the down-facing cam remains the same in terms of framerate but the front-facing has around 12 fps and at least 2 cores max out with the rest at above 60%. At this point, I'm not currently transferring the videos through the network. I'm still conducting some tests on the local computer, but you would be right that it will be much worse on a slower network. I will try to use a different camera driver like cv_camera but I think I skipped it since it seemed older than usb_camera and even less supported | {
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[GRAPH]ig9m08odka[/GRAPH]
It should not be hard to prove algebraically that every $\varphi$ except the first one increases unboundedly after $x=4$, and so the sequence that starts with $x_0=5$ tends to infinity.
#### evinda
##### Well-known member
MHB Site Helper
We have the algorithm:
$$x_{i+1} = \varphi(x_i)$$
Now suppose that $\varepsilon_i$ is the error in $x_i$ with respect to the actual root $x$.
That means that $x_i = x + \varepsilon_i$ and $x_{i+1} = x + \varepsilon_{i+1}$.
Then substituting gives us:
$$x + \varepsilon_{i+1} = \varphi(x + \varepsilon_i)$$
From Taylor's theorem we know that:
$$\varphi(x + \varepsilon_i) = \varphi(x) + \varphi'(x) \varepsilon_i + \mathcal O(\varepsilon_i^2)$$
And since $x$ is the root, we also know that:
$$\varphi(x) = x$$
In other words:
$$\varepsilon_{i+1} = \varphi'(x) \varepsilon_i + \mathcal O(\varepsilon_i^2)$$ | {
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"url": "https://mathhelpboards.com/threads/how-can-i-choose-which-is-the-most-suitable-method.8784/"
} |
The Product Rule is evident from the visual representation of all possible outcomes of tossing two coins shown above. The possible outcomes will be HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. Suppose we have 3 unbiased coins and we have to find the probability of getting at least 2 heads, so there are 2 3 = 8 ways to toss these coins, i. Click here👆to get an answer to your question ️ 1. So, P (A) = 4/8 = 0. So the probability of all 5 tosses coming up heads is (1/2)^5 = 1/32, about 3%. The left-hand column shows all the possible outcomes for 3 tosses of a coin. What is the probability of getting 2 heads in 3 tosses. This curve represents the probability that at most x heads will be thrown from the 100 tosses. If 50 people did this on average 0. So, to compute the p-value in this situation, you need only compute the probability of 8 or more heads in 10 tosses assuming the coin is fair. This row shows the number of combinations 5 tosses can make. Created by Sal Khan. P(H on 11th toss) = P(T | {
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"tags": null,
"url": "http://fleckundzimmermann.de/probability-of-getting-2-heads-in-5-tosses.html"
} |
ros2
[1663092212.790100] debug | SerialAgentLinux.cpp | send_message | [** <<SER>> **] | client_key: 0x77247145, len: 13, data:
0000: 81 00 00 00 0A 01 05 00 9B 01 00 00 80
[1663092212.992004] debug | SerialAgentLinux.cpp | recv_message | [==>> SER <<==] | client_key: 0x77247145, len: 32, data:
0000: 81 80 9B 01 07 01 18 00 01 A5 00 05 0B 00 00 00 74 65 73 74 5F 6C 61 62 65 6C 00 00 00 00 00 3F
[1663092212.992291] debug | DataWriter.cpp | write | [** <<DDS>> **] | client_key: 0x00000000, len: 20, data:
0000: 0B 00 00 00 74 65 73 74 5F 6C 61 62 65 6C 00 00 00 00 00 3F
[1663092212.992429] debug | SerialAgentLinux.cpp | send_message | [** <<SER>> **] | client_key: 0x77247145, len: 13, data:
0000: 81 00 00 00 0A 01 05 00 9C 01 00 00 80
[1663092213.194446] debug | SerialAgentLinux.cpp | recv_message | [==>> SER <<==] | client_key: 0x77247145, len: 32, data: | {
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} |
general-relativity, curvature, tensor-calculus, stress-energy-momentum-tensor, trace
Title: Flat space Solution of Einstein Field Equation Does a trace-free energy-momentum tensor $T_{\mu}^{\mu} = 0$ ensure that the Einstein's field equations have a flat space solution? The Einstein field equations
$$
R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu}
$$
for zero stress energy means that the Ricci Curvature $R_{\mu\nu}$ is proportional to the metric with $R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar $R~=~R_{\mu\nu}g^{\mu\nu}$ this is a spacetime of constant curvature, such as a 4-sphere.
By taking the trace of this Stress energy it is not hard to show that
$$
8\pi G\left(T_{\mu\nu}~-~\frac{1}{2}Tg_{\mu\nu}\right)~=~R_{\mu\nu}.
$$
The traceless condition $T~=~0$ just means the Ricci tensor is propotional to the stress energy, but the Ricci scalar is zero. | {
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ros, ros-fuerte, topic, openni-launch, openni-camera
Originally posted by Thomas with karma: 4478 on 2013-10-23
This answer was ACCEPTED on the original site
Post score: 2
Original comments
Comment by Athoesen on 2013-11-01:
Is there a way to get topics from more than one Kinect?
Comment by Thomas on 2013-11-01:
I guess if you launch the nodes twice in different namespaces it should work. You may need to connect your two kinects on two different USB controllers though.
Comment by Athoesen on 2013-11-03:
Well as I explained in my question here: http://answers.ros.org/question/96071/accessing-multiple-kinects-in-ros-hydro/ I'm not sure exactly how to modify the launch file to launch separate nodes. Any ideas?
Edit: Thanks for the suggestion Thomas, it helped me figure it out. | {
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probability-theory, entropy
Title: Conditional entropies of sum relations Let $(X_1,Y_1)$ and $(X_2,Y_2)$ be identically and independently distributed. Also consider $Z=X_1+X_2$. I am trying to prove the following inequality:
$$ H(X_2 \vert Y_1 Y_2 Z) \leq H (X_1 \vert Y_1)\leq H(Z \vert Y_1 Y_2).$$
I think I can show that middle term is the average of the other two. So now if I can show
$$ H(X_2 \vert Y_1 Y_2 Z) \leq H(Z \vert Y_1 Y_2),$$ I should be done. Can someone give a hint? Since $X_1,X_2$ are independent,
$$
H(X_1+X_2) \geq H(X_1+X_2|X_1) = H(X_2),
$$
and similarly $H(X_1+X_2) \geq H(X_1)$.
Therefore
$$
H(X_2|X_1+X_2) + H(X_1+X_2) = H(X_2,X_1+X_2) = H(X_1,X_2) \leq \\ H(X_1) + H(X_2) \leq 2H(X_1+X_2),
$$
which implies that
$$
H(X_2|X_1+X_2) \leq H(X_1+X_2).
$$
Since we have only used the independence of $X_1,X_2$ in this argument, we also have
$$
H(X_2|X_1+X_2,Y_1Y_2=y_1y_2) \leq H(X_1+X_2,Y_1Y_2=y_1y_2),
$$ | {
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cc.complexity-theory, np-hardness, cg.comp-geom, polygon
I am interested in the complexity of deciding the existence of a polygonization of a set of 2D lattice points (in Euclidian plane) with prescribed perimeter. I suspect that it is NP-complete. What is known about its complexity? Your problem is NP-hard, since it contains the Hamiltonian cycle problem on grid graphs as special case: Given a set of lattice points in the plane, is there a cycle in which all edges have lengths $1$ (and hence connect two horizontally or vertically adjacent lattice points.
The grid graph has a Hamiltonian cycle, if and only if the corresponding $n$ lattice points (interpreted as geometric points) have a polygonization with perimeter $n$. | {
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Could you please say me if (i) is correct and help me with (ii)?
With regards
math12
• You could take the infinite matrix of $T$ with respect to the canonical orthonormal basis. This will show you immediately that $T$ is not self-adjoint. – Julien Feb 17 '13 at 14:12
• I am always bad in finding examples or counterexamples, therefore I determined the adjoint operator explicitly. To my calculation, the adjoint operator is here given by $T^*\colon\ell^2\to\ell^2, (a_n)_{n\in\mathbb{N}}\mapsto\left(\frac{a_n+a_{n-1}}{2}\right)$ for $n\geq 2$ and so it is another operator and different from $T$ itself. Therefore $T\neq T^*$ and so $T$ is not selfadjoint. – math12 Feb 17 '13 at 14:35 | {
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} |
4.3k views
Compute the post fix equivalent of the following expression $3^*\log(x+1)-\frac{a}{2}$
in DS
retagged | 4.3k views
+1
postfix of log(x+1) will be ??
+1
it will be x1+log
+1
how ?
+1
in uninary operator we do firstly takes the value i.e x+1 and then apply what opration we want here log so firstly x+1 convert into psotfix then applly log on it other example is (a+1)++ for that we firstly convert a+1 into postfix then apply operation increment
+1
so unary opern of log(x1+) == x1+log ??
+1
yes u can say
+3
just construct expression tree and do post order traversal.....no need to worry about priorities | {
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"url": "https://gateoverflow.in/15708/gate1998-19b"
} |
waves, interference, geometric-optics
Thus, the path difference between any pair of adjacent rings is $\pm \lambda$ (depending on whether the adjacent ring is on the inside or outside relative to its partner).
The rings themselves are formed via thin film interference, by a minuscule layer of air between a curved glass surface and the flat glass surface where it resides. Due to the curvature of the glass surface, the thickness of the air layer does not increase linearly. The farther a point is from the center, the smaller the horizontal distance, which means an increase in vertical thickness of $\lambda$. Thus, the horizontal distance between two neighboring rings decreases as we observe radially outward from the center.
Here is a visual: | {
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c#, design-patterns, stored-procedure
private void CreateCommandWithParametersCommandTimeoutAndTransaction()
{
_command = StoredProcedureDbCommandCreator
.CreateStoredProcedureDbCommandCreator(_connection, _procedureName)
.WithParameters(_procedureParameters)
.WithCommandTimeout(_commandTimeoutOverride)
.WithTransaction(_transaction)
.BuildCommand()
.Command;
}
private void DisposeCommand()
{
if (_command != null)
{
_command.Dispose();
_command = null;
}
}
private void ExecuteCommand()
{
if (HasNoReturnType)
{
ExecuteCommandWithNoReturnType();
return;
}
ExecuteCommandWithResultSet();
}
private void ExecuteCommandWithNoReturnType()
{
_command.ExecuteNonQuery();
} | {
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java, beginner, quiz
}
}
class SetUp {
public void chooseNumberRange(UserChoices user) {
//These are the possible number ranges for a math exercise.
int min1 = 0;
int max1 = 9;
int min2 = 10;
int max2 = 100;
Scanner scan = new Scanner(System.in);
//Here we prompt the user to choose his type of math exercise.
System.out.println("Welcome to Math Work! Please select your "
+ "choice of operation: \n" + " 1.Addition"
+ "\n 2.Subtraction" + "\n 3.Multiplicaiton"
+ "\n 4.Division");
int userOperatorChoice = scan.nextInt();
switch (userOperatorChoice) {
case 1:
user.setUserOperator('+');
break;
case 2:
user.setUserOperator('-');
break;
case 3:
user.setUserOperator('*');
break;
case 4:
user.setUserOperator('/'); | {
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c, image, library, postscript
http://code.google.com/p/xpost/source/browse/src/lib/xpost_interpreter.h
Note: I have also posted this code to comp.lang.postscript for a more discussion-type discussion.
Also note: The interpreter produces temporary files in the current directory named gmemXXXXXX lmemXXXXXX xdumpXXXXXX (where XXXXXX is a system-generated unique filename sequence) which may needlessly accumulate diskspace. They periodically must be manually removed. And be careful with code that may do an infinite series of pushes to the stack, lmemXXXXXX may grow very large to hold a huge stack. Based on these two lines… | {
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"url": null
} |
python, beginner, web-scraping, beautifulsoup
The goal is to have cleaner code, with a #comment on each line (unless extremely basic) so as to improve my annotation habits.
Comments on each line could overall decrease readability of the code and they are, in essence, extra information and weight you need to make sure stay up to date with the actual code. Self-documented code is something you should strive to achieve, using comments as an extra measure used to explain the "why" some decisions were made and not the "how". | {
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classical-mechanics, torque, equilibrium
Title: Solving fulcrum / seesaw problem with multiple stacked weights I came across an article that 6 year olds in the UK will be asked questions like this in maths class: | {
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thermodynamics, thermal-conductivity
I can follow up to this point, but my problem arises when it’s said that,
Q1=Q2
My argument is that, Q1 is the heat transferred along the d displacement, not the entire rod. If T1 and T2 was placed at the beginning and end of the rod respectively, only then we could say Q1=Q2. But this is not specified. All the sources say that d can be any known distance on the rod. But clearly anything other than d=the entire length of the rod wouldn't give the correct value, right? Should this clarification be made? Consider a proof by contradiction: $Q_1\neq Q_2$, such that $Q_1=Q_2+Q_3$, where $Q_3$ must be stored on the right side of the rod. (We can show this with an energy balance of the right side of the rod.) We consider the sensible heating of that region: $Q_3=mc\frac{dT}{dt}$, where $m$ is the mass, $c$ is the specific heat capacity, $T$, is the temperature, and $t$ is time. At steady state, the temperature must be constant ($\frac{dT}{dt}=0$), so $Q_3=0$ and we have in fact $Q_1=Q_2$. | {
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python, performance, algorithm, pandas
# Rearrange the dataframe so that values which appear in both
# columns appear first in the second column
c1_in_c0 = df[1].isin(df[0])
df = pd.concat([df.loc[c1_in_c0], df.loc[~c1_in_c0]])
c0_in_c1 = df[0].isin(df[1])
df = pd.concat([df.loc[~c0_in_c1], df.loc[c0_in_c1]])
# In some cases the above algorithm will loop indefinitely if there
# are values that occur in both columns but there are no repeated
# values in the second column. Avoid this finding rows with values
# in the first column that are present in the second column and, if
# any exist, swapping values in the first row found.
c0_in_c1 = df[0].isin(df[1])
if c0_in_c1.any():
to_swap = df.loc[c0_in_c1].index[0]
df.loc[to_swap, :] = df.loc[to_swap, [1, 0]].to_numpy()
df.columns = input_columns
# Remap the modified dataframe
return make_bidirectional_map_one_to_many(df) | {
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ros, dynamic-reconfigure, image-pipeline, prosilica-camera
Just to be clear, I know that you can crop the images using dynamic_reconfigure but I'm wondering whether it can down sample the entire image somehow. I'm a vision noob so I think this is probably feasible - I just don't know how to do it. | {
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java
getPropertiesOfNumberSequence(num1);
i++;
}
break;
case "even":
if (isDuckNumber(num1) && isEven(num1)) {
getPropertiesOfNumberSequence(num1);
i++;
}
break;
case "odd":
if (isDuckNumber(num1) && isOdd(num1)) {
getPropertiesOfNumberSequence(num1);
i++;
}
break; | {
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python, python-3.x, raspberry-pi
while True:
if measure_temp() < 8:
GPIO.setmode(GPIO.BCM)
RELAIS_1_GPIO = 17
GPIO.setup(RELAIS_1_GPIO, GPIO.OUT)
GPIO.output(RELAIS_1_GPIO, GPIO.HIGH)
print('{:%d.%m.%Y %H:%M:%S}'.format(datetime.datetime.now()))
print('Heating on')
print()
time.sleep(900)
GPIO.cleanup()
else:
GPIO.setmode(GPIO.BCM)
RELAIS_1_GPIO = 17
GPIO.setup(RELAIS_1_GPIO, GPIO.OUT)
GPIO.output(RELAIS_1_GPIO, GPIO.LOW)
print('{:%d.%m.%Y %H:%M:%S}'.format(datetime.datetime.now()))
print('Heating off')
time.sleep(900) | {
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"openwebmath_score": null,
"tags": "python, python-3.x, raspberry-pi",
"url": null
} |
ecology, population-dynamics, population-biology
I would assume λ is under ideal conditions or as an average of whatever species you're working with. Your modified lifetime reproductive output still leaves out many variables which can affect λ. By your own words it leaves out mutations and daughter cells that are dead-on-arrival. There could also be predation or conditions where the Mother-cell produces more than one Daughter-cell, resulting in R = 3 instead of 2 and your subsequent λ also being 3. UCLA Bioengineers have actually observed up to 5 daughter cells in cancer lines.
I'm not sure what you mean by "... or it depends on the natural situation?", but ultimately reality is always more complicated than basic equations usually describe. Either B or D can increase or decrease, depending on the species, environment, and how you define t. | {
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} |
php, mysqli
if ($types === null)
{
$types = array();
foreach ($params as $param)
$types[] = substr(gettype($param)), 0, 1);//boolean -> b, integer -> i, string -> s
//float or double both yield "double" type -> d, which is what you need
}
$typeString = implode('', $types);//make string
//optional validation of typeString:
if (preg_match('/[^bsdi]/', $typeString))
throw new InvalidArgumentException('unknown data types used');
//add to params:
array_unshift($typeString, $params);
$stmt = $this->mysqli->prepare($query);
call_user_func_array(
array($stmt, 'bind_param'),
$param
);
//this call_user_func_array equates to:
$stmt->bind_param($typeString, $param[0], $param[1], $param[2],...);
}
//usage
$indb->queryParam(
"SELECT COUNT(*) counted_column_test
FROM users
WHERE userid != ?
AND disabled = ?
AND email = ?",
array(1, "this@email.net", false)
);
//or | {
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"openwebmath_score": null,
"tags": "php, mysqli",
"url": null
} |
general-relativity, gravity, spacetime, curvature, time-dilation
Title: Why would spacetime curvature cause gravity given that the value of gravitational time dilation is so small? In this question, mpv provides the clearest explanation of the operation of gravity in his answer:
The apple moving first only in the time direction (i.e. at rest in space) starts accelerating in space thanks to the curvature (the "mixing" of the space and time axes) - the velocity in time becomes velocity in space. The acceleration happens because the time flows slower when the gravitational potential is decreasing. Apple is moving deeper into the gravitational field, thus its velocity in the "time direction" is changing (as time gets slower and slower). The four-velocity is conserved (always equal to the speed of light), so the object must accelerate in space. | {
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"tags": "general-relativity, gravity, spacetime, curvature, time-dilation",
"url": null
} |
http://MathMeeting.com. This law considers ASA, AAS, or SSA. D The only prob… It is also applicable when two sides and one unenclosed side angle are given. Law of sines may be used in the technique of triangulation to find out the unknown sides when two angles and a side are provided. Below is a short proof. Two values of C that is less than 180° can ensure sin(C)=0.9509, which are C≈72° or 108°. ′ We also know nothing about angle-A and nothing about side-a. on plane To use the law of sines to find a missing side, you need to know at least two angles of the triangle and one side length. The absolute value of the polar sine of the normal vectors to the three facets that share a vertex, divided by the area of the fourth facet will not depend upon the choice of the vertex: This article is about the law of sines in trigonometry. That is, when a, b, and c are the sides and A, B, and C are the opposite angles. E As you drag the vertices (vectors) the magnitude of the cross product of the 2 | {
"domain": "fanthai.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9752018376422666,
"lm_q1q2_score": 0.822968054788499,
"lm_q2_score": 0.8438951025545426,
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"openwebmath_score": 0.7975123524665833,
"tags": null,
"url": "https://www.fanthai.com/matte-pink-uoy/e5934b-law-of-sines"
} |
formal-languages, turing-machines, reductions, undecidability, halting-problem
Title: Is a language of some deciders decidable? Is
$$L = \{ \langle M \rangle \mid M = (\{Q_1, Q_2, . . . , Q_{100}\}, \{0, 1\}, \{0, 1, \_\}, δ, Q_1, Q_2, Q_3) \text{ is a decider}\}$$
decidable?
I know
$$HALT_{TM}= \{ \langle M \rangle \mid M \text{ is a decider}\}$$
is not decidable, but in the case we are given a specific Turing Machine that has 100 states, alphabet $\{0,1\}$, tape alphabet $\{0, 1, \_\}$ (where _ is space), a transition function $\delta$, a start state $Q_1$, an accept state $Q_2$ and a reject state $Q_3$.
Checking that M is in the correct format is easy but is it possible to build a Turing Machine that decides $L$? I assume I can use the fact that $HALT_{TM}$ is undecidable to show that $L$ also is, but I am not sure how to proceed.
And I don't see how I could reduce $L$ to $A_{TM}$ to prove by contradiction that it is not decidable?
How should I proceed? I don't fully understand the details of your question (especially the notation that you've used). | {
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"openwebmath_score": null,
"tags": "formal-languages, turing-machines, reductions, undecidability, halting-problem",
"url": null
} |
c#, stream
/// <summary>
/// Constructor. Specifies an output encoding, and a byte order mark detection option for the input stream
///
/// </summary>
/// <param name="strInput">Input data, that will be decoded and re-encode into the specified output encoding</param>
/// <param name="encodingOut">Output encoding</param>
/// <param name="bDetectInputEncodingFromByteOrderMarks">Indicates whether to look for byte order marks at the beginning of the input stream.</param>
/// <remarks>
/// This constructor initializes the encoding to UTF8Encoding
/// The detectEncodingFromByteOrderMarks parameter, if true, detects the encoding by looking at the first three bytes of the stream. It automatically recognizes UTF-8, little-endian Unicode, and big-endian Unicode text if the file starts with the appropriate byte order marks. Otherwise, the UTF8Encoding is used. See the Encoding.GetPreamble method for more information.
/// </remarks> | {
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"tags": "c#, stream",
"url": null
} |
And that's it. If there are two Matrix with dimension A (2 x 3 ) and B (3 x 2). Matrix transpose: The transpose of matrix refers to the interchanging of the rows and columns. The number $$4$$ was in the first row and the second column and it ended up in the second row and first column. In practical terms, the matrix transpose is usually thought of as either (a) flipping along the diagonal entries or (b) “switching” the rows for columns. We can find its transpose by swapping the column and row elements as follows. Definition. The transpose will also be of dimension (2x2). Here are a couple of ways to accomplish this in Python. The transpose of a matrix by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. Menu. The conjugate transpose of a matrix is the matrix defined by where denotes transposition and the over-line denotes complex conjugation. [noun] In matrix mathematics, the process of rearranging elements in a matrix, by | {
"domain": "nubrace.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9773707966712549,
"lm_q1q2_score": 0.8247984344465039,
"lm_q2_score": 0.8438951084436077,
"openwebmath_perplexity": 629.0233137196188,
"openwebmath_score": 0.7180114388465881,
"tags": null,
"url": "http://nubrace.com/usda-direct-kwbnnc/transpose-matrix-definition-62cf45"
} |
the amount of variance that needs to be explained is greater than the percentage specified by n_components. You can see matrices as linear transformation in space. A second difference to svd is that fast. For comparison, a solver that applies the normal equations is included. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. However, the 'svd' solver cannot be used with shrinkage. So, i want to use SVD to solve orthogonal Procrustes problem with respect to that hyperparameter and then minimize the solution to find the best value for that hyperparameter. Solve the lower triangular system Ly = b for y by forward substitution. SVD also does this for all the data points, that is, dividing every data point with the value we divided the eigenvector with. K-SVD:Step1-SparseCoding Fix D, solve for X in minimize X k Y − D X k2 F subject to ∀i, k x ik 0 ≤ T, (14) Note that k Y − D X k2 F = XN i=1 k y i − D x ik 2 2. This causes a problem as the size of the | {
"domain": "cesenaticoopen.it",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109485172559,
"lm_q1q2_score": 0.8227933346453876,
"lm_q2_score": 0.8354835371034368,
"openwebmath_perplexity": 1043.0111490718846,
"openwebmath_score": 0.6158665418624878,
"tags": null,
"url": "http://yjcg.cesenaticoopen.it/svd-solver.html"
} |
beginner, c, http, socket
int type is guaranteed to store [−32767, +32767], but port numbers are [1, 65535]
atoi converts initial portion of the string, so garbage like 8080garbage is allowed
it does not detect errors
I suggest you to use strtol:
if (argv[2] < '0' || argv[2] > '9') {
/* Port argument must start with digits */
usage(stderr);
return EXIT_FAILURE;
}
char *end;
long int port = strtol(argv[2], &end, 10);
if (end != NULL || port < 0 || port > 65535) {
/* Port argument must not contain garbage and must represent a valid port number */
usage(stderr);
return EXIT_FAILURE;
} | {
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"id": 30199,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "beginner, c, http, socket",
"url": null
} |
java, object-oriented, minesweeper
break;
}
default: break;
}
if(currentCellImage!=null)
graphics.drawImage(currentCellImage, j*CELL_WIDTH, i*CELL_HEIGHT, CELL_WIDTH, CELL_HEIGHT,null); // draw current cell
}
}
@Override
public Dimension getPreferredSize()
{
// report grid dimensions in order to adjust parent's size
return new Dimension(CELL_WIDTH*nCols, CELL_HEIGHT*nRows);
};
public GameGrid(int cols, int rows, int mines) {
super(cols, rows, mines);
icons_init();
}
private void icons_init()
{
// load icons that represent different cell types and states
icons = new HashMap<String,Image>();
try {
for(int i=0; i<=8; i++)
{
// Current cell is open and has N (0-8) mines around (horiz., vert. and diag.) | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "java, object-oriented, minesweeper",
"url": null
} |
classical-mechanics, lagrangian-formalism, conservation-laws, symmetry, noethers-theorem
$$\frac{d}{dt} Q(q_i, \dot q_i) = 0.$$
Note that $Q(q_i, \dot q_i)$ does not depend on $t$ explicitly; it only depends on $t$ insofar as $q_i$ and $\dot q_i$ do. However, an initial condition depends on $q_i$, $\dot q_i$, and $t$. You need to know $t$ in order to know "how far to turn back the clock" to find the initial position and velocity.
A slick "proof" of Noether's theorem goes as follows. Say you have some differentiable group of transformations that leave your Lagrangian invariant. Imagine changing a path in configuration space by an infinitesimal group action, using a tiny number $\varepsilon$. For example, an infinitesimal translation in the $x$-direction in 3D space ($i = 1, 2, 3$) would be given by
$$q_1 \to q_1 + \varepsilon$$
$$q_2 \to q_2$$
$$q_3 \to q_3$$
$$\dot q_i \to \dot q_i$$
and an infinitesimal rotation in the $xy$-plane would be given by
$$q_1 \to q_1 + \varepsilon q_2$$
$$q_2 \to q_2 - \varepsilon q_1$$
$$\dot q_1 \to \dot q_1 + \varepsilon \dot q_2$$ | {
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"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "classical-mechanics, lagrangian-formalism, conservation-laws, symmetry, noethers-theorem",
"url": null
} |
c++, performance, algorithm, backtracking
using namespace std;
bool found;
/* This function check, for every time a letter is added, if the sequence
* is going to repeat itself. For each pair of substrings with length i
* they are compared to check if they are equal.
* ex:
* The sequence NONPNOPN will check if N = P, PN = NO, OPN = NPN, NOPN = NONP
*
* The function returns True if it should reject the sequence and False otherwise.
*/
bool reject(string str){
int i = 1, len = str.size();
while(i <= len/2){
if(str.substr(len-2*i,i) == str.substr(len-i,i)) return true;
i++;
}
return false;
}
/* This function uses backtracking to generate the answers. It recursively adds
* a letter to the string and checks if it is a probable candidate, printing the answer if
* it's the right answer and returning otherwise
*/
void Thue(int n, string str){
if(found) return;
if(reject(str)) return;
if(str.size() == n){
cout << str << endl;
found = true;
return;
} | {
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"id": 24171,
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"openwebmath_score": null,
"tags": "c++, performance, algorithm, backtracking",
"url": null
} |
hydrology, statistics
There are many examples of calculation and software packages to actually build these curves, which are used often for engineering design to help decide the storm a given piece of infrastructure is being designed to (e.g. 5 year storm, 25 year storm, etc.). Most municipalities will have developed their own IDF curves to set standards for design.
There are also examples of online tools to help generate IDF curves for a given location, such as this one in Ontario. This tool is neat as it calculates the IDFs for a given year, and uses trend analysis to predict future changes to the IDF curve, allowing you to also pick the target year for generating the IDF curve. | {
"domain": "earthscience.stackexchange",
"id": 1916,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "hydrology, statistics",
"url": null
} |
angular-momentum, quantum-spin, atomic-physics, notation, orbitals
For $^1D$ and $^1S$ it is pretty simple because $S$ is zero, which means that for $^1D$ $J=2$ only, and for $^1S$ $J=0$ only. Note that in a magnetic or electric field the $^1D$ $J=2$ state will be split into five sublevels with $M_J=+2,+1,0,-1,-2$.
By contrast the $^3P$ state has three different $J$ values; $J=2,1,0$ depending on how the $L$ and the $S$ combine. We write these as $^3P_2$, $^3P_1$ and $^3P_0$. Note that Note that in a magnetic or electric field the $J=2$ state will be split into five sublevels with $M_J=+2,+1,0,-1,-2$, the $J=1$ state will be split into three sublevels with $M_J=+1,0,-1$ and the $J=0$ state will have a single sublevel with $M_J=0$ - Thus there are a total of 9 different sublevels of the different $J$ states of $^3P$. | {
"domain": "physics.stackexchange",
"id": 47243,
"lm_label": null,
"lm_name": null,
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"openwebmath_score": null,
"tags": "angular-momentum, quantum-spin, atomic-physics, notation, orbitals",
"url": null
} |
waves, harmonic-oscillator, speed
$$f_n=\dfrac{n}{2L} \sqrt{\dfrac{T}{\mu}}$$
The problem is, this formula is derived by assuming that the wave speed is $$v=f\lambda$$ or $$v=\sqrt{\dfrac{T}{\mu}}$$
But if the wave is actually standing, what does this speed actually mean? Does it mean twice the length of the string times the frequency of the individual particles performing harmonic oscillations? And if that is so, who is to say that it will be equal to $\sqrt{T/\mu}$? What makes a standing wave is having waves in a single medium propagating in opposite directions at exactly the same speed, and in phase. This requires the waves to possess a unique speed through the medium in order to establish the "standing wave" condition (and it also requires the medium to have a finite length into which an integer number of those waves can fit). That's why the standing wave condition must account for the wave speed. | {
"domain": "physics.stackexchange",
"id": 70956,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "waves, harmonic-oscillator, speed",
"url": null
} |
general-relativity, classical-mechanics, metric-tensor, space-expansion
Title: Deriving Friedmann Equations without General Relativity Can we derive the analytic Friedmann Equations without general relativity, starting from completely classical/nonrelativistic arguments? (If we consider sufficiently small volumes.) A Newtonian approach is possible but of of course it is not rigorous.
We're in a newtonian approximation now, and we want to describe the motion of a unit mass at a point P on the surface of a sphere.
So let's take this spherical distribution of matter, take a point at a distance $l$ from the origin. The equation of motion of that point is
$$
\frac{d^2l}{dt^2} = - \frac{GM}{l^2} \tag{1}
$$
$$
\Rightarrow \, \, \, \ddot{l}\, \dot{l} = \frac{d}{dt} \frac{\dot{l}^2}{2}= - \frac{GM}{l^2} \dot{l} = \frac{d}{dt} \frac{GM}{l} \tag{2}
$$
from which you can find
$$
\frac{d}{dt} \Big(\frac{\dot{l}^2}{2} - \frac{GM}{l} \Big) =0 \tag {3}
$$
$$
\Rightarrow \, \, \dot{l}^2 = \frac{2GM}{l} + constant \tag{4}
$$ | {
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"tags": "general-relativity, classical-mechanics, metric-tensor, space-expansion",
"url": null
} |
algorithms, graphs, minimum-spanning-tree
(Hint to) Proof: One approach is to verify the condition 5 of lemma 1. Another approach is to verify the condition 2 of lemma 1 by showing that in any set of edges, a lightest edge by $w_2$ is also a lightest edge by $w_1$,
A comparison-based algorithm on a graph $G$ is defined as comparison-compatible if for any two (weakly) comparison-compatible weight functions on all edges, we can run the algorithm with one weight function in a way that can be repeated without any change with the other weight function. In particular, those two runs of the algorithm will have the same output.
Please note that most if not all MST algorithms can be stated in two flavours. The first flavour assumes that distinct edges of $G$ have | {
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"openwebmath_score": null,
"tags": "algorithms, graphs, minimum-spanning-tree",
"url": null
} |
audio, frequency-response, impulse-response
Regarding the decibel scaling, an easy thing would be to play a signal at a known level (for example 94 dB SPL) and record it with the microphone. Then check what is the dBFS level of that recording. This will allow you to map the dB SPL to dB FS. Once it's done, you can convolve your signals and analyse the spectra to see what level they simulate. If any gain adjustment is necessary then you can do it afterwards. This, of course, assumes that you know the original level of those recordings, which is generally not true. If you had the same sort of measurement for the device that was used to capture those recordings, then it's easy. You know what is the sensitivity of that device, you know how much it differs from your microphone, so you can correct for that either by adjusting the level of impulse response or applying gain post convolution. Finally, you can simply ignore all of that and set the gain of your impulse response so the frequency response peaks at 0 dBFS. This means no | {
"domain": "dsp.stackexchange",
"id": 8276,
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "audio, frequency-response, impulse-response",
"url": null
} |
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