Why does Poisson regression not have a closed form solution?

Question: I am trying to understand GLMs by trying to run them on my own on some Poisson data. My understanding is that if I have my Y values and X values, then using the log-link function, $log[Y] = mX+b$ for some $m, b$. Now the question is to find the coefficients which best explain my Y variable. When I read about it online, they will exponentiate this expression to optimize the likelihood over $m, b$. My question is, why can this not be done through simple linear regression on the $log[Y]$ values and then exponentiated after fitting a simple linear curve? Does this not result in a simple closed-form solution? Is there something wrong about this approach?
Comment: GLMs consider $\log \mu$, where $\mu$ is the expected value of $Y$. That's different than considering $\mathbb{E}[\log Y]$, which has the expectation and the logarithm interchanged.
Comment: What would you do with the logarithm upon observing a count of zero, which always has a positive probability for any Poisson distribution?  For various accounts of what Poisson regression actually is, [please search this site](https://stats.stackexchange.com/search?tab=votes&q=glm%20poisson%20link%20function).  https://stats.stackexchange.com/questions/69820 answers the same question for logistic regression.
Answer: You could use a linearized form of the equation and apply ordinary linear regression. But, this makes two errors:

You ignore non-homogeneity of the error distribution. The variance in $Y$, and also the variance in the transformed variable $\log(Y)$, is not the same for different values of $X$. You need to perform the linear regression with weights $w_i =\hat{Y}_i$ to correct for this, where $\hat{Y}_i$ is the estimated value (more about that later).

A non-linearity in the effect of errors on the response when a link function is applied to the response. (ie. the model of the error distribution is for $Y$ and not for $\log(Y)$ and that needs to be corrected).
This is a bit the same as the lognormal distribution (the exponetiation of a normal distribution with mean $\mu$ and deviation $\sigma$) not having a mean $\exp(\mu)$ but instead a mean $\exp(\mu + 0.5 \sigma^2)$.
That is an additional correction, and instead of using $Y_i^\prime = \log(Y_i)$ you use $Y_i^\prime = \log(Y_i) + (Y_i-\hat{Y_i})/\hat{Y_i}$.

So to find the solution $\hat{Y}_i$ with a linear regression, you need to use two corrections that depend on the solution $\hat{Y}_i$ itselve. That is why there is no closed form and an itterative procedure is used. We use a starting point, assume some initial value as solution and use that to compute a new value. Based on that new value we compute a new value, and so on.
See also What is the objective function to optimize in glm with gaussian and poisson family?
Comment: I understand. I was assuming log[Y] = mx + b, where errors in the log[Y] term are typical gaussian noise (as in linear regression). In fact it is log[E[Y]] = mx+b .


According to Valmiki Ramayana, was Hanuman a human being or a monkey (vanara)?

Question: Hanuman is one of the central characters of Ramayana. He's revered as a great devotee of Shree Rama and worshiped as an avatar of Shiva. He's also known as the son of Vayu. We see his images, statues, photos, sculptures, temples, etc. all over India and he's popularly depicted as a monkey. But what does Valmiki Ramayana say about Hanuman? Is he a vanara or a human being?
Comment: @SwiftPushkar how is it even duplicate
Comment: Again I pose that it is NOT clear if Hanuman is an avatar of Shiva.
Comment: @SwiftPushkar How is it duplicate? I didn't get the answer about the Vanaras in the above question.
Comment: @ProudKafir You can edit and remove this and any other lines which are a duplicate of the other: "If he was a monkey, how could he have a yagyopavit (sacred thread) and master all the four Vedas?" Also, change the title to: "According to Valmiki Ramayana, was Hanuman a human being or a monkey (vanara)?"
Comment: @sv. Okay, done.
Comment: Also all the answers are present in above Post.
Answer:
But what does Valmiki Ramayana say about Hanuman? Is he a vanara or a human being?

Valmiki Ramayana says that he is a Monkey - The most powerful divine monkey. Following are 3 verses which address Hanuman as kapi and vAnara - meaning a monkey.

Here in Ramayana, Baalkaand, Sarga 1, Verse 77. Narada gives a summary of the Ramayana to Valmiki and addresses Hanuman as MahAkapiH = The Great Monkey

अस्त्रेणोन्मुक्तमात्मानं ज्ञात्वा पैतामहाद्वरात् |
  मर्षयन् राक्षसान् वीरो यन्त्रिणस्तान्यदृच्छया || १-१-७६ ||
  ततो दग्ध्वा पुरीं लङ्कामृते सीतां च मैथिलीम् |
  रामाय प्रियमाख्यातुं पुनरायान्महाकपिः || १-१-७७ ||
"Knowing himself to be free from the captivity of the weapon according to Brahma's boon, the valiant great monkey Hanuma enduring those demons with self-will who tied his tail with ropes with the intention of burning it, there after burning the city of Lanka while leaving from fire the Seetha of Mithila, got back again to convey the pleasant news to Rama ... [1-1-76, 77]

And here again in Bala-kanda, Sarga-17, verse-16-17, mention Hanuman as the celestial vanarah, the son of Vayudeva, with an indestructible body and with the speed of Garuda:

मारुतस्यौरसः श्रीमान् हनुमान्नाम वानरः |
  वज्रसंहननोपेतो वैनतेयसमो जवे || १-१७-१६
  सर्ववानरमुख्येषु बुद्धिमान् बलवानपि |
  ते सृष्टा बहुसाहस्रा दशग्रीववधोद्यताः || १-१७-१७
The opulent/marvellous monkey son of Vayudeva known by the name Hanuman. Whose body is as indestructible as the Vajra of Indra and in speed only Garuda, the son of Vinita can match him. Among other vanara, he is the most intelligent and powerful!

And here in Kishkindakand, Sarga-3, Verse-2. Hanuman himself is thinking that my monkey form (kapi-rupam) no one would trust so I should change into a ascetic/sannyasi.

कपि रूपम् परित्यज्य हनुमान् मारुतात्मजः |
  भिक्षु रूपम् ततो भेजे शठबुद्धितया कपिः || ४-३-२
Casting off his monkey's semblance Air-god's son Hanuma attained the persona of an ascetic, for that monkey is incredulous in mind about Raghava-s. [4-3-2]

There are many other verses but I think this much should suffice to answer the question.
Hope this Helps.
Comment: @SwiftPushkar I like to stick to the point of answering the question. But I certainly agree with you and the sentiments behind it, and hence will try to add a few of those verses relevant to question as well as showing his divinity. Kindly forgive if at all the way I answered is not proper.
Comment: I think adding some references of divinity of Hanuman  from VR will make  the .answer more accurate . Because he is Chiranjeevi and also a divine monkey , he is not a normal ape , we pray to him as god.
Comment: Haha , there is no issue of sentiments here. But my point is adding the ref. from VR where Haumuna is mentioned as divine monkey will be more accurate than simply quoting the shlokas where he is just called as monkey plainly. Your answer is correct and proper. But a bit less accurate with the current shlokas . Just changing the shlokas will make it the Best. And after all it all depends on you . I was just pointing , so do not take it seriously.  : - )


How to take screenshot on Ubuntu-guest (Virtualbox)

Question: I am running Ubuntu 10.10 amd64 in a Virtualbox on windows host. whenever i try to take a screenshot by pressing the print screen or by launching the gnome-screenshot option the screenshot appears completely black.
any idea how to get a screenshot inside the Virtualbox?
Answer: Ok guys! I have found out the problem and the solution. :P
The problem is Compiz and the solution is to disable it using the command:
<code>metacity --replace &
</code>
Now, I can take screenshot of any window inside the VirtualBox.
Comment: Well... this kills all my icons and everything... so partial credit...
Comment: This does not work for me on Mint 17.1 inside Win 8.1. If I use this to try to take a screenshot, I just get a screenshot of my background. It does not matter which application I try to take a picture of (Terminal, Firefox, folder explorer, ...).
Answer: VirtualBox provides a menu option to take screenshots of the guest, View --> Take Screenshot (Host+E). Alternatively, just Host + E (that's usually Right Ctrl + E).
Comment: These screenshots from View_>Take Screenshot are currently blank(black empty pictures) for me, and doing a Print Screen not even toggled into the Host results in nothing in the copy paste buffer  :(
Answer: In Ubuntu guest go to System -> Preferences -> Keyboard Shortcuts. On the following window define a new shortcut to take a screenshot (e.g. in this example ShiftAltP):

Thenafter a screenshot will be taken whenever you press your keyboard shortcut. The default setting using Print is not available in the guest (host only).
Taking screenhots will not work with desktop effects activated, i.e. when Compiz is running (see here for reference).
Comment: Hi, Thanks for the idea, but even after this , i am unable to take a screenshot with any image. everything is dark in the screenshot. please see the attached screenshot image. http://i.imgur.com/RQkCP.png
Comment: o.k. my fault: I should have read your question better. Your shot is convincing ;). See Edit.
Answer: A workaround would be to switch to windowed mode and take a screenshot of the VM window contents from the host OS.
Comment: thanks. yes thats a good idea. i discovered it 2 days back. +1 to u :)
Comment: I've been using this for a long time. It has some limitations. For instance, today I had to take a screenshot of a mouseover effect. Impossible with this method.
Answer: I'm on windows 7 and you can use the snipping tool to take screenshots on ubuntu guest. I'm on a dual monitor setup though.
Answer: I'm on Windows 7 running Ubuntu 11.10 on VirtualBox. Taking screenshots doesn't work for me either, but here's a workaround:
Press [Host Key] + Alt + Print Screen.
You can now access the screenshot from the host's (Windows) clipboard. The Host Key is by default Right Ctrl.
Answer: In VirtualBox 4.2.6 with Windows7-host and Ubuntu12.10-guest, I took correct screen shots before installing Guest Addition tools. After this, I get only black screens and desktop background image.
After disabling VirtualBox 3D acceleration in VM configuration, the problem seems to be gone.
Now I can use PrintScreen to take the full screen and Alt + PrintScreen to take a window screen shot through the gnome default capture utility.
Comment: Please detail precisely *what* you did and *how* you did it to solve the problem.
Answer: for linux host
Disable "Auto capture keyboard" from virtualbox preferences.
Main virtualbox window (not running guests) -> File -> Preferences -> Input -> Unselect "Auto capture keyboard"


Should edits that only add links to Wikipedia be allowed?

Question: Edits such as this have received both accept and reject votes. What's our consensus on such edits?
Comment: That particular example looks like somebody manually emulating the (often pointless) automatic link generation done internally by Wikipedia itself. If that is the specific type of edit you are referring to, I would definitely vote "no" - but on the other hand the *intelligent* use of links can be valuable.
Answer: Yes, but only when the information is not widely known or reasonably obvious to current-day users, and the explanation is not given by a question or answer on a StackExchange site.
One heuristic is whether or not we have a tag for the thing in question. If we have a tag for it, it's probably (but not always) well known enough that a link to an explanation isn't necessary.
Another is whether someone with a broad general knowledge of the era or area in question but not knowledge of that specific thing could develop an answer given that knowledge. There are a lot of questions I see here that I (and I suspect others) have no difficulty answering with a small amount of research; the correct links make that research faster and make it more likely that I'll do it to develop an answer.
Examples:

What does a Nintendo Game Boy do when turned on without a game cartridge inserted? The Game Boy is so widely known amongst consumers even today that there's no need for link that explains what it is.

What happened to ZIP RAM? ZIP packaging for RAM is long obsolete and was used for only a relatively short time (late '80s and early '90s); most people who didn't use certain computers within that period would not be expected to know about it. Further, not all popular search engines give good results for "ZIP RAM"; though Google's first result is the wikipedia page, both Bing and DuckDuckGo have no relevant results on their first result page except for the question itself. This should have a link. (I'd like to see my proposed edit to add this restored, but it was rejected once so someone else who agrees with me should do that.)

What was the point of Apple Pascal having its own file system?, in particular the edit suggestion linked in the OP.

A link should be added for "Apple's UCSD Pascal" because this particular product isn't widely known even amongst people who've heard of the computers that could run it. But it should be to the Apple Pascal Wikipedia page, since that it's a derivative of UCSD Pascal is somewhat clear from context, and very clear from the first paragraph of the page, which itself provides a further link.
Links for "Apple II" and "Pascal" are unnecessary; even today most people know what these are.
Links for "DOS 3.3" and "ProDOS" aren't necessary since all you need to know about those for the purposes of this question (that they are other Apple II DOSs) is already clear from context.

In a hypothetical question that mentions the Game Boy instruction set being different from the standard 8080 or Z80 instruction set, the "different instruction set" text should be linked to Is the Game Boy Sharp LR35902 object-compatible with the 8080/Z-80?. And it's perfectly reasonable to add a new question and answer to a StackExchange so you can link to it, if that question is appropriate for the StackExchange site. This would not be the case for something like "Apple's UCSD Pascal" in the example above, because "What is Apple's UCSD Pascal" is too general a question for an SE site, as explained below.

This differs from wizzwizz4's answer below because the RetroComputing SE is not an attempt to be a comprehensive archive of all information about retrocomputing, but is merely a set of (generally) narrowly-focused questions with specific answers. According to the help center, we should have "only...practical, answerable questions based on actual problems that [we] face" that are "reasonably scoped." As it says, "If you can imagine an entire book that answers your question, you're asking too much," which is why we should link externally to general information and large, detailed resources.
Answer: Yes
Links to Wikipedia help to give extra context to posts, providing information about the topics mentioned that is far in excess of what can be provided in posts. This can help the understanding of the casual reader.
Comment: I tend to add such links to many of my questions and answers as I write them, often (but not always) to Wikipedia. The audience on RC.SE will have varying levels of familiarity with specific topics, and the presence of a link can help add context to what is written (where needed). The answer should still be complete if the link were moved/removed, so quotations still need to be inserted into the post. There is also, of course, a certain balance to be achieved: not *everything* needs to be hyperlinked.
Answer: No
We already have tag wikis for providing extra information about topics - this question was tagged with apple-ii, file-system and apple-pascal at the time that the edit was suggested. Relying on links to external sites for information is frowned upon because this site should aim to be self-sufficient - that's one reason why we don't have link-only answers.
Comment: The purpose of tag wikis is to explain how to use the tag, not necessarily to give additional information about the subject.  Although additional information can be useful in explaining how to use the tag.
Comment: Agreed.  Gratuitous links to an external site, especially one that isn't as reliable a source of information as our own site aims to be, are at best irrelevant, at worst harmful.  The occasional link to Wikipaedia for background can be helpful but we should not be editing links into posts for the sake of them.
Comment: @Chenmunka I admire the sentiment, but what this site (or any other site) "aims to be" and what it actually is are often two different things. (I'm tempted to add Wikipedia links to "hubris" and "nemesis." but humour and irony are dangerous on the internet!)
Answer: No, and when the information is not widely known or reasonably obvious to current-day users we should document that in a Q&A pair and link to that instead. (See Curt J. Sampson's answer for examples.)
Pros

We're keeping all the necessary information on-site, not linking elsewhere for important information.
It's fulfilling our mission to document and preserve information about retrocomputing for future readers.
The Q&A pair could form a sort of thing index or thing hub, where an overview is given and further questions are linked for details.

Cons

If they're not just giving a general overview, such questions could be too broad.
There's a temptation to produce these Q&A pairs for things that aren't on-topic, such as alternating current or semiconductors. (Or is that just me?)
Comment: I don't think that the on-topic thing is an issue for examples like the one you gave: the question and answer should simply go in another more appropriate StackExchange (probably [electronics.se], in this case) where it will be preserved just as well as if it were in this SE.
Comment: @CurtJ.Sampson I nicked your answer, and then you nicked mine. It has come full circle. ☺ I agree with everything you've said in these two comments, and kind of want to downvote my answers now.
Comment: Making and maintaining Q&A pairs feels like an awful lot of work to me, given the wide scope of RC.SE. Some of these may just duplicate information available on Wikipedia and elsewhere.
Comment: @Kaz Is that a problem? All information is available elsewhere. And the "making and maintaining" argument applies to every single one of our 8500 odd posts.
Comment: I may have misunderstood. I thought the suggestion was to have "Q&As" defining and describing everything, such as "What was the Manchester Small Scale Experimental Machine?". That feels to me like it's straying toward being a reference book or wiki rather than a question and answer model.
Comment: @Kaz Yeah, that is what I meant. But how many of those will there be? And "what's an NES" is more useful, since it can link to stuff about the I/O and graphics hardware…
Comment: Hm. I suspect it could work, and work well, though I'm having a hard time pictuing it at the moment. I guess it feels a bit removed from what (I understand) a SE site is intended to be. Maybe the proof of the pudding would be in the eating, for me at least.
Comment: However, while I agree that explanatory links should be to SE questions or answers wherever those exist, I think that there are still plenty of cases where those won't exist because the questions would just be too broad. I've updated my answer above to give an example and explain my opinion that we _should_ often link elsewhere for important information that's not in "specific question with specific answer" format.
Comment: So cool to see SE working.


Would it be possible to generate muons using neutrinos and electrons?

Question: The decay products for a muon are an electron, a muon neutrino, and a electron antineutrino. As the decay products for a neutron (electron, proton, neutrino) can combine together to form a neutron again, would it be possible to generate muons using electrons and neutrinos as long as they have sufficient energy?
Comment: Yes, trivially, but with very tiny chances of happening.
Comment: It almost impossible collide two neutrinos
Comment: But you can collide one electron with one electron antineutrino and get one muon with one muon antineutrino
Comment: why would the chance of it happening be very low?
Answer: Yes, there are two possibilities for inverse muon decay: $\nu_\mu e^- \rightarrow \nu_e \mu^-$ and $\bar{\nu}_e e^- \rightarrow \bar{\nu}_\mu \mu^-$.  The first process has been observed, but not yet the second.
The standard model $\nu_\mu e^- \rightarrow \nu_e \mu^-$ lowest-order cross-section for a muon neutrino - electron interaction with centre-of-mass energy squared $s$ is (for $M_W^2\gg s \gg m_\mu^2$):
$$
\sigma(\nu_\mu e^- \rightarrow \nu_e \mu^-) \approx \frac{G^2_F}{\pi} s
$$
I believe this process was first observed by the CHARM experiment at CERN and later by the CCFR and other experiments at Fermilab. The subsequent CHARM II experiment observed $15758 \pm 324$ $\nu_\mu e^- \rightarrow \nu_e \mu^-$ events, in agreement with the expected number.
The cross-section for $\bar{\nu}_e e^- \rightarrow \bar{\nu}_\mu \mu^-$ is comparable (for $M_W^2\gg s \gg m_\mu^2$):
$$
\sigma(\bar{\nu}_e e^- \rightarrow \bar{\nu}_\mu \mu^-) \approx \frac{G^2_F}{3\pi} s
$$
but it has not yet been observed because it is almost impossible to make a pure enough high-energy anti-electron-neutrino beam.  For electrons at rest, the minimum neutrino energy threshold for $\bar{\nu}_e e^- \rightarrow \bar{\nu}_\mu \mu^-$ is about $m_\mu^2/2E_\nu = 10.9$ GeV. Such high energy neutrinos are produced by smashing protons into targets which produce lots of charged pions and kaons whose decays mostly produce muon neutrinos and anti-neutrinos. For example, the NuTeV "anti-neutrino" beam was $98\%$ $\bar{\nu}_\mu$ and only $1.6\%$ $\bar{\nu}_e$, and although they observed 24 events consistent with $\bar{\nu} e^- \rightarrow \bar{\nu} \mu^-$, this number was consistent with backgrounds such as beam impurities and muon charge misidentification. These backgrounds were about $7$ times greater than the expected $\bar{\nu}_e e^- \rightarrow \bar{\nu}_\mu \mu^-$ signal.


Sci fi short, about birds that flew through soil and rock as if it was air

Question: Does anyone know the title of, or where to (re-) find a short story where, once a day, the birds of the planet would fly as a flock, but were able to continue on through solid objects. This caused problems such as eroding the foundations of buildings etc! I believe it was set on a single planet in a binary star system. Story must be 45 years old minimum, as I borrowed the collection from  the local library as a child. Ring any bells for anyone?
Comment: Was the story aimed at children? Also what language/nationality was it? Does you remember any of the other stories in the collection?
Comment: Birds came flying from the underground? If I could see it, then I'd understand.
Answer: "Placet is a Crazy Place" by Fredric Brown, first published in Astounding Science Fiction, May 1946, available at the Internet Archive. If you read it about 50 years ago in England, it could have been in this anthology. Here is a description of the setting from Wikipedia:

Placet is a fictional planet that appeared in the science fiction short story "Placet is a Crazy Place" by Fredric Brown, first published in the May 1946 issue of Astounding Science Fiction.

Placet is composed of both ordinary and heavy matter. The core is made of the collapsed, heavy matter, giving the planet a gravity three-fourths that of Earth. Bird-like animals made of heavy matter fly through the earth above the heavy matter core.

Placet revolves faster than the speed of sound in a figure eight pattern around two stars, Argyle I and Argyle II. Argyle I is normal matter and Argyle II is of contraterrene or negative matter. Half way between them is a field where light slows to the speed of sound. This field allows Placet to be able to eclipse itself twice at the same time, run into itself every 40 hours and chase itself out of sight.

This field also affects the brains of human beings, resulting in extremely vivid visual hallucinations. The sense of touch is unaffected by the hallucinations.

And here is a passage from the story:

You can close your eyes, of course, but you don't—because even at the height of the effect, your eyesight gives you the relative size and distance of things and if you stay in familiar territory your memory and your reason tell you what they are.So when the door opened and a two-headed monster walked in, I knew it was Reagan. Reagan isn't a two-headed monster, but I could recognize the sound of his walk.I said, "Yes, Reagan?"The two-headed monster said, "Chief, the machine shop is wobbling. We may have to break the rule not to do any work in mid-period.""Birds?" I asked.Both of his heads nodded. "The underground part of those walls must be like sieves from the birds flying through 'em, and we'd better pour concrete quick. Do you think those new alloy reinforcing bars the Ark'll bring will stop them?"
Comment: A classic. Thanks for reminding us of it.
Comment: User14111. This is definitely the right story - I was convinced the title started 'The birds of …" but you've found the right one, and given me an author to look up . Thank you.


How to use value under cursor in the shell command?

Question: Is it possible to use word under cursor in shell command? Say:
Elvis
^ cursor here

then
:r! echo <word_under_cursor>

and get "Elvis" read in.
Answer: If you are typing interactively in the vim command line you can hit CtrlrCtrlw to include the current word or CtrlrCtrla to include the current WORD.
If you are writing a function or a command you can use expand('<cword>') or expand('<cWORD>') for the current WORD version.
Like in:
function! Hello()
  echo 'The current word is: ' . expand('<cword>')
  echo 'The current WORD is: ' . expand('<cWORD>')
endfunction

Comment: Yes, using `_` solves it. Using expand() is not possible in command line mode as far as I see.
Answer: There is something to remember: Vim treats the commands differently. Some of them accept literals and cmdline-specials, so you can do
:r !echo <cword>

(or maybe :.!echo <cword>)
While others accept VimScript expressions, so it could look as, for example,
:put=expand('<cword>')

(or maybe :put=system('echo ' .. expand('<cword>'))
So the syntax of VimScript is heavily context-dependent.


Not Able to watch my Cardano faucets when i query the node

Question: I am using following command to query node.
<code>./cardano-cli query utxo --address $(cat payment.addr)    --testnet-magic 1097911063</code>
Getting the Faucets in preview testnet by the following site
https://docs.cardano.org/cardano-testnet/tools/faucet
The funding gets successful, also i can see the transaction hash on https://preview.cexplorer.io/tx ,But not getting faucets on node query.
Although my node is also synced to 64% with Alonzo era. Kindly guide me what i am missing.
Also i am using the latest cardano-node version ie. 1.35.4
Answer: First, in order to be able to query your address after you just used the faucet, the node must be in sync (100%). The version 1.35.4 is correct, it is the only one that will sync your node to 100%.
Second, if you are on the preview testnet, the testnet magic is 2. The testnet magic 1097911063 is for the legacy testnet.
Comment: Hi thanks for your answer, ill check it again when it gets 100% synced. Also i checked my testnet magic from db created which shows me 1097911063. And if i query the node with magic 2 then it gives me error of "cardano-cli: HandshakeError (Refused NodeToClientV_13 "version data mismatch: NodeToClientVersionData {networkMagic = NetworkMagic {unNetworkMagic = 1097911063}} /= NodeToClientVersionData {networkMagic = NetworkMagic {unNetworkMagic = 2}}")"
Comment: If your magic in the db folder is 1097911063, then you are on the legacy testnet, not on preview.
Comment: To switch to preview, you need to stop the node, download the configuration files for preview from here: https://book.world.dev.cardano.org/environments.html, delete the db folder and start your node.
Also, 1.35.4 will never sync the legacy testnet to 100%. Only 1.35.2 will sync it to 100%.
Comment: Yah, I have tried the prepod testnet also, but its syncing doesn't go beyond 0.01% even after a whole day. With that syncing speed i am unable to get any faucets. Can you please recommend its solution. I have also tried cardano 1.35.3 (preview and also preprod), but again its not syncing enough then how can i get faucets and proceed with that?
Comment: Syncing on preview and preprod should be a matter of minutes, you are probably doing something wrong.
For preview and preprod use 1.35.4, 1.35.3 will still go for a while for preprod, but not for preview.
You should check the logs to see what the problem is. If you are not starting your node as a systemd service, take a look at the tutorial I wrote here: https://apexpool.gitbook.io/stake-pool-scripts/running-a-cardano-node
To see the logs if you set it like I described, run:
```journalctl -f -n 50 -u cardano-node.service```
Comment: Ok after alot of research, finally i got the solution. I checked my log, it was giving error of  `TrConnectError (Just 127.0.0.1:1337) 220.127.116.11:30000 Network.Socket.connect: : invalid argument (Invalid argument)` . So i resolved this error by running the node with **host-addr 0.0.0.0** So final command is ```cardano-node run --config $HOME/cardano/preprod/config.json
--database-path $HOME/cardano/db
--socket-path $HOME/cardano/db/node.socket
--host-addr 0.0.0.0
--port 1337
--topology $HOME/cardano/preprod/topology.json```


completly removing a package of this kind

Question: hi I was trying to configure my laptop and went online trying to figure out about things to do after installing Ubuntu 14.04. I came across this.

Fix overheating and extend battery life
There is a good chance Ubuntu is not handling your fan, battery, etc. optimally but there is a nifty utility that probably will. Do this:
sudo add-apt-repository ppa:linrunner/tlp
sudo apt-get update
sudo apt-get install tlp tlp-rdw
sudo tlp start

but it seems to have made my laptop behaving weirdly but shutting down regularly after booting an entering password. how do I remove the whole package? please.
Also my graphics seem to turn a bit greyish for a mn or two before returning to normal.
I am using a Lenovo w510 Intel core i7 with 4gb ram and 1tb hdd.
many thanks.
Answer: Installs PPA Purge
sudo apt-get install ppa-purge
Removes the PPA
sudo ppa-purge ppa:linrunner/tlp
Uninstalls the packages
sudo apt-get purge tlp tlp-rdw
Restart system. Optionally, if there are no open documents etc, shutdown -r 0 for a quick reboot.


Resetting a hijacked router

Question: If you reset a router that's been hijacked to factory defaults with the inbuilt reset button, will that take care of the problem?
Comment: It depends on how it's been hijacked. It might also let the intruder in again if the initial attack vector has not been closed
Comment: It can also depend on the type of router, I could normally be of more help but there is little information given.
Comment: Do you know how the router was compromised in the first place?
Answer: If your only problem is that the router is compromised than a reset to factory defaults will probably fix this problem. But make sure that you are really doing a factory reset and not a simple reset (restart) of the router.
If your problem is also that an attacker was able to compromise the router in the first place then this problem has not been solved by a factory reset. Maybe it can be resolved by updating the firmware, but if a fixed firmware is out there depends on the vendor. Also you cannot be sure that the firmware really fixes the problem even if it claims so, because some vendors just fix the one bug which got reported or seen in the wild and leave other unreported bugs open.
Answer: In short, probably not.
Resetting the router to default settings will likely remove the back-door or the privileged access gained while high-jacking the device. What resetting doesn't do is stop the initial vector that cause the high-jacking in the first place.
As a rule there are a few things you should do with new or recently reset home routers to mitigate security issues:

Change the administrator password
Enable security (WPA2) and disable broadcast beacons
Disable remote administrator access (make sure an administrator can only login from the local network)
Disable UPnP which is a common vector in high-jacking routers.
Update your firmware to the latest available.
Disable NetUSB if it's an available feature, it's another vector.

Basically, if the hole is still there you're open to another attack - internet scans are perpetually looking for vulnerable devices so it'll just be a matter of time because the router gets re-high-jacked.

Resetting a hijacked router


Point feature changes over time?

Question: QGIS strategy for denoting point feature changes over time?
My project file has a shapefile layer (point symbols) geolocating 50 historic street name mosaics embedded in concrete sidewalks throughout an urban neighborhood. Label for each point displays the historic street name. Contemporary street names and curb outlines are displayed on background shapefile layers from City GIS portal.
As sidewalks are gradually replaced, tiles at some locations will be removed, then reset at same locations in the new sidewalks. Tiles at some locations may never be removed, others may be removed/reset several times. Attribute table for my original shapefile includes separate columns for creation date and subsequent reset date.
How to indicate on map that a tile was present in original sidewalk, and now is reset in new walk, or even was reset several times? Label text needs to show historic street name.
My ideas:

New shapefile layer with larger point symbol, e.g. a circle with no color fill. Circle symbol on new layer would surround but not obscure original symbol on original shapefile layer. New layer table includes only one attribute column = reset date. For tiles that are removed/reset several times, new shapefile layers can be added with bigger concentric circle symbols.  Example, any mosaics that that are removed/reset 2 times would have a symbol on 3 separate layers (orig layer, reset #1 layer, reset #2 layer) and would display 2 concentric circles around the original symbol.
Multiple, nested symbols, or label colors, or other differentiation on original layer. No clue how to place several different symbols on same layer.  If this is a good strategy, please tell me what this process is called and point me to where to learn about it.
Other strategy?
Answer: I would recommend checking out rule based styling (e.g. here and here).  Using rules you can setup various rule/expressions against the layer field(s) values to evaluate your various conditions.  You may also stack rule symbols on-top of each other based on conditional priority using the Rendering order button.  This may eliminate the step of having additional shps to track data change over time.
Comment: OK, thanks.  I will check out rule based styling.  End product is file(s) for submittal to City so they can add/update these features in City portal.

Point feature changes over time?


Core data, how to get NSManagedObject's ObjectId when NSFetchRequest returns NSDictionaryResultType?

Question: I have an NSFetchRequest which is returning the objects' properties in an NSDictionaryResultType. Is it possible to also get the objects' ObjectId within this dictionary? Otherwise I will need to run the query with a return type of NSManagedObjectResultType which is much slower for a large number of returned items.
Answer: Yes you can, using the very nifty but badly-documented NSExpressionDescription class. You need to add a properly-configured NSExpressionDescription object to the array of NSPropertyDescription objects you set via setPropertiesToFetch: for your NSFetchRequest.
For example:
NSExpressionDescription* objectIdDesc = [[NSExpressionDescription new] autorelease];
objectIdDesc.name = @"objectID";
objectIdDesc.expression = [NSExpression expressionForEvaluatedObject];
objectIdDesc.expressionResultType = NSObjectIDAttributeType;

myFetchRequest.propertiesToFetch = [NSArray arrayWithObjects:objectIdDesc, anotherPropertyDesc, yetAnotherPropertyDesc, nil];
NSArray* fetchResults = [myContext executeFetchRequest:myFetchRequest error:&fetchError];

You should then have a @"objectID" key in the the dictionaries you get back from your fetch request.
Comment: Nice find Nick, and your documenting the example was very much appreciated.
Comment: This is a freaking answer of the year :)
Answer:  NSFetchRequest *request = [[NSFetchRequest alloc] init];
    request.entity = [NSEntityDescription entityForName:@"yourEntity" inManagedObjectContext:context];
    request.sortDescriptors = [NSArray arrayWithObjects:[NSSortDescriptor sortDescriptorWithKey:@"title" ascending:YES], nil];
    request.predicate = nil;
    request.fetchLimit = 20;

    NSError *error = nil;
    NSArray fetchedResults = [context executeFetchRequest:request error:&error];

    NSLog(@"%@", [fetchedResults valueForKey:@"objectID"]);

Since your fetched results are already in an array why not pull them out with the valueForKey:@"objectID" ? Clean, simple only need one fetch request so you can pull all other data you need as well.
Comment: Jason wants results in dictionary format, while your approach returns a faulted managed object
Answer: Swift Version of the accepted answer
    let objectIDExpression = NSExpressionDescription()
    objectIDExpression.name = "objectID"
    objectIDExpression.expression = NSExpression.expressionForEvaluatedObject()
    objectIDExpression.expressionResultType = .objectIDAttributeType
    let fetchRequest = NSFetchRequest<NSManagedObject>(entityName: entityName)
    fetchRequest.resultType = .dictionaryResultType
    //
    var propertiesToFetch: [Any] = [objectIDExpression]
    propertiesToFetch.append(contentsOf: entity.properties)
    fetchRequest.propertiesToFetch = propertiesToFetch

Comment: thanks for that last bit to append the remaining fields from the entity
Answer: Nick Hutchinson's answer in Swift:
    let idDescription = NSExpressionDescription()
    idDescription.name = "objectID"
    idDescription.expression = NSExpression.expressionForEvaluatedObject()
    idDescription.expressionResultType = .objectIDAttributeType

I can't comment on it because I don't have enough rep :(
Answer: The only solution I have found so far is executing a second fetch request, that is similar to the initial fetch request except the following differences:
[fetchRequest setReturnsObjectsAsFaults:YES];
[fetchRequest setPropertiesToFetch:nil];
[fetchRequest setFetchLimit:1];
[fetchRequest setFetchOffset:index]; // The index for which the objectID is needed
[request setResultType:NSManagedObjectIDResultType];

This will cause the fetch request to return an array with exactly one object, the wanted objectID. Performance seems good, even when the initial fetch request's result contains 10000 objects.
If there are any better ways to handle this I would be glad if someone could post them here.


How do I log data from Courier .mailfilter rule sets?

Question: Background
I'm running Ubuntu 16.04 LTS on a virtual machine. It is the mail server for a small medical clinic. We are trying to move mail filtering from the client-side to the server-side, and I am having difficulty with $HOME/.mailfilter rule files.
I've been reading various copies of the man-page at multiple locations, but there is not much information about turning on logging. An example of the data that I am working from can be found here: Maildrop Filter Man Page
Current Setup
At the moment, my directory structure looks like this:

The three files that I created to contain the filter rules contain very simple instructions:
logfile "/var/log/maildropfilter.log"
log "Executing code in {file_path}"

Naturally, I change "{file_path}" to be whatever file I'm currently in. My hope is that if the rule file is executed that the logfile will have a statement that says "Executing code in {file_path}", and I will know which of the three files to start putting control logic into.
Expectation
According to the documentation, or rather my interpretation of it, after receiving an email to my email account for user jaredclemence, I should see the following in my log file:
Executing code in /etc/courier/maildroprc
Executing code in /home/jaredclemence/.mailfilter

I should not see any reference to the third file in the Maildir, because this file is not $HOME/.mailfilter, which is the file that the documentation points to.  I created this file just in-case $HOME references the root of the Maildir instead the root of the user account's home folder.
Current Result
The trouble is, when I send an email to the user jaredclemence from outside the network, the email does get received, but the log file is empty.
What am I doing wrong?

Is there something special that I have to do in order to register my filter rules, or is having the file in the correct location all that is required?

If having the file in the correct location is all that is required, then what is wrong with the code that I have written, which sets the logfile and outputs a message to the log with the log command?

All help is appreciated.
Update 2018-01-04 10:17
I read the maildrop man-file and learned more about the various modes that it will run in, and identified that one of the three modes does not use the .mailfilter rules.
Noting that there is a manual mode, I executed the maildrop command in manual mode to see if the logfile were activated and utilized.
After running the maildrop command manually, I verified the following output:
Executing commands from /home/jaredclemence/.mailfilter
Date: Thu Jan  4 09:37:05 2018
From: Synegi Inc <firstname.lastname@example.com>
Subj: Wishing You a Happy New Year from Synegi!
File: /var/mail/jaredclemence                                          (29594)

This is interesting, because the global rule file is never activated. It is also interesting because it delivers the mail to an mbox at /var/mail/jaredclemence, which is not the Maildir.
This suggests one of two things. Either the maildrop program is not used as the local delivery agent, or a configuration file changes the delivery location when the maildrop program is run in the 'Embedded Mode', which is the only mode of the three that does not execute the .mailfilter file instructions.
From here, I have started reading man-pages for trace(8), postfix(1), and local(8) to see if I can track down which program is being used by the mail delivery agent.
Answer: After much research and searching, I have discovered that my system is not using maildrop for delivery.
If you are having the same question, first identify which mail delivery agent your system is using.
Start with your SMTP server. Mine is Postfix.
Postfix uses it's own local delivery agent in a process called local(8).  In the file /etc/postfix/main.cf, look for a variety of references to any of the local(8) handoff parameters. The two most prominent would probably be mailbox_transport or virtual_transport (the difference depends on whether the people who receive mail have accounts with home directories or if an account is maintained virtually).
If these parameters are not set, then postfix is probably handling its own local delivery, and as far as I can tell, postfix does not do inspection based redirection and delivery to specific mailboxes, only from one email address to another based on a series of forwarding or alias addresses.
So, if you are trying to implement advanced filtering with Courier or Dovecot-LDA or Dovecot-LMTP, then you will need to tell Postfix using the mailbox_transfer or virtual_transfer settings which program to call.


Energy change during electron transitions

Question: I was studying Bohr's atomic model and came to know that when electrons make transitions in between the orbits they lose or gain energy in the form of electromagnetic radiations. I understand why they lose or gain energy. But why is it only in the form of electromagnetic radiations and in not any other form?
Comment: What other forms do you have in mind?
Comment: What other forms do you expect?
Answer: Actually other forms of energy could be emitted or absorbed. These are the Raman lines. Sometimes a raman line corresponds to transfer of part the energy into the internal rotational/vibrational states. In a solid this would correspond to part of the energy bring emitted as a phonon. Sometimes a phonon energy is added to the energy if the incoming photon and one gets a higher energy photon as the output
Ofcourse phonons are bosons just like photons
Answer: The issue of emission or absorption of electromagnetic radiation by an atom, within Bohr's atomic model, goes the other way around.
The model was introduced by Bohr to provide a unifying picture where stability of motion of an electron moving around the nucleus (thus being an accelerated charge which should emit e.m. radiation, according to Maxwell's equations) and the experimental evidence of line spectra of atoms could coexist. Bohr's model explicitly eliminates e.m. radiation of the orbiting electrons, with the only exception of transition from an orbit to another.
Other conversion of energy when electrons jump, at least in the case of an isolated atom are hard to imagine.


AC Condenser Not Starting, New Capacitor, New Switch

Question: I am troubleshooting my home ac unit model GSX130361EC. It was not turning on with the contactor repeatedly clicking and the fan not turning, however trying to start then giving up. I had replaced the contactor and the capacitor on the unit since I had these parts on hand. When testing yesterday, it turned on and ran no problem. I went today and tested, and the unit will not start. It is not a blow fuse, I can manually depress the contactor and the unit starts. I verified all the measurements on the capacitor before installing. While researching I read this could indicate a low freon amount, my unit uses 410a. I borrowed a neighbors gauge, surrounding air temperature is 65 degree, and as I read this I have high suction pressure thus a low freon with likely leak  do you concur dr.?
Comment: you were right! thanks!
Comment: the relay is switched by 24 volt coming from thermostat, check that. The compressor will fail to start if the pressure is to high
Comment: Uou mean the wires going to the relay with a multimeter?
Comment: the really has low voltage (24 V) for the coil, and high voltage (120V) so set your multimeter appropriately.
Comment: You need licensure, recovery equipment and upstream supply chain to deal with recovered refrigerant, in order to mess with R410. Of course, some equipment is designed from the ground up for DIY installation that doesn't require refrigerant handling - Pioneer etc. overcharges the outside unit somewhat and fills the rest of the system from that; and MrCool uses special self-sealing quick-connect line sets.
Comment: licensure smishensure
Answer: Wow, the fins of the condenser are really furry. That debris should be rinsed off so the unit can function better.
Each refrigerant has a definite relationship between temperature and pressure. The blue and red scales on these gauges show PSI; the pink scale is calibrated to temperature. We see that both gauges are showing the same pressure, so we can conclude that the compressor has not been running for some little while and the system has reached a non-running equilibrium. The pink scale reads about 66 °F which matches your reported temperature.
We can't determine whether the refrigerant charge is correct when the unit isn't running.
It sounds like you'll need to do more electrical troubleshooting. Trace the wires and compare with the wiring diagram to confirm that you've re-connected everything correctly after those parts were replaced. Use a volt meter to confirm whether the correct voltage is reaching both the compressor and the fan. Wiggle the connectors and confirm that they're clean and snug. Check all the connections: at the capacitor, input and output of the contactor, input and output of the disconnect, the disconnect itself, the fuses inside the disconnect (if any), the circuit breaker.
Answer: The first thing to check is the low voltage side of the contactor. If the new connection came loose, it could obviously cause the contactor to fail to close.
Assuming that is not the problem, it looks like this unit has a high pressure switch in line with the contactor. This is on the 24V signaling line. You can try briefly jumpering it out by attaching a jumper to both sides. If it is the switch preventing the contactor from closing, then it will close one jumpered out and compressor will start. (Obviously I'm assuming the thermostat is calling for cooling).
Assuming it is indeed the pressure switch causing preventing the contactor from closing, we now need to figure out why. This switch on your model requires 610 PSIG to go open circuit (cutting off the contactor), and should go closed circuit again once the liquid line pressure drops below 410 PSIG. Assuming you connected the gauges correctly, and that image was the system not running, then your pressure is nowhere near high enough for the pressure switch to be cutting off, at least not before the compressor starts.
Especially since the pressure has equalized on high and low sides. (And matches the temperature, suggesting that the refrigerant is not crazy low or anything).
This would suggest a faulty or worn out high pressure switch, which needs to be replaced. Of course this may not be the only issue with the system. For example if the switch was worn out, that would suggest another issue causing high head pressure, which needs to be addressed, lest the new switch get worn out.


Embedding content in external website

Question: I'm new to Drupal (and SE) and am looking for a way to embed Drupal 7 content in an external website.
D6 had modules for that (web widget and embed widget).
How can I do it in D7?
Thank you!
Answer: Web Widgets now has a Drupal 7 release, so it might be worth trying it out.
Answer: You can implement the hook_preprocess_page() in your theme's template.php file to unset the regions you don't need (if your don't do this it will generate a performance issue. see http://www.lullabot.com/articles/drupal-performance-tip-block-visibility) and then you can set a custom template file for the page using the $variables['theme_hook_suggestions'] variable. This will give you the ability to show just the content region on the desired path on your Drupal site using a custom HTML structure for instance.
/**
 * Implements hook_preprocess_page().
 */
function mytheme_preprocess_page(&$variables) {
  $path = request_path();

  if ($path = 'the/path/you/want/to/customize') {
    unset($variables['page']['header']);
    unset($variables['page']['footer']);
    unset($variables['page']['sidebar_right']);
    $variables['theme_hook_suggestions'][] = 'page__my_custom_template_file';
  }
}

Then you need to create the file page--my-custom-template-file.tpl.php on your theme's templates directory to set the HTML that you need in the page.

    <?php print drupal_render($page['content']); ?>

For more info about Drupal theme files, see http://drupal.org/node/171194
After that you will just need to add an iframe tag in the place you need that content on the external website following the example:

For more info. about the iframe tag. See http://www.w3schools.com/tags/tag_iframe.asp
Answer: Not sure about a Drupal 7 Module that is out yet. As an interim solution maybe PHP scrapping from your Drupal 7 Installation?
http://net.tutsplus.com/tutorials/php/html-parsing-and-screen-scraping-with-the-simple-html-dom-library/
The solution is not ideal but might help until one of the projects you posted is ported to 7/
Comment: thank you. Do you think it's possible to create a view and assign a custom template that's more or less blank. Then embed the page with an iframe?
Comment: Web Widgets module does it for you

Embedding content in external website


A woman who prays yom tov shemona esrei on chol hamoed

Question: Does a woman who prays yom tov shemona esrei on chol hamoed have to repeat?
Comment: Do you have any reason to suspect a women and man would be different in this regard?
Comment: Phillip, welcome to Mi Yodeya! I recommend that you [edit] this question to explain more about what you think the relevant issues are and what motivates this question. That will make it much more likely that you'll get answers that address your point of curiosity. Please take a look at our [How to Ask](http://yodeya.com/help/how-to-ask) page for tips on how to pose an especially answerable and valuable question. Finally, please consider [registering](http://yodeya.com/users/login) your account, which will give you access to more of the site's features.
Answer: Rabbi Ovadiah Yosef rules (Yechavveh Da'at 5:37) that she should not repeat it, but must recite the entire 19 blessings including Ya'aleh veYavo that she should have said to begin with. He reasons that since Chazal enacted to recite a full 19-blessing-prayer on Chol haMoed, one does not fulfill one's obligation with only 7 blessings. Thus the initial mistaken prayer is void.
His ruling that the initial mistaken prayer is void appears to apply to anyone who said such a prayer, whatever their gender may be.
Rabbi Moshe Feinstein has a similar argument in Igrot Moshe OC 4:21:3, explicitly applying the ruling to days of Chol haMoed which fall on Shabbat as well.


print function in python3.x?

Question: I want to know how print function works. But I can't find print function in python3.2's source code!
In python2.5 print an integer(print 100, for example), will call int_print(). Because when create an int object, initialize PyIntObject's function pointer - tp_print that point to int_print():
printfunc tp_print;
/* We can't provide a full compile-time check that limited-API
   users won't implement tp_print. However, not defining printfunc
   and making tp_print of a different function pointer type
   should at least cause a warning in most cases. */
typedef int (*printfunc)(PyObject *, FILE *, int);

In python3.x, all integer is an object of PyLongObject, but it initialize tp_print=NULL. And print is a function not like python2.x.
Please tell me when call print, how it works! Thanks a lot!
Answer: I have already known! When print a builtin object, print() will call builtin_print() function.
builtin_print(PyObject *self, PyObject *args, PyObject *kwds)

print function in python3.x?


Can you "fear doing something"?

Question: Could I say sentences like : "I fear skydiving" or "I fear being enclosed in small spaces" or are such sentences wrong? Would I have to use "scared of skydiving/being enclosed etc" or do I need to add the preposition "of" to fear? "Fear of skydiving"? (so use fear as a noun and not a verb)
Comment: "I fear" and "I have a fear of" or "I am afraid of" are all fine, as is "I am scared of".  It would be incorrect to say "I scare skydiving" because skydiving cannot be made afraid.
Answer: "I fear skydiving." is a complete and correct sentence. In this sentence, fear is the verb.
"fear of skydiving" is a noun phrase, not a complete sentence. Fear is a noun in this phrase. You could say "I have a fear of skydiving"
In the phrase "scared of skydiving", the word "scared" is an adjective. You could also use "afraid of skydiving"
Using "I fear skydiving" instead of "I'm scared of skydiving" is a little more formal, and less common, but both are correct.
The reason for all these synonyms is the multiple sources of English. "fear" is Anglo-saxon, as is "fright" but from a different source, "afraid" is Anglo-french, and "scared" is from Old Norse. The stronger "terrify" is from Latin.
Comment: You're obviously correct.  I was always afraid of French but terrified of Latin.
Answer: The idiomatic expression is "I have fear of ...", probably because the fear is latent and abstract, so have is used as Auxiliary verb

Can you "fear doing something"?


Classification of recursively enumerable sets

Question: It's quite easy to show that there are only three recursive sets up to m-equivalence, namely $\emptyset,\{1\},\mathbb N.$ Can we state something "similarly short" for recursively enumerable sets, like there are only 4 recursively enumerable sets up to m-equivalence – $\emptyset,\{1\},K,\mathbb N$ (where $K=\{i\mid\varphi_i(i)\text{ is def.}\}$)? More specifically, is every non-recursive recursively enumerable set m-equivalent to $K$?
Comment: Since Turing reducibility extends many-one reducibility, yes it does: if $A\not\ge_TB$ then a fortiori $A\not\ge_m B$.
Comment: See https://en.wikipedia.org/wiki/Turing_degree#Recursively_enumerable_Turing_degrees. Recursively enumerable Turing degrees are dense.
Comment: I am not sure if that answers my question, because I don't know much about Turing equivalence. Does it imply that there is a non-recursive r.e. set $A$ such that $K\nleq_m A$ ?
Answer: There are lots of these. Even with respect to Turing reducibility (which extends $m$-reducibility: if $A\not\ge_TB$ then $A\not\ge_mB$), the structure of c.e. sets is extremely rich.
That said, constructing an infinite non-recursive r.e. $A$ with $A<_mK$ is significantly easier than constructing one with $A<_T K$. The former was done by Post in 1944 who introduced the notions of creative and simple sets (each of which are necessarily non-recursive), proved that a simple set exists and that $K$ is creative, and proved that no creative set is $m$-reducibile to any simple set; the latter wasn't achieved until a decade later by Friedberg and Muchnik independently, and required a fundamentally new technique (the priority method).


stuck trying to fix bad Android 6.0 update

Question: When manually updating my Nexus 9 (Wifi) to Android 6.0, I forgot to flash the cache.img and vendor.img images. Now my Nexus 9 is stuck at the Google boot screen (just says Google, no animation). How do I get the Nexus 9 back to the state where I can re-flash images manually again (i.e., "flashboot flash...")?
Answer: The solution was simply booting in fastboot mode (I already was there). Apparently, booting in fastboot → plugging in the USB cable → waiting a couple of minutes for the device to be detected by ADB (adb devices) works best (otherwise it won't list the device). I was then able to install Android 6.0 successfully via fastboot commands (USB debugging and OEM unlocking were previously enabled and the bootloader unlocked).

fastboot flash bootloader bootloader-flounder-3.48.0.0135.img
fastboot reboot-bootloader
fastboot flash recovery recovery.img
fastboot flash boot boot.img
mfastboot flash system system.img
fastboot flash vendor vendor.img
fastboot flash cache cache.img
fastboot reboot

mfastboot was used instead of fastboot to flash the system.img file because the latter gave an "could not allocate -1995847632" error message (it can't handle big image files).
I will re-lock the bootloader later.


Does the absence of a region restriction notice mean a game is region free?

Question: Steam has begun to display region restriction and language info for some games, as seen in the picture. Does the abscence of such a notice indicate that the game on offer is completely region free and not in any way different from releases in other regions (except for those that are region restricted)?
Comment: Related: Some games may have region locked versions in some regions but a general 'for everyone else' version. What happens in this case if you happen to be in the 'everyone else' category?
Answer: This is what Steam support had to say on the matter. Hopefully it is true.

Region restrictions will be noted on the store page of the product. If there are none listed, then there are no region restrictions for the game.
Answer: It looks like people have been buying games in other countries, and then playing them locally on Steam.  It's not hard to do, a simple VPN from the UK to the US, and I can buy my games in USD, and have access to the US inventory, then play them in the UK.
From this post and it's comments I'm guessing people were using similar ideas to VPN into Russia, get cheap games, and then play them in their native country/language.
Long story short, this looks/feels/smells like game manufacturers and Steam trying to make sure people pay full price for games. (no judgement, just stating they're trying to maximize revenues)
I would imagine in most cases the games are the same, though in certain locations (Germany) I would guess that the content might be edited based on Germany's watchdog restrictions.  This might also be to broaden the market into countries that have more restrictive content managed systems.
All best guess, and IMHO, hope that helps!
Comment: You might want to edit your answer to avoid sounding like the Publishers and Steam getting full price for a game is a bad thing.  Games sold in Russia are cheaper for a reason, because the at the normal price, they wouldn't sell a single copy.
Comment: @Ramhound - I'm not saying anything like that.  Have added the caveat, but was simply stating that looks like what's going on, whether good or bad.


Permanently disable Apple's OpenSSH?

Question: I have an older OS X 10.5 PowerMac (PowerPC) that needs to be online for testing software under the processor. It needs to be online because others want to use it for testing, too. I've updated OpenSSL and OpenSSH with new binaries placed in /usr/local. I also added a new Launchd plist at /System/LaunchDaemons to enable the new OpenSSH daemon on port 1522.
Now I want to permanently disable the existing Apple provided OpenSSH located at /usr/sbin/sshd. I don't want to delete it or replace it because that often causes more trouble than its worth. A search with http://www.google.com/search?q=apple+disable+sshd+site:apple.com appears to reveal no support documents on the subject. I found lots of discussions about updating OpenSSH and disabling it through the UI, but nothing about permanently disabling it.
I tried removing the execute bit, but that did not work. After reboot, Launchd still managed to start sshd. Other thoughts include a filesystem extended attribute to deny access to it.
How do I safely and permanently disable the existing OpenSSH daemon?
Answer: The most simple way to disable the Apple provided sshd without removing or replacing it is just renaming it:
sudo mv /usr/sbin/sshd /usr/sbin/sshd.old

Additionally you may add a symbolic link at /usr/sbin/ to the /usr/local/sbin/sshd with
sudo ln -s /usr/local/sbin/sshd /usr/sbin/sshd

Permanently disable Apple's OpenSSH?


How to assign defined keyword as value of variable in Robot Framework? How to run a Variable as Keyword?

Question: I'm currently using Robot Framework for creating UI automation for our project. And I would like to know how to assign defined keyword as a variable value and how to run variable as keyword?
Comment: For me it's unclear what you are asking for. You want to know how you can store a value from a keyword which have [Return] statement or you are asking something else? Please update you question with more details.
Comment: You should share the code to we understand and perfaps help....
Answer: Robot framework has for this prepared keyword Run Keyword, where you supply actual keyword to run as a parametr - stored in variable.
Answer: You can assign a defined keyword as a value of a variable in Robot Framework using the Set Variable keyword.
For example:
*** Variables ***
${KEYWORD_NAME} My Keyword

*** Keywords ***
My Keyword
Log To Console This is my keyword.

*** Test Cases ***
Test Keyword as Variable
Set Variable ${KEYWORD} Evaluate ${KEYWORD_NAME}
Run Keyword If '${KEYWORD}' == 'My Keyword' ${KEYWORD}

In the above example, the value of the KEYWORD_NAME variable is assigned to the KEYWORD variable using the Set Variable keyword. And then the Run Keyword If keyword is used to run the KEYWORD as a keyword, if it's equal to My Keyword.
You can also use the Evaluate keyword to run a variable as a keyword, as shown in the example.


Long-haired cat has poop caked on butt

Question: I have a 10 year old long-haired cat and for the very first time.
Now it is an absolute mess!  She Will NOT tolerate a bath.
I've always had dogs. So this is new to me.
PLEASE HELP!
Answer: Will she let you wipe her with unscented baby wipes? They sell "pet wipes" too, but they're the same thing only more expensive. Unscented is the key. If she won't let you do that, either,  you may have to take her to a groomer or the vet to trim the fur on her hind-quarters.  When my long haired cat was older, she didn't have the flexibility or energy to keep herself clean, and trimming the fur was a big help to her.
Answer:
She Will NOT tolerate a bath. I've always had dogs.

Cats generally don't like water - especially not when it's not their own decision.
Dogs are able to defer to humans and do something that the human tells them to do. Cats generally don't and rely on their own decisions.
Cats generally also don't need baths, they are self-cleaning animals. The poop getting stuck in their fur is something that needs to be addressed, but I suggest not using baths as a solution as it will lead to an unhappy cat, it may start to avoid coming near you.

Dealing with the poops getting stuck in her fur is an annoying problem.
In the short term, I suggest cutting the hair in which the poop is stuck to fix the issue. Just focus on the hard to get bits. Simple cleaning can be done by the cat herself (and I'm pretty sure it would prefer doing it herself).
In the long term, if your cat reasonably allows you to (or you can get her to endure it), I suggest trimming/cutting the hair on her posterior to prevent poop getting stuck in it. It doesn't need to be shaven off but keeping it to a reasonable length (like short haired cats) will prevent the issue.
Our long-haired cat used to have this issue once in a while, and her immediate reaction would be to drag her butt across surfaces to get it off. I think you can imagine the poop smears we can home to in those cases.
So what we did was get a very bristly floor mat (example) and put it next to the litter box. We had to replace the mat once it a while as it would get disgusting but it was considerably better than having to clean poop streaks off the floor and furniture.

Long-haired cat has poop caked on butt


Critical effect sizes and power for paired t test

Question: Is there some sort of easy way to calculate the required sample size for a paired t-test with an effect size of 2 standard deviations? In my line of work, we often use an effect size of 2SD from the mean for power and sample size calculation with independent t-tests but I'm not sure how to adapt that to a paired t-test. I'm not even sure if that typed of effect size makes sense for a dependent t-test. Thanks for the help.
Comment: With a paired t-test, the actual test is a one-sample t-test on the pair-differences.
Answer: Yes, this is possible and even fairly easy, but additional information is required. Specifically, we have to make an assumption about what the correlation between the observations from each pair are.
The effect size as a difference in standard deviation units is usually referred to as $d$. We can apply a correction factor to $d$ to incorporate the information about the aforementioned correlation, and then we can use our standard power formulae with this corrected $d$ (making sure to also mind the change in degrees of freedom associated with moving to the paired design) to compute power. The corrected $d$ is
$$
d_o = \frac{d}{\sqrt{1-r}},
$$
where $r$ is the correlation. I have called this $d_o$ because this is sometimes referred to as the "operative effect size."
Here is a little R routine that computes a table of minimum number of PAIRS as a function of the assumed correlation and the desired power level, with $d=2$ assumed.
library(pwr) # package for pwr.t.test() function
             # may need to install first with install.packages()

# define a function to get the minimum number of pairs
# for a given correlation and desired power level
getN <- function(r,p){
  unlist(mapply(pwr.t.test, d=2/sqrt(1-r), power=p,
    MoreArgs=list(n=NULL, sig.level=.05, type="paired"))["n",])
}

# apply this function to all combinations of the parameters below
tab <- outer(seq(0,.95,.05), c(.7,.8,.9,.95,.99,.999), "getN")
dimnames(tab) <- list("Correlation"=seq(0,.95,.05),
                  "DesiredPower"=c(.7,.8,.9,.95,.99,.999))
tab

Which returns the following:
           DesiredPower
Correlation      0.7      0.8      0.9     0.95     0.99    0.999
       0    3.767546 4.220731 4.912411 5.544223 6.888820 8.656788
       0.05 3.691858 4.126240 4.787326 5.389850 6.669683 8.350091
       0.1  3.615930 4.031562 4.662220 5.235637 6.451021 8.044096
       0.15 3.539645 3.936653 4.537050 5.081483 6.232774 7.738792
       0.2  3.462940 3.841433 4.411750 4.927417 6.014903 7.434270
       0.25 3.385708 3.745774 4.286234 4.773338 5.797404 7.130529
       0.3  3.307922 3.649640 4.160447 4.619143 5.580267 6.827580
       0.35 3.229382 3.552889 4.034209 4.464751 5.363362 6.525430
       0.4  3.149970 3.455310 3.907393 4.309986 5.146613 6.224026
       0.45 3.069435 3.356743 3.779777 4.154653 4.929824 5.923282
       0.5  2.987581 3.256903 3.651065 3.998456 4.712773 5.623032
       0.55 2.904079 3.155423 3.520841 3.841020 4.495111 5.323066
       0.6  2.818472 3.051834 3.388672 3.681805 4.276260 5.022875
       0.65 2.730145 2.945449 3.253781 3.520048 4.055501 4.721751
       0.7  2.638237 2.835369 3.115118 3.354639 3.831565 4.418560
       0.75 2.541442 2.720152 2.971074 3.183823 3.602697 4.111397
       0.8  2.437713 2.597460 2.819127 3.004879 3.365682 3.796879
       0.85 2.323340 2.463226 2.654597 2.812710 3.114890 3.468745
       0.9  2.190677 2.309002 2.467897 2.596901 2.838233 3.113596
       0.95 2.018024 2.110699 2.231866 2.327720 2.501567 2.692358

Note that $d=2$ is considered in many fields quite a large effect size, so the resulting minimum numbers of pairs are all quite low.
Comment: @Glen_b Yes, indeed. My assumption is that specifying $d$ and the correlation is a little more intuitive, but obviously this is a matter of opinion.
Comment: Alternatively, one could (equivalently to specifying the correlation and the 2-sd difference), directly specify the standard deviation of the distribution of the differences.
Comment: I think it depends on the situation one is in. In many cases the correlation will be the more natural way to do it, but in some other situations, you can have either experience (or perhaps intuition) of the typical variability in differences themselves.
Comment: should that formula be $d_o = \frac{d}{\sqrt{1-r^2}}\,,$ rather than $d_o = \frac{d}{\sqrt{1-r}}$?
Comment: @Glen_b No, the $\sqrt{1-r^2}$ form does seem to arise a lot, but in this context the correct denominator really is $\sqrt{1-r}$. I originally pulled the formula from Cohen's (1988) textbook, but I have also verified it by deriving it myself (I think at one point I had the same suspicion as you).
Comment: If you think about it, it makes sense for the $r$ not to be squared, because this means that when $r<0$, the paired $t$ is smaller than the independent-sample $t$, due to increased variance of the differences. If it was $r^2$ then the paired $t$ would always be $\ge$ the independent $t$, which we know is not the case.
Comment: Thanks. Quite right; I had an error in my calculation of the variance of the difference.
Answer: Any effect size is possible, it's hard to tell what you mean by "makes sense".
There are many power calculators on the internet found with a simple Google search. You'll see that G*Power is often recommended here.
Conceptually you calculate power more or less the same way. In the independent case you're probably using the pooled variance of your conditions to get an SD. That's a Cohen's D. However, with the repeated measures case it will be the SD of the effect, which incorporates the covariance between the conditions. If you have a well justified repeated measures design the covariance should be fairly large, which will result in a small effect SD. Therefore, you'll require a smaller N. You need to somehow find this effect SD or covariance (or correlation). If there are no repeated measures studies from which to derive it or you don't have prior data you might better off taking a different approach (at least for your first repeated measures study).
You don't have to calculate power before conducting a study. You could take the following approach. You could decide how much of an effect you care about. Let's say it's 4 (not 4SDs, 4 on your scale, whatever it is...it could even be your original 2 SDs, although that's rather large). Effects smaller than that wouldn't really mean much even if the test was significant and you want to be able to detect effects at least that size. Run subjects until you can detect an effect of that size (calculate the width of an effect confidence interval). Stop once you can detect the desired value. I've previously posted the simulations that show that this does not inflate alpha appreciably, and gives you the expected power and N on average. (Keep in mind that you are not allowed to make decisions on significance of a test, only sensitivity. That requires discipline.)


Why is my front room ceiling fan switch setting off my smoke alarms

Question: When I come down in the morning and switch my ceiling fan on it sets off the fire alarms. It shouldn't be set off by a surge in the circuit as far as I know as the fan is downstairs and the smoke alarms are connected to the upstairs lighting circuit. I have got a box for the alarms on the wall with the option to locate alarm, test alarm and silence alarm but when I press the silence alarm button it just seems to reset the alarm and it goes off again. It only happens in the morning when the fan is first switched on.
Comment: Hello, and welcome to Home Improvement. When did this start, and what changed? And, who's responsible for the fire alarms? And, you should probably [take our tour](//diy.stackexchange.com/tour) so you'll know you'll know the details of contributing here.
Comment: What make and model is your fire alarm panel?
Comment: What happns if you leave the ceiling fan on all night?   What if you then turn it off when you get up,  wait 3 or 4 hours,  then turn it on?
Answer: There are two types of smoke detectors, Ionization and photoelectric. If you have the latter, it might be that they have some particulate around the photocell and when the fan goes off it disturbs the dust enough to simulate smoke. If you know your type and manufacturer you can check your manual (many are online) and follow the cleaning procedures.
If you don't know the manufacturer sometimes you can bag the smoke detectors with a "shower cap" or a plastic bag and a rubber band and see if turning on the fans will set of the alarm.
After that it gets a little complicated and you might need to consult an alarm specialist in your area.
Good luck.
Comment: Also check if your smoke alarms are approaching or over 10 years old. If so they need to be replaced anyway due to end of life of the detector.  If they are dirty/dusty inside they are likely to be very old so do check.
Comment: The house was rewired by the council about 4 years ago and all brand new alarms was put in, there is a circuit breaker switch for the alarms but it is the same one as the upstairs light circuit breaker. There is a wall in between the fan and the alarm. It has also done it at night when I have turned the fan off but they only beep once when it happens at night and it ha only done it about 5 times at night. I have had to have the council out since and they found out that the screws in the back of the stitches and lights were loose.
Comment: Also about a year ago I had a problem where the landing and kitchen lights would just go off and after a while just come back on randomly. We checked the bulbs in other sockets and they worked fine, it only played up in the landing and kitchen light sockets. I dont really trust the new wiring as the electricians that rewired the house didnt have trunking big enough for the wires going into the main circuit board so they just used a hammer to hit them in.


How do I check the texts of New York State laws online?

Question: I want to read up the text of laws online. I am specifically interested in New York State laws. Is there a webpage where I can read up on the text of laws in New York State, or perhaps some other state in the USA?
Answer: There's an FAQ on the Meta site for Online Legal References which includes New York Law from the FindLaw.com site, with the other states here.


Are force vectors members of a vector space?

Question: Vectors in vector spaces depend only on their size and direction.
Force vectors, for example, depend also on their location. Opposite force at different locations, for example, do not annihilate each other but form torque.
Why this anomaly?
Answer: The answer is that force is a vector, you just have to be careful about what the vector space is. As a starting point, I think it would be useful to review what exactly a force is. Imagine we've chosen some inertial reference frame $O$ and are using it to describe a universe which consists of nothing but a single particle, which has coordinates $\mathbf{x}(t)$. As I'm sure you're aware, Newton's second law for this system just asserts that if we have some force law for the particle $\mathbf{F}(\mathbf{x}, \mathbf{\dot{x}},t)$, the particle's motion satisfies
$$\frac{d^2\mathbf{x}}{dt^2} = \mathbf{F}(\mathbf{x}, \mathbf{\dot{x}},t)/m$$
where m is the mass of the particle. The important thing to notice about this is that force isn't tied to random points in space-- it's tied to the movement of particles within space. I can't apply a force on an empty piece of vaccuum-- I can only do so to a particle. In this vein, we should then not be thinking of the force as acting on the vector space describing space, but rather the vector space describing the location of our particles.
Now, for one particle, this is somewhat of a moot point because both of these vector spaces are just $\mathbb{R}^3$. But what about if we have two particles? There's no "nice" (ie continuous, linear) way to describe the location of two particles in space using one vector in $\mathbb{R}^3$, which we need if we're gonna be using vector algebra and calculus to solve stuff. But there is a nice way if we expand our vector space to $\mathbb{R}^6$-- just let the first three components describe the location of particle 1, and the second three components describe the location of particle 2! In this formalism, it's clear how a force acting on particle 1 is different from a force acting on particle 2-- they're completely different vectors in $\mathbb{R}^6$.
Hopefully the generalization to arbitrary numbers of particles is obvious-- we can describe the location of $N$ particles using a vector in $\mathbb{R}^{3N}$, and it's in this vector space-- known as configuration space-- that force lives. Using this concept of configuration space is very useful in physics and guides us to the incredibly important concept of lagrangian mechanics. Note-- when I say force "lives" in configuration space, I'm using that as shorthand for saying that it's the co-domain of the force.
To recap-- force isn't tied to arbitrary locations in physical space, it's tied to the location of particles inhabiting that physical space. And these locations are described by the $3N$ dimensional configuration space, which is the vector space that force formally lives in.
An alternate viewpoint
If my description of force as a vector in $\mathbb{R}^{3N}$ seems contrived, be aware that another, perhaps more intuitive way of looking at the situation is by saying that each particle has its own force vector associated with it that depends on time and the locations/velocities of itself and all the other particles. In this viewpoint (which is the one taken in JiK's answer), the reason that applying a force to two different particles results in different dynamics of your system is because although both forces are members of $\mathbb{R}^3$, they're two completely independent vectors since they're describing the dynamics of different particles. It's this independence that lets us combine the vector spaces of all the different forces in a natural way using the direct sum.
Aside from just doing it because we can, this combination of vector spaces is useful for three main reasons:

As mentioned before, it helps lead us to Lagrangian mechanics.
On the aesthetic front, it combines $N$ different vector differential equations into one.
It helps clarify the conditions we need to impose on $\mathbf{F}$ to guarantee existence and uniqueness of trajectories through configuration space via the  Picard-Lindelöf theorem. Strictly speaking to do so we must first turn the force equation from a $3N$ dimensional 2nd order ODE to a $6N$ dimensional 1st order ODE, but this is trivial.

Further discussion
In the comments, Aloizio Macedo brought up a good point, which is that if you read my answer in a certain way it can seem like it's saying that force should only be described in terms of what particles are actually doing. Yet, I also seem to implicitly define force as a function on all of configuration space which is a contradiction because the system doesn't occupy every point of configuration space along its trajectory. So what gives?
Well, in my mind, this comes down to a chicken and the egg type situation-- is the path of the particle through configuration space what's given and a force law deduced from that? Or is it the other way around and the force law is what's fundamental and what determines the path through configuration space? When I was writing my answer, I had the second one in mind, with force being given as some function $\mathbf{F}: \mathbb{R}^{3N} \times \mathbb{R}^{3N}  \times \mathbb{R} \rightarrow \mathbb{R}^{3N}$ satisfying some continuity requirements.
However, as Aloizio said, it's possible to read my second paragraph as saying that forces should only be defined when acting on particles, which necessitates the first viewpoint and a different formalism that makes the path through configuration space $\gamma$ the fundamental thing describing our universe and derives force through that.
This wasn't my intention with the second paragraph. I just wanted to emphasize for the OP that force is something that fundamentally has to do with how particles move in space, rather than with space itself-- a statement that's true regardless of which of the above viewpoints you subscribe to. Hopefully that clears things up and addresses the concerns people may have had with my description of force.
Comment: @BlueRaja-DannyPflughoeft I think your comment makes a valuable contribution, but there's some issue with the terminology. A *vector space* is a set together with some operations (like $\mathbb{R}^3$ with $+$ and multiplication by scalars). The concept you're describing, an association of a vector to each point in space, is a *vector field* (hence the shortening to just "field"). Don't mean to be pedantic, but since the OP and this answer are both particular about what vector space they're in, it's worth distinguishing.
Comment: This answer seems confused. If you require forces to be defined only where they are acting upon something, then forces don't live in the configuration space, they should probably live in something like the space of sections of the pull-back bundle $\gamma^*TM$, where $M$ is the configuration space and $\gamma$ is the path of the particle(s). But your conclusion and final paragraphs imply that forces are vector fields on the configuration space, directly contradicting the first part.
Comment: Another way to look at it: for many forces, such as gravitational force, you can describe the vector space of what the force would be if there were a particle _(of a certain mass/potential/etc)_ there.  We call that vector space a [field](https://en.wikipedia.org/wiki/Field_(physics)).
Comment: @bemjanim The force vectors are now six dimensional-- the first 3 components correspond to the force acting on the 1st particle, and the second 3 components correspond to the force acting on the second particle. If you prefer, you can think of them as two different forces acting on each of the particles, but combining them into one vector has some mathematical and aesthetic advantages.
Comment: @AloizioMacedo After thinking about it for a while, I think I understand what you were getting at and it's a valid point. I think the confusion stems from my second paragraph coming across differently than I intended. I added a note at the bottom discussing this-- hopefully it adequately addresses your concerns.
Comment: @AloizioMacedo Given that our configuration space $M$ in my answer is the trivial manifold $\mathbb{R}^{3N}$, wouldn't the fiber of $TM$ for any $p \in M$ and thus of $\gamma^*TM$ for any $t$ still be $\mathbb{R}^{3N}\cong M$? I didn't want to get bogged down with unnecessary formalism, although I can see how the distinction might be important for more general coordinates/theories. Also, I was considering forces at a single time since it's easier to visualize, which is why I'm more concerned about the fibers of $\gamma^*TM$ than sections of it. I'll clarify this in my answer.
Comment: I don't understand how three dimensional vectors can live within a six dimensional vector space. Am I missing something? Shouldn't the correct vector space just be $\Bbb R^3$ as the set of vectors, $\Bbb R$ as the set of scalars, along with just regular vector addition and scalar multiplication? Or are the force vectors now six dimensional?
Comment: @el duderino ah ok thank you for clarifying
Answer: Is mass a scalar? If I spread $10^{57}$ hydrogen atoms around the universe, it barely has any effect on anything. But if I concentrate them into a small volume, it creates distinguishable gravity and planets have closed orbits around them. So clearly, the position of a mass also matters, not only the amount of the mass.
Is the amount of money a scalar? If I have 1000 € in my bank account and pay a 100 € restaurant bill, I end up having 1000€+(-100€) on my bank account. But if instead John pays the bill, I end up having 1000€ in the bank account. So clearly the owner of the money also matters.
Are real numbers scalars? If a question has a $2$ and a $3$, then you can't know whether $5$ is the correct answer to the question without knowing other things.

Force vectors, for example, depend also on their location. Opposite force at different locations, for example, do not annihilate each other but form torque.
Why this anomaly?

There is no anomaly. Force vectors are vectors. But when you want to find out what happens in a physical system involving forces, you can't just take the vectors and do stuff with them without thinking about where they apply.
Like you can't just take masses and sum them to find gravity without thinking about where the masses are. Like you can't just take amounts of money and sum them without thinking about who owns the money. Like you can't just take real numbers and sum them without thinking about what the problem actually is.
Answer: In addition to @elduderino's very good answer, there is a second way to address your question.  It can be hard to see what I'm talking about in a Euclidean space (like $\Bbb{R}^2$ or $\Bbb{R}^3$) so let's talk about point particles constrained to move on the surface of a sphere.
A particle constrained to move on the surface of a sphere cannot react to the component of a force vector along a radius of the sphere, so we might as well restrict our force vectors to have no component along the radius.  So the set of allowed force vectors lie in a 2-dimensional real vector space (equivalent to a plane) where we imagine the origin of the vector space is at the particle and the plane is tangent to the sphere.  (When we start moving the origins of vector spaces around, we get affine spaces.)  At each point of the sphere, we have a different vector space because we have a different tangent plane.  (Many pairs of tangent planes seem to intersect in lines -- these are not physically meaningful.  We should think of each plane as being tagged by its point on the sphere and we should think of that tag as being an extra coordinate, so these intersections are projection artifacts of ignoring that extra coordinate.)
Because these spaces are tangent to our sphere, they are called tangent spaces.  The collection of all of these on our sphere is called the tangent bundle of our sphere.  There is much math of tangent spaces and tangent bundles.  One thing I want to make very clear is that we have not gone far from @elduderino's answer -- these tangent spaces also have straightforward descriptions in $\Bbb{R}^{3n}$-dimensional configuration spaces.
Notice that these tangent spaces capture something we know about motion on a sphere -- velocities are parallel to the surface of the sphere, accelerations are parallel to the surface of the sphere, and (as $F = ma$ suggests) so are force vectors.  (If you imagine some sort of generalized force vector that does not lie in the tangent plane for its particle, then project it into the component that does and discard the other component -- the other component is attempting to violate or geometric constraint, so does nothing.)
Let's circle back to my claim about it being hard to see what's going on in the plane or in 3-space.  At each point of the plane, the tangent space is a 2-dimensional real affine space, so it looks exactly like the plane we started with.  This can be confusing.  At each point in space, the tangent space is a 3-dimensional real affine space, so it looks exactly like the space we start with.  This can be confusing.  So to explain this, start with a space where the tangent spaces to various points are not just the space you start out with -- many people are familiar with moving around on the surface of an approximate sphere, so start there.
Answer: Yes, they are. The vector space is isomorphic to $\mathbb{R}^3$, as you might expect.
In at least the Euclidean case, I find that the most helpful way to understand the relationship between points and vectors is to have the idea of an affine space along with a vector space. An affine space is a space where that points are all "homogeneous" in that you cannot discern, using the operations available to you in the space, any point from any other point. This contrasts with a vector space, where you can always obtain the zero vector $\mathbf{0}$ by multiplying any given vector by the scalar 0.
In affine space, you can do two things:

Subtract two points to get a vector: if $P$ and $Q$ are two points, then $Q - P$ is the vector representing the displacement from $P$ to $Q$,
Add a vector to a point to get another point: if $P$ is a point and $\mathbf{r}$ a displacement vector, then $P + \mathbf{r}$ is the point you get to by following that displacement.

Note that in neither of these does any particular point in the space have any significance as none is referenced, while the zero vector is both explicitly referenced and obtainable in the axiom set for vector spaces. And it is the second idea above - adding a vector to a point - that relates to the idea of forces "having a point of application". In particular, the force $\mathbf{F}$ generates an acceleration $\mathbf{a}$ which describes change in the rate of displacement $\mathbf{v}$, and the object's position finally undergoes small displacements
$$P' = P + (\mathbf{v}\ dt)$$
at each moment of time: here is where we have the affine vector-to-point addition function at work, but the force vector $\mathbf{F}$ does not directly add to a point.
However, I believe the notion of "point of application" most properly belongs to the study of spatially extended objects, where the above discussion applies to point-like objects only, because for a point-like object, you cannot intuitively "apply a force" anywhere else but at the single point it occupies, otherwise you're just "pushing at empty space". Describing extended-body movements is a little different. A solid body can be mathematically represented by a suitably-"nice" set of points drawn from space, representing all the locations it occupies, which we may call $I$ (for "image"). Then we describe the forces slightly differently, as a function
$$\mathbf{F}(P)$$
that tags each point $P \in I$ with a force vector which may be zero. These points on the object where force vectors are associated is are "points of application" for those forces. The "application point" is not a property of the vector, again, but a property of the mapping that tags the points with vectors: it is the point that was tagged with that given vector. (Note that this case subsumes the previous one by allowing $I$ to consist of only the single point that a point particle occupies.)
So yes, force vectors are members of a vector space, but the "point of application" is not a property of the force vector itself, so much as it is a property of the relation between the force vector and the object.
Answer: tl;dr–  A force vector at a location can be described as part of a vector field.  You probably mean to refer to a vector field rather than a vector space.

The tl;dr probably answers the question, if I understand it correctly.
So, just to clarify some more vocabulary...
Reference:  Vocabulary terms.
A temperature can be quantified with a scalar.  We can describe scalar values at different points in space as a scalar field, e.g. a temperature field.
A force can be quantified with a vector.
We can describe vector values at different points in space as a vector field, e.g. a force field.
Scalars and vectors are both tensors.  Scalar fields and vector fields are both tensor fields.
Answer: You do add the forces applied at different locations in order to calculate the net force and determine, whether the body is being accelerated as a whole. The difference is that in this case one thinks of the object as one point with the forces applied to it, whereas when analyzing the rotation, one looks at different parts of the body as if they were different points/objects.
Answer:
Opposite force at different locations, for example, do not annihilate
  each other but form torque.

Maybe. Maybe not.
For example, you can have two equal and opposite forces applied at different locations that do not form a torque if the forces have the same line of action. Such forces are called "collinear forces". An example is two people pulling on each end of a rope (tug of war) with equal forces. No torque is involved.
On the other hand, a system of two forces that are equal in magnitude, opposite in direction, and parallel to each other acting on an extended (as opposed to a point mass) object will result in a moment, or torque. Such as system is called a "couple" (a.k.a force-couple).
You can also have a system of forces whose lines of action (the direction in which a force acts) meet at a common point. Such forces are called "concurrent force" systems. Such forces can be added up to get a resultant force at their common point. Since they intersect at a common point, there is no torque involved. An example is a point mass suspended by two strings. Three concurrent forces act on the mass: gravity and the tensions in the two strings. No torque is involved.
For a general system of $n$ forces (non concurrent, non parallel system) applied to an extended object, equilibrium (no net force and no net torque) will depend on whether or not $\sum F_{n}=0$ and $\sum M_{n}=0$.
Hope this helps.
Comment: I know. It still means that locations CAN matter. That's not something that happens on normal vectors
Comment: @user4951 Yes it means locations can matter. I'm just saying it's not ONLY the location that matters but also the nature of the force systems involved.
Answer: Force vectors do not depend on their location. What depends on the location of a force vector is the effect to which that vector contributes to the motion of a body, or to else to considerations of its equilibrium.
Of course, mathematically, vectors are always part of a vector field, regardless of their interpretation in a physical theory!
We consider a body in space, and locate some arbitrary point as the origin. A given force acting on a point on the body creates a moment with regard to the chosen origin. The moment is a combination (dot product) of two vectors: the position vector and the force vector. Moment is a generalization of torque; torque is the magnitude of moment, which is a vector quantity.
The moment is not the only effect of the vector, of course, and it is arbitrarily determined with regard to a chosen point.
We can look at all the forces acting on an body, and consider whether their moments about a given point add to zero. If they add to zero about some point, they add to zero about any other point: the moments are balanced, and the object will not rotate.   But the forces might not be balanced, and so the object might accelerate.
Answer: It's not that the two forces don't "annihilate" but that the two torques don't cancel out, and that's a different thing.
There are rules on how to move around and add up vectors on Euclidean space and are of course consistent with general rules for vector spaces. One generally can't just simply add together vectors from different points because they usually belong to a strictly different vector space. But if you can move one vector to another point(different vector space)and  while doing so being true to its original direction and magnitude, than you can add it up normally. So forces of course are members of a vector space, but not all vector spaces can be used to represent forces.