Can a gapless system be a topological state?

Question: For a gapless system without boundary (i.e. in the bulk there is gapless excitation while no clear meaning of boundary excitations like QFT), can it be a topological state? What is the property of EE in these systems? Thanks!
Answer: One first needs to define what it means for a state to be topological when the system is gapless. This is not an easy question! I don't think there is a unified answer at the moment.
I'll give some examples of exotic gapless states which are "proximate" to gapped topological states. Here I take "topological" to mean exotic low-energy excitations (much like anyons). But of course there are other notions of topological gapless states, such as Weyl semimetals in 3D which can be characterized by exotic boundary states.

Emergent Fermi liquid. By this I mean the following: starting from a gapped topological phase, which contains emergent fermionic quasiparticles (not the electrons). Now imagine the system is tuned in such a way that the gap to this fermionic excitation is reduced, while the gap to all other types of quasiparticles do not close. Eventually, the gap to the fermionic quasiparticles close, and we imagine those fermions form some kind of Fermi liquid. A well-known example of this type is Kitaev's honeycomb lattice model, at the isotropic point. The gapless emergent fermions are still coupled to a $\mathbb{Z}_2$ gauge field. A related example is the $\nu=1/2$ composite Fermi liquid in FQH, but it has a different gauge structure ($\mathrm{U}(1)$ gauge field).
Fluctuating order parameters. Some topological states, especially those that occur in free fermion band systems, are sometimes associated with symmetry breaking. For example, topological superconductors break $\mathrm{U}(1)$ symmetry. One can ask what if the superconducting order parameter has phase fluctuations (gapless Goldstone mode). An example that is studied very thoroughly is 1D class D superconductor, with such phase fluctuations, see http://arxiv.org/abs/1106.2598. The phase fluctuations change the exponential splitting between the degenerate ground states to power law.

I think in both cases people would have no issue calling the gapless states "topological", since they still inherit some of the nontrivial low-energy excitations from the nearby topological states. But a general characterization of gapless states does not exist yet.
I recommend the very inspiring paper http://arxiv.org/abs/1212.6395, where the authors defined "quasi-topological phases".
Comment: Thanks for your helpful comment! More naive questions: Is Weyl semi-metal a topological system with gapless excitations? Is it possible that for such system (e.g. translational invariant ideal WSM in some strongly coupled case) there is no Fermi arc like boundary excitations? Thanks!
Comment: Ha, I forgot to mention this most popular example. Weyl semi-metal is topological in the sense that the Weyl point is a monopole in the momentum space, although they are not stable to translation-breaking perturbations. Although I don't know what "strong coupled case" you have in mind, it is unlikely that the interactions can get rid of Fermi arc without a phase transition in the bulk.
Comment: I was curious about the role of boundary excitations in WSM. Is it possible to realise a similar topological system without any boundary excitations? Maybe I am wrong, since the effective electromagnetic interaction strength in WSM systems is or order 1, WsM should be a strongly coupled system. The free Weyl fermion picture (with Weyl nodes) might be broken down in some strongly coupled WSM systems (thus no boundary state?), but it might be still possible to be in a topological state.
Comment: Well, our understanding of Weyl semimetals is based on non-interacting electrons, so is the material realizations. Electromagnetic field is not part of the system, we only turn on a background EM field to probe the EM response of the system. You can definitely imagine some exotic states obtained from coupling Weyl fermions to dynamical U(1) gauge fields, and free fermion picture could break down, but it does not mean that there is no boundary state.
Comment: I am totally from outside:) Is there any material example in which all the transport physics are very similar to WsM (can be explained by chiral anomaly, e.g. negative magneto resistance, anomalous Hall effect etc.) but there is no similar ARPES observations (Weyl nodes + Fermi Arc)? If one realise such material, would it be topological matter?
Comment: No, no such examples are known. People are not looking for materials just blindly, one needs to have some guide from theory(first principle calculations, etc), and the most reliable theories we have are non-interacting (or weakly interacting) fermions.
Comment: Thanks for your patient clarifications! But I am still curious if a topological state can be defined via a response property since with interactions the single particle picture may breaks down.             A related question: recently `@Xiao-Gang Wen` is trying to classify the topological order for topological state from gapped systems. What would be the topological order parameter for this WsM like gapless topological state?


Let $f(x) = a_nx^n + a_{n−1} x^{n−1} + · · · + a_1x + a_0$ be a polynomial with integer coefficients. Prove that it has no rational roots.

Question:
Let $$f(x) = a_nx^n + a_{n−1}  x^{n−1} + · · · + a_1x + a_0$$ be a polynomial with integer coefficients. If $a_0$, $a_n$ and $f(1)$ are odd, how to prove that $f(x)$ has no rational roots.
Answer: From the rational root theorem the root is of form $p/q$ where $p$ and $q$ are odd.
So:
$$0=f(x) = a_n(p/q)^n + a_{n−1}  (p/q)^{n−1} + · · · + a_1(p/q) + a_0$$
by multiplying both sides by $q^n$:
$$0= a_np^n + a_{n−1}  p^{n−1}q + · · · + a_1pq^{n-1} + a_0q^n$$
but this sum should be odd as the $f(1)$, $p$ and $q$ are odd (easy to prove).
While the $0$ is even.
Comment: i dont follow your last sentence,  what should be odd? how did you get that
Comment: If $f(1)$ is odd and the $p^nq^m$ is odd  so the $a_ip^nq^m$ has the same  parity as $a_i$ this means that the sum of $a_ip^nq^m$ has the same parity as the $f(1)$ which means it can not be 0.
Comment: Proof that $p,q$ are odd for completeness' sake $q$ can either be even or odd. Assume $gcd(p,q)=1$ and suppose $q$ is even. Then $0$ is even, and so is $a_i p^i q^n-i$ for $0\leq i<n$. So $a_np_n$ is even, which implies $p$ is even, which is a contradiction since we assumed $gcd(p,q)=1$.

Now suppose $q$ was odd. Then $p$ can't be even, as if it were, all of the terms in the RHS would be even except for $a_0q^n$, but the $LHS=0$ is even, so this implies that $q$ is even, a contradiction.

Thus $p,q$ are both odd. (After typing this up, I realise you can just use the Rational Root Theorem)
Answer: If $\frac{p}{q}$ is a rational root, then $f(x)=(x-\frac{p}{q})(a_{n}x^{n-1}+...+a_{0}\frac{q}{p})$.
Since $p|a_{0}$ and $q|a_{n}$, we conclude that both $p,q$ are odd.
Now, $f(1)=\frac{(q-p)\big(p(a_{n}+...+a_{1})+a_{0}q\big)}{pq}$, and thus $f(1)$ has an even factor $(q-p)$, which is a contradiction.
Comment: how do you know (p(an+...+a1)+a0q) / pq is an interger
Comment: pq which is odd, has to divide one of the two factors on top, since $f(1)$ is an integer. So it can only cancel out odd factors from the numerator. It can still divide $pq$ but only odd terms in $pq$.


How can I lower the water intake on my booster pump from the main water pump?

Question: I have a 25 foot shallow well for water sprinkler with 2 horse power barracuda pump. Low pressure so added 3/4 hp barracuda booster pump. Booster is dry running for lack of water from main pump. How can I lower the booster intake too stop dry run?
Comment: where is the air getting in?
Comment: What do you consider low water pressure, I have had both simple suction and jet pumps on wells that could easily pump 80 psi. Is the primary pump or the line(s) to the water the real problem?
Comment: My main pump has 1 1/4 inch in and out of pump. I don't believe their is no air leak. Took everything apart and resealed piping. My sprinkler system has was too many misters and rotating heads on each zone. My cut on and off pressure is 40/20 with pressure take attached. Put muriatic acid in piping to well and let set for a day before hooking back up.  With booster off. The main pump builds pressure and shuts off in less than minute so I have water. I sank well my self 24 years ago. Please help. Getting fed up.
Answer: @Ed Beal is right in his comments so you should not need a booster pump. I think it is time to check out the primary pump and piping to and from the pump since a 2 HP pump will move a lot of water at a very high pressure. A booster pump should not be necessary.


How straightforward is in vitro compartmentalization?

Question: in vitro compartmentalization (IVC) is one of those technologies that everyone knows about, talks about, but never actual does due to the rather technical difficulties in setting the system up. I was wonder if anyone not named Tawfik, Griffiths or someone who did a post-doc with them have been successful in making IVC work? Details in how stable emulsions and a homogeneous mixture of cell-free components in the droplet are made would be wonderful.
Comment: *one of those technologies that everyone knows about*. Everyone but me! Nice link, always good to learn something new.
Comment: Thanks for pointing that out, it'll save me some reading, too.  Sequencing technologies like 454, SOLiD, IonTorrent use emulsion PCR on a large scale.  The SOLiD system even has an sample prep machine that automates the creation of these emulsions.  I think these emulsions end up a bit smaller, but they're at least in the right ballpark.
Comment: You might want to look up synthesis techniques  artificial cells, like artificial red blood cells.  A resource that has led me to believe this stuff isn't impossible to set up is at www.artcell.mcgill.ca
Comment: @ThomasIngalls, while certainly neat, the dimensions of artificial cells are completely different from the dimensions of IVC which prevents the concepts learned in this body of work from being translated to smaller emulsions. This will still be interesting reading material.
Comment: @ThomasIngalls i think you should expand into an answer, eh?
Answer: It is actually surprisingly easy and reproducible to make emulsions with defined components. We do it routinely in our lab:
http://www.sciencemag.org/content/332/6026/209.full - CBT protocols
http://www.sciencemag.org/content/336/6079/341.full - CST protocols
Comment: Thanks! Since it has been awhile since I've asked this question, I now get the impression that certain reactions like transcription and smPCR work well. When concentrations and size become more important or translation is important things go south.


how to hide Access to media files in a directory, if we have the full URL for a file

Question: how to hide Access to media files in a directory, if we have the full URL for a file.
How to,
I make an online store, so to download one of the files in my store, I have to log in first, then it can be downloaded, for example https://mystore.com/product/widia/ they must log in first,
but if we access it from the url of the file, for example https://mystore.com/wp-content/uploads/2019/01/widia.ttf
then the file is downloaded immediately,
Please help
Comment: Not a graphic design question. Check [ask] and [help/on-topic] to learn more.
Answer: Just in general.....
Put an index.html file in the directory. Leave this index file completely blank (no markup or text whatsoever - although you can style it or add content if you desire).
When a user targets the URL the index.html file is called an they merely see a blank page (or the content of the index.html file).
This does not prevent direct access to items within the directory if the full path and file name is known. However, it does prevent "browsing" of the directory itself.
I place a blank index file in every directory which is not meant to be viewed by anyone over the web - even if the access is limited via server configurations. The blank file hurts nothing if it is not necessary, but will prevent browsing if it is needed.

But...
If this is a "store" and you are selling items in a directory and want to prevent access to those files, even if the file path is known, you need much more robust protection via some server-side scripting (php/asp) and authentication to control the access to downloads. It's not an easy answer especially if you aren't familiar with server-side scripting.
In general you'd probably set an authorization cookie after purchase. When the user accesses the download directory, the cookie is checked to see if it is valid. If the cookie is not valid, has timed out, or is simply not present, the user is rejected from the directory and unable to download.
StackOverflow.com is site in the stack network more directed at web site configuration, creation, and assistance with those matters. this stack isn't really geared towards assisting with server-side scripting.
Comment: right, this is for the shop,
where can I find help for my shop, so that we can avoid being accessed in the url of one of the files directly, because if that happens it is very detrimental to the shop owner, people who know the url of one of these files can be downloaded without having to buy,thanks
Comment: You really should hire a developer.


What attributes of a surface are relevant to light sensors?

Question: I'm doing a study on surfaces and I would like to describe all of the surfaces that I expect to encounter in the field and determine which ones will be the most difficult to detect. To do this, I need to determine what attributes in a surface I need to consider. I have no experience in optics so I'm asking this question here in hopes that someone could save me from overlooking important factors.
Comment: I think you need to add some of what you wrote in the comment to Carl's answer in your question, to give your question more detail. From that comment, the light sensor you use will not be the main issue. Your problem will be analyzing/interpreting the data from the sensor. Some things you may need to consider will be contrasts of light, shadows & edge effects produced by 3D surfaces, like jagged cliff faces or buildings. There will also be the issue of dealing with glare - you may may need to use a polarization filter.
Answer: Surface roughness, absorptivity, transmittivity, absorption spectrum, etc might be some attributes that may help you.
Answer: Your question lacks important parameters: in addition to the qualities of the surface or material that Pranav mentions, you need to be able to define what "detect" means, what sort of light sensors you think you want to use, what the interfering noise sources are, and more.
If your target surface is not active (like a lightbulb, generating its own output) then you need to know as a minimum what the illuminating sources are.  Then you can calculate, based on surface qualities such as BRDF the expected signal magnitude at your sensor location.   But there's a huge difference between saying "Look out! you're going to hit that thing!"  and saying "Hey, there's a chunk of rock with 20% silicates and 1% manganese at about 240 Kelvins over there"
Comment: I'm unsure yet about what types of light sensors I will use--that is one of the main purposes of this study. I would say that the surfaces I am looking at are not active, but there may be a case when there are active surfaces--I am aiming to detect cliffs within safe stopping distance for mobile robots. My application is therefore closer to the former, "Look out! You're going to hit that thing!". At a higher level, my goal is to improve safety.
Comment: @Klik If you're using it for collision detection you could perhaps augment your system with sonar as well to at least help detect outliers missed by your primary system


Should I automatically delete accounts if they have not verified their email address?

Question: I am launching a new site soon and in my previous experience I know many bots will create accounts, I'm trying to avoid this so before a user can sign in they need to verify their email address. If they email address doesn't get verified within X amount of time should I automatically remove the account? What's best practice in this area?
Comment: What can a user with an unverified account do on your site?   Usually users can't do much of anything until they have created and verified an account.  That leaves little difference between an unverified account and a deleted account.
Comment: @davidgo I think there's some value in the above question, though. For one thing, it's not something people inquire about every day, so existing resources may be limited. For another, it touches on best practices for account creation flow. There may be more than one answer that makes sense, but the OP's considerations are a very real topic for marketers and web managers.
Comment: I voted to close as opinion based. The answer really depends on why people are signing up,  where you are - which influences data retention laws etc. My opinion is get rid of the data.
Comment: I doubt there is "best practice" - you can make your own.  "This link will be valid for x days after which the account will be deleted if not verified"
Answer: To add to Stephen's comment, it's unlikely that absolutely nothing goes into any database at the point of unverified account creation. Therefore, the question is what you want to do with the information you do have. Leaving the inactive info in there indefinitely will not do any immediate harm, but it will create noise in your data and add unnecessary weight to your tables.
If the user's email and whatever other information is saved somewhere, but the user doesn't take the final step to verify, you may want to send a reminder email the next day. You may want to follow up in a week or so with a second reminder, as people get busy, forget, or procrastinate.
If they don't verify after a couple of weeks, I'd delete the info to keep your data tables clean. You can play with the timeframes in which you do all this, but that's the general approach I'd recommend.
Comment: Many thanks for clarifying what I was thinking. I'll go down the step of a follow up email a day and one a week later and if no response remove the account. Thanks
Comment: @JTheBiker You're welcome - glad it was helpful!


How low should the surface water table be in order to be considered well draining?

Question: My yard has a high water table (creek, and 4 foot elevation about 4 foot elevation over almost all of the yard), where tree roots often just sit at the surface, so I was wondering how low the water table should be in order to consider something well draining.
Answer: Not remotely related. The level of the water table has nothing to do with how well the soil drains.
High water table can be in clay that doesn't drain very well at all, or in a gravel pit that drains exceedingly well.
If you filled a bathtub with each soil and set the water level to 4 inches from the top, it would be the same place for ether type. If you pulled the plug on the bathtub, they would drain very differently.
There are places where you have to drill hundreds or thousands of feet to find water, but the surface turns to a slimy clay mess when it rains.
Comment: So how far off the water table is best for most plants then?
Comment: As @Ecnerwal said -- it's irrelevant. Well draining means how quickly water flows through the soil. Sand, rocks are well draining. Composted leaf mulch is slow draining. Clay is very poor draining (almost none!).
Answer: If your water table is 4 feet down from the surface of the soil, that should be fine for most garden plants. But if you have soil that has a lot of clay, that retains too much moisture for too long, you will have to mix in some sand where you plant your garden.
Most plants don't like feet that are wet all the time. Water logged roots = dead roots = dead plant.
Example: My normal soil is very sandy in most places, so I have to water more often as almost no water is retained there. But I got some potting soil and started some seeds in there in pots, and this soil keeps moist for much longer so I had to water much less or the roots would get diseased from too much moisture.
Does that help?
Comment: that's the area the house is on, and the part that I want to plant on is <1 foot above the surface water, and almost never floods as it's probably richer than potting soil.
Comment: If the water table is 1 foot below the soil surface you might have some issues with the water wicking even higher in the soil. Does the soil look like it dries out down to 6 inches, ever? My best suggestion is just to try several different plants, or use a raised bed made with cedar or bricks.


Nautilus does not show the left bar, how do I turn it back on?

Question: When I open nautilus does not open the left bar where it says computer and down markers such as music, documents, videos, downloads. I reinstalled nautilus from the synaptic package manager but nothing happened. What I can do to make that bar visible again?
Answer: Just press F9 when you open nautilus.
Another way to do this is
Click on 'view' then check the 'side pane' option.
Answer: You can activate or deactivate the sidebar simply by pressing F9. You can also go up to the global menu, then select View → Sidebar → Show sidebar.
Answer: Just press F9 or if you like it the 'mouse' way, jut select 'Side Pane' from the 'View' menu.
Answer: Open Files and then in the left side of the top bar select Files which will present a dropdown menu from which you can select Sidebar.

An alternative (reminder courtesy of @PRATAP) is to press F9 on opening Nautilus,


Length of an email list in sent mail

Question: I want to send an email to a large list, using Mac Mail. How many email addresses can I have in To: or cc: or bcc:? I recall there is a limit.
Answer: I'm not sure if there is a hard limit in Mail for the number of recipients in any of the particular fields. The larger obstacle may be your mail service blocking the message because of the number of recipients: their spam filters tend to flag messages as spam when the number of recipients goes over a certain number.
What you can try - and I have had this work - is to use an Address Group. Put the contacts you are sending to in a Group in your Address Book, then enter then name of the Group in the To: field. It masquerades the email addresses to the recipients and may get it past the spam filters.
Comment: Technically there is no difference between adding all addresses manually to a mail or using an Address Group. In the later case Apple Mail replaces the group with the individual addresses before handing the mail to the mail server.

Length of an email list in sent mail


Windows cannot be installed to this disk. Your computer hardware may not support booting to this disk

Question: Trying to install Window 8 on a separate partition alongside Ubuntu, I got the message "Windows cannot be installed to this disk. Your computer hardware may not support booting to this disk. Ensure that the disk controller is enabled in the computer's BIOS menu." My system doesn't use EFI or GPT, which cause known issues with Windows installation.
Answer: Windows must be installed to a primary partition, not an extended or logical partition.
I checked the error log (in X:\Windows\panther\ I believe), and it says that the partition is large enough but it is not a primary partition. Yet in Windows installation, even when I delete the partition and recreate it, it's created as a logical not a primary partition. Also,in GParted, I can't change the partition type from logical to primary, though I flagged it as "boot".
Finally I discovered that no, primary partitions can't be created inside an extended partition. So using GParted, I shrank my extended partition and moved it to the right/end of the drive, and created a NTFS partition to the left/at the beginning of the drive. I could install Windows 8 on that partition.
Answer: As (you) indicated, Windows needed a PRImary partition. More specifically; (modern) Windows wants either an (U)EFI setup, or a disk with an MBR layout. But all is not lost. Given you indicate an Ubuntu install. I suspect your disk has a GPT layout. Which, strictly speaking, is incompatible with MBR. But it's still possible to create what's often called a "hybrid" layout, that incorporates a MBR partition.
At the point where Windows indicates an error condition that prevents Windows from being installed. (Disk, partition selection). Press the keyboard combo SHFT+F10. This will open a so-called "dos box", command prompt. Execute DISKPART:
Microsoft DiskPart version 6.1.7100
Copyright (C) 1999-2008 Microsoft Corporation.
On computer: SomeComputer

DISKPART>

Then lis vol
  Volume ###  Ltr  Label        Fs     Type        Size     Status     Info
  ----------  ---  -----------  -----  ----------  -------  ---------  --------
  Volume 0     F                       DVD-ROM         0 B  No Media
  Volume 1     E                       DVD-ROM         0 B  No Media
  Volume 2     C   8            NTFS   Partition     16 GB  Healthy    System
  Volume 3     D   Data         NTFS   Partition    263 GB  Healthy
  Volume 4     G                       Removable       0 B  No Media

DISKPART>

Select the volume you want to modify
DISKPART> sel vol 3

Volume 3 is the selected volume.

DISKPART>

What happens from here is completely dependent upon your disk layout, and what you have. If you already have a Volume (as above). You could delete it. Then create a PRImary Volume/Partition. Make it Active, and largely be done with it.
Thusly:
DISKPART> del vol 3

Volume 3 has been deleted

DISKPART>

Create a Volume -- you can create it with size; see help diskpart for greater detail:
DISKPART> cre par pri

...

DISKPART>

See what just happened:

DISKPART>list vol

YOUR NEW LIST WILL BE SHOWN (with the others)

DISKPART>

Now select your newly created volume:
DISKPART>sel vol (the volume you're interested in)

(your selected volume) is the selected volume.

DISKPART>

Format it, and Label it (I will use Win10 for this example)
DISKPART> for fs=ntfs quick label="Win10"

  100 percent completed

DISKPART>

Now all you need is to make it Active (bootable)
DISKPART> act

DiskPart marked the current partition as active.

DISKPART>

Done. EXIT (close the CLI)
DISKPART> exi

At this point you have a Primary, Active partition you can install Windows to.
Now that the "dos box" is gone, you'll have the Disk Selection dialog in front of you. Choose "Refresh". The ERROR should now be gone, and you can select your newly created Partition, and install.
I could have been even more specific, and provided even more example dialog. But you didn't provide any indication of your current disk layout. So I improvised as best I could, based on what I imagined you might have.
At the worst. You'll have enough clues to proceed in a way that gets you to where you need to be.
Answer: After reading so many forums and answers:
Played a lot with BIOS options
Released an entire HD space in order to convert it to GPT
whatever ...
Nothing worked out in my case.
I have an Asus M5A78L-M LX mobo (just fyi) but the tricky here is that I also have an IDE 2 PCI controller in order to connect an old CD Recorder.
It was just a matter of removing the board from the slot, configuring @ BIOS the USB Disk as HDD and adding it as the 1st HD so it will be the primary boot option (these steps I had done before without success).
Just that and it took me many hours until I figure it out that Windows Installation or something @ BIOS could be looking to the IDE controller as primary boot option and since my HD has on SATA controller Windows kept saying that it may not be bootable.
That's it


Spoiled garlic ok to use?

Question: I bought a 3 pound bag of peeled garlic. Some of them have white mold on them. Should I throw the lot away or pick through the bad ones & discard?
Comment: When you can see mold, you are seeing the fruiting body.   The mold tendrils penetrate the substrate on which the mold appears.

I would discard the entire bag, on the grounds that I am paranoid, and garlic is not that expensive.
Answer: Throw the lot away. Mold is not always visible but when it is, the mycelium may be present in the entire substance, especially in foods with high moisture content. Eating the mycelium can make you sick and it can continue to develop mold after you've cut the moldy parts away.
Answer: Pick through.  I even cut out, then use the good part.  It is most likely Penicillium hirsutum, "blue mold of garlic".  Sporulation and germination are delayed at low temps (i.e. fridge)[1], therefore you see only white which are the mycelia.
[1] "Low-temperature biology and pathogenicity of Penicillium hirsutum on garlic in storage", P. Bertolini et. al, 1996, https://doi.org/10.1016/0925-5214(95)00025-9

Spoiled garlic ok to use?


Op-amp circuits with capacitor in negative feedback loop instead of resistor

Question: When the circuit is in steady state, a capacitor can be replaced by an open. However, I have seen an example where a capacitor was used in place of a resistor in a negative feedback loop, and the condition that the input voltages were equal was used in solving for the output voltage of the op-amp. How is this possible?
Comment: Do you have more context (and a schematic) for that example? Was it analyzing the behavior of an op amp when exposed to high-frequency signals? Nothing so far seems like a contradiction, but it's difficult to write up an answer without having the context.
Comment: Probably using complex-valued AC analysis.
Answer: I really like this question, in spite of its brevity and lack of context. It shows that at least you've identified issues with capacitive feedback and quiescent state.
Start with the classic op-amp integrator circuit, assuming that all components and conditions are ideal:

simulate this circuit – Schematic created using CircuitLab
If this were a perfectly ideal operational amplifier, and the capacitor had no idiosyncrasies that might complicate matters, and that there's no such thing as noise, then behaviour is really easy to explain. Start with the fact that C1 provides negative feedback, which has the effect of causing the op-amp to adjust its output potential to whatever value is necessary to equalise the potentials at its inputs Q and P, \$V_Q = V_P = 0V\$.
Secondly, the op-amp input draws no current, so by KCL and Ohm's law:
$$ I_3 = I_1 = \frac{V_{IN}}{R_1} $$
Then there's the relationship between current \$I_3\$ through C1 and the voltage across C1 \$V_{C1} = V_Q-V_{OUT}\$:
$$ I_3 = C_1 \frac{dV_{C1}}{dt} $$
which translates to a few interesting behavioral facts:

If \$I_3\$ is constant and non-zero, the voltage across it will change at a constant rate (in volts-per-second), a rate proportional to \$I_3\$. Since node Q is held at a virtual-ground, zero volts, \$V_{OUT}\$ is what changes, and where you will see this "integrating" behaviour.

There will be an accumulation over time of charge on C1 (and the corresponding voltage across it), if the input potential \$V_{IN}\$ is even the tiniest bit non-zero. This is usually very problematic, because it's super difficult to produce conditions that result in exactly zero quiescent current through C1.

If \$I_3\$ can be maintained at exactly zero, the output potential stays where it is, which is not necessarily zero.

In reality of course there are input offset voltages, input bias currents, and all manner of imperfections that complicate matters, but the fact remains, if you can establish a condition at IN that results in no current flowing via R1 through C1, then \$V_{OUT}\$ will remain steady, unchanging.
Under those conditions, with all imperfections accounted for, with no further perturbations in temperature, input voltage or anything, and \$I_3 = 0\$, then since no current is flowing through capacitor C1, it's no longer playing any role in this "quiescent state ", and it can be removed without influencing anything. Conditions would remain as they are even if you repeatedly removed and re-inserted C1, or even replaced it with a cell of exactly the same potential difference.
However to return to the vague context of your question, what is the probability that the DC operating point of the entire circuit that contains this module provides exactly the right potential \$V_{IN}\$, to push exactly the right current through R1, that will be exactly the right amount to bias the op-amp input so that \$V_P - V_Q\$ multiplied by some ridiculously large open-loop gain produces a conveniently small \$V_{OUT}\$, well within the op-amp's power supply potentials (i.e without saturating the op-amp output)? I would say that probability is vanishingly small.
In other words, this integrator circuit is not really suitable for DC operating point analysis, except in the context of ideal components and conditions. If indeed you were to remove C1, to establish the quiescent state  of the circuit, and if you were to account for all imperfections of the op-amp, such as input offset voltage, output voltage limits and so on, you'd probably find the op-amp to be saturated before you even start transient analysis.
You are right to ask this question, and I think you can give yourself a pat on the back for identifying the absurd idea that this circuit has any such thing as a "steady state". It does not. Without C1 in place the op-amp is open-loop, and all bets are off. You are at the mercy of any variance in the op-amp's parameters from device to device. Without feedback, it is now a digital comparator, not a linear amplifier.
Even with C1 in place, the integrating nature of the circuit means that the output is always changing, unless you can establish \$I_3 = 0\$ somehow during your DC operating point analysis. Steady state here is very difficult to achieve.
The solution is usually to place a large resistance in parallel with C1, to slowly "bleed away" any charge accumulated, due to non-zero quiescent current through C1, and to provide a tiny amount of negative feedback at DC, bringing DC gain down to something more predictable and reasonable than 100,000 or a million, or whatever the op-amp has.
Comment: So under ideal conditions, in the quiescent state the input voltage difference can equal 0 only when Vin is 0. Then is there no difference between using a capacitor as negative feedback versus no feedback?
Comment: @nova that is correct, though I reiterate that the condition for quiescence is that I3 be 0A, which in practice will *not* correspond to Vin being exactly zero. Close, but non-zero none-the-less.
Answer: Those circuits are incomplete.
If it is an integrator you're talking about (or anything with capacitive feedback), then they'll normally take feedback resistor in parallel with that capacitor so that the input and output of the amplifiers are well-defined at DC.
I believe standard electronic books do not, usually, include the parallel feedback resistor into the schematic when introducing the circuit to a new student as it might clutter the concept and schematic itself... something I totally understand.
There are cases in which the resistors are not needed, but that's usually when the input and output nodes are defined at DC elsewhere. This happens, typically, when an integrator is used in the feedback loop (as a "DC servo").
Comment: Provided the integrator *is within an overall DC feedback loop* (such as is often the case in voltage regulators), a resistor in the feedback loop of the error amplifier (which can also be the loop integrator) is not necessary.
Comment: Indeed. I meant to say that with "input and output nodes are defined elsewhere".
Comment: Another case when there is no need of parallel feedback resistor is if the capacitor is periodically discharged.
Answer: These are my thoughts.
If you are solving for the DC operating point t->inf and you know that at t->inf the circuit goes to DC values and the AC goes away (sometimes with prior knowledge you can do this with visual inspection). The input must also not be changing.
If you know that it does go to a DC value at t->inf and you are solving for DC then the current through the capacitor must be zero at that time. You can then omit it from the DC analysis.
Answer: Although the moderators declared that this question needs details of clarity and, for this reason, the question is closed, the question title unambiguously defines the class of circuits so distinctly that you do not have to see the circuit drawing in order to fully appreciate the problem.
The question body discloses the context of the OP confusion that resulted in this question. That the OP accepted the meticulous answer that reiterates considerations of many textbooks and lecture notes containing basic math of the ideal op amp analysis at the beginner level, leaves no doubt that this derivation is what the OP sought, and they got it with the accepted answer.
Still, the question title is attracting the attention of visitors via internet searches, and, for the students at the more advanced level, it might be of interest that the opamp circuits with pure capacitive feedback exist and are very useful. @ErnestoG mentions one example in his answer, with @PeterSmith's apt comment. I am going to elaborate on this subject with a reference to the Renesas application note AN551 Recommended Test Procedures for Operational Amplifiers.
For visitor's convenience, I use here the circuit drawing and few excerpts from this document. The circuit is developed for the industry-accepted NULL method for operational amplifier testing used in mass production. I learnt of this circuit while looking for an answer to @kenearl's question What is the best way to simulate the open-loop gain of an op-amp in LTspice.
The circuit TEST CIRCUIT 1 from the Renesas application note 551, page 5

is used to measure offset voltage, input bias current, input offset current, common mode rejection ratio, and open loop gain.
When measuring the offset voltage, the gain of this circuit with RF = 50K (RF = 5M) requires the output to be driven to 1000 (100,000) times the offset voltage necessary to maintain the output of the AUT at 0V. Overall circuit stability is maintained by the adjustable feed-back capacitor \$C_A\$. Notice that this adjustable feed-back capacitor \$C_A\$ is the only component of the path connecting the BUFF opamp output and the BUFF inverting input, and this is the circuit of the topology class identified in the OP question title.


Customise Views PDF export

Question: I am using Views PDF to export some of my pages as PDF files. So far I can get it to export the files .
I am however getting challenges in trying to adjust the styling. I have looked at the module for any tpl files and they are not there. Anybody with a headstart on this?
Answer: In the pdf view under "PDF SETTINGS" you can add a css file for styling. Note that it only supports basic css.

Customise Views PDF export


Different fonts on close reason popup

Question: There're different fonts for items in the list on the popup for closing a question with "A community-specific reason":
Comment: No bug. Some items have title, and some don't, and each "type" has its own design. When there is title, it is bold and the body with smaller font, which I find rather good.
Comment: @ShadowWizardChasingStars looks like the first two should have their own titles as well.
Comment: That would make it consistent, however will also clutter the already big popup, with something that isn't really required.
Answer: The reason some close reasons have small text is because they have been updated to use the new format, which includes a "brief description" (bigger text) in addition to "usage guidance" (smaller text).
Apparently there are some active discussions about changing the rest, such as Let's update our blatantly off-topic close reason to use the 5 available close fields!.

Different fonts on close reason popup


Solution for the coexistence of overpopulated human species with nature

Question: Is there any solution to deal with the overpopulation of our species at the same time protect the natural environment of ours? (other than finding a inhabitable planet?)
Comment: I think we need some more details on this: Which issue coming with population growth are you talking about? How do you define overpopulation? How do you define environmental protection?
Answer: It's important to understand that this problem is a bit more nuanced.
Population growth in every species is exponential (as much as exponential growth can exist, in reality almost nothing is proven to grow infinitely, therefore at some point stops growing exponentially).
Check out this for a description of carrying capacity. In a lot of ways it shows that human population growth is not much different than any other species... but we do it on such a grand scale, and that is something to consider. https://en.wikipedia.org/wiki/Carrying_capacity#:~:text=The%20carrying%20capacity%20of%20an,water%2C%20and%20other%20resources%20available.
The biggest problem in the future may not be our exponential growth, but our increase in carrying capacity for our specific environment (which for humans is the entire Earth). The most significant cause of this is the Green Revolution (agricultural improvements, not environment).
https://en.wikipedia.org/wiki/Green_Revolution
We certainly don't want to see that carrying capacity drop.
Comment: Population growth in most species is not exponential: typical populations are *stable* as they have reached limits imposed by their environment. The human population is even more limited and will start declining in a few decades if current family sizes don't grow. we are already past peak children.
Comment: I should clarify a new successful species in a new ecosystem often grows at an exponential rate until an inflection point where population grows starts to slow towards stability (population capacity, k).

But as I mention in brackets above, exponential growth is actually not common. As most things have a stable cap. Thus if you see the entire event from introduction of a species to it's stable population, it looks sigmoidal.  But cut the curve in half (temporally), then it "looks" exponential, which is where we are in human population.
Answer: Measures to control human population that have been tried so far have been unpalatable and very unpopular. Everyone want's their own "tribe" to survive.
Wars and other significant conflicts caused millions of deaths & the world is still in a precarious situation. World War 2 resulted in the death of 75 million people. It had a particularly devastating effect on the population of Russia and its economy has suffered since. Deaths resulting from Stalin's purges and the Ukraine Famine added to the devastating result.
During the 1970s, the Emergency in India, lead to nearly 11 million people being sterilized, mostly forcibly sterilized. A 1970 law lead to the mass sterilization of native American women.
In 1980, China introduced its One Child policy, the results of which are still being felt.


Integrating rational functions of sine using residues

Question: Can anyone point me to a book that discusses integrals of the following type?
$$\int_{\mathbb{R}}^{}\frac{\cos(ax)dx}{1+x^2}$$
Comment: Not really, but at least now the integrals are not trivially zero. My answer to the question would be: about every introductory textbook to complex analysis.
Comment: Could you explain more precisely the type of integrals you are interested in? (Those in your question all equal zero.)
Comment: @Didier, sorry, I've edited. Is it clear now?
Comment: You want some version of Jordan's lemma: http://en.wikipedia.org/wiki/Jordan's_lemma.
Comment: @Micah, thanks a lot for the answer(I will accept if you write it below). At Didier, thanks for the correction.
Answer: Actually, Jordan's lemma is not needed here, at least if assuming $a$ is real. Since the degree of the denominator is two, we can just write $\cos ax = \operatorname{Re} e^{iax}$, and if $a > 0$ and $z$ is in the upper halfplane, $$|e^{iaz}| = |e^{ia(x+iy)}| = e^{-ay} < 1.$$
Hence, choosing $f(z) = \dfrac{e^{iaz}}{1+z^2}$ and a contour $\gamma = [-R,R] \cup C_R^+$, where $C_R^+$ is a semi-circle in the upper halfplane of radius $R$. By the residue theorem it follows that
$$ \int_{-R}^R \dfrac{e^{iaz}}{1+z^2}\,dz + \int_{C_R^+} \dfrac{e^{iaz}}{1+z^2}\,dz = 2\pi i \operatorname{Res}_{z=i}(f(z)).$$
By the estimate above,
$$\left|\dfrac{e^{iaz}}{1+z^2}\right| \le \frac{1}{R^2-1}$$
on $C_R^+$, so the integral over $C_R^+$ tends to $0$ as $R\to\infty$ by the "ML-inequality". Summing up, you will see that
$$\int_{-\infty}^\infty \dfrac{\cos ax}{1+x^2}\,dx = \operatorname{Re}(2\pi i \operatorname{Res}_{z=i}(f(z))).$$
If $a < 0$, you can do the same with a semicircle in the lower halfplane (or just use the fact that $\cos(ax) = \cos(-ax)$).
Similar estimates work for integrals of the type
$$\int_{-\infty}^\infty \cos ax \frac{P(x)}{Q(x)}\,dx
\quad\text{or}\quad
\int_{-\infty}^\infty \sin ax \frac{P(x)}{Q(x)}\,dx
$$
where $P$ and $Q$ are polynomials and $\deg Q \ge \deg P + 2$. If $\deg Q = \deg P + 1$ you do need some variant of Jordan's lemma, this should be covered in any textbook on complex analysis.


MCP23017 INT Pin always high - INT cannot be cleared

Question: trying to enable INT on MCP23017 so that, in case of a state change on any PIN from Port A (e.g from low to high) an interrupt should be triggered on INTA.
Unfortunately the INTA pin remains high as long the pin on port A is high. So I can not clear the INTA pin
What I have done so far. Requirement is to use pure CMD and no framework for it.
I took i2cset and i2cget tools from raspberry pi 4.
wrote a simple script for configuration and added an oscilloscope to INTA.
#!/bin/bash
i2cset -y 1 0x20 0x00 0xff # all ports bank A to inputs
sleep 0.5
i2cset -y 1 0x20 0x0a 0x02 # set Bit 1(INTPOL) to HIGH
sleep 0.5
i2cset -y 1 0x20 0x08 0xff # set Bit 1(INTCONA) to HIGH
sleep 0.5
i2cset -y 1 0x20 0x04 0xff # High on all pins to enable all pins to interrupt on a change of polarity (i.e switch change)

After executing the code, the level on INTA is low.
Adding a high level via cable (so connecting +3.3V to any pin of the port A ==> all pins are wired via 10k to GND) to a pin of PORT A triggers a high level signal on INTA.
It goes back to low if also the high level on the pin from PORT A disapears.
No way of resetting (clearing) the interrupt via CMD like
i2cget -y 1 0x20 0x10  # ==> this is always showing the pin who was triggering the INT
or
i2cget -y 1 0x20 0x12 # ==> this is always showing the right state of the inputs (so which pin is high and which is low... )
Any ideas what's wrong?

I am using IOCON.BANK = 0 therefore IOCON Register for Port A should be according to table 3-5 IOCON Address=0x0a.
According to datasheet the bit 1 should be set to 1 or 0.
In my top post I think I must have a bug.
i2cset -y 1 0x20 0x0a 0x02 # set Bit 1(INTPOL) to HIGH

I must set it instead of 0x02 to 0x01 but how can I define to set exactly the bit 1 to value 1?
i2cset -y 1 0x20 0x0a 0x01 1 # ????? that does not work :-(

Answer: Have you setup the IOCON register (address 0x05)?
The default is that the interrupt output is active low - INT being high means it is already cleared.
MCP23017 daatsheet
Comment: see my answer (new posting)  below
Comment: @D.S - Hi, Your "answer" (which wasn't an answer to your original question, and therefore should not have been posted in the "Your Answer" box) has been added to the question as an edit i.e. an update. Unless you are writing the full answer to your own question (i.e. you have solved the problem yourself, with no further help needed), please don't use the "Your Answer" box below. Thanks!


Implications of an Archipelago Planet

Question: If an entire planet was made up of nothing but small archipelagos, what would come of it? Would living things be able to exist? Would said living things ever form cities or even nations? How would I explain why the world is nothing but small islands in a massive ocean?
Answer: Continents are formed by tectonic activity, so you would need a planet with little or no tectonic activity. No large moons either. The planet would need to be very old and/or having formed from a nebula poor in radioactives, which means a Population II or (more rare) Population III star. Elements beyond iron would be rarer than on Earth.
Life would be different from Earth's (I believe that several heavier-than-iron elements are required by Earth metabolisms, even if in trace amounts), but CHONP would be there, and the basic organic molecules could form. So, I don't see why there shouldn't be life on such a planet.
Of course it would reman aquatic for way longer than on Earth.
But at some point, islands could be formed through biogenesis - you'd have a world of shallow oceans and coral reefs.
Comment: When thought about deeply, it really is something that seems improbably to exist. Tectonic plates would not have to exist or they would produce land and cluster to become a continent. However without underwater volcanoes water organisms would find it harder to exist and evolve. Difficult.
Answer: There is always the possibility that the planet once had continents.  At one point Earth had 1 big continent and shallow seas.  The archipelago planet could be a several million year phase the planet is going through.
Another possibility is that it is a water world with or without plate tectonics.  It is theorized that before Earth got hit by the planetoid that created the Moon, Earth was a water world.  Volcanoes would likely happen even without plate tectonics, just no ware near the rate we have on Earth.  And coral reefs are also a possibility.
I would bet that in the second scenario, any intelligent live would be amphibians.  Too much tech involves fire.  Having islands would allow for smelting and most of the tech that we know.
In the first scenario, I don't know that there would be time for intelligent life to form unless it represents the survivors of whatever caused the continents to disappear.
Answer: There are different theories regarding plate tectonics and why it exists. These theories try to explain why Earth has plate tectonics and Venus, and likely Mars don't have it. Consensus seems to be that an Earth-like planet should have planet tectonics, and consequently, continents, but we shouldn't take it as settled science. For Earth itself plate tectonics took more than a billion years to establish itself, and before that Earth is thought to be pretty much what you have described - an Archipelago Planet.
Aquatic life should have no problem developing on such a planet. However, terrestrial life would have major problems. Not only the amount of dry land would be small - this land would be split among large number of isolated pockets, and any species that are not good swimmers or flyers would have to die out when an island goes under. I propose that your planet would have a civilization of mermaids :)

Implications of an Archipelago Planet


Using Hamachi to develop XNA on one PC

Question: I'm currently trying to develop a network game for XNA (LAN only as no desire to release), however I only own 1 PC.
I've been trying many different things but all seem to have failed. I'm running VMWare Player with a Win7 Prof image (with vs2008 + XNA 3.1 installed) from my main PC (Win 7, everything installed).
I've installed Hamachi on both but Games For Windows Live seems to refuse to use the hamachi connection (even though it's at the top of the preferred list).
If anyone has any advice on what I could try next, or how to get Hamachi working, I'd be extremely grateful.
Many thanks,
Chris
Answer: You'll have to get either another PC or an Xbox 360. You can't do multiple multiplayer game instances on 1 PC.

Using Hamachi to develop XNA on one PC


Confusion about subjectivity

Question: I am having some confusions about subjectivism. I am a pretty new to this subject. Here is my question:
Subjectivists believe that everything is subjective, including philosophy and morals. So, how does criticizing, or calling some philosophy 'bad' make sense? (in the perspective of a subjectivist).
Comment: Does this answer your question? [Is absolute moral relativism impossible?](https://philosophy.stackexchange.com/questions/78872/is-absolute-moral-relativism-impossible)
Comment: Indeed strictly speaking the often heard speech of calling someone not making sense doesn't make sense on its own as any epistemology as a web of all kinds of *beliefs* unavoidably contains its own subjective element in the Cartesian dualist's terms which is usually defined in the first chapter or introduction section of modern western philosophy. But later you'll learn other different ideologies and a branch called moral philosophy (ethics), so you'll have more perspectives for what *ought* to be the norm or the "right" way of thinking per social convention or specific expert domain...
Comment: Mathematics and logic are also part of metaphysics, like morals or aesthetics, and can be _bad_ as well. Any dialog implies common agreements about subjectivities: some say that _objectivism is just a shared subjectivism_. So, 1=2 is a "bad mathematical operation" because you and me have common agreements about mathematical objects: we share the same subjective mathematical notions. In the same line, saying that "fascism is good" is probably a bad philosophy for most, or "being egoist" is a "bad" moral rule for most.
Answer: It's easy to oversimplify the concept of subjectivism. For instance — on the most superficial level — 'subjectivism' might refer to (using non-standard terminology):

Solipsism: The belief that one's subjective experience is the only truth, and that all other people are manifestations of one's own subjective experience, without subjective experiences of their own. This is one way of interpreting Descartes' 'cogito ergo sum'.
Intersubjectivity (linguistic/symbolic): The belief that reality is a composite of the subjective experiences of a community of people, carried primarily in language and symbolic interaction. Thus something like a traffic law has no ontological existence — it only exists within the minds of people — but still has significant 'real' effects in the world.
Intersubjectivity (scientific/ontic): The belief that reality is a composite of the collective subjective experience of some underlying objective/ontological world. E.g. scientific inference, where subjective experience is collated, abstracted, and formalized into a subjective concept that seems to have objective reach.

Any form of intersubjectivity carries the suggestion that collective knowledge can be (not necessarily is) superior to individual knowledge on purely pragmatic grounds. Collective knowledge integrates greater quantities of subjective experience, opening the possibility that individual misperceptions and biases can be balanced out by a weight of external experience. Thus as individuals assimilate broader sources of collective knowledge, their own (individual) subjectivity becomes more sophisticated, nuanced, and developed: aka 'better'. Some people assume that there is a 'best' that this bettering aims for; others assume an unfixed, process-oriented system of communal adaptation. The jury is still out on that one...
Comment: Isn't "cogito ergo sum" from Descartes?
Comment: @Barmar: Yeaaaaah. Total brain fart.
Comment: I might not have noticed except that it was the answer to a Jeopardy question a day or 2 ago.
Answer: From E.E. Sleinis' Nietzsche's Revaluation of Values: A Study in Strategies, page 59

[Nietzsche's] position is not that any view is as good as any other
view either in regard to value judgements or to judgements about the
world.  In considering values, we cannot do better than perspectival
truth, but we can certainly do worse. There are still better perspectives and worse perspectives, and these need to be sorted by
subjecting them to rational scrutiny. Provided only that one desires
to maximise value, the ground is set for an inquiry into how such
desire can be met.
Comment: In this perspective truth is what you think is true.  In a more forceful form it is justice, as quoted [here](https://philosophy.stackexchange.com/a/84530/5154).
Comment: Very nice (+1) but I wonder about this formulation: is perspective truth? Whose perspective is truth? The truth is "in" perspectivalism, certainly according to N it would seem -- but "the truth" reflects back our own desire for it, our own way of understanding it. The Truth doesn't belong to either relative forms or absolute principles etc, and Nietzsche would emphasize it is a useful fiction etc

Confusion about subjectivity


What's the difference between generateblock and generatetoaddress

Question: Both commands generate blocks, and one specifies an address, what is the usage for each? Thanks!
Answer: generatetoaddress generates a block using transactions from the mempool. It uses the given address for the coinbase output.
generateblock generates a block with the given transactions in the given order. It does not automatically select transactions from the mempool like generatetoaddress does.
Both RPCs are only relevant on regtest and are used for testing. generatetoaddress is generally used when any unconfirmed transactions can be included in the block. generateblock is used when only some specific transactions should be included. generatetoaddress is also used to generate coins via coinbase transactions.


add open face chinese

Question: I posted a question about pineapple OFC and was told that it was not within the scope of the poker stackexchange. I was wondering if we could add Open Face Chinese (OFC) and its pineapple variant POFC to be valid.
https://en.wikipedia.org/wiki/Open-face_Chinese_poker
Comment: Could you add links to examples of each game so myself & others can take a look at the games you're asking about, thanks.
Comment: beyond a link to wikipedia? what do you want exactly?
https://en.wikipedia.org/wiki/Open-face_Chinese_poker
Answer: I've looked into it more deeply & id suggest that OFC and its variants are On-Topic.
My reasoning being in relation to another game that was suggested, "Pandante", which id suggest is off-topic.
To be a poker variant there are a couple of constraints more important than others that need fulfilling. Those are use of a standardized deck, and use of a common hierarchy of hand rankings. OFC and variants of it have those characteristics.
An argument used with Pandante was that similarities in skill sets (eg. Counting outs, psychology, etc.) were important. I wouldn't call baseball and tennis variants of the same game because they both requires ball striking skills.
I propose reopening the question.
Comment: Thank you a lot for looking into this. I really really wanted to do some research on my own on the subject in order to make a more informed decision, but the past few days were just packed for me, so there was no way I could find some time to do it. Tomorrow (Sunday) I have time, so I promise I will look into it; I will probably come to the same conclusion: it should be on-topic.
Comment: Mate no probs ;) It was worth looking in to. I'll leave it to you to re-open if you come to that decision.
Comment: I reopened the question.

add open face chinese


how to combine different tex files to a single file and generate pdf using R

Question: I have three data frames which I would like to put to pdf. I generate tex files for the three dataframes. So I have three tex files. I would like to combine it and generate pdf in R. Right now I use some online tool manually copy paste all the tex and generate the pdf. Is it possible using R to combine and generate the pdf.
Comment: Welcome to TeX.SX! You should look at [`knitr`](http://yihui.name/knitr/). It does what you want.
Answer: With the  Sweave function of R (or with its successor knitr) you do not need make one .tex file per each data frame. Instead, you can make a LaTeX file with the .Rnw extension  and insert one/several chunk/s of R code (between lines with <<>>= and @) containing all the data frames and the R code for LaTeX/verbatim printing. One example with Sweave:
\documentclass{article}
\begin{document}
<<Data,echo=F,results=tex>>=
library(xtable)
# Making 3 data frames
df1 <- data.frame(Foo=c("Foo","Bar","Baz"),Bar=sample(1:49,3,replace=F))
df2 <- data.frame(Stuff=round(rnorm(10),2) ,Thing=round(rnorm(10),3))
df3 <- t(df1)

# show three data frames
xtable(df1, caption=names(df1))
xtable(df2, caption="Data frame 2")
xtable(df3, caption="Trasposed data frame")
@
\end{document}

Sweave can convert this is a true .tex file. One way is from the terminal:
R CMD Sweave my3dataframes.Rnw

Then you can compile as any tex file, for example:
pdflatex  my3dataframes.tex

And read with your pdf viewer:
acroread my3dataframes.pdf

If you do not need run pdflatex again for solve cross-references, you can make your PDF with a single command:
R CMD Sweave -pdf my3dataframes.Rnw; acroread my3dataframes.pdf

N.B.: With the editor RStudio you only need to remember where are the buttons to compile and open the PDF.
In any case, the result of the above MWE will be:
Comment: Thanks a lot .. :D This is exactly what i was looking for. I will try this out.
Comment: @ArunRaja You're welcome, but note that the usual  way of say thanks in this site is upvote any good answer (of any question)  and  [accept  the best answer](http://meta.tex.stackexchange.com/questions/1852/how-do-you-accept-an-answer)  (only one) to  your own question (of course, if solve your problem o at least deserves more  acknowledgment that a simple  upvote). Accept answers increase also (a little) your reputation in this site.
Comment: Yeah I am regular user of stackoverflow and have points on that. I dont have enough points to upvote here.


Hyperplane section of embedded curve

Question: Let $X$ be a smooth projective curve over $\mathbb C$ and take a point $p \in X$. For some very large $k$, $\mathcal O(kp)$ is very ample, so gives an embedding $i: C \to \mathbb P^n$. How can I find a hyperplane $H \subset \mathbb P^n$ with hyperplane section divisor equal to $kp$? Or at the bare minimum, I just want a hyperplane divisor intersecting $X$ at only $p$. For reference, I am trying to understand why a smooth projective curve minus a finite number of points is affine, and there is a proof here. Thank you.
Answer: I don't know if you are still interested, but this is an application of the Riemann-Roch theorem. Indeed, as you said the divisor $D = (2g+1)P$ is very ample by Riemann-Roch, so $|D|$ gives an embedding of $\phi : X \to \Bbb P^{2g}$.
On the other hand, the elements in the linear system $L(D)$ corresponds exactly to the set of hyperplane sections $H \cap \phi(X)$, so in particular there is an hyperplane $H$ with $H \cap X = \{P\}$ and in particular $X \backslash \{P\}$ is affine.
This correspondence holds because if $f_0, \dots, f_n$ is a basis of $L(D)$, an divisor is an element of the linear system $E = D + \text{div}(g)$, with $g =\sum a_i f_i$, will corresponds to $X$ intersected with the hyperplane $H : \sum a_i x_i = 0$. For more details see the book by Rick Miranda, Riemann surfaces and algebraic curves, page 159.

Hyperplane section of embedded curve


Ambari scaling memory for all services

Question: Initially I had two machines to setup hadoop, spark, hbase, kafka, zookeeper, MR2. Each of those machines had 16GB of RAM. I used Apache Ambari to setup the two machines with the above mentioned services.
Now I have upgraded the RAM of each of those machines to 128GB.
How can I now tell Ambari to scale up all its services to make use of the additional memory?
Do I need to understand how the memory is configured for each of these services?
Is this part covered in Ambari documentation somewhere?
Answer: Ambari calculates recommended settings for memory usage of each service at install time. So a change in memory post install will not scale up. You would have to edit these settings manually for each service. In order to do that yes you would need an understanding of how memory should be configured for each service. I don't know of any Ambari documentation that recommends memory configuration values for each service. I would suggest one of the following routes:
1) Take a look at each services documentation (YARN, Oozie, Spark, etc.) and take a look at what they recommend for memory related parameter configurations.
2) Take a look at the Ambari code that calculates recommended values for these memory parameters and use those equations to come up with new values that account for your increased memory.
Answer: I used this https://docs.hortonworks.com/HDPDocuments/HDP2/HDP-2.4.0/bk_installing_manually_book/content/determine-hdp-memory-config.html
Also, Smartsense is must http://docs.hortonworks.com/HDPDocuments/SS1/SmartSense-1.2.0/index.html
We need to define cores, memory, Disks and if we use Hbase or not then script will provide the memory settings for yarn and mapreduce.
root@ttsv-lab-vmdb-01 scripts]# python yarn-utils.py -c 8 -m 128 -d 3 -k True
     Using cores=8 memory=128GB disks=3 hbase=True
     Profile: cores=8 memory=81920MB reserved=48GB usableMem=80GB disks=3
     Num Container=6
     Container Ram=13312MB
     Used Ram=78GB
     Unused Ram=48GB
     yarn.scheduler.minimum-allocation-mb=13312
     yarn.scheduler.maximum-allocation-mb=79872
     yarn.nodemanager.resource.memory-mb=79872
     mapreduce.map.memory.mb=13312
     mapreduce.map.java.opts=-Xmx10649m
     mapreduce.reduce.memory.mb=13312
     mapreduce.reduce.java.opts=-Xmx10649m
     yarn.app.mapreduce.am.resource.mb=13312
     yarn.app.mapreduce.am.command-opts=-Xmx10649m
     mapreduce.task.io.sort.mb=5324

Apart from this, we have formulas there to do calculate it manually. I tried with this settings and it was working for me.

Ambari scaling memory for all services


Retrieving and using Permissions from a Service Provider in WSO2 IS

Question: I am using OIDC Authentication with WSO2 IS (5.7.0) in an Angular Application, and passing the OAuth2 Access Token (JWT) through to the backend REST APIs for Authentication (Identity Propagation) and Authorization.
I have configured the Service Provider in WSO2 so that the Roles assigned to a User are included in the Groups Claim (groups) in the Access Token (via the oidc Scope), but I'm not sure how I can determine which Permissions the User has inherited from those Roles so that I can apply some coarse-grained Authorization (RBAC) logic within my Angular App / REST APIs. Specifically, I am interested in custom Permissions added to the Service Provider.
I don't particularly want to use WSO2 IS as an XACML PDP for this coarse-grained Authorization at the boundary, but am considering using it for more fine-grained Authorization (ABAC) nearer to data access / manipulation - where we want to take attributes of the data into account. However, I'm not even sure if these Permissions can be used in XACML policies as they are not direct attributes of a User, which is the same reason they are not returned as Claims in an OAuth2 Access Token.
Is there a way to include the inherited Permissions as Claims in the OAuth2 Access Token?
Alternatively, is there a separate WSO2 IS Endpoint I can call with the Access Token to retrieve a list of Permissions - possibly even as Claims returned in a separate Token (JWT)?
From reading the documentation and searching online, there appears to be a complete disconnect between OAuth2 (Scopes/Claims) and RBAC (Roles/Permissions) in WSO2 IS. There's detail on how to Configure Users, Roles and Permissions (RBAC) in WSO2 IS, but nothing on how to then access and use that data to enforce Authorization.
Answer: This is possible with an extension.
Check https://github.com/wso2-incubator/samples-is/tree/master/custom-permission-claim-handler out.
I guess this is exactly what you are looking for.


Graph API calls to SharePoint Online is returning 504 responses and is taking more than usual time to complete the requests

Question: I have Http triggered Azure functions that interacts with SharePoint Online using Microsoft Graph SDK, the solution has been in production for couple of years - it creates documentsets/uploads files/updates metadata etc.,.  In past month or so it appears that the http calls are taking more than usual time to complete (and failing).  Below is one such example,

Few observations,

The Graph api calls are retried multiple times (exponential back-off) due to gateway timeout errors (504 response).
The first 3 Graph calls terminates around 30 seconds due to 504.
The last Graph call errors out with 409 - eTag mismatch error,

resourceModified Message: The resource has changed since the caller
last read it; usually an eTag mismatch
Inner error:
AdditionalData:
date: 2023-03-01T11:21:53
request-id: 570d092f-8980-44f1-a1a1-8d5185e6006a
client-request-id: 570d092f-8980-44f1-a1a1-8d5185e6006a
ClientRequestId: 570d092f-8980-44f1-a1a1-8d5185e6006a
4) We believe the number of calls are within the throttling threshold.

Questions:

Why are Graph API calls are terminating at 30th second?  Is this a default setting of the Graph SDK?  If so, is it recommended to change?
How to avoid eTag mismatch error while updating metadata, or, uploading files?  In our use case, all operations are replace of existing data.  (Note:  We have observed this error in scenarios where there is only one call targeting an item in SharePoint i.e., no real reason for mismatch.)
Is there a way to determine which applications is contributing to hitting throttling threshold?  As per my analysis there is no reason for SPO to throttle request (no. of requests is within the allowed limit as per the user licenses that our org. has), which team from Microsoft/SharePoint can assist us in troubleshooting this issue.
Answer: I can't reproduce your issue locally. For service issue, I recommend you raise a support case with Microsoft Graph, a Support Engineer will be able to assist you better. You can raise support ticket from:
http://aad.portal.azure.com or https://admin.microsoft.com/#/support/requests.
Hope this helps.
If the answer is helpful, please click Accept Answer and kindly upvote. If you have any further questions about this answer, please click Comment.


Why doesn't Darth Vader's breath rate increase while he is fighting?

Question: Is it because he is a Jedi? Or does it and it is simply inaudible?
Comment: Do you have some evidence of the fact that it does not increase?
Comment: I'm assuming you're mostly referring to episodes 4-6 where his breathing is regulated by artificial means. There are a few scenes where his breathing does not match up with his dialog, so I don't know that his breathing rate is ever really accurate in those movies.
Answer: It's because he's not actually breathing himself.  His breathing is controlled by his suit, which is a mobile iron lung.  See the question "Why was the cyborg Darth Vader wheezy?" for more info.
Comment: Makes sense - this would also mean his V02 max could never increase.  Perhaps his heart rate is always constant, or if not, his suit would monitor when more blood is required in the blood vessels he has left, as well as managing his breathing and oxygen flow for him.
Answer: Given he has no arms and legs I don't think he would require similar levels of oxygen. Also, I would assume anyone who has mastered the force to the level of Vader will actually expend very little energy during a battle and therefore would have a constant metabolism.
Comment: Interesting, although other Jedi knights have displayed shortness of breath / tiredness during physical combat. Perhaps these knights were not as powerful as Vader and had not yet mastered complete control of their own mind / bodies, which encompasses metabolism.
Comment: I don't think Vader ever ran, which might explain it.
Comment: Maybe he cant run. Wonder what is 100m time is.


Proving the inequality $e^{-2x}\leq 1-x$

Question: How do I prove the inequality $e^{-2x}\leq1-x$ for $0\leq x\leq1/2$?
Comment: @Tim: Nowhere. I downvoted this.
Comment: @TheChaz, Sarcasm does not translate well on the internet.  It would have been much more helpful and less confusing had you given your actually reason why you downvoted this question.
Comment: @TheChaz: Out of curiosity: where did you see any research effort?
Answer: For example, $(1-x)(1+2x)=1+x-2x^2\geq 1$ so
$$ e^{2x}\geq 1+2x \geq \frac{1}{1-x} $$
Answer: A slightly ugly but standard beginning calculus approach is to let
$$f(x)=1-x-e^{-2x}.$$
We want to show that $f(x)\ge 0$ in the interval $[0,1/2]$.
A first experimental step might be to use software to graph $y=f(x)$ as $x$ ranges over our interval.  If we have a high degree of trust in the graphing software, the picture tells us that $f(x)$ is quite likely to be $\ge 0$ in our interval.
Certainty is better. Use the derivative to study the behaviour of $f(x)$. Note that $f(0)=0$, and
$$f'(x)=2e^{-2x} -1.$$
The derivative is positive at $x=0$, and is clearly decreasing. It reaches $0$ at $x=(\ln 2)/2\approx 0.3465$.  So $f(x)$ is increasing in the interval $[0,(\ln 2)/2]$, and decreasing from $(\ln 2)/2$ on. By $x=1/2$, $f(x)$ is about $0.13212$, and in particular still positive.
Thus $f(x)$ is $\ge 0$ for $0 \le x\le (\ln 2)/2$, and $f(x) >0$ for $(\ln 2)/2 \le x\le 1/2$. It follows that $f(x)\ge 0$ on the whole interval $[0,1/2]$ (and somewhat beyond $1/2$).
Comment: There are far better ways to prove the inequality. But let's stick to calculusy approaches. To cut down on the negatives, note that equivalently we want to show that $e^{2x}(1-x) \ge 1$ on our interval (we multiplied both sides by the positive number $e^{2x}$). Let $g(x)=e^{2x}(1-x)-1$. Then $g(0)=0$. Also, $g'(x)=(1-2x)e^{2x}$, so $g$ is increasing on the interval $[0,1/2]$, and we are finished.
Comment: @leftaroundabout: The SE software seems to have mangled your link.  I think you mean something like [this](http://www.wolframalpha.com/input/?i=plot+%28tanh%5E2%28%28x-2%29200%29-exp%28-x%29%29+from+0+to+4) instead (with `^` encoded as `%5E`).
Comment: Perhaps it's good to give [an example](http://www.wolframalpha.com/input/?i=plot+%28tanh^2%28%28x-2%29*200%29-e^%28-x%29%29+from+0+to+4) where a software plot is misleading: this function has a value $<0$ at $x=2$, yet the plot seems to be positive for all positive $x$. – Actually, I'm a bit surprised Wolfram|Alpha does this, a good CAS certainly should be able to _proove_ that the plotted function is well-behaved between two evaluation points, and only then simply connect them.
Comment: @IlmariKaronen yeah that's what I meant.
Answer: Let $f(x)=e^{-2x}-(1-x)$. We have:

$f''(x)=4e^{-2x}$, which is always positive. Hence, $f$ is convex.
$f(0)=0$
$f\left(\dfrac{1}{2}\right)=\dfrac{1}{e}-\dfrac{1}{2}\leq0$.

Since $f$ is convex, its curve is always below $0$ on $\left[0,\dfrac{1}{2}\right]$. This proves that $e^{-2x}\leq 1-x$.
Answer: By Taylor expansion, we have $$\ln \frac{1}{1-x} = \sum_{n=1}^{\infty} \frac{x^n}{n},$$ whose convergence radius is $R=1$.
The equation above can also be achieved by integrating both sides of $$\frac{1}{1-x} = \sum_{n=0}^{\infty} {x^n} .$$
For $0\leq x\leq 1/2$ we have the following upper bound
$$
\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq \sum_{n=1}^{\infty }\frac{1}{
n\left( 2^{n}\right) }=\ln 2 \leq 2.
$$
Therefore
$$
-\ln \left( 1-x\right) =\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq
2x.
$$
The given inequality follows.
Answer: We may start out in an elementary manner with a very known result that comes from W. Sierpiński inequality or $(e^x)$ Taylor expansion, namely:
$$e^x\geq 1+x $$
We get that:
$$e^{-x}\leq\frac1{x+1};  \space e^{-2x}\leq\frac1{2x+1}\leq1-x,\space \space \space 0\leq{x}\leq\frac{1}{2}$$
The proof is complete.
Comment: In fact your proofs works for $0\le x\le 2$. We have (for $x\ge 0$) the following: $\frac1{1+2x}\le 1-x$ $\Leftrightarrow$ $1\le (1+2x)(1-x)=1+x-2x^2$ $\Leftrightarrow$ $0\le x-x^2=x(x-2)$ $\Leftrightarrow$ $x\in[0,2]$.
Comment: @Martin Sleziak: thanks for your comment!
Answer: For $x > 0$, the Taylor series for $e^{-2x}$ is the alternating series $\sum_n {(-2x)^n \over n!} = 1 - 2x + 2x^2 -...$ If $x < {1 \over 2}$, the terms decrease in magnitude.  It's a fact about alternating series that in an alternating series with terms of decreasing magnitude, the overall sum is going to be less than what you get if you add finitely many terms, stopping with a positive term. So we have
$$e^{-2x} < 1 - 2x + 2x^2$$
But after a little simple algebra, for $x > 0$ the statement $1 - 2x + 2x^2 < 1 - x$ is equivalent to the statement that $x < {1 \over 2}$. So the inequality follows. (The endpoint case $x = {1 \over 2}$ can be gotten by just plugging in the value).


Stimulated emission of radiation

Question: Laser light is monochromatic, coherent and (in many cases) collimated. Are these properties due to stimulated emission of radiation? Also, is it true that a photon emitted as a result of the stimulated emission is always emitted to the same mode as the original stimulating photon? In other words, is it true that the stimulated photon is exactly the same as the stimulating photon? What is strange to me is that spontaneous emission does not seem to have the above properties although we can think about the spontaneous emission as a result of emission stimulated by vacuum fluctuations.
Comment: It can be a little tricky to make this precise, but generally the phase coherence of a laser is indeed due to the fact that the beam is made up of stimulated emission photons.  In fact, you can basically figure out how long it will take for the phase to meander far away from its initial value by looking at the fraction of photons $\approx 1/n_{k}$ that were emitted spontaneously rather via stimulation.
Comment: Yes, yes and yes. State transition in spontaneous emission is affected by _all_ vacuum modes, while in stimulated emission,- just of resonant frequency and "in phase" photon is able to "knock out" atom from a population inversion to the ground state thereby producing sibling collimated photon. Good analogy of this collimation is the snow avalanche in the mountains, where new parcels of snow is also roughly "collimated" to the same direction. Old snow parcels generates new ones and the process repeats until high-power flow of snow is generated at the bottom.
Answer: Coherence is due to stimulated emission. But collimation and monochromaticity are more largely due to the cavity.
A typical cavity has two slightly concave mirrors. One is perhaps 99% reflective and 1% transmissive. If you solve the Maxwell's equations for that boundary condition, you get a number of modes where the phase change on a round trip back and forth between the mirrors is an integer number of wavelengths. In this case each successive reflection interferes constructively with the last.
There are many such modes. If one frequency has N wavelengths in a round trip, another slightly higher frequency has N+1. All of these can coexist. The laser medium has a band of wavelengths that it can amplify. Wavelengths near the center of the band have a higher gain. A wavelength at the center will grow at the expense of lower gain modes, and soon it will be the only mode.
The best shape of the solution is a Gaussian Beam. To a first approximation, a laser beam is cylindrical and the rays are parallel straight lines. This ignores diffraction, which is another way of saying the wave properties of light. In a Gaussian beam, rays are hyperbolic. Far from the laser, they approximate a cone with a typical divergence angle of a few milliradians.
Comment: Excellent answer, but I would like to rephrase your 3rd paragraph in other words, in order to emphasize the selection the cavity does, to address OPs concern's regarding spontaneous emission not having specific rules, and why those arent present. The cavity is extremely highly selective to which modes are allowed, and which aren't: non-allowed modes are extremely lossy, which in turn makes them less likely to get amplified, while allowed modes are so much more efficient that they "eat away available gain". These two get rid of non-allowed modes from spontaneous emission and stabilize output

Stimulated emission of radiation


New SQL Server with slow Index Spool (Eager Spool) and old server used index

Question: So yesterday I posted this question regarding that we have upgraded from SQL Server 2008 to SQL Server 2017 and that we excperience some performance problems.
I got the answer to try updating statistics which worked for one database, but not the others. Should I set compability level from 2008 to 2017?  At the current time, compatibility level has been left at 100 (SQL Server 2008).
Old server was a physical server, the new server is a VM (Hyper-V).
After the upgrade the execution plan looked different from the old one
and it seems to use Index Spool (Eager Spool) instead of using index. Which means that execution time went from 3 seconds on the old server to 1 minute 50 seconds on the new server.
The only thing I notice that is different from the database that does not have Eager Spool with same query is that the table rows are less, it's 13.000.000 rows on the non-working vs 70.000 rows on the working with correct execution plan.
The plans below are anonymized but I hope there's enough information.
Here is the plan with Eager Spool vs The other databas with Index
I have tried

Updating statistics
Changing cost threshold for parallelism and max degree of parallelism accordingly.
Double check indexes if they're as before.
I've looked at READ_COMMITTED_SNAPSHOT.
I've checked the NUMA configuration.

Another problem is that wait types are high
LCK_M_IS,
CXPACKET,
LCK_M_X,
ASYNC_IO_COMPLETION,
LCK_M_IX,
PAGELATCH_EX,
BACKBUFFER,
TRACEWRITE.

I am starting to run out of ideas.

Max memory is set to leave 4GB for OS, I noticed yesterday that power plan on the host machine was set to balanced but changed it to high perf, but still same problem.
On received advice to update statistics:
Seems to have worked on one of the databases, however I still have a problem on all the other databases.
The old server was a physical server with 8 cores and 32GB RAM.  The new server is a VM running on Hyper-V with 4 cores and 64GB Ram.
Answer: It's possible having double the amount of RAM in the new server is causing the query optimizer to choose a plan that performs worse than expected.
You could try using DBCC OPTIMIZER_WHATIF1 to see what happens if you have 20 cores and only 32 GB of RAM, like the old system.  I wrote a blog post showing how that works on sqlserverscience.com.
To see the effect on plan choice with 32GB of RAM, you could use this command:
dbcc optimizer_whatif ('MemoryMBs', 32768);

You'll need to evict the plan for the query from the plan cache, or force a new plan to be compiled using WITH RECOMPILE or OPTION (RECOMPILE) to see if that causes SQL Server to choose the "better" plan.
If you see a better plan after using dbcc optimizer_whatif, you might consider either re-factoring the query to allow the optimizer to make a better choice, or you could create a plan guide.
You may be interested in this question, and the answers about the differences in execution plans on two servers with different resources.
FYI for future visitors: the question states that most of the databases have been kept at compatibility level 100; this rules out an array of potential issues related to cardinality estimator changes in SQL Server 2014+.

1 - be aware, this is an unsupported DBCC command, and should only be used for testing purposes.
Answer: When upgrading SQL Server there are a number of tasks that should be completed, to ensure a "smooth"(er) transition.
A list that I use can be found here. This list includes tasks like running CHECKDB, checking compatibility levels, refreshing views, and updating statistics.
In your case, stale statistics appears to have been your problem. The result of this is that the cardinality estimator is unable to make accurate estimations for predicting data distribution and rows returned, and thus produces a different and potentially sub-optimal execution plan.
Edit based on updated question
You also mention about updating the Compatibility Level. The above links provide information with regards to how, and why the Compatibility Level should or could be adjusted. One of the main reasons for changing it is that v12 (SQL Server 2014) provided a major upgrade in the Cardinality Estimator. The best way to determine if this will make a difference for you is to test it, preferably before your upgrade has gone live. This may help you in generating similar execution plans, post-upgrade.


TALEND : while moving employee data from source to target I want to remove duplicates and move them to separate file

Question: TALEND :  while moving employee data from source to target I want to remove duplicates and move them to separate file
Comment: Hi - whats your definition of duplicate? Same employee id or identical values in every field of the employee records? When there is a duplicate do you want to move all the records for that employee to a separate file or 1 record to the main target and any others to the separate file; if so, which of the records goes to which target?
Answer: Exactly if the full record is same its a duplicate
like there could be 2 scenarios

duplicate records in the source file
Duplicate records in the source file when compared with   existing table.

duplicate in source file :
if there are 3 duplicate records
catch the first record and move it to target and other 2 move them to another file (may be u shd add a surrogate key to identify them uniquely)
duplicate rec compared with table data :
compare each rec with table data and if its duplicate move it to another file


How to calculate a limit without using L'Hôpital's Rule

Question: I have to calculate the limits $$\lim_{x\rightarrow 0}\frac{\sin(x\sqrt{x})}{x}\;\;\;\text{and}\;\;\; \lim_{x\rightarrow0}\frac{\sin\sqrt{x}}{x}$$ without using L'Hôpital's Rule, and I can't see how. Calculating those derivatives would help me determine if the two functions $x\mapsto\sin\sqrt{x}$ and $x\mapsto\sin(x\sqrt{x})$ are differentiable at $x=0$.
Comment: You can use Taylor's theorem to get $\sin(x) = x + O(x^3)$ as $x \to 0$. So $$\frac{\sin(x\sqrt{x})}{x} = \sqrt{x} + O(x^{3.5})$$ and $$\frac{\sin(\sqrt{x})}{x} = x^{-1/2} + O(x^{1/2}).$$
Comment: @Acccumulation Taylor's theorem does not use L'Hopital's rule. The method of proof is different. Admittedly, the proof of Taylor's theorem is a bit more elaborate than the proof of the version of L'Hopital's rule needed here, but I like Taylor's theorem because it gives detailed quantitative information.
Comment: @Mason Taylor series use derivatives, so kinda are using L'Hopital implicitly.
Comment: @Mason. I like your comment (I am in love with Taylor series for 65+ years). Moreover, we could say that L'Hopital's rule is a byproduct of Taylor expansion. Cheers ;-)
Answer: HINT
The result of the first limit is given by
\begin{align*}
\lim_{x\to 0^{+}}\frac{\sin(x\sqrt{x})}{x} & = \lim_{x\to 0^{+}}\frac{\sqrt{x}\sin(x\sqrt{x})}{x\sqrt{x}} = 0
\end{align*}
Can you justify why it is so?
On the other hand, we can rewrite the second limit as follows
\begin{align*}
\lim_{x\to 0^{+}}\frac{\sin(\sqrt{x})}{x} & = \lim_{x\to 0^{+}}\frac{1}{\sqrt{x}}\left(\frac{\sin(\sqrt{x})}{\sqrt{x}}\right)
\end{align*}
As $x$ approaches $0$ from the right, the second term tends to one and the first converges to $+\infty$.
Since $1\times +\infty = +\infty$, we conclude it converges to $+\infty$.
Answer: You should look at how $\lim \frac{\sin x}x$ is calculated, and go from there. Also, it's probably easier to use the substitution $u=\sqrt x$. Then you have $\lim_{u \rightarrow 0} \frac{\sin u^3}{u^2}$.


Retrieve a deleted trojan by Kaspersky for analysis (on windows 7)?

Question: After installation of Kaspersky on a windows machine, several trojan viruses where discovered and deleted.
Now, I'm looking for a way to retreive those trojans deleted by kasperky. (to export them to a VM for further analysis)
Answer: If Kaspersky has given you logs of what it deleted you may be able to undelete the files. The company formerly known as diskeeper has an undelete program which should allow you to find it.
Be sure to temporarily disable Kaspersky or designate a directory that it should ignore to put the files into. Once the files are undeleted you'll need to password zip them up otherwise any time you try to move them around antivirus will want to purge them again.
All that said, unless you are an expert in reverse engineering/assembly hacking this doesn't seem like a safe thing to do.


How do particles in a wave make the next particles move?

Question: I only know few things about waves, but this question just popped up in my mind.
I already know that particles transfer energy to each other,but I do not know HOW they do it!
The truth is that I really want an explanation using energy transfer and an explanation using the forces that one particle applies to the one next to it.
I am talking about an analysis for each type of wave. Longitudinal,transverse and surface waves.  Thank you in advance :)
Comment: Waves in what medium? On a guitar string? In water?
Comment: Sure, the motion of the wave is the same. But the question of "how the particles exert the force on each other" can be quite different. The interactions of molecules in air don't necessarily look very much like the interaction of molecules in a crystal lattice, even though both support transverse pressure waves.
Comment: I do not think that this is important.Because if you think about it,the important thing here is the type of wave.In each type of wave,the wave moves in the same way in one medium as in another.Difference comes from the different types.For example,longitudinal waves travel in the same way in water(under the surface) as in air with the only difference being that different particles are moving(water and air in each example respectively).
Comment: Transverse and longitudinal waves are the two different types. Surface waves are not a new type, but could be any of these two.
Comment: ok,i agree but i just want a more general explanation of each type.We can't go into specifics on each and every situation.
Answer: A rope is the best example.
Think of a rope as a chain of particles attached to each other.

You now grab the particle in the end and lift it up. Your hand applies the force that causes the acceleration of the particle.
As that particle is starting to move up, it will pull the next particle beside it by exerting the same force as the hand applies.
This second particle replies by starting to accelerate, and as it starts moving, it will pull the third particle along.
And so on

Whatever you exert of force on the end particle will therefore be sent all the way through to each particle in turn.
When your hand reaches the top of its movement (the amplitude of the wave), it will stop moving and thus it will stop the motion of the first particle also. It now exerts a force downwards, so the first particle will stop and starts a downwards acceleration.
As before these same forces are sent all the way through all particles in turn.

This is for a transverse wave. For a longitudinal wave, the same reaction chain of forces will happen, but it would rather be a compression force.
During an earthquake, the inner particles close to the point of expansion are being compressed. Think of a bath of elastic balls.

If you push so fast on the first ball, that it will not have time to move or anything, it will get squeezed a bit.
this first ball will now try to "unsqueeze" and return to original size. Therefor is will push in the next ball.
this second ball now gets a bit squeezed. In the same way, this ball will try to return to the relaxed state and release it's stored elastic energy. To do that it must push on its neighbour, the third ball in the row.
and so on.

I hope this makes it clearer.
Comment: Great answer, but I still have a question.You explained that for transverse waves, the particles in the rope will pull the next particle with it..but how about surface waves (water surface waves are transversal waves).the water molecules are not connected with each other.so how does a particle pull the next particle with it?
Comment: Well, why would you say that water molecules are not connected? In fact they are - just with weaker bonds (Van der Waals bonds, if I remember correctly) than inside a solid. They pull and push in each other in the same manner. And these bonds will easily break when the wave is too high (at shadow water or so), and you see the white splashing of the water at
Comment: Yes, you are right!
Comment: One should add that matter binds and interacts with electric and magnetic forces, in the spill over fields of the atoms and molecules .
Answer: Electrostatic forces, a.k.a attractive forces,
between particles.
(for a transverse wave) On a micro scale, when the wave is initiated, a particle is displaced perpendicular to its neighbouring particle. Because the displaced particle and the neighbouring particle are electrostatically attracted to each other, the neighbouring particle is pulled perpendicular to the next particle along.
The original particle is moved the opposite direction, and so the neighbouring particle is pulled in this direction.
You were correct that the medium in which the wave propagates is irrelevant/unnecessary to give an answer.


Briefly stated, why does the function N(x) appear in the European call option pricing model?

Question: I'm aware of the the mathematical formula for the price of a European call option on a stock however I'd like to think about it in an intuitive way.
Answer: It is because Black and Scholes assume that the stock follows a geometric brownian motion, i.e. under the historical probability $\mathbb{P}$ the stock moves according to:
$$
\frac{dS(t)}{S(t)} = \mu dt + \sigma dW^{\mathbb{P}}(t)
$$
Solving this SDE we obtain that
$$
S(t)=S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{P}}(t)} = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}
$$
where $z \sim N(0,1)$.
Thanks to Girsanov's theorem we can show that under the risk neutral measure, the stock still follows a Brownian Motion, in particular it can be shown that under $\mathbb{Q}$ the stock price follows
$$
S(t)=S(0)e^{(r-\frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{Q}}(t)} = S(0)e^{(r - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}
$$
It can be shown that the price of an option should be equal to the expected value under the risk neutral measure ($\mathbb{Q}$) of the discounted payoff at expiration, i.e.
$$
c(S,t) = \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}]
$$
Now, take the expected value:
$$
c(s,t)= \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}] = \int_{S(T)>K}(S(T)-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =
$$
$$
= \int_{S(T)>K}(S(t)e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \sqrt{T-t} z}-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =$$
$$= S(t)N(d_1) - Ke^{-r(T-t)}N(d_2)
$$
where
$$
d_1=\left[ln\frac{S(t)}{K} + (r-\frac{\sigma^2}{2}(T-t))\right]\frac{1}{\sigma\sqrt{T-t}}
$$
$$
d_2= d_1 - \sigma\sqrt{T-t}
$$
Notice that $d_1$ just indicates the cutoff for z s.t. S(T) > K, i.e. where the option closes in the money.
Summarizing, the normal cdf simply derives from the assumption of a stock diffusing like a geometric brownian motion both under the historical and the risk-neutral probability measure.
Comment: Thanks Franic, that's just what I was looking for. Did you solve the SDE for the geometric brownian motion model of a stock using Ito's lemma?
Comment: Yes, just use Ito's lemma on $d\log S(t)$. You have something like $$d\log S(t) = S(t)\left[\mu dt + \sigma dW(t) -\frac{1}{2}\sigma^2 dt\right]$$ which can be solved diving by $S(t)$ and integrating, i.e. $$\int_0^t \frac{d\log S(t)}{S(t)} = \int_0^t (\mu-0.5\sigma^2)dt + \int_0^t \sigma^2 dW(t)$$ $$\log S(t) = (\mu - 0.5\sigma^2)t + \sigma W(t)$$ then exponentiate to obtain $$S(t) = e^{(\mu - 0.5\sigma^2)t + \sigma W(t)}$$


What the difference to update Android App Widget between two approach?

Question: I read some example in internet/book about the App Widget, a normal example to update the widget is in onUpdate(Context context, AppWidgetManager appWidgetManager,int[] appWidgetIds) method of AppWidgetProvider like this:
final int N = appWidgetIds.length;
for (int i=0; i<N; i++) {
 int appWidgetId = appWidgetIds[i];
 RemoteViews updateViews = new RemoteViews(context.getPackageName(), R.layout.mywidget_layout);
 updateViews.setTextViewText(R.id.mytext, "updated text");

 appWidgetManager.updateAppWidget(appWidgetId, updateViews);
}

It update each Widgets in a loop.
But now, I have to implement an App Widget, it is updated in BroadcastReceiver, onReceive(Context context, Intent intent) method since there are no int[] appWidgetIds passed in. So I implemented the code like this:
RemoteViews updateViews = new RemoteViews(context.getPackageName(), R.layout.mywidget_layout);
updateViews.setTextViewText(R.id.mytext, "updated text");

ComponentName myComponentName = new ComponentName(context, AndroidBatteryWidgetProvider.class);
AppWidgetManager manager = AppWidgetManager.getInstance(context);
manager.updateAppWidget(myComponentName, updateViews);

It didn't update widget one-by-one, but actually all widgets were updated at once. Even though it worked as I want, but I got confused as to why there is no need to update all widgets one-by-one as before.
What's the difference between two methods?
Can I send another broadcast from BroadcastReceiver.onReceive() to trigger AppWidgetProvider.onUpdate()? And how to?
Answer: It's the same thing. Update with ComponentName loops through all the ids like your first code block.
You can see it in the Android code here:
http://grepcode.com/file/repository.grepcode.com/java/ext/com.google.android/android/4.0.4_r1.2/com/android/server/AppWidgetService.java#AppWidgetService.updateAppWidgetProvider%28android.content.ComponentName%2Candroid.widget.RemoteViews%29