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How does gravity cause nuclear fusion in stars?

I am a bit confused regarding the nuclear fusion that occurs during star formation. For example, suppose there is a huge hydrogen cloud. It gets more mass and therefore pulls in more and more hydrogen, but it could just do this endlessly and get endlessly big. What causes it to stop growing? Is it because of the heat generated by the huge mass?

This is not meant as a detailed description of how fusion starts in stars: I just want to convince you that it can start, and where the energy comes from to start it.

Let's start with a big ball of hydrogen (and let's assume it's not spinning very fast). There are two things which determine what happens to it:

• it has a lot of mass, and so gravity wants it to get smaller;
• pressure wants it to get bigger.

If we were very careful and built this ball very slowly and carefully we could get to a state where it just sat in equilibrium (so long as it was not too big when bad things famously happen) with pressure just counteracting gravity.

But in fact what happens is that it starts off with quite a low pressure, so gravity wins, and starts making it smaller. What this means is that all the hydrogen atoms start moving down the gravitational potential gradient: they are losing gravitational potential energy. But energy is conserved, so they must be gaining some other kind of energy. And that's kinetic energy: they start moving faster, and bouncing around off each other as you'd expect.

Well, a thing where the atoms have a lot of kinetic energy like this is hot: that's what being hot is. So the ball of gas bargains gravitational potential energy into heat, and starts getting really hot.

Some of this heat -- in fact half of it, thanks to the virial theorem -- gets radiated away into space, but it's a big ball of gas, so its surface area is relatively small compared to its volume, so the rate at which it can radiate stuff away is relatively low. So as it continues to shrink it gets really hot, especially in the centre.

Really hot means that the hydrogen atoms in the centre are moving really fast (they also get ionized so they end up as a plasma). The centre of the ball of the gas also gets really dense as the thing falls in, so the atoms (really, the protons) start being more likely to crash into other protons, and to do so with lots of energy.

If the ball is too small, then eventually the pressure becomes high enough to halt the collapse and then the thing sits there radiating away heat and slowly cooling.

But if the ball is big enough then the centre gets hot and dense enough that some of these collisions have high enough energy that fusion starts. This releases more energy, so the temperature in the centre climbs further, and the protons have more kinetic energy and more fusion happens. It's kind of astonishing that this process settles down at some point to being a star which is stable for millions or billions of yeas rather than causing what you'd naively expect, which is a huge explosion as fusion runs away and blows the thing to bits.

• It's really astonishing. – Ray Luxembourg Apr 26 '18 at 11:40
• Half the gravitational potential is radiated away as it contracts. – ProfRob Apr 30 '18 at 17:49
• @RobJeffries: but only half... – user107153 May 2 '18 at 7:37
• @tfb Just suggesting you replace "some of" with half. The virial theorem is quite specific on that. Doesn't affect your qualitative answer. – ProfRob May 2 '18 at 7:39
• @RobJeffries Oh, sorry, I thought you were picking nits! I have edited it to say that (I'd also have been very happy if you had just unilaterally done so!). – user107153 May 2 '18 at 7:51

a huge hydrogen cloud, it gets more mass therefor it pulls more and more hydrogen but it could just do this endlessly and get endlessly big.

Yes, it could. This is related to the questions "why is the unviverse homogeneous down to $10^-5$ as seen in the the microwave background radiation?"

Graph of cosmic microwave background spectrum measured by the FIRAS instrument on the COBE, the most precisely measured black body spectrum in nature. The error bars are too small to be seen even in an enlarged image, and it is impossible to distinguish the observed data from the theoretical curve.

The curve is the best fit to the black body distribution, BUT, and it is a big BUT, the graph is the same in every direction of the universe ,where radiation could not travel due to the light cones at that early time, (the time of photon decoupling, see plot below) to set up a thermodynamic equilibrium.

Nine Year Microwave Sky The detailed, all-sky picture of the infant universe created from nine years of WMAP data. The image reveals 13.77 billion year old temperature fluctuations (shown as color differences) that correspond to the seeds that grew to become the galaxies. The signal from our galaxy was subtracted using the multi-frequency data. This image shows a temperature range of ± 200 microKelvin. Credit: NASA / WMAP Science Team WMAP

The second question related to yours is "why there are inhomogeneities below that level?".

The uniformity is modeled by proposing quntization of gravity and and an inflationary period where quantum mechanics homogenizes the universe, with small statistical inhomogeneity, the colered regions in the plot, which become the seeds of concentration of matter.

So your question, within the Big Bang model, is answered by positing that the original energy distribution of the universe was homogenized quantum mechanically. The uncertainty principle in quantum mechanics allows for inhomogeneities which as the universe progresses to hydrogen creation have regions in space with more concentrated mas, which will then attract gravitational more and more mass and generate the clusters of galaxies, galaxies and the present day universe.

I think that the reason it doesn't keep getting bigger and bigger is that once it has enough mass, it starts fusion so it's making photons. Now those photons have momentum so they can collide with the surrounding gas particles and keep them from reaching the star. It's just a wild guess tho.. could be wrong.

• So this is just natural behavior. Once object gains enough mass it will start fusion process? – Ray Luxembourg May 1 '18 at 8:31
• If you're asking whether a star will naturally start fusion when it has enough mass, then I think the answer is yes. When a star gains mass, the pressure inside it also increases, which pushes the hydrogen atoms together with more and more force. At one point, when the star has enough mass, the electrostatic repulsion can no longer keep the H atoms apart and they fuse to form helium; I'm no expert in nuclear fusion but that's what my intuition tells me. – Anurag B. May 1 '18 at 10:33


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SEARCH HOME
 Math Central Quandaries & Queries
 Question from maria, a student: I've read your solution to the problem of how to derive the formula of the surface area of the sphere. But i've got a question that initially you demenstrate the area is apprximately equal to that of the lateral area of the cylinder. Afterwards you simply say that the two areas are exactly equal. I'm very confused, how you connect your two statements together. Can you explain to me how they are exactly equal? Your quick reply would be appreciated. Thanks for your help!

Hi there.

If you take a finite strip of the cylinder and sphere, they are approximately equal - I don't think you are questioning this. The leap to exactness is due to taking a smaller and smaller strip height hc and hs. The smaller these are, the more exact the comparison becomes. When you use calculus, you make this height infinitely small, so the distortion (comparison of approximation to exact value) becomes infinitely small.

It's a bit like comparing 0.9 to 1. That's approximately the same, but if we make it closer, then 0.99 is approximately 1, but certainly nearer than the first one.

Similarly, 0.99999....repeating forever is infinitely close to 1, so we say it is equal to 1. Thus it is exact.
(1/9) = 0.1111...., 9/9 = 0.9999.... = 1.

Stephen La Rocque.>

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.

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Dividing by Powers of Ten (C): 1/10^2

This Dividing by Powers of Ten (C): 1/10^2 worksheet also includes:

In this math activity, students are given 26 numbers which they divide by 10 to the -2 power. Once the procedure is understood, the solution is easy to find.

Concepts

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with Tarki's axioms of geometry what is a plane?

Reading about Tarski's axioms of geometry https://en.wikipedia.org/wiki/Tarski%27s_axioms I was puzzeling how they would look for 3 dimensional geometry.

the big problem then is the upper dimension Axiom

For 2 dimensions it is (copied from wikipedia)

$(xu \equiv xv \land yu \equiv yv \land zu \equiv zv \land u \ne v) \rightarrow (Bxyz \lor Byzx \lor Bzxy)$

in 3 dimensional geometry it should be something like:

For every points P , Q , X, Y and Z if PX = QX, PY=QY and PZ= QZ then there is a plane A and X , Y and Z all belong to plane A .

But how to formulate plane A in Tarski's axioms?

The Upper n-Dimensional Axiom for n = 2, 3, . . . asserts that any three points a, b, c which are equidistant from each of $n$ distinct points $p_1$, $p_2$, . . . , $p_n$ must be collinear.


text:
SEARCH HOME
 Math Central Quandaries & Queries
 Question from Awudu, a student: Given the roots of the quadratic equation ax^2+5x+c=0 are 5 and -2/3,find the values of a,b and c.

Hi,

I am going to assume the equation be $a x^2 + b x + c = 0.$

If $5$ and $\large \frac{-2}{3}$ are the roots of a quadratic equation then you know that

$p x^2 + q x + r = (x - 5)\left(x + \frac{2}{3}\right)$

for some real numbers $p, q$ and $r.$

Expand the right side and equate the like coefficients. This will give you values of $p, q$ and $r$ with $p = 1$ and $q$ and $r$ fractions. I expect you want $a, b$ and $c$ to be integers. How do you modify the equation $p x^2 + q x + r = 0$ so that the coefficients are integers and the roots are still $5$ and $\large \frac{-2}{3} \normalsize ?$

Penny

Math Central is supported by the University of Regina and the Imperial Oil Foundation.


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# True randomness? [duplicate]

I am a physics high-school student so my knowledge is not very deep on the subject. We started learnning about quantum mechanics and on some processes that my teacher described as random. I began to think on the concept of randomness and question it, thinking how can a process or an outcome be determined without any cause, how an outcome be determined at all in complete randomness?

I searched the internet and figured the scientific community does not agree with me. I'd really like to understand how can true randomness exist? Why the scientific community rejects the idea that ''random'' events may just have a cause we are not aware of? What am I missing?

## marked as duplicate by ACuriousMind♦, Kyle Kanos, Ryan Unger, John Rennie, Kyle OmanMay 29 '15 at 16:15

• But isn't it different? Saying that we can't measure something precisely and saying the state is random? I'll tell you where exactly I began to think about it so you could maybe explain why I am wrong in those spesific things and then in the general sense. One time is when we learned about matter-decay, we were taught the process is random. Second time (I hope my description will be understandable with my vocabulary) is with the process that photon destabilize an electron and ''bounce'' it to other level, then the electron got back to the original level through sevral ways in a random manner. – user3917631 May 28 '15 at 15:51
• "Random" means "practically unpredictable." An individual tritium atom will undergo beta decay if you wait long enough, but we have no theory that can predict when it will happen (i.e., it is "random" as far as we are concerned). That doesn't mean that there is no cause: It only means that if there is a cause, the cause is beyond our present knowledge. – Solomon Slow May 28 '15 at 16:15
• @ james large : according to Bohr, there is no hidden story , nor hidden variable , for pure random. Not only unknown, it can't be known if any – user46925 May 28 '15 at 19:10
• "True randomness" doesn't exist. A finite sequence of numbers does not have that property and an infinite sequence is not physically realizable. Like with every other philosophical/mathematical idea one has to check first if it satisfies trivial existence requirements before wasting any time on it in the context of physics. This one does not. Having said that, so far we have not seen any violations of the laws of quantum mechanics, but that has nothing to do with "true randomness". Quantum mechanics is about uncertainty, not randomness. – CuriousOne May 28 '15 at 19:36

First of all, welcome to the site. What quantum mechanics tells us is not that processes are random; but that there is a "fundamental" (at least from the QM point of view) impossibility of having certain informations (for example the trajectory of a particle) with perfect precision. That means that the maximal informations we can have of the state of a system is of probabilistic nature, id est we are only able to say "the outcome $a$ of a measure about the quantity $A$ have a $x\%$ probability of being measured".

That does not mean, however, that the processes happen randomly, they are governed anyways by physical laws; nevertheless these same laws forbid a completely precise "classical" description of the system. So I may say that randomness, in this particular context, is not a good word, also because it may be misleading since the study of random processes is an important branch of mathematics and physics (but not strictly related to basic quantum mechanics).

Related to the OP comment above: those processes are not random from a QM point of view. Simply QM introduces, as I said, the concept of probability of a measure as opposed to exact knowledge. Take the specific example of the photon interacting with an electron. What we can say, very roughly speaking, is that there is only a certain probability that the photon would interact with the electron (and that probability depends on the state of the system photon+electron); and that this causes the electron to absorb the energy of the photon. After that, there is a certain probability that the electron would lose energy and emit a photon, perhaps of different energy. That does not mean that the events are random, they are dictated by the laws of quantum electrodynamics; it simply means that the maximal information that the QM description can give is this, i.e. which is the probability of a certain event.

It is a different type of prediction, that changes radically the point of view from the classical one: if in classical mechanics we would have predicted, given the initial conditions, exactly if the photon and the electron (seen as classical particles) would collide (as in a billiard); on the contrary in quantum mechanics we do not know a priori if the collision happens, we know only which is the probability of the collision. However there is nothing random about that probability, it is determined exactly by the quantum theory (as the trajectory was in classical mechanics).

• How do you define randomness in the first place because wikipedia clearly says "During a measurement, on the other hand, the change of the initial wavefunction into another, later wavefunction is not deterministic, it is unpredictable (i.e., random)." Of course there might be non-local hidden-variables that determine the events but that is not the issue here. – Gonenc Mogol May 28 '15 at 16:52
• @gonenc There is a very long dispute about the quantum measurement process; there are mathematical models of it that have nothing unpredictable (at least in principle). Nevertheless, also from a naïve point of view, when you do a measurement you know a priori the probability of the possible outcomes, and given the effective outcome, it is postulated by QM that the original state is projected on the eigensubspace corresponding to the obtained measurement. This seems a quite predictable description, within the indeterminacy proper of quantum measurements (that is again predictable). – yuggib May 28 '15 at 17:09
• This bit "those processes are not random from a QM point of view. Simply QM introduces, as I said, the concept of probability of a measure as opposed to exact knowledge." is interpretive and not backed up by clear experimental data. The data tell us that it is either actually random or practically random because we can't know the hidden variables, but it doesn't distinguish the two cases. You can have either a local truly random theory or a non-local hidden-variables theory. You just can't have a local hidden-variable theory. – dmckee May 28 '15 at 17:15
• Hmmm ... on further reading I see that you are trying to draw a strong denotative difference in the works "random" and "probabilistic". I see where you're coming from on that, but I don't believe that this distinction is universally recognized. – dmckee May 28 '15 at 17:18
• @dmckee exactly, I think that there is a strong semantic difference between the two terms, however as you may see also in chat now there is not much consensus on that...;-) – yuggib May 28 '15 at 17:27

Bell proved that, if there exists such an unknown cause, then it surely must violate special relativity (Information must travel faster than the Speed of Light).

Taken from Wikipedia

"Realist interpretations of quantum mechanics are possible, although, such interpretations must reject either locality or counter-factual definiteness."

See


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# How to predict colour of precipitate? [duplicate]

I am currently preparing for Junior science competitions and I am coming across many questions where colour of precipitate plays important role... Like

When compound $$\ce{Hg2(NO3)2}$$ reacts with $$\ce{Na2CrO4}$$ solution, the colour of the compound formed is--

The question is a bit long and the compound can be determined from a series of experiments given in question. the product is $$\ce{Hg2CrO4}$$ but guessing the colour is impossible. Is there any way to determine colour of compound i.e. precipitate.??

• Actually the question is a a bit long and from the question the compound is determined. Let's assume the product can be determined easily. Is there any way to know it's colour?? Oct 2 '21 at 17:46
• May be the solution is in the details (I am looking for the hints there). Please post the entire question. Oct 3 '21 at 0:34
• @Nilay Ghosh Looks perfect, Brilliant, Thanks Oct 3 '21 at 3:30

• the mercury(I) chromate $$\ce{Hg2CrO4}$$ is brown at room temperature, and turns red at $$100$$°C. It changes crystallization system at high temperature.
• the mercury(II) chromate $$\ce{HgCrO4}$$ is yellow at room temperature, and becomes red at $$100$$°C. This change is due to the formation of a basic salt.
• The dichromate ion $\ce{Cr2O7^{2-}}$ does not give a precipitate with $\ce{HgCl2}$, but is gives a brownish yellow precipitate with $\ce{Hg(NO3)2}$, according to Treadwell's Qualitative Analysis. Oct 3 '21 at 9:38


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Roman Yurchak
@rth
@sameshl See https://github.com/pypa/pip/issues/5247#issuecomment-381550610 probably best to reinstall in a new virtual environment.
Andreas Mueller
@amueller
@thomasjpfan you like puzzles, right? scikit-learn/scikit-learn#14704
PLT
@Pakigya

I can't seem to get to make the virtual environment with sphinxgallery

conda create -n sklearndev numpy scipy matplotlib pytest sphinx cython ipykernel sphinxgallery

or

conda create -n sklearndev numpy scipy matplotlib pytest sphinx cython ipykernel sphinx-gallery

never mind! the solution is in the other gitter chat!
Bin Wang
@biwa7636

Hi team, I am new to Cpython but really wants to play with the internals of sklearn. I want to test out some of the cdef classes in the pyx file but looks like the methods are inaccessible within Python. Any thought?

For example:

from sklearn.tree import _utils
ph = _utils.PriorityHeap(100)
dir(ph)

And I cannot find call methods like pop, push.
Usually how does the workflow look like if I want to play with the internals of sklearn within Jupyter notebook.

FritzPeleke
@FritzPeleke
hello everyone. I'm really new to Machine learning in general and i have been working with some sklearn Regressors. I need some help :). My question is how do i know if the RMSE i have is minimum enough for good predictions. To what do i compare this RMSE to?
enoch-sun
@enoch-sun
I was able to create a model by curve fitting a set of data that has 5 variables using GaussianProcessRegressor. The problem is I am unable to export/load this model into an older version of python (version 2.5.2). Is there a way to dump the equation/formula into mathematical terms in relations to these 5 variables so that I can use this prediction on the older python? Thanks
@enoch-sun We don't really support those Python versions anymore. You can try and figure it out with some other persisting models such as ONNX or PMML, but you'll be mostly on your own
Thomas J. Fan
@thomasjpfan
@biwa7636 The PriorityHeap functions pop and push are cdef, which means they are not available in python.
Jesse Leigh Patsolic
@MrAE
Is there a scikit-learn preferred way to store a vector using Cython? I've seen libcpp.vector, array.array and numpy used in the code base. @NicolasHug @amueller
Nicolas Hug
@NicolasHug
The way we do it now is to allocate numpy arrays (in python or in cython), and then use a memory view for pure cython parts. You can take a look at how we do it in e.g. ensemble/_hist_gradient_boosting
Sakitha
@Sakitha
Hi, does apply in df.apply(fun) iterate over each columns in 'df' data-frame and pass them to 'fun' function as a series?
Bin Wang
@biwa7636
@thomasjpfan, you are right, however, I also tried to execute the above code too using %%cython magic also from sklearn.tree cimport _utils but still did not work. Was it supposed to be like that?
%%cython
from sklearn.tree._utils cimport Stack
s = Stack(10)
print(s.top)
>>> AttributeError: 'sklearn.tree._utils.Stack' object has no attribute 'top'
I found the source code so well written, fascinating and really want to be able to get the development environment up and running.
Bin Wang
@biwa7636
Weird, the above code will work if I replace s = Stack(10) with cdef Stack s = Stack(10), I believe this must have something to do with static type declaration.
Jesse Leigh Patsolic
@MrAE
Does anyone know why the base estimator for ExtraTreesClassifier is ExtraTreeClassifier, instead of DecisionTreeClassifier with splitter='random'? I am working on adding a new type of tree. @NicolasHug @amueller
Nicolas Hug
@NicolasHug
No idea. It doesn't make much sense for ExtraTreeClassifier to allow for a splitter that isn't 'random' IMO.
Would you want to submit a PR to deprecate the parameter?
motmoti
@motmoti

Hi All, Im getting the following error while executing the python setup.py install
error: Command "cl.exe /c /nologo /Ox /W3 /GL /DNDEBUG /MT -IC:\Users\Moti\Anaconda3\envs\motidevs\lib\site-packages\numpy\core\include /EHsc /Tpsklearn\svm\src\libsvm\libsvm_template.cpp /Fobuild\temp.win-amd64-3.7\sklearn\svm\src\libsvm\libsvm_template.obj" failed with exit status 127

Do you have any idea? Thanks!

kirk86
@kirk86
Any scikit devs who can shed some light on why calibration_curve is only for binary estimators?
Anjali Singh
@Anj-ali
how can i start committing to the open source
@Anj-ali you can start by going through our contributing guides: https://scikit-learn.org/dev/developers/contributing.html#contributing
Anjali Singh
@Anj-ali
thank you Sir, surely i will do that
Olivier Grisel
@ogrisel
Heads up: if you use conda and upgrade your env, you might get a crash when using n_jobs>=2. This is caused by an updated version of intel-openmp in the default channel of conda. I reported the issue upstream as ContinuumIO/anaconda-issues#11294 and the problem is tracked in this PR on the scikit-learn side: scikit-learn/scikit-learn#15020
The error message is OMP: Error #13: Assertion failure at z_Linux_util.cpp(2361) reported by the dying worker process.
Which in turns causes loky to raise: TerminatedWorkerError: A worker process managed by the executor was unexpectedly terminated. This could be caused by a segmentation fault while calling the function or by an excessive memory usage causing the Operating System to kill the worker. The exit codes of the workers are {SIGABRT(-6)}.
Samesh Lakhotia
@sameshl
If someone is free to review, please take a look at scikit-learn/scikit-learn#14993 and scikit-learn/scikit-learn#15045.
Andreas Mueller
@amueller
hm is there a pandas gitter? Or is @jorisvandenbossche around lol? For a pandas dtype, how do I get the closest numpy dtype to cast to?
Joris Van den Bossche
@jorisvandenbossche
yep
there is pandas gitter actually (pydata/pandas)
I don't think there is a typical way to do it
If I remember correctly, there is an issue about it
Basically, you would like to know the dtype of np.asarray(obj).dtype right? (but without needing to do the actual conversion?)
Andreas Mueller
@amueller
indeed
it's for scikit-learn/scikit-learn#15094 which is currently failing because np.result_type(pd.CategoricalDType) raises an error
Joris Van den Bossche
@jorisvandenbossche
the issue that I rememered is pandas-dev/pandas#22791
Andreas Mueller
@amueller
ok. so no solution :-/ is there a work-around?
like what does actually happen when you do the conversion?
is it from the pd.DataFrame.__array__ method or something?
Andreas Mueller
@amueller
yeah it is, no way to figure that one out :-/
Jesse Leigh Patsolic
@MrAE

Hello all (I am new to Cython),

I am currently working on adding an augmented version of Brieman's forest-RC (similar to RandomForest) algorithm into my fork of scikit-learn: In short, the algorithm takes linear combinations of features and projects them with weights randomly selected in {-1,1} to form a new feature to split on. The number of features combined at each split is a random variable.

The current SplitRecord only holds one feature, I need something to store a vector of features and a vector to hold weights.

1. I tried initializing an np.ndarray and using memoryviews, but ran into GIL issues.
2. I tried to make an ObliqueSplitRecord class, but that can't be passed as a pointer into functions because it is a Python object.
3. I tried to augment the SplitRecord struct in _splitter.pxd but that didn't seem to work because vectors would then be of fixed length.
4. I tried to use something similar to the tree/_utils:Stack but fell into the same problem as it was a class and couldn't be passed as a pointer into a function.

I am looking into using cppclass, but am not sure if that will fix solve my problem.

Does anyone have suggestions on how to best implement this in a Cythonic way? i.e. storing a vector of things while avoiding the GIL and not using python objects?

@MrAE you can use a cpp vector in cython. But since you're changing the splitrecord struct, you'll need to change the code in quite a lot of places.
Mateusz Sokół
@mtsokol
Hi, I have some basic question about local docs build for scikit. I've been trying to modify docs inside API for some file in sklearn/linear_model and followed instructions in Contributors Guide. But after few attempts the make command inside /docs does not seem to modify local docs build inside _build. In the browser, API docs didn't change although I modified the sources. Am I missing something?
Nicolas Hug
@NicolasHug
@mtsokol it seems that you're doing it right... maybe double check that 1. you're actually changing the sources, i.e. not anything in the _build folder, 2. the doc that you're changing is about a public estimators/tools (private tools aren't rendered in the doc anyway) and 3. that you're looking at the generated html in doc/_build/html/stable/
Nicolas Hug
@NicolasHug

@MrAE

re 1. you can't use (let alone allocate) numpy arrays when the GIL is released because these are Python objects. Is there a way for you to allocate the arrays somewhere where the GIL is held, and use memory views when the GIL is released? Memory views are safe to use without the GIL

re 2. is it still considered a Python object if you use a cdefed class and all the attributes are cdef`ed as well?

re 3. what vectors? can't you use a view as a field of the struct?

Nicolas Hug
@NicolasHug
Also @MrAE I happen to have been writing about Cython over the weekend... maybe that could help http://nicolas-hug.com/blog/cython_notes
Jan-Benedikt Jagusch
@janjagusch
could somebody share a good example for class docstrings in scikit-learn that we could use as a sort of template? thanks!
Kristiyan Katsarov
@katsar0v
Alessandro Surace
@zioalex
Here is the issue search string "is:issue is:open examples class docs involves:adrinjalali"


text:
Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
2017.
The collaboration

Abstract (data abstract)
Measuring K$^0_{\rm S}$K$^0_{\rm S}$ and K$^{\rm \pm}$K$^{\rm \pm}$ three-dimensional femtoscopic correlations using Pb-Pb collisions at $\mathbf {\sqrt{s_{\rm NN}}=2.76}$ TeV LHC ALICE collaboration KchKch and K0sK0s femtoscopy from 2.76 TeV/nucleon Pb-Pb collision 0-10%, 10-30% and 30-50% centrality, -0.8 < eta < 0.8 We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. The $m_{\mathrm{T}}$ scaling predicted by pure hydrodynamic models appears to be broken in our data, which indicates the importance of the hadronic rescattering phase at LHC energies. A $k_{\mathrm{T}}$ scaling of pion and kaon source radii is observed instead. The time of maximal emission of the system is estimated using the three-dimensional femtoscopic analysis for kaons. The measured emission time is larger than that of pions. Our observations are well supported by the hydrokinetic model predictions. For correlation functions (Figure 2, Figure 3) the errors are statistical only, systematic uncertainties are equal to or less than the statistical errors. For all other figures total errors are presented.

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# Stirling's Approximation for binomial coefficient

In this proof, it is assumed that, for $$k \ll n$$, $${n \choose k} \approx \frac{n^k}{k!}$$, given Stirling's approximation.

How does Stirling's Approximation, in either form $$\ln n! \approx n\ln{n} -n + (\ln(n))$$ or $$n! \approx \sqrt{2\pi n} (\frac{n}{e})^n$$, give this result?

• Stirling's approximation isn't needed here. You just need to show that $\frac{n!}{(n-k)!}\approx n^k$. Sep 23, 2015 at 0:19
• Ok, but for $k <<n$ I get $\frac{n!}{(n-k)!} \approx \frac{n!}{n!} = 1$ which is where I have been stuck.
– user157818
Sep 23, 2015 at 0:20
• Oh I see so can we say $\frac{n!}{(n-k)!} = \frac{n(n-1)(n-2) ... (n-k)(n-k-1) ... (1)}{(n-k)(n-k-1)...(1)}=n(n-1)(n-2)...(n-k+1)=n^k + O(n^{k -1})$
– user157818
Sep 23, 2015 at 0:23
• For fixed $k\ge 1$, $\lim_{n\to\infty} \frac{n!}{(n-k)!}=\infty$, so the $\approx 1$ a few comments above is very wrong. Sep 23, 2015 at 0:26

Although certainly not the most efficient approach, we can use Stirling's Formula to arrive at the asymptotic expression for $$\binom{n}{k}$$. We have

$$\frac{\sqrt{2\pi n}(n/e)^n}{k!\sqrt{2\pi(n-k)}((n-k)/e)^{n-k}}=(1-k/n)^{k-1/2}\frac{e^{-k}}{k!(1-k/n)^{n}}n^k \tag 1$$

Note that the first term on the right-hand side of $$(1)$$ tends to $$1$$ as $$n\to \infty$$. Note also that the denominator of the middle term on the right-hand side goes to $$k!e^{-k}$$. Therefore the middle term also tends to $$1/k!$$. We are left with $$n^k/k!$$ as expected.

Note that the notation $$k\ll n$$ is nebulous (See THIS note's discussion on asymptotics of the binomial coefficient). Herein, we have tacitly assumed that $$k$$ is fixed and that $$k=o(\sqrt n)$$.

• Isn't the denominator in the LHS supposed to be $\sqrt{2\pi(n-k)}$ instead of $\sqrt{\pi(n-k)}$. Feb 6, 2017 at 19:01
• @Taylor Yes. I've edited. Thank you for caching the typo. Feb 6, 2017 at 19:45
• I know this is quite old, but whey do you say that the first term in $(1)$ tends to $1$? The question is for $k \ll n$. But $k = n / \log n$ satisfies this. So the term actually tends to $\exp(-k^2/n)$, which may well diverge. Apr 15, 2020 at 8:54
• @SamT The notation $k<<n$ is nebulous. Herein, we assumed that $k$ is fixed and that $k=o(\sqrt n)$. I added a note at the end of the post to state this assumption. Apr 15, 2020 at 17:39
• @SamT You raised a good point. See this short note on the binomial coefficient. In particular, the sections on Bounds and Asymptotics are of interest here. Apr 16, 2020 at 16:13

The approximation $n! \approx (n/e)^n$ suffices. As $n \to \infty$ and $k/n \to 0$ we have

$$(n-k)! \approx \left(\frac{n-k}{e}\right)^{n-k}= \left(\frac{n}{e}\right)^{n-k} (1-k/n)^{n-k} \approx\left(\frac{n}{e}\right)^{n-k} e^{-k}$$

Hence

$${n \choose k} \approx \frac{ \left(\frac{n}{e}\right)^{n}} {\left(\frac{n}{e}\right)^{n-k} e^{-k} \, k!} = \frac{n^k}{k!}$$

• This is nice but $n! \approx \frac{n^n}{e^n}$ is wrong, unless I am not interpreting $\approx$ correctly. To me, $a_n\approx b_n$ means $\frac{a_n}{b_n}\to 1$. By Stirling's theorem your approximation is off by a factor of $\sqrt{n}$, (which later cancels in the fraction expressing the binomial coefficients). Sep 30, 2015 at 18:21
• I may be wrong but that double twidle sign stands for "approximately equal to". I believe you are referring to only one twidle sign when it comes to asymptotc equivalence. Jun 4, 2019 at 13:59
• @Juan123 Good point, I'm being a little sloppy in that regard. Actually the first "approximation" is just that, it's not an asympotic equivalence (we'd need the factor $\sqrt{2 \pi n}$ for that). (This also answers the previous comment). Jun 4, 2019 at 14:42
• I know this is old, but as with the other answer, this one also seems to be very slightly wrong. Of course, "$\approx$" can mean basically whatever you like. If "$a_n \approx b_n$" means that "$\log a_n / \log b_n \to 1$", then it's true. But you're missing a factor $\exp(-k^2/n)$, coming from $(1 - k/n)^k$. (Note that you only assume that $k/n \to 0$, not that $k^2/n \to 0$.) Apr 15, 2020 at 8:56
• @SamT I wrote the first order (logarithmical) term for the approximation. Even if $k^2/n$ diverges, this is still correct, just like the usual $n! \approx (n/e)^n$ Apr 15, 2020 at 11:12


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# Thread: test n varience unkown

1. ## test n varience unkown

Im not sure how to do this question. A chemical test on a compoun gave the following results(grams of a particular element in 1kg of compound):1.60, 1.56 ,1.59, 1.53, 1.58 , 1.52, 1.59 ,1.64, 1.55. From these figures make an estimate of the size of the sample required so that the standard error of the mean will be about 0.005. Explain what assumptions have to be made, and what sources of error there could be in the estimate.

I dont quite understand the structure of the working required. I believe the population mean to be 1(unsure) sample variece to be 0.0122, sample mean 1.57 and from here whether i use a chi squared test or t test and how inco-operate an n value to estimate the standard error??? Somewhere along the line do i use a null hypothesis of mean = 1, alternative mean < 1 then use a type 1 error? The answer is 54 using the sample varience, pop varience unkown, small sample so sample varience may not be very good estimate of pop varience. If somebody could show me the working that for 54 id appreciate it very much thanks.

2. Originally Posted by oxrigby
Im not sure how to do this question. A chemical test on a compoun gave the following results(grams of a particular element in 1kg of compound):1.60, 1.56 ,1.59, 1.53, 1.58 , 1.52, 1.59 ,1.64, 1.55. From these figures make an estimate of the size of the sample required so that the standard error of the mean will be about 0.005. Explain what assumptions have to be made, and what sources of error there could be in the estimate.

I dont quite understand the structure of the working required. I believe the population mean to be 1(unsure) sample variece to be 0.0122, sample mean 1.57 and from here whether i use a chi squared test or t test and how inco-operate an n value to estimate the standard error??? Somewhere along the line do i use a null hypothesis of mean = 1, alternative mean < 1 then use a type 1 error? The answer is 54 using the sample varience, pop varience unkown, small sample so sample varience may not be very good estimate of pop varience. If somebody could show me the working that for 54 id appreciate it very much thanks.
The standard error of the mean of a sample of size $n$ is:

$se=\frac{\sigma}{\sqrt{n}}$

so the required sample size to give a $se$ of $0.005$ would be:

${n}=\frac{\sigma^2}{0.005^2}$

we will estimate this by:

$\widehat{n}=\frac{s^2}{0.005^2}$

where $s^2$ is the unbiased estimator of the population variance derived from the sample.

The source of error in this estimate is the sampling error in the sample based variance estimate.


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# How To Calculate The Value Of X In Angles References

How To Calculate The Value Of X In Angles. (1) we know that in the triangle aec exterior angle is equal to the sum of two opposite interior angles. //calculate the total accelerometer vector //57.296 = 1 / (3.142 / 180) the arduino asin function is in radians angle_pitch_acc = asin((float)acc_y / acc_total_vector) * 57.296;

A common system of quantifying angles is degrees; A transversal is a line that crosses at least two other lines.

### 7 Inch Metric Aluminum Alloy Speed Square Roofing Triangle

Ab = ac and bd = cd. Acc_total_vector = sqrt((acc_x * acc_x) + (acc_y * acc_y) + (acc_z * acc_z));

### How To Calculate The Value Of X In Angles

Another practical formula is the sine of 3 times an angle:B) to find angle x, subtract 41° and the value of a right angle from 180°.Below are some example angles with their associated degree measures.C) as angle z is one of 5 angles around a point, you can calculate angle z by dividing 360° by 5.

Calculate the angle x and angle y value in the given figure?Calculate the value of $$p$$.Calculate the value of $$r$$.Calculate the value of $$t$$.

Calculate the value of $$y$$.Calculate the value of $$z$$.Calculate the value of angle x and complete the sentence below to describe the relationship between the two angles.Calculate the value of n.

Calculate the values of angles a, b, and c and complete the sentence that describes their relationship.Calculate the x and y value in the given figure.Cos(x) is a trigonometric function which is used to calculate the value of x angle.D) find the missing angles x , y and z.

Do not round this answer yet.Eugene brennan (author) from ireland on november 23, 2017:For example, the sine of 9 degrees is the sine of (3×3°).Formulas to calculate the angles of the right triangles are given as, sin a = x/z cos a = y/z tan a = x/y formula to calculate the sides of the right.

Hi jeetendra, this is called a scalene triangle.If a = 40 o, find boc.If the bisectors of b and c of triangle of a abc intersect at o, then prove that boc = 90 o +.If the sum of two angles adds up to 180°, then the angles are said to.

In the figure given below, in triangle aob and triangle aoc, ab = ac (given) ⇒ ∠abo = ∠aco (angles opposite to equal sides.Justify your statement in each case.Let as consider x, y and z are the three sides of the right angle triangles and x is the angles of the right angle triangles.Let x = 3k, y = 4k and z = 5k ∠adb = x + z = 8k and ∠abc = x + y = 7k (exterior angle of a ∆ is equal to the sum of pair of interior opposite angles) also, ∠abc + ∠adc = 180° (pair of opposite angles in a cyclic quadrilateral are supplementary) ⇒ 8k + 7k = 180° ⇒ 15k = 180° `∴ k.

Let’s calculate ∠x and ∠y.Lines intersected by a transversal pairs of angles formed by a transversal.Notice that the angles start from zero on the left and increase to the point at which the rays have returned to their initial coincident positions.Opposite angles of a cyclic quadrilateral sum 180° , so acb.

Setting your miter saw to 0 degrees is actually cutting a 90 degree angle on the material.Simply enter your desired height, width, and material width, and the calculator will display the x angles and lengths.Sin 3a = 3 sin a − 4 sin 3 a.Sines of other angles many angles can be computed exactly by many methods.

So we can write it as.So we can write it as.So, the value of x is 136º.So, the value of x is 18°.

So, with a = 3, we arrive atThe bisectors of cbd and bce meet at o.The leftmost diagram above would be an angle of 0 degrees (or 0°), and the rightmost diagram above would be an angle of 360 degrees (or 360°).The longest edge of any triangle is opposite the largest angle.

The right angle triangles are shown below:The sides ab and ac of a triangle are produced to d and e respectively.The value of angle x is 37º and the value of angle y is 70º.The value of the angle x and angle y can be calculated using the congruence property of the triangle.

Then pythagoras theorem is used to calculate the value of angles in right triangles.Then the diagonally opposite angles are equal.This means that when two (or more lines) create an x the angles in the opposite corners are equal to each other.To determine the angle to set your miter saw, simply subtract the angle on the material from 90.

Two sides of each of the regular pentagons and two sides of a meet at the point p.Use the information given in the diagram to find x.Use the information given in the diagram to find x.We can calculate the values of the sines of 18° and 72° from the above expression.

We know that in the triangle bde exterior angle is equal to the sum of two opposite interior angles.X = _____ the 130° angle and angle x are _____ angles.X = _____ the 60° angle and angle x are _____ angles.X o = ∠ceb + 45 o.

$$\cos{x} = \frac{a}{h}$$ $\cos{x} = \frac{4}{6}$ \(\cos{x} = 0.∠abd = ∠acd or ∠bdc = ∠bac.∠bac = ∠bad + ∠cad.∠bdc = ∠adb + ∠adc and.

∠cdb = ∠ceb + ∠dbe.

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# Standing wave and its formation

1. Aug 15, 2012

### pcsx22

I came across a question about wave from univeristy physics (sears n zemansky) and the question was
'can a standing wave.be produced on a string by by superposing two waves travelling in opposite direction with same frequency but different amplitudes? And same amplitude but different frequencies?

I have no idea about this question but i believe two opposite waves of same frequency but different amplitudes can form standing wave..

Last edited by a moderator: Aug 16, 2012
2. Aug 16, 2012

### Jolb

Well, let me show you some equations and maybe you can answer the question for yourself.

How do we represent a wave with wavenumber k moving to the right at a velocity c? The equation would be:
$$\sin(kx-ct)$$
This means we have a sine wave that is periodic in space with a width of one period equal to 2∏/k. However, it's moving to the right at the velocity c.

What if we superpose two waves of equal amplitude and wavenumber, one moving to the right and one moving to the left, both at a velocity c? The combined wave would be:
$$\sin(kx-ct)+\sin(kx+ct)$$
If we apply the angle addition formula:
$$\sin(kx-ct)+\sin(kx+ct)=[\sin(kx)\cos(ct)-\cos(kx)\sin(ct)]+[\sin(kx)\cos(ct)+\cos(kx)\sin(ct)]=2\sin(kx)\cos(ct)$$
Thus the sum of these two waves is just a sine wave of wavenumber k moving at velocity=0, modulated by the factor cos(ct). It is not moving, the only thing changing is its overall amplitude.

Would the same things hold if the two superposed waves had differing amplitudes or frequencies (wavenumbers)?

Last edited: Aug 16, 2012
3. Aug 16, 2012

### pcsx22

So you're saying if two opposite waves of different amplitudes and frequency superimpose, velocity of the resulting wave will not be zero therefore standing wave is not formed?? Correct me if I'm wrong

Last edited by a moderator: Aug 16, 2012
4. Aug 16, 2012

### sophiecentaur

Any reflection at an interface will produce a standing wave - it's just that the anti-nodes will not have zero amplitude because a small reflected wave will not produce perfect cancellation of the forward wave. This sort of thing is easiest to observe with EM waves on an RF feeder with a mis-matched load. The 'Voltage Standing Wave Ratio' (VSWR) gets nearer to unity as you improve the match of the load. It used to be the most convenient way of measuring the match but nowadays, the reflection coefficient is often easier to measure. (You can calculate one from the other.)

5. Aug 16, 2012

### Jolb

A standing wave, by definition, is a wave with zero velocity. For a wave to have a well-defined velocity (in the usual sense), you have to be able to write it in the form $A(t)\sin(kx-ct)$, where A(t) is a possibly time-dependent amplitude. In this case, the constant c is the velocity.

The only reason I was able to write the superposition of the two waves in the following way
$$\sin(kx-ct)+\sin(kx+ct)=[\sin(kx)\cos(ct)-\cos(kx)\sin(ct)]+[\sin(kx)\cos(ct)+\cos(kx)\sin(ct)]=2\sin(kx)\cos(ct)$$
is a lucky coincidence of the fact that there is exact cancellation between the two terms of the form $\cos(kx)\sin(ct)$. If we had waves with arbitrary amplitudes A and B and arbitrary wavenumbers j and k moving in opposite directions, then their superposition $A\sin(jx-ct)+B\sin(kx+ct)$ doesn't in general have the exact cancellation and so the wave cannot be written in the form $A(t)\sin(kx-ct)$ required to identify the wave's velocity.

Last edited: Aug 16, 2012
6. Aug 22, 2012

### alextx

7. Aug 22, 2012

### Alvydas

For you can be interesting standing wave of this type:
sin(ω*(x/(c+v) +t) + sin(ω*(x/(c-v) -t)
It is interesting because visually (if you animate) it will not look like standing wave.
But if you look at any particular x point you will find it waves with the same amplitude
like classical standing wave
sin(ω*k*(x/(c) +t) + sin(ω*k*(x/(c) -t), where k = 1/(1-v^2/c^2)


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# How can you use trigonometric functions to simplify  12 e^( ( 19 pi)/12 i )  into a non-exponential complex number?

Apr 19, 2018

$3 \sqrt{6} - 3 \sqrt{2} - i \left(3 \sqrt{6} + 3 \sqrt{2}\right)$

#### Explanation:

We can turn into $r {e}^{i \theta}$ into a complex number by doing: $r \left(\cos \theta + i \sin \theta\right)$

$r = 12$, $\theta = \frac{19 \pi}{12}$

$12 \left(\cos \left(\frac{19 \pi}{12}\right) + i \sin \left(\frac{19 \pi}{12}\right)\right)$

$3 \sqrt{6} - 3 \sqrt{2} - i \left(3 \sqrt{6} + 3 \sqrt{2}\right)$


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4.6.2: Surface Tension

The surface tension is one of the mathematically complex topic and related to many phenomena like boiling, coating, etc. In this section, only simplified topics like constant value will be discussed. In one of the early studies of the surface tension/pressure was done by Torricelli. In this study he suggest construction of the early barometer. In barometer is made from a tube sealed on one side. The tube is filled with a liquid and turned upside down into the liquid container. The main effect is the pressure difference between the two surfaces (in the tube and out side the tune). However, the surface tension affects the high. This effect is large for very small diameters.

Example 4.31:

In interaction of the molecules shown in Figure ? describe the existence of surface tension. Explain why this description is erroneous?

Solution 4.31:

The upper layer of the molecules have unbalanced force towards the liquid phase. Newton's law states when there is unbalanced force, the body should be accelerate. However, in this case, the liquid is not in motion. Thus, the common explanation is wrong.

Fig. 4.47. A heavy needle is floating on a liquid.

Example 4.32:

Needle is made of steel and is heavier than water and many other liquids. However, the surface tension between the needle and the liquid hold the needle above the liquid. After certain diameter, the needle cannot be held by the liquid. Calculate the maximum diameter needle that can be inserted into liquid without drowning.

Solution 4.32:

Under Construction


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## Back To School '18: An FFT Problem

View as PDF

Points: 12 (partial)
Time limit: 2.0s
Memory limit: 128M

Author:
Problem type

To prepare for the upcoming school year, Richard has bought books for his English class. Each book is assigned a value, , Richard's willingness to read that book.

Richard wants to choose of the books and calculate his willingness to read those books. The willingness to read those books is the product of the willingness to read each individual book. For example, if he bought books of value , and he chose books with indices , the willingness to read those books would be .

Richard wants the sum of the willingness of all distinct combinations of books for all values of .

However, since Richard does not like large numbers, he wants each sum modulo .

Two combinations are considered distinct if the indices of the books chosen are different, regardless of the values of the books.

#### Input Specification

The first line of input will contain a single integer , the number of books that Richard bought.

The second line of input will contain space-separated integers, the integer representing , the value of each book.

#### Output Specification

On one line, print space-separated integers, the integer representing the sum of the willingness of all distinct combinations of choosing books, modulo .

We define modulo in the 2 equivalent ways:

1. Taking the remainder of , adding if the result is negative.
2. Subtracting from , or adding to , until is in the interval .

It may or may not help to know that .

No further constraints.

#### Sample Input 1

4
1 2 2 3

#### Sample Output 1

8 23 28 12

#### Explanation for Sample Input 1

, .

There are distinct combinations to read book:

Their sum is .

There are distinct combinations to read books.

Their sum is .

There are distinct combinations of reading books.

Their sum is .

The only distinct combination of reading books is .

#### Sample Input 2

3
-1 -1 -1

#### Sample Output 2

998244350 3 998244352


text:
Example 70.6.11. Let $k$ be a field. Let

$U = \mathop{\mathrm{Spec}}(k[x, y]/(xy))$

be the union of the coordinate axes in $\mathbf{A}^2_ k$. Denote $\Delta : U \to U \times _ k U$ the diagonal and $\Delta ' : U \to U \times _ k U$ the map $u \mapsto (u, \sigma (u))$ where $\sigma : U \to U$, $(x, y) \mapsto (y, x)$ is the automorphism flipping the coordinate axes. Set

$R = \Delta (U) \amalg \Delta '(U \setminus \{ 0_ U\} )$

where $0_ U \in U$ is the origin. It is easy to see that $R$ is an étale equivalence relation on $U$. The quotient $X = U/R$ is an algebraic space. The morphism $U \to \mathbf{A}^1_ k$, $(x, y) \mapsto x + y$ is $R$-invariant and hence defines a morphism

$X \longrightarrow \mathbf{A}^1_ k$

This morphism is a universal homeomorphism and an isomorphism over $\mathbf{A}^1_ k \setminus \{ 0\}$. It follows that $X$ is integral and Noetherian. Exactly as in Example 70.6.10 the reader shows that $\text{Cl}(X) = \mathbf{Z}/2\mathbf{Z}$ with generator corresponding to the unique closed point $x \in |X|$ mapping to $0 \in \mathbf{A}^1_ k$. However, in this case the henselian local ring of $X$ at $x$ isn't a domain, as it is the henselization of $\mathcal{O}_{U, 0_ U}$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach space. I think that everyone who tries to study "classical" operator spaces like $\mathcal{K}$, $p$-Schatten class operators etc. immediately discovers the similarity with "commutative" counterparts, i.e. $c_0$ and $\ell^p$. This phenomenon is visible when one uses (generalised) singular numbers for certain classes of operators. Again, I have got plenty of questions concerning this stuff, let me list at least two of them:
1) what are the complemented subspaces of $\mathcal{K}$? Is $\mathcal{K}$ complemented in $\mathcal{B}(\ell^2)$? Recently, Haydon and Argyros constructed an HI-space $E$ such that $\mathcal{K}(E)$ has codimension 1 $\mathcal{B}(E)$, thus complemented.
2) is every bounded operator from $p$-Schatten class to $\mathcal{K}$ compact?
What other properties $\mathcal{K}$ shares with $c_0$?


text:
# Source Panel method for flow pressure distribution over cylinder

1. Feb 19, 2012

### frozenguy

1. The problem statement, all variables and given/known data
Develop a source-panel computer program to numerically calculate the potential flow pressure distribution on a right circular cylinder with and without circulation.

1.) Tabulate and plot your numerical results for the pressure coefficient distribution on the cylinder together with the corresponding exact analytic solution as a function of the angular variable omega, measured clockwise from the negative x-axis to the center of panels, from 0-360 degrees. Tabulate the following results:
i, omega, q, numerical Cp, theoretical Cp, % error.

2.) Do your calculations for the following values of circulation: K =$\Gamma/\left(2\pi\right)$ = 0, -1, -2, and -3. For K=0 do your calculations for N=4 and 250. For all other values of K, use only N=250.

2. Relevant equations
w=omega
Theoretical Cp=1-4*sin(w)^2
w=(360/N)(i-0.5)
θi= ARCtan((yi+1-yi)/(xi+1-xi))
ζi=0.5(xi+xi+1)
ηi=0.5(yi+yi+1)
yi=sin(wi)
xi=cos(wi)

U0sin(θi - α) =(qi)/2 - SUM^{N}_{j=1}qjcij j does NOT equal i
cij=sin(θi-$\Phi$ij)/2$\pi$rijΔS
rij=[(ζij)2+(ηij)2]1/2
$\Phiij$=ARCtan(ηji)/(ζji)
Sorry but I am making some mistake with the complex symbol program on this forum.

3. The attempt at a solution

I have calculated i, ωi, ηi, ζi, xi, yi, θi for N=250. I can't find the j components of those needed.

Also I'm having trouble finding out the length of each panel. If a square was inside the unit circle, each length would be 1.414 right? But N=250 not 4.

Last edited: Feb 19, 2012


text:
## Results (1-50 of at least 1000)

Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin
2-567-63.52-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.52 $$0.0 0 0.0909 0 1.72107 Modular form 567.1.t.a.136.1 2-567-7.6-c0-0-3 0.531 0.282 2 3^{4} \cdot 7 7.6$$ $0.0$ $0$ $0$ $0$ $2.52899$ Modular form 567.1.d.c.244.2
2-567-63.34-c0-0-2 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.34 $$0.0 0 -0.0555 0 1.67920 Modular form 567.1.l.b.55.1 2-567-63.32-c0-0-0 0.531 0.282 2 3^{4} \cdot 7 63.32$$ $0.0$ $0$ $0.232$ $0$ $2.25104$ Artin representation 2.567.6t5.b.a Modular form 567.1.n.a.53.1
2-567-63.34-c0-0-1 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.34 $$0.0 0 -0.138 0 1.65885 Modular form 567.1.l.d.55.1 2-567-63.34-c0-0-3 0.531 0.282 2 3^{4} \cdot 7 63.34$$ $0.0$ $0$ $0.111$ $0$ $1.80881$ Artin representation 2.567.6t5.d.b Modular form 567.1.l.c.55.1
2-567-63.13-c0-0-3 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.13 $$0.0 0 -0.111 0 1.89819 Artin representation 2.567.6t5.d.a Modular form 567.1.l.c.433.1 2-567-63.13-c0-0-2 0.531 0.282 2 3^{4} \cdot 7 63.13$$ $0.0$ $0$ $0.0555$ $0$ $1.86503$ Modular form 567.1.l.b.433.1
2-567-63.34-c0-0-4 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.34 $$0.0 0 0.361 0 2.71344 Modular form 567.1.l.d.55.2 2-567-63.13-c0-0-4 0.531 0.282 2 3^{4} \cdot 7 63.13$$ $0.0$ $0$ $0.388$ $0$ $2.35058$ Modular form 567.1.l.a.433.1
2-567-63.23-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.23 $$0.0 0 0.0454 0 2.02089 Artin representation 2.567.6t5.c.b Modular form 567.1.j.a.296.1 2-567-63.34-c0-0-0 0.531 0.282 2 3^{4} \cdot 7 63.34$$ $0.0$ $0$ $-0.388$ $0$ $0.866612$ Modular form 567.1.l.a.55.1
2-567-63.13-c0-0-1 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.13 $$0.0 0 -0.361 0 1.54709 Modular form 567.1.l.d.433.2 2-567-7.6-c0-0-2 0.531 0.282 2 3^{4} \cdot 7 7.6$$ $0.0$ $0$ $0$ $0$ $1.91754$ Artin representation 2.567.6t3.a Artin representation 2.567.6t3.a.a Modular form 567.1.d.b Modular form 567.1.d.b.244.1
2-567-7.6-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 7.6 $$0.0 0 0 0 1.17964 Modular form 567.1.d.c.244.1 2-567-7.6-c0-0-1 0.531 0.282 2 3^{4} \cdot 7 7.6$$ $0.0$ $0$ $0$ $0$ $1.30083$ Artin representation 2.567.3t2.b Artin representation 2.567.3t2.b.a Modular form 567.1.d.a Modular form 567.1.d.a.244.1
2-567-63.2-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.2 $$0.0 0 -0.232 0 0.950768 Artin representation 2.567.6t5.b.b Modular form 567.1.n.a.107.1 2-567-63.13-c0-0-0 0.531 0.282 2 3^{4} \cdot 7 63.13$$ $0.0$ $0$ $0.138$ $0$ $1.11999$ Modular form 567.1.l.d.433.1
2-567-63.61-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.61 $$0.0 0 0.131 0 2.24789 Modular form 567.1.k.a.460.1 2-567-63.40-c0-0-0 0.531 0.282 2 3^{4} \cdot 7 63.40$$ $0.0$ $0$ $-0.0909$ $0$ $1.20133$ Modular form 567.1.t.a.271.1
2-567-63.11-c0-0-0 $0.531$ $0.282$ $2$ $3^{4} \cdot 7$ 63.11 $$0.0 0 -0.0454 0 1.63236 Artin representation 2.567.6t5.c.a Modular form 567.1.j.a.431.1 2-567-63.31-c0-0-0 0.531 0.282 2 3^{4} \cdot 7 63.31$$ $0.0$ $0$ $-0.131$ $0$ $1.43296$ Modular form 567.1.k.a.514.1
2-567-567.104-c1-0-11 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 0.362 0 0.774803 Modular form 567.2.bm.a.104.5 2-567-567.101-c1-0-54 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $0.314$ $0$ $1.97505$ Modular form 567.2.br.a.101.53
2-567-567.104-c1-0-12 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 -0.261 0 0.782479 Modular form 567.2.bm.a.104.37 2-567-567.101-c1-0-33 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $-0.0311$ $0$ $1.16148$ Modular form 567.2.br.a.101.54
2-567-567.101-c1-0-44 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 0.193 0 1.50588 Modular form 567.2.br.a.101.8 2-567-567.101-c1-0-53 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $0.358$ $0$ $1.94767$ Modular form 567.2.br.a.101.31
2-567-567.101-c1-0-63 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 0.244 0 2.46216 Modular form 567.2.br.a.101.57 2-567-567.104-c1-0-10 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.447$ $0$ $0.692688$ Modular form 567.2.bm.a.104.65
2-567-567.101-c1-0-29 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 -0.143 0 1.08882 Modular form 567.2.br.a.101.14 2-567-567.101-c1-0-34 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $-0.239$ $0$ $1.17063$ Modular form 567.2.br.a.101.36
2-567-567.101-c1-0-26 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 -0.303 0 0.918684 Modular form 567.2.br.a.101.45 2-567-567.101-c1-0-23 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $-0.229$ $0$ $0.877061$ Modular form 567.2.br.a.101.39
2-567-567.101-c1-0-27 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 -0.204 0 0.926451 Modular form 567.2.br.a.101.17 2-567-567.101-c1-0-35 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $0.0685$ $0$ $1.18124$ Modular form 567.2.br.a.101.12
2-567-567.104-c1-0-13 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 -0.450 0 0.827111 Modular form 567.2.bm.a.104.67 2-567-567.104-c1-0-54 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.359$ $0$ $1.95659$ Modular form 567.2.bm.a.104.24
2-567-567.104-c1-0-42 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 -0.255 0 1.69880 Modular form 567.2.bm.a.104.68 2-567-567.104-c1-0-55 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.0510$ $0$ $1.96428$ Modular form 567.2.bm.a.104.52
2-567-567.104-c1-0-28 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 0.236 0 1.36959 Modular form 567.2.bm.a.104.35 2-567-567.104-c1-0-25 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.0944$ $0$ $1.33278$ Modular form 567.2.bm.a.104.21
2-567-567.104-c1-0-32 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 -0.134 0 1.49722 Modular form 567.2.bm.a.104.51 2-567-567.104-c1-0-41 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.00507$ $0$ $1.68689$ Modular form 567.2.bm.a.104.63
2-567-567.104-c1-0-50 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.104 $$1.0 1 0.451 0 1.87990 Modular form 567.2.bm.a.104.20 2-567-567.104-c1-0-53 2.12 4.52 2 3^{4} \cdot 7 567.104$$ $1.0$ $1$ $0.387$ $0$ $1.94531$ Modular form 567.2.bm.a.104.39
2-567-567.101-c1-0-20 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 -0.375 0 0.816816 Modular form 567.2.br.a.101.42 2-567-567.101-c1-0-22 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $-0.107$ $0$ $0.853819$ Modular form 567.2.br.a.101.26
2-567-567.101-c1-0-24 $2.12$ $4.52$ $2$ $3^{4} \cdot 7$ 567.101 $$1.0 1 -0.146 0 0.882477 Modular form 567.2.br.a.101.3 2-567-567.101-c1-0-25 2.12 4.52 2 3^{4} \cdot 7 567.101$$ $1.0$ $1$ $-0.269$ $0$ $0.887226$ Modular form 567.2.br.a.101.23
Next


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# Homework Help: Getting a rhombus vertex from one vertex and its area

1. Jun 9, 2010

### Telemachus

Hi there. Im tryin to solve this one, I know that if I find the way to get one more vertex i'd have it solved.

1. The problem statement, all variables and given/known data
The point A(-1,8) is the vertex of a rhombus which minor diagonal is situated on the line L: $$L=\begin{Bmatrix} x=3\mu & \mbox{ }& \\y=1+4\mu & \mbox{}&\end{matrix}$$, $$\mu\in{R}$$. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.

2. Relevant equations
$$A=\displaystyle\frac{dD}{2}$$

3. The attempt at a solution
Well, I haven't done too much. Actually I did some, but then I realized that I had confused something, cause I got the line L' where would be located the major D, but I've used for it the point A, and then I was trying to get the point of intersection between L and L', so then I doubled it, and I was going to get my second vertex, but then I realized that it was wrong, cause I couln't use point A, cause A don't belongs to L', it belongs to L, so...

$$30=\displaystyle\frac{dD}{2}\Rightarrow{60=dD}$$

So, I don't know much about L', but that its perpendicular to L.

$$L'=\begin{Bmatrix} x=x_0+4/3\lambda & \mbox{ }& \\y=y_0-\lambda & \mbox{}&\end{matrix}$$

Last edited: Jun 9, 2010
2. Jun 9, 2010

### tiny-tim

Hi Telemachus!
That's right , but from then on you're making it too complicated.

Hint: Where is the midpoint of the rhombus?

3. Jun 9, 2010

### Telemachus

On the intersection between L and L', but how should I get L'? I thought maybe using pythagoric equation I could find another vertex, but im not too sure. I know it gives enough data, but I don't know how to use it.

4. Jun 9, 2010

### Telemachus

I had a misunderstood, I think I was on the right way, cause actually I didn't corroborate if A belongs to L, and the sentence don't say so. I'll corroborate, if it doesn't I was on the right way, and it will be easy to solve, cause if A don't belongs to L, it must belongs to L'.

Bye there, and thanks.

5. Jun 9, 2010

### tiny-tim

L' is perpendicular to L, and goes through A.

(and remember, you don't need to find L', you only need the midpoint )

6. Jun 9, 2010

### Telemachus

How could I get the midpoint without L'? I thought of it as the intersection between L and L'.

7. Jun 9, 2010

### tiny-tim

AM is on L', so it's perpendicular to L.

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Thread: Condition number of simple eigenvalues

1. Condition number of simple eigenvalues

Hi, there's a step in the derivation of the condition number that I do not understand.

Let $A\in\mathbb{C}^{n\times n}$ and let $\lambda$ be a simple eigenvalue of $A$.
Let $u$ be a right eigenvector and perturbe such that

$(A+\delta A)(u +\delta u) = (\lambda +\delta\lambda)(u + \delta u).$

I multiply this thing and use the fact that $Au=\lambda u$ to get,

$A(\delta u) +(\delta A)u = \lambda(\delta u) + (\delta\lambda)u.$

Now let $v^*$ be a left eigenvector of $A$ and multiply the above equation from the left. Using the fact that $v^*A=v^*\lambda$ we can write

$v^*(\delta A)u = v^*(\delta\lambda)u,$

hence,

$|\delta\lambda| = \frac{|v^*(\delta A)u|}{|v^*u|}.$

By assuming that $||u||_2=||v||_2=1$ we can bound this as

$|\delta\lambda| \leq \frac{||\delta A||_2}{|v^* u|}.$

I hope you are still awake What happens next is confusing me. If $\lambda\neq 0$ then

$\frac{|\delta\lambda|}{|\lambda|}\leq \frac{||A||_2}{|\lambda||v^*u|}\cdot\frac{||\delta A||_2}{||A||_2},$

and therefore

$cond(\lambda) = \frac{||A||_2}{|\lambda| |v^*u|}.$

I do not understand the (and therefore) part Is this some alternative definition of condition number? Hope someone has the patience to help me out, thanks!

2. This doesn't look quite right to me either. The condition number of a simple eigenvalue is $1/|v^*u|$. What's the reference here?
Now that I think about it, the quantities given are relative errors. This might be a kind of relative condition number, if such a notion exists.


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## Problem of the Day #81: Reversing digitsJune 8, 2011

Posted by Sreenath in : potd , trackback

Suppose that a two digit integer is equal to the sum of its digits times some integer $k$. Let $a$ be the integer formed by reversing the digits of this number, and let $b$ be the sum of its digits. Find an expression for $\frac{a}{b}$ in terms of $k$.

## Comments»

no comments yet - be the first?


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# Comparing divergent and convergent sums

Let $$(x_n)$$ be a monotonically decreasing sequence of positive real numbers that is also summable.

Let $$(y_n)$$ be a sequence of positive real numbers such that $$\sum_n x_n y_n$$ converges.

Let $$(z_n)$$ be a monotonically increasing sequence of positive real numbers such that $$\sum_n x_n z_n =\infty.$$

Assume that the sequences $$y_n$$ and $$z_n$$ are such that $$2^{-\varepsilon y_n}$$ and $$2^{-\varepsilon z_n}$$ are summable for every $$\varepsilon>0.$$ Does it follow that there is some $$\delta>0$$ such that

$$\sum_n \Big(2^{-\varepsilon y_n}-2^{-\varepsilon z_n}\Big) \ge 0 \text{ for all } \varepsilon \in (0,\delta)?$$

The motivation for this statement to be true is that $$z_n$$ should be larger most of the time than $$y_n$$ and we capture this most of the time by taking $$\varepsilon$$ small.

Please let me know if you have any comments, questions or remarks.

No. For example $$x_n=2^{-n}$$, $$y_n=n$$, and we now keep $$z_n$$ constant on long intervals. More precisely, consider first $$\epsilon=1$$, and set $$z_1= \ldots = z_{N_1}=c_1$$, with $$N_1$$ taken so large that $$N_1 2^{-1\cdot 1}>\sum 2^{-1\cdot n}$$.
We then continue in the same way: let $$z_{N_1+1}=\ldots = z_{N_2}= c_2$$, and again take $$N_2$$ so large that the inequality you want fails for $$\epsilon=1/2$$ etc.
Here we must choose the $$c_n$$ such that $$\sum 2^{-n}z_n=\infty$$ and $$\sum 2^{-\epsilon z_n}<\infty$$. I'll take $$c_n=2^{N_{n-1}}$$ to guarantee the first property. If the $$n$$th step deals with $$\epsilon=1/n$$, then, since $$\sum n2^{-\epsilon n} \simeq 1/\epsilon^2$$, I can take $$N_n=n^3 2^{N_{n-1}/n}$$ to make sure your inequality fails.
With these choices in place, the final condition becomes $$\sum N_n 2^{-\epsilon 2^{N_{n-1}}} =\sum n^3 2^{(-\epsilon+1/n)2^{N_{n-1}}}< \infty$$ for all $$\epsilon >0$$, and this holds because the $$N_n$$ increase rapidly.
• thanks a lot for this wonderfully worked out example. If I understand things correctly, then what matters here of course is that I am looking for this delta neighbourhood. I could have also asked if $\liminf_{\epsilon \to 0} \frac{\sum_n 2^{-\varepsilon y_n}}{\sum_n 2^{-\varepsilon z_n}}>0$ holds true to better quantify their relative size to one another? I am not entirely sure how the series of your constructed $z_n$ scales with $\varepsilon$ so I don't see it right away? Mar 20, 2021 at 19:11
• @PritamBemis: I think the basic issue here is that $\sum 2^{-\epsilon u_n}$ (for $u=y$ or $u=z$) only depends on what values $u_n$ takes, not on where exactly these occur, while the other series $\sum x_nu_n$ can of course be quite sensitive to rearrangements (for given $x$). This suggests that we shouldn't really expect a very close connection between the two, and I think the situation is still essentially the same in the modification you suggested. Mar 20, 2021 at 20:41


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# The figure shows a rectangular conducting frame MNOP of resistance R placed partly in a perpendicular magnetic field $\vec{B}$ and moved with velocity $\vec{v}$ as shown in the figure.Obtain the expressions for the(a) force acting on the arm ‘ON’ and its direction, and(b) power required to move the frame to get a steady emf induced between the arms MN and PO.

(a) The induced emf in the moving conductor MNOP

$e=BlV$

where B = Magnetic Field

l= length of the conductor

The induced current, $i=\frac{e}{r}=\frac{BlV}{r}$

Hence, the force on the arm 'ON' $(F)=Bil$

On putting the value of 'i' we have force,

$(F)=\frac{B\times B\times l\times V\times l}{r}$

The force is directed in the direction opposite to Velocity of the rod (v).

(b) We know that,

$Power:P=F\times V$

Where F = Force

V = velocity
We know, $F=\frac{B^{2}l^{2}V}{r}$

therefore, $P=\frac{B^{2}l^{2}V^2}{r}$

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text:
# Statistical mechanics and ocean circulation

We apply equilibrium statistical mechanics, based upon the conservation of energy and potential enstrophy, to the mass density distribution within the ocean, using a Monte Carlo method that conserves the buoyancy of each fluid particle. The equilibrium state resembles the buoyancy structure actually observed.

Atmospheric Physics Seminars
Event series  Atmospheric Physics SeminarsNoble Seminar Series


text:
## Sunday, April 26, 2009

### Elbow power: Things going *bump* in the night

Whenever some form of hardware doesn't work, many people bash it or elbow it, hoping and praying that this desperate measure might solve the problem. So, I figured, what if you could harness this form of kinetic input to control your computer

What if you could set the computer up so that when you slam it with your elbow, and like yours was the elbow of the Fonze, it kills Firefox (porn surfers rejoice), or renews your network connection, switches keyboard layout, switches desktop background, the possibilities are limited only by your imagination.

Disclaimer: Your computer will stand up to some degree of slapping and bashing about, it does have a breaking point. Use this at your own peril. I do not take responsibility for whatever damage you inflict upon your computer with this program.

Instead of buying expensive gyroscopes, accelerometers or hacking your wiimote, there's a very cheap way of making this a reality. The poor man's accelerometer is nothing but a computer mouse. Place it on top of your computer (make a casing for it if you want it to look neat). Now, the problem is how to read the data from the mouse. In Linux, this is a trivial matter. All hardware in Linux is a file, and you can simply open the mouse like any other old file.

Poor man's accelerometer.

What protocol the mouse device uses is a different pony. There's several mouse protocols, and they are all slightly different. The basic PS/2 protocol has 3 byte packets, where the first byte is buttons and some flags, and the second is relative motion in the X direction, and the third is relative motion in the Y direction. Older 2 button mice with no scroll wheel is generally PS/2 mice. Newer mice allow for longer packets, but they are generally backwards compatible with PS/2. There is more information on the protocol on Wikipedia if anyone is interested.

Great. So let's write a program that we can use to track jolts. The algorithm is going to be like this:

1. Open the device.
2. Read data from the file descriptor.
3. If the magnitude of the relative motion exceeds some threshold value, return from the program.

Add some red carpet code to allow customization, and you'll end up with this:

kinput.c
Makefile

Compile with make, and you're (hopefully) go.

Now that this helper program is done, what you need to do is write a shell script that performs some action when it detects a bump.

#!/bin/bash# DISCLAIMER! Your computer will stand up to some degree of # slapping and bashing about, it does have a breaking # point. Use this at your own peril. I do not take# responsibility for whatever damage you inflict upon# your computer with this program.KINPUT=/path/to/kinput # Change this to where you put kinputMDEVICE=/dev/input/mouse1 # Change this to some appropriate deviceTHRESHOLD=10 # Increase this if it goes off too easilyTIMEOUT=1 # Change this to how long you want to sleep # before waiting for another motionWHAT="killall firefox-bin" # Change this to what you want performed on a bumpwhile [ 1 ]; do if "$KINPUT" -d "$MDEVICE" -t "$THRESHOLD"; then echo "Bump detected!" eval "$WHAT" else exit; fi # Exit if something has gone wrong sleep "\$TIMEOUT"; # Sleep a momentdone

... and that's it. Running this in the background will allow you to solve your problems with your elbow. Correctly configured, you should only need to nudge your computer slightly. Bashing your computer too hard may harm your hard disks and cause problems in the wiring, so you shouldn't do that.

You could conceivably also use this program to detect mouse motion for other purposes. Running it in the background in the login manager and tracking the regular mouse could allow you to set up your computer to grab snapshots with your webcam of everyone who tries to log into your computer (which may or may not be legal where you live/work) and then email them to you.

There are some problems that might arise with this though, the most likely is that you don't have permission to read from the mouse device. Instead of running the script as root (are you insane?), it's a lot better to allow the mouse device to be read by all users, or add trusted users to the same user group as the mouse.


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# Grashof number ground-air

The Grashof number is a dimensionless number in fluid dynamics and heat transfer which approximates the ratio of the buoyancy to viscous force acting on a fluid. It frequently arises in the study of situations involving natural convection. The transition to turbulent flow occurs in the range $10^8 < Gr < 10^9$ for natural convection from vertical flat plates. At higher Grashof numbers, the boundary layer is turbulent; at lower Grashof numbers, the boundary layer is laminar.

Symbol
$\mathrm{Gr}_{L}$
Formulae
 $\frac{L_{cm}^{3} \beta_{gas} g \left(T_{gas} - T_{\infty}\right)}{\nu_{gas}^{2}}$ for vertical flat planes $\frac{D_{o}^{3} \beta_{gas} g \left(T_{gas} - T_{\infty}\right)}{\nu_{gas}^{2}}$ for pipes
Related
$\beta_{gas}$
$D_{o}$
$L_{cm}$
$\nu_{gas}$
$T_{gas}$
$T_{\infty}$
Used in
$\mathrm{Ra}_{L}$


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# Computing eigenvalues and eigenvectors

I need a little help with this question:

A is this self-adjoint matrix $$\pmatrix{ 1 & 0 & i \\ 0 & 2 & 0\\ −i & 0 & 1}$$

Compute the eigenvalues and a set of orthonormal eigenvectors of A.

I know how to compute the eigenvalues but my eigenvectors seem to be different. They normalise them for some reason

Any help would be appreciated, thanks

• Wouldn't that be because they want ortho'normal' eigenvectors rather than just orthogonal eigenvectors? – jgon May 11 '15 at 19:59
• oh i see. but for example (for the first eigenvalue of 0): any vector of the form (x,0,ix) is an eigenvector. But then they choose 1/root2 (1,0,i). how come? if you find the unit vector of (1,0,i): root of 1^2 + 0^2 + i^2 = 0 – jdhokia May 11 '15 at 20:04
• $||(1,0,i)|| = \sqrt{(1,0,i)\cdot\overline{(1,0,i)}} = \sqrt{(1,0,i)\cdot(1,0,-i)} = \sqrt{2}$ – uranix May 11 '15 at 20:14
• argh thanks :), i was using the dot product – jdhokia May 11 '15 at 21:00

To find the eigenvalues you have to solve the equation: $$\det\pmatrix{ 1-\lambda & 0 & i \\ 0 & 2-\lambda & 0\\ −i & 0 & 1-\lambda}=0$$ and developing the determinant you find: $\lambda(2-\lambda)(\lambda-2)=0$, so the eigenvalues are: $\lambda_1=0$,$\lambda_2=2$ and $\lambda_2=2$.
The eigenspace of $\lambda_1$ is the subspace of vectors that verify the equation: $$\pmatrix{ 1-\lambda_1 & 0 & i \\ 0 & 2-\lambda_1 & 0\\ −i & 0 & 1-\lambda_1}\pmatrix{x\\y\\z}=\pmatrix{0\\0\\0}$$
and thay have the form $$\pmatrix{-iz\\0\\z}$$ so you can chose : $$v_1= \pmatrix{-i\\0\\1}$$ In the same way, the eigenspace of $\lambda_2=\lambda_3$ is: $$\pmatrix{iz\\y\\z}$$ and you can chose the two orthogonal vectors:
$$v_2= \pmatrix{i\\0\\1} \qquad v_3= \pmatrix{0\\1\\0}$$
Now you can normalize these vectors with: $u_i=\dfrac{v_i}{||v_i||}$ (where the norm is defined as $||v||=\sqrt{\langle v,\bar v\rangle}$ in a complex vector space). So you find:
$$||v_1||=\sqrt{2}\qquad ||v_2||=\sqrt{2}\qquad ||v_3||=\sqrt{1}$$


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# Automatically fix outstanding mathjax issue involving backticks

After the transition to the stack exchange network, I have occasionally noticed posts being edited to replace occurrences of the following code (note the backticks):

$some math here$

with

$$some math here$$

Most recently, such an edit was made to the question Homotopy type of tensors of Moore spectra. These edits are infrequent, but still recurrent.

Can someone with access to the database find all occurrences of such code (and perhaps some variations of it) and automatically make the replacement?

• I took the liberty of changing this from a discussion to a feature-request. – Scott Morrison Jul 30 '13 at 5:24
• @Scott: Thanks. – Ricardo Andrade Jul 30 '13 at 6:25
• @ScottMorrison: I have a couple of questions. Would this question warrant the tag 'bug', even if tangentially? Also, would tagging the question 'bug' get a quicker developer reaction on this matter? – Ricardo Andrade Aug 15 '13 at 23:45
• Good question. I can email the stackexchange folks to remind them of this. More generally, I think we need better triaging of feature-requests and bugs, so everyone knows what's happening. – Scott Morrison Aug 16 '13 at 3:16
• Thanks, @Scott. Please let me know if you learn something useful. The feedback on feature-requests is very scarce. Any suggestions on how we should perform this triaging? – Ricardo Andrade Aug 16 '13 at 8:12


text:
Project 3a: Graphs Checkpoint

## Introduction

There will be a checkpoint due for Project 3 that will be due Friday, 11/30/2018 by 11:59pm. Please tag your checkpoint submissions as proj3a-x where x is the submission number. This checkpoint will have two different parts that are both based on the graph representation portion of the graph package (specifically just the non-traversal portion). Each of these parts will be for 1 point of extra credit for a possible total of 2 points of extra credit. The relevant skeleton files are Graph.java, GraphObj.java, GraphFilter.java, LabeledGraph.java, DirectedGraph.java, and UndirectedGraph.java.

As you are completing the graph package you should be writing unit tests. For this part of the checkpoint we will check that you have made sufficient effort in the creation of unit tests for the graph package. Make sure that the tests you have added run when "make check" is called. (Tests added to GraphTest.java should automatically run with make check). If you have trouble running make locally, remember you can always SSH into your instructional account and run it there.

## Our Tests

We will also be testing you on a number of unit tests that we have written (some subset of the final autograder tests) in order to test for correctness of your graph package. You will have to pass a portion of these to receive full credit.

## Project Parties!

We plan to have two project parties for you on the following dates and times:

• Wednesday, 28 November 2018, 7-9 PM (Soda 271 and 277).
• Monday, 3 December 2018, 6-8 PM

The parties will likely take place in the Soda labs. but check back for confirmation soon.

Project parties are primarily there for you to work with other students on the project. We’ll split you all up by the portion of the project you’re working on: graph representation / graph traversals / make + trip clients. We will have staff there but we won’t have queues like we do in office hours and labs.


text:
# Exercise C.3.4

Argue that if $X$ and $Y$ are nonnegative random variables, then

$$\E[\max(X, Y)] \le \E[X] + \E[Y]$$

There is a hidden assumption that $X$ and $Y$ are on the same domain (otherwise, you can't define $\max(X, Y)$). The expectations can be expanded as this:

$$\E[\max(X, Y)] = \sum n \cdot \max(\Pr\{X = n\}, \Pr\{Y = n\}) \\ \E[X] = \sum n \Pr\{X = n\} \qquad \E[X] = \sum n \Pr\{Y = n\}$$

Each of the summands in the first formula appears in either $\E[X]$ or $\E[Y]$. Their sum contains twice as many, all nonnegative. That makes it equal or greater.


text:
# Talk:Thomas Young (scientist)

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## German wiki

The German wikipedia mentions a Young's formula, which supposedly governs wetting to some extent. Yet I can find no mention of it here. Could someone please check to see if Thomas Young did in fact formulate this formula:

${\displaystyle \cos \Theta ={\frac {\sigma _{S}-\sigma _{LS}}{\sigma _{L}}}}$

-- Ec5618 10:43, 11 October 2005 (UTC)

The Wikipedia article on Surface Energy credits Thomas Young and links to the article on Contact Angle which appears to be angle Theta in your formula. I think it is an accomplishment worth adding to the Thomas Young article. Also there should be mention of the Young temperament and possibly Young's Rule for determining child medicine doses.

Harold14370 16:01, 10 April 2007 (UTC)

## "Last Person to Know Everything"?

How can he be considered "the last person to know everything" when two of the others claiming the same "title" (Coolerige, Goethe) died after Young? 208.201.250.2 04:26, 29 December 2006 (UTC)

Not meaning any disrespect to the two persons mentioned, but I don't think they really qualify. Coleridge was pretty much just a poet. I don't even know why he was mentioned. Goethe made an impact as poet, dramatist, and novelist which are closely related. As a scientist, he really made no impact. His theory of light was a crock. As a painter, he just dabbled. A great man surely, but not really a polymath on the same order as Young. Harold14370 03:48, 14 April 2007 (UTC)

## This article needs work

Please compare this article to the Answers.com entries and improve this article. For instance, Young was the first to use the term energy in its modern, physical sense.--76.220.203.60 06:30, 18 July 2007 (UTC)

## F = kx (formula attributed to Newton)

In the section titled "Young's modulus", the formula F = kx is attributed to Newton, but I always thought that was (Robert) Hooke's law of elasticity. Does anyone know for certain who devised this "law"? Did Newton also devise this relationship/law? TimProof 19:18, 21 July 2007 (UTC)

All the books I've consulted say Hooke. The only Newton connection is that Newton prevented Hooke's ideas from being given any play while Newton was alive. Dicklyon 23:27, 21 July 2007 (UTC)

## John Zachary Young

I'm sure that J. Z. Young was a stalwart fellow, but surely the place to mention that he's a direct line descendant of Thomas Young is on his own (large and lengthy already) wikipedia article, and not on Thomas Young's. arkuat (talk) 06:46, 19 May 2008 (UTC)

Why not mention it in both? Dicklyon (talk) 14:18, 19 May 2008 (UTC)

Because then we'd be expected to do the research to make sure that all of Thomas Young's other descendants who were at least as notable as JZ were also mentioned in Thomas Young's article. arkuat (talk) 21:23, 20 May 2008 (UTC)

That's a lame excuse; there is no requirement to be complete or balanced in wikipedia. Dicklyon (talk) 21:48, 20 May 2008 (UTC)

I think in this case an excuse is required for leaving this irrelevancy in Thomas Young's article, and no excuse is required for removing it. arkuat (talk) 03:02, 8 June 2008 (UTC)

## Doctor of Physics?

German wikipedia says that he obtained a decree as doctor of medicine. Considering physics' reputation back then, this seems to be more credible. Possibly change the article in this regard? Christoph Hahn (talk) 11:37, 1 November 2008 (UTC)

Check and cite a source or two in the process. Apparently the Göttingen degree and later degrees at Cambridge were in "physic", which is what physician studies, i.e. what we now call medicine. See this book and this book. Dicklyon (talk) 16:46, 1 November 2008 (UTC)
D'accord. I was confused by a today's statement of my professor. Christoph Hahn (talk) 17:59, 1 November 2008 (UTC)

## Pendulum

The text says: "In 1816 he was secretary of a commission charged with ascertaining the length of the second's pendulum" There have been amendments concerning the apostrophe - but what does a "second's pendulum" mean? Myrvin (talk) 14:08, 13 November 2009 (UTC)

It had been editted without comment to "second pendulum", which is clearly not what was intended. You can find hudreds of books that talk about a "second's pendulum" or a "seconds pendulum" on google book search. Dicklyon (talk) 02:40, 14 November 2009 (UTC)
You make it seem that what it is is obvious - it isn't. There is an article in Wiki on the "seconds pendulum", so there should be a link. Also perhaps a short explanation. Myrvin (talk) 13:50, 14 November 2009 (UTC)
Did it myself Myrvin (talk) 09:43, 9 December 2009 (UTC)

## References

--Stone (talk) 08:20, 11 February 2011 (UTC)

## Latest edits

There have been some strange edits by Cajun80.I have reverted them. Myrvin (talk) 19:52, 16 February 2011 (UTC)

## Last person to read everything?

Does anyone know if it's true that he was the last person to read all books (in English, presumably) that were published up to and including his lifetime? (I'm sure that can't be strictly true since I'm sure some books were published just before his death, but it could be true up until a certain point in his life.) If true, should it be mentioned? M-1 (talk) 00:41, 19 September 2013 (UTC)

## Young on animals?

I found this work called "An Essay on Humanity to Animals" (1798) at Googlebooks, and I wonder whether this is the Thomas Young of this article or another. Hope somebody may help.--Goose friend (talk) 07:00, 31 December 2014 (UTC)

Read this [1]; and this [2]. Myrvin (talk) 11:28, 31 December 2014 (UTC)

## boxhorn and IE

Boxhorn's name for the group was Scythian because he did not know about Sanskrit. The information about Boxhorn here is irrelevant (Pamour (talk) 15:08, 25 January 2015 (UTC)).


text:
# Tag Info

## New answers tagged reference-request

0

When you talk about 1x3 or 3x1, we do not know if you are talking about COLxROW or ROWxCOL. The typical way of representing a vector in a matrix is ROWxCOL terms, so 3x1 is what you are looking for The matrix multiplication uses a single column with $N$ entries multiplied by an $N*N$ matrix. Since this is multiplication the order does not matter, $P*K$ is ...

7

Curve25519 was designed to take advantage of the Montgomery ladder, which combined with Montgomery curves forgoes the $Y$ coordinates, is side-channel resistant, and enables public keys to be any 255-bit string. The ladder looks something like this (pseudocode): Q[0] = P; Q[1] = 2*P; for(int i = log2(exponent) - 2; i >= 0; --i) { Q[ bit(exponent, i)] ...

6

Did you take a look at DjB's paper? One of his design criterias in order to improve performance is "Use a fixed position for the leading 1 in the secret key". The set of secret keys is defined to be $\{\underline{n} : n \in 2^{254} + 8\{0, 1, 2, 3,\ldots, 2^{251}-1\}\}$.

4

Full disclosure — I'm a Skein/Threefish co-author. Also, when I mention Skein/Threefish without any other qualification, I mean Skein/Threefish-512. The security proofs we did for Skein prove that if there's a weakness in Skein, it implies an underlying weakness in its components (Threefish or UBI). As Dmitry says above, Threefish is very strong, and there ...

7

Pure Threefish has received less attention than Skein. Shortly speaking, it has a large security margin, and can be safely used for encryption. In more details, Threefish has been tweaked twice. The first two versions were vulnerable to rotational cryptanalysis in weak models (related-key attacks or distinguishers) up to 57 rounds. All these attacks are ...

Top 50 recent answers are included


text:
# Midterm 1

General Chemistry 3

is over.

Check the solutions on the questions and answer that you have decided correctly

You can revisit questions later in your account.

Here is an equilibrium: C (s) + CO2 (g) $⇌$ 2 CO (g)     [ΔH0rxn = 65 kJ]

Predict and explain the direction that the equilibrium will shift when each of the following changes occurs:

- CO is removed at constant temperature

- the total pressure is increased by addition of N2 (g) at constant temperature

- the volume is increased at constant temperature

- the temperature is decreased

- CO is removed at constant temperature:

direction: toward product

explanation: when product is removed, the reaction shifts to make more product (Le Chatelier’s principle)

- the total pressure is increased by addition of N2 (g) at constant temperature

direction: no change

explanation: volume did not change à partial pressure did not change à Q did not change à no shift

- the volume is increased at constant temperature

direction: toward product

explanation: side with most moles of gas

- the temperature is decreased

direction: toward reactants

explanation: reaction is endothermic

[SO3]0 = 2.00 M, [O2]0 = 0.0850 M, [SO2]0 = 0 M and [SO2]eq = 0.120 M

Calculate Kc for the following reaction:

2 SO3 (g) $⇌$ 2 SO2 (g) + O2 (g)

Kc =

From the stoichiometry of the reaction:

[SO3]eq = [SO3]0 – 2X

[O2]eq = [O2]0 + X

[SO2]eq = [SO2]0 + 2X = 2X = 0.120 M

⇒ X = 6.00 x 10-2 M

At the equilibrium:

[SO3]eq = [SO3]0 – 2X = 2.00 – 0.120 = 1.88 M

[O2]eq = [O2]0 + X = 0.0850 + 0.0600 = 0.145 M

[SO2]eq = 0.120 M

Kc =

Kc =

Kc = 5.91 x 10-4

Calculate the pH of a 0.100 M aqueous solution of ethylamine CH3CH2NH2.

Data: Kb [CH3CH2NH2] = 4.30 x 10-4

Acido-basic reaction:

CH3CH2NH2 (aq) + H2O (l) $⇌$ CH3CH2NH3+ (aq) + HO- (aq)     [Kb]

A t = 0:

[CH3CH2NH2] = 0.100 M and [CH3CH2NH3+] = [HO-] = 0 M

At the equilibrium:

[CH3CH2NH2] = 0.100 - X M and [CH3CH2NH3+] = [HO-] = X M

Kb = $\frac{\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}{{\mathrm{NH}}_{3}}^{+}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}{\mathrm{NH}}_{2}\right]}$  =

Let’s suppose 0.100 >> X

Kb = 4.30 x 10-4 = $\frac{{\mathrm{X}}^{2}}{0.100}$

⇒ X = 6.56 x 10-3 (X cannot be < 0)

⇒ the hypothesis 0.100 >> X is verified

pOH = - log [HO-] = 2.18

pH = 14 – pOH = 11.8

A 7.50 x 10-3 M solution of a weak base has a pH of 11.3.

Determine the Kb for this weak base.

A- + H2O $⇌$ AH + HO-     [Kb]

A t = 0:

[A-] = 7.50 x 10-3 M and [AH] = [HO-] = 0 M

At the equilibrium:

[A-] = 7.50 x 10-3 - X M and [AH] = [HO-] = X M

pH = 14 – pOH = 14 + log [HO-]

⇒ [HO-] = X = 10pH – 14 = 10-2.7

⇒ [HO-] = 2.00 x 10-3 M

Kb = $\frac{\left[\mathrm{AH}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{A}}^{-}\right]}$

K=

Kb = 7.23 x 10-4

Write the relative reaction rate equation for each component in the combustion of ethanol CH3CH2OH.

Combustion of ethanol:

CH3CH2OH + O2 → CO2 + H2O     [unbalanced equation]

CH3CH2OH + 3 O2 → 2 CO2 + 3 H2O     [balanced equation]

Reaction rate:

rate = - $\frac{∆\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{OH}\right]}{∆\mathrm{t}}$ = - $\frac{1}{3}$ $\frac{∆\left[{\mathrm{O}}_{2}\right]}{∆\mathrm{t}}$ = $\frac{1}{2}$ $\frac{∆\left[{\mathrm{CO}}_{2}\right]}{∆\mathrm{t}}$ = $\frac{1}{3}$ $\frac{∆\left[{\mathrm{H}}_{2}\mathrm{O}\right]}{∆\mathrm{t}}$