Proof of the Newton Raphson method

Question: I know only a little about this method.
I get about the big picture. You draw a tangent from an appropriate point on function f, find the spot where it meets the x-axis, and draw another tangent from the point which has the same x coordinate on the function f. the meet converges toward the root.
I get the big picture, I can see it on the graphs, but I can't prove it.
Can you please elaborate a proof for me? I'm in a hurry, plz
Comment: [here you can find it](https://en.wikipedia.org/wiki/Newton%27s_method#Proof_of_quadratic_convergence_for_Newton's_iterative_method)
Answer: Consider that you look for the zero of $f(x)$. At a starting point $x_0$, make a Taylor series limited to first order
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+ O\left((x-x_0)^2\right)$$ Ignore the higher order terms and solve for $x$ and let us call $x_1$ the solution of $$0=f(x_0)+f'(x_0)(x-x_0)\implies x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
Repeating
$$f(x)=f(x_1)+f'(x_1)(x-x_1)+ O\left((x-x_1)^2\right)$$
Ignore the higher order terms and solve for $x$ and let us call $x_2$ the solution of $$0=f(x_1)+f'(x_1)(x-x_1)\implies x_2=x_1-\frac{f(x_1)}{f'(x_1)}$$
Repeat again and again to get the famous
$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
Comment: @user247327. For sure I totally agree with all your statements. The only thing I tried to do is to formulize the steps using the tangent since he/she wrote *I get the big picture, I can see it on the graphs*. This was my very modest contribution. Cheers.
Comment: What, exactly, do you want proved?  That the method converges to x such that f(x)= 0, **if it converges**, is pretty straight forward since if f(x) did not go to 0, the method would not converge.  If you are talking about showing that the method always converges, there is no such proof because that is not always true.

Proof of the Newton Raphson method


How we distinguish process model and process framework?

Question: How do we distinguish between process model and process framework (like CMMI). How they relate each other?
Answer: Your question can be answered by distinguishing the terms model and framework. Do you know the meaning in other contexts? For example, do you know the difference between programming model and programming framework?
To clarify, let's first answer my own question. A model is a more tangible way for describing a concept. But a framework, is a collection of tools, metrics, etc. to make sth. In the context of programming, a programming model is a way you describe your program code. For example, using the imperative programming model, you can describe someone how your program is run. On the other hand, a programming framework, is a collection of debuggers, profilers, compilers, testers, etc. (e.g. .Net Framework) to help you make your programs easier.
The same story is true for processes. A process model is a way you can describe the execution of your process. It may contains its activities, transitions, roles, etc. On the other hand, a process framework, e.g. CMMI, is a collection of goals, practices, sub-practices, CLs, appraisal methods, etc to help you implement a mature process.


Windows 11 and Ubuntu, Different Drives

Question: I want to install ubuntu on a different drive. Do I have to unplug Windows 11 drive in order for Ubuntu not to install stuff on it? Or maybe ubuntu does not install anything on different drives when they are connected, but only to the drive which is choosen? I just want to have two isolated systems. One for work, one for private stuff and I don't want for my work drive to install or touch anything on a private one and vice versa.
Comment: It'll use the preexisting ESP (EFI System Partition) for the bootloader, that's all. When installing or using it won't write to the Windows partitions unless you tell it to do that. And this isn't something new, it always has been since dual-boot was invented.
Comment: Be aware of launchpad bug 1396379, installer ignores your bootloader location information and uses first ESP it finds.  Is your second disk removable?  Do you ever want to boot without the second disk?  The suggestion to remove the first disk during install to the second disk is to avoid having grub wired to both disk, needing both to boot successfully.
Answer: If you want to install Ubuntu on another drive, at installation select "Erase disk and install Ubuntu" and it will ask you, what is the drive you want to install it.
So you just have to select the drive you want to install Ubuntu, and Ubuntu will not install anything on the Windows 11 one.
Yes, Ubuntu will only install on the chosen one, but if you feel safer, you can unplug the Windows one.

Windows 11 and Ubuntu, Different Drives


Could you explain meaning of "getting hit on something" & "laid back" in this passage,please?

Question:
In this week's Fashion Q&A, Cosmopolitan.com's style editor, Charles Manning, answers your questions about high heels versus flats, the best underwear for avoiding panty lines, and how to make denim on denim work for you.
Q: I usually wear heels because I think they're sexier, but lately I've been going for casual drinks in flats and getting hit on. Is there something to a flat shoe that's enticing, and which ones should I be wearing?
A: It's possible that guys see your flats and think, "Hey, she seems chill and low maintenance." Or that they are less intimidated than they might be if you had the added height a pair of heels would normally give you. It's also possible that they never noticed your shoes in the first place and were responding to something you were not even aware of. There's really no way of knowing.
As sexy as high heels are, their real value lies in how they make you feel and how you project that feeling out into the world. If you feel cool, laid back, and confident, it doesn't really matter how high your heels are. And although there are not nearly as many fabulous flats in the world as there are fabulous heels, there is usually an appropriate flat equivalent for any high heels.
Answer: To "get hit on" means to have guys expressing a sexual interest in you. You are the target of their "sexual advances".
To be "laid back" means to be calm and easygoing, not nervous or agitated; "centered" in your being; a synonymous slang term would be "mellow" or "chill".


Opening Multiple Accounts In Outlook 2016

Question: I'm trying to setup another account in Outlook, besides my primary, (this is another account through Account Settings > Email > New, not a shared mailbox) and I cannot get it to work. I can add the account just fine (autodiscover works and I get all the green checks) but when I try to open it I get an error "The set of folders cannot be opened. The file  is not an Outlook data file (.ost)". If I remove the account and add it back again with cached mode disabled I get the error "The set of folders cannot be opened. Network problems are preventing connection to Microsoft Exchange"
I'm using the C2R version of Outlook but I've tried it with the MSI version and it's the same thing.
Both accounts are Exchange Online accounts. Is this supported? Are other people able to add multiple EXO accounts to a single Outlook profile?
Comment: I don't deal with EXO but if I were working with previous versions I would make sure I had given myself full access permissions so that I could add the extra account.
Comment: This is for adding another account (using the account owner credentials) not adding a shared mailbox. Thanks for the reply though.
Comment: We have different definitions of "shared mailbox". Exchange has the term for a type of mailbox called shared mailbox. Then there's granting yourself access to someone else's mailbox. If you have userA with mailbox and userB with another mailbox. You want to be able to add userB to your outlook so you can see its contents. You give yourself permissions to view his mailbox while logged in with your own creds. This is what I'm referring to and it's not a shared mailbox. It remains userB's mailbox but you can view it. Without full access permissions you get the error message you've mentioned.
Comment: Fair. Yes I'm referring to accessing a mailbox using the owners account. Just having both in a single Outlook profile.
Answer: It works for me: adding two different Exchange Online mailboxes in the profile. Both of them work as expected.
Please ensure the second mailbox is working when setting up it in a single profile.


How do I find out the IP Address for all print jobs to a printer

Question: Our network printer is spitting out unwanted print jobs which no one seems to collect. I want to know from which machine the jobs comes to the printer.
We have PaperCut in our server but these specific jobs are printing without any entry in Papercut.
Comment: The Printer (Kyocera Taskalfa7500) does come with a recent job list, but the user info is left blank. All it tells is the time it started and ended printing and a document name (doc2134xxx12)
Comment: If someone is able to send print jobs to the printer and PaperCut is installed on your server then that suggests the computer that is creating the print jobs does not have the PaperCut service installed or running. Have you ruled out someone printing from a rogue computer?
Comment: put a computer with 2 network cards in front of the printer and log the traffic - or use a passive hub and sniff the traffic.
Answer: Check the printer integrated log on the admin web interface, if your printer come with a log viewer.
Edited: As your printer does not have a log viewer, please use network monitor like wireshark to capture the packet from where they come. You can put a mirror port on your switch to capture or you can use a old hub, plugged in your printer and on a laptop to capture the packet.
The setup would look that way:
<code>[switch] ----- [hub] ---- printer
                     ---- laptop
</code>
After the packet capture, please check your DHCP to find some detail about who print.
A guide for wireshark there
Comment: Printer does have a log viewer but it only says socument name(doc0213xx)and accepted time for the print job. the rest of the feilds are blank.
Comment: @vinayakvijayan Will update my answer, as you will have to capture some data manually
Answer: It's a bit of an open question depending on your network... but some kind of network traffic monitoring might be useful... https://en.wikipedia.org/wiki/Network_traffic_measurement
What kind of printer is it?  There may be a function to print out the IP address of the sender, or better yet, print recent logs.  You might need to enable logging first: http://smallbusiness.chron.com/check-printer-logs-54952.html
Is your printer and/or network behind a firewall?  There are almost certainly firewall logs on your network... but that all depends on your network...
Alternatively, you could try unplugging the printer and pluggin a laptop, with fixed IP the same as the printer's (if your network allows this).  Then you can monitor IP-based requests directly.  (not a great idea, but probably one that I would try just for fun and if nothing else works).  This might help if you're on a windows system: apparently I need more reputation to paste this link... google "nextofwindows tracking firewall activity with a windows 8 firewall log"
Let us know:
1) Printer model
2) Network type
3) Firewall status
Maybe we can help more?
Good luck!
Comment: These are all valid relevant points. Are you trying to answer the question or asking for more info?
Comment: Printer Model-Kyocera Taskalfa, Network Type-LAN with both DNS and DHCP. Windows 2012 server, No Firewall between the printer and network and yes the network is behind a firewall but i dont have access to it.
Answer: How are jobs submitted not through Papercut? If possible you could force authentication and then you would be able to see who submitted via the accounting interface.


Are non-speaking characters in a screenplay introduced in all caps?

Question: For example, if a guard enters a room and watches someone but does not speak, is this character introduced in all caps as GUARD? What if the character's role is even more limited: a little girl hands flowers to someone? Also, how does this apply to large groups, such as spectators, protesters, or crowds? If the character(s) requires casting, should all caps be used?
Answer: Yes - if the character is going to be featured (ie obviously doing something) then the director and casting director are going to need to know who they are, so they should be called out in caps.
If they're merely background colour, then they stay lower case.
Example.
INT. MUSEUM. DAY.

Smith enters the museum, pushing her way through the crowd of parents and
children. A GUARD spots her, and draws his gun, firing into the crowd.

If we wanted to call out someone who gets shot, or is otherwise important to the action in the crowd, then they would be capped. So the little girl handing out flowers in your example would be too.
Some writers like to say upfront that a character won't be speaking later. This is usually done via N/S (for non-speaking) or ND (no dialogue). Pick whichever you prefer and use it consistently.
INT. OPERATING THEATRE - DAY

DOCTOR BOB is elbow deep in a patient. An N/S NURSE wipes the sweat from his
brow.


Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$?

Question: Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$?  I get the feeling it isn't compact, but I can't figure out a way to prove this.  I understand that,  in $\Bbb Q$, any open set $U = \Bbb Q \cap O_i$ such that $O_i \subseteq R$.  But I can't figure out a way to construct a finite set that doesn't contain the intersection set.
Answer: Compact spaces have the property that they are closed in any Hausdorff superspace. Since ${\bf R}$ is Hausdorff and $[0,1]\cap {\bf Q}$ is not closed in ${\bf R}$, it cannot be compact.
To see that directly, you do the same as you would with, say $[0,1]\setminus \{\sqrt{1/2}\}$.
Comment: I have another reasoning for this question could you please tell me whether I am correct;we know that if Y is subspace of Z and X is subspace of Y then X is Subspace of Z.Now for X to be compact in Y,it has to be compact under subspace topology induced by Y but then X is also subspace of Z and hence it has to be compact under subspace topology induced by Z.So for $[0,1]Π\mathbb{Q}$ to be compact in $\mathbb{Q}$ it is equivalent to say that it is compact subset of $\mathbb{R}$ and hence we can used Heine Borel theorem now.
Comment: @lbs: This is basically the argument I used here.
Answer: A set in a metric space is compact iff it's complete and totally bounded. $\Bbb{Q}\cap [0,1]$ is not complete.
Answer: Let $r$ be an irrational number in $[0,1]$. There exists $N \in \mathbb{N}$ such that
$$ \frac{1}{N} \leq \min\{r, 1-r\} = \text{dist}\big(r, \mathbb{R}\backslash [0,1]\big).$$
The sets
$$U_n := \left(-1, r-\frac{1}{n}\right) \cup \left( r+ \frac{1}{n}, 2\right),~~~~~n \geq N$$ are open. The collection of $U_n$ forms an open covering of $[0,1] \cap \mathbb{Q}$ with no finite sub-covering. Hence $[0,1] \cap \mathbb{Q}$ is not compact in $\mathbb{Q}$.
Answer: A set is sequentially compact if every sequence contained in the set have a subsequence that converges to a point of the set.
Because the set proposed is only composed of rational numbers then any sequence that converges to some irrational number doesn't have any subsequence that converges to a point of the set (because for convergent sequences every subsequence converges to the same point).
Answer: Well, let's figure it out knowing it can't be true.
In R a set is compact iff it is closed and bounded.  [0,1] restricted to Q is bounded but it isn't closed.  (Which is enough... we are done.  But ...yeah, I know, it doesn't feel we are done.  It'd feel better if we could find an open cover that has no finite cover.)
Well, let's find a limit point of [0,1] restricted that isn't in the set.  Well, that'd be any irrational $x; 0 <x <1$.  Let's make an open cover that "tends" toward x in a "limit pointy sort of way".
Oh, all right ... I'll out and say it.  Let's use the 1/n to get open sets that come within 1/n of x.  As x is a limit point, these will get arbitrarily close to x and cover all but x.  But as x isn't in the set it will cover the set.
Okay, let's formalize that.
Let $U_n = (-\infty, x -1/n)\cup (x+1/n,\infty) $.  This collection $\{U_n\} $ is an open cover as it covers everything but $x$ which wasn't in the set to begin with.  Any finite subcollection, $ \{U_{n_i}\} $ will have an $m=\max (n_i) $ and so $ [x-1/m,x+1/m]\cap [0,1]\cap \mathbb Q \subset [0,1]\cap \mathbb Q$ will not be covered.
So that's an open cover with no finite subcover.
Comment: Oh, this is basically what Masacroso posted while I was typing.
Answer: This is a particular case of a more general affirmation, namely: Every compact set in $\mathbb{Q}$ has an empty interior.
To prove this use the fact that in any open interval $(a,b)\subseteq\mathbb{R}$ there is a sequence of rationals converging to an irrational (which will of course contain no subsequence converging to a rational), and recall that in any compact space every sequence contains a convergent subsquence.


Load different data source for same charts in Google Charts

Question: Overall Problem: I have year-wise data in different JSON files. For each JSON file, I generate 3 charts that are displayed on the webpage. Now the user should be able to select a different year, and the charts should update based on that particular year's data.
What I have Done: I wrote a function called initialize_charts() which parses the given JSON, prepares data as per google chart requirements and draws those charts. I am calling this function from google.setOnLoadCallback. (as of now I have hardcoded the json file name).
Main Issue: Now, I am not sure how do I re-render the charts when the user selects some other year's JSON file? Somehow, I should be able to pass-on filename parameter to google.setOnLoadCallback or google.load, such that my initialize_chart reads this new data and re-renders the charts. But I am not sure how.
What I tried: I tried writing a wrapper function called initialize like this:
google.setOnLoadCallback(initialize);

function initialize()
{
  initialize_charts("year1.json");
}

So, when it loads first, it loads year1.json data. Then for onclick event i called onclick="initialize_charts("year2.json"). But that doesn't work. I suppose that the placeholders for the charts (div elements) are replaced by the charts itself. So, it can't find those placeholders.
I would greatly appreciate any sort of help with this. If you need more information or my code, I would be happy to provide. I am stuck on this from the last 2 days :-(
Answer: Solved it. Using this piece of code:
google.setOnLoadCallback(function() {
    initialize_charts('year1.json');
});

function initialize_charts(filename)
{
   .....
}

After the first load, if you have to change the data source, just call initialize_charts with some other json file (for example: year2.json)


How to make it dark inside water?

Question: I am creating a map that simulates subnautica in minecraft, and i´m having problems because no matter how deep i build the water is not dark at night time, i tried using the blindness effect but it does not you regulate the range as the level of the effect makes no difference at all.
Answer: There is no feature to do what you want (I assume you want some kind of "super-blindness" with which you see nothing), but you can abuse a bug for this: https://bugs.mojang.com/browse/MC-4647
Simply give the player both blindness and night vision at once and everything will turn black.
You could also achieve a similar result by setting your brightness setting extremely low, but that would require editing options.txt and restarting the game every time you want to go outside the normal range that is possible in the game settings. In case you still want to do it, the entry is "gamma".
Comment: doesn´t need to be completely black but the water is too clear
Comment: Then your only option is to play 1.12 or earlier. :D

How to make it dark inside water?


Why are minifigure head studs now hollow?

Question: Comparing my childhood minifigure heads to my post dark age ones, there is now a hollow stud at the top vs the solid ones I grew up with. At what point did this change happen and what was  the reason for the change? Up until recently I didn't think it had a functional difference, but I have seen at least one hair piece that takes advantage of the hollow stud to hold it in place (CMF4 punk rocker). Was that a motivating factor for the change?
Answer: Part (or mould) changes have numerous reasons, the main ones being:

decrease mould complexity and thus production cost (for example  to )
decrease plastic quantity used and thus production cost (for example, look at the bottom of several 1 x n bricks — the new ones have hollow tubes)
increase part sturdiness (for example  to  — honestly, how many of these did you break as kid?)
add functionality (I have no example in mind, but considering the cost of new moulds, they probably do that only when it has to be replaced, or they treat it as a new part altogether).

Making sure a part fits better into the system is a huge factor in new part creation nowadays, and designers will get a new part produced easier if the part plays nice with the other parts, mainly in terms of connection possibilities, but also of existing similar parts. In the past, things were way less rational.
I'm not so sure it influences part redesign, but it could. At the time the hollow stud appeared on the minifig heads, however, functionality was certainly not the main interest, so I would bet on mould complexity. Of course, since it added functionality anyway, you can count on designer to use it. So it's a bit of a chicken-and-egg problem: did the designer ask for the new functionality first, or did the mould change first? In this case, I would say the latter.
EDIT: Even if I answered the "full stud" against "hollow stud" part, I was intrigued by the "anti-choking" argument of having not only a hollow stud, but even one which was fully hollow, that is, you could see through it. So I asked Jan Beyer, LEGO community manager, about it and it turns out the "non-choking" reason was indeed correct. Here is his answer:

Fast answer:
This is correct, previous the hole was made in the element due to product safety if the element was swallowed.
  These regulations changes from time to time, and it has been evaluated and decided that the hole is no longer needed due to safety regulations.
Comment: The bar with a claw on the end was changed, and it added functionality.  The edges of the claw were rounded and it can now be put into more positions.
Answer: One theory I've read online is that this change was made to increase safety in case of choking.
This reason appears in a Gizmodo article:

We added this hole on the top of the head just in case any kids got one of the heads stuck in their throat. That way they would be able to keep breathing.

The article, though formatted as an interview, is a compilation of reader questions asked by the writer on a tour, so it is not clear if these are the exact words of Lego staff.
The same rationale is discussed in a Brickset post:

...there's a book that LEGO came out with in Japan, called "LEGO Book Museum" (volume 1). This guy gave a review of it, which is unfortunately lost to the internet ether. The response indicates that his review indicated an "anti-choking" reference relating to minifigs, but apparently wasn't specific, leading people on LUGNET to speculate that they were referring to the hollow-stud head.

The writer provides a translation and concludes that the book seems to just point out the design change itself:

So, I haven't found anything about choking on the page...
Comment: The question was about full versus hollow stud, not about the fact that the hollow stud itself is pierced. That said, since LEGO is currently removing that particularity (going to a not-pierced hollow stud), I doubt it was done for safety. If it was, they wouldn't go back.
Comment: I remember it happening around 1991. Seems to correlate with growth of toy safety concerns, and trial lawyers.
Comment: Well, I've seen enough references to this that it sounds likely, but then I don't get why LEGO would go back. I've sent the question to someone there in the hope of settling it once and for all.
Comment: Well, it was correct, actually — I'll add that to my answer.
Answer: A hollow head stud allows to connect a 4140303 lightsaber blade sticking out the top. (in case you ever wanted to do that)
It's probably used to save plastic. For every 100 hollow-studded heads, Lego probably saves several heads worth of plastic. In the long run, that saves much material.
Comment: Four words: orange transparent chainsaw mohawk. Can't have a haircut more awesome than that :)
Comment: Not necessarily just lightsabers, I think some of the minifigures use it to hold hair pieces too
Answer: More a comment as I don't know the real answer, but I wonder if it in part had to do with the small vacuum created when you connect to a body piece. Pulling apart tight heads usually made an audible popping noise, especially if the pieces were wet or dirty.
However, I'm sure the friction from grimy pieces had more to do with the tight fit than the vacuum.


Swapping transmission 98-99 possible

Question: I'm wondering if my 99 Bmw 528i transmission with a bad engine can be a potential doner for a 98 bmw 528i with a good engine bad transmission help so I jump on the purchase before it's to late
Comment: Welcome to Motor Vehicle Maintenance & Repair! While I'm not a BMW expert, I understand parts are very swapable between years and models. Since the 98 and 99 are both 4th Generation 5-Series (E39) and out of the same model (528i), I doubt there'd be any issue with the swap. Hopefully someone on here could tell you for absolute sure for you to make the decision.
Answer: Looking at Wikipedia, it appears there were two different automatic transmissions used in 1999. Depending on when your car was manufactured.
The two autos are the:

GM 4L30E - Used up until 06/98 manufacturing date.
GM 5L40E - Used from 07/98 to 03/00 manufacturing dates (inclusive).

There was only one standard shift tranny for either of the years, the S5D 250G.
If you have an automatic transmission and it falls before the 07/98 manufacturing date, you should be fine. If after that, then the two would most likely be incompatible. It looks as though, the standard version shouldn't have an issue.
Note: Some information was pulled from this source.

Swapping transmission 98-99 possible


Problem with rebooting Linux Mint 18 Cinnamon 32-bit

Question: I was browsing on FireFox in Linux mint 18 cinnamon and I couldn't log in my google account, so I closed the browser, but after I tried to turn it back on, it showed me a notification:

"Firefox is already running, but it is not responding. To open a new window, you must first close the existing Firefox process, or restart your system."

Firstly I restarted it, it worked fine for a while, but I still couldn't log in the google account, so I closed the browser again and when I tried to open the browser, the same notification appeared. I logged out my profile, but then was written on the screen (right after I wrote the Security key):

BusyBox v1.22.1 (Ubuntu 1:1.22.0-15ubuntu1) built-in shell (ash) Enter 'help' for a list of built-in commands.

I wrote exit and this is what appeared:

(initramfs) exit /dev/mapper/mint--vg-root contains a file system with errors, check forced. Inodes that were part of a corrupted orphan linked list found.
/dev/mapper/mint--vg-root: UNEXPECTED INCOSISTENCY; RUN fsck MANUALLY (i.e., without -a or -p options) fsck exited with status code 4 The root filesystem on /dev/mapper/mint--vg-root requires a manual fsck**

Does anyone know, why this occured and how to solve it?
Answer: You have encountered a serious harddrive problem.
Boot from install media and save your data to another disk - ASAP


Combining prediction intervals

Question: I'm using ML regressors (neural networks and random forests) to predict some numbers. I can put in my inputs and get out a value and its prediction interval. The inputs to my regressors are noisy, however, so I vary the inputs within their uncertainties in a monte carlo fashion and obtain a sample of prediction intervals. How can I combine them to get one big prediction interval?
Answer: If your "raw" prediction intervals are simply quantiles of Monte Carlo outcomes of the separate methods, the simplest way would be to just throw all the MC outcomes from all methods together, then take quantiles of this "total" set of outcomes.
You could even weight the outcomes from separate methods, if you happen to trust one method more than another, then take quantiles from weighted data.
Apart from that, there is quite some literature on "combining interval forecasts", e.g. this. Alternatively, you could look at combining density forecasts (which is really what I am proposing above, and what you seem to be doing), which is somewhat different. Here is a relevant paper by Hall & Mitchell (2007) in the International Journal of Forecasting.

Combining prediction intervals


How do you load test a wing rib?

Question: When builders say they load tested a wing, most of the time what they are saying is that they load tested the wing spar(s) by turning the plane upside down and putting weights on the wing. This procedure doesn't load test the ribs, so how do you load test a single rib?
Comment: It seems like loading the wing would load up the whole wing, not just the spars.
Comment: Testing a rib by itself may not give proper results. If the rib is kept straight as by a metal skin when assembled into a wing, testing without the skin will cause a failure at much lower load.
Comment: What country are you asking about? As far as I know, there is no requirement to test single ribs, just the assembly as a whole. Testing a single rib may be meaningless anyway since it obtains (and contributes) strength in regards to the entire structure, not just the single structural member.
Comment: @RonBeyer based on OP's track record, they are asking for the purposes of a homebuilt ultralight, and their country is Nigeria.
Comment: @DavidTeahay I'd say proctoring it the SAT or GED would give it considerable amounts of stress, so if you do that you'll have to be careful it doesn't crack. If it does well, then good luck to your new wing going off to a wonderful university! :P
Answer:
Rib is load tested upside down with weights on its flat bottom side to simulate air deflection during flight,and two ropes holding the rib up at positions where the spars will sit.

How do you load test a wing rib?


Boolean Modifyer dissapears if I delete object

Question: So after watching many tutorials on how to carve a 6 into the side of a cube (making dice), I followed steps to use the boolean modifyer to use a "6" object to carve into one face of the cube. All the tutorials say I should be able to delete the 6 object now. But when I do, the boolean modifyer deletes also. I end up with a blank cube again. Can someone help me? using the 2.8 blender. Also when I export as a stl file so i can put it in my 3d cutting software for the printer, it exports with all the numbers popping out of it, instead of boolean differences cut out of the cube.
Comment: I am not sure what tutorials you are watching, but you cannot delete the second object if you're using a Boolean modifier unless you apply the modifier first.
Comment: Did you apply the boolean modifier?
Answer: Blender keeps data of all modifier in temporary state until the modifier is applied. Simply putting at this stage you can do whatever you want with modifiers. So, blender is not sure at this point. You can delete the modifier and object will return back to it's base form. So, modifier data and object data are dependent to each other. If one is affected then other one will be affected too.
Coming to your scenario, Assume there is a cylinder going through the cube. Boolean modifier is configured to affect the cube. So, if cylinder is going through the cube it will make a hole in cube. When you delete the cylinder you will loose hole too. Because object data and cylinder data are dependent to each other. So, to save your hole apply the modifier. This will now make modifier data independent to modifier object and your target object will be modified. After you can delete the modifier object. in this case after applying modifier you can delete cylinder and you will have hole in cube.


Dimension of a vector space below

Question: I have to prove that the dimension of the vector space of real numbers over Q (rational numbers) is infinity. How can I prove? I have no idea.
Comment: Do you know that the cardinal of $\;\Bbb R\;$ is really bigger than that of $\;\Bbb Q\;$ ? In fact, we have that $$|\Bbb Q|=\aleph_0\;,\;\;|\Bbb R|=2^{\aleph_0}=\mathcal c$$ If you know the above, the reult follows at once from cardinality considerations.
Comment: Is there a finite linear combination with coefficients in $\mathbb Q$ that gives $\pi$? I guess if you have to prove the answer though it might take you some time...
Answer: Hint: Show that all $\log(p)$ for $p$ prime are linearly independent over $\mathbb{Q}$.
Apply the exponential function to the equation $\lambda_1\log(p_1)+\cdots +\lambda_n\log(p_n)=0$, and conclude that all $\lambda_i$ are zero. It follows that $\dim_{\mathbb{Q}}\mathbb{R}\ge n$ for all $n\ge 1$.
Answer: Remember that the dimension of a vector space is finite means that there is a finite basis.
In your case, the "vectors" are real numbers and the "scalars" are rational numbers. So the dimension being finite means there is a finite set $r_1,\ldots,r_n$ of real numbers such that every real number $r$ can be written in the form $r=q_1r_1+q_2r_2+\cdots q_nr_n,$ for some rational numbers $q_1,\ldots,q_n$. Can you see the contradiction here by thinking about countability vs uncountability?

Dimension of a vector space below


Find the IP address of expired domains

Question: I want to know the previous hosting IP address of an expired domain. Is there any way find this?
Answer: The well named DNS History tool should give you what you are after.
Comment: I like the idea of this, however, the history seems to be surprisingly sparse/non-existent for the domains I've tried?!
Answer: Other useful sources besides dnshistory.org:

https://viewdns.info/reverseip/ 250 queries per free account
2013 DNS census: https://opendata.stackexchange.com/questions/1951/dataset-of-domain-names/21077#21077

Both of these tend to contain more data than dnshistory.org. The data of these three sources is disjoint however, so searching all of them can be fruitful.
Another one is: https://securitytrails.com/dns-trails appears to be the most complete one I've seen so far. I believe without great evidence that they have pieced up other databases together. Signup blacklists common email providers like gmail to force professional accounts. Query limit appears to be 100/day.
Related: https://superuser.com/questions/686195/how-to-find-the-last-ip-used-for-an-expired-domain-name

Find the IP address of expired domains


How to create dex pallet?

Question: I want to create dex pallet(without using pallet-contracts) with functionality like uniswap with pools of liquidity, this dex gonna work with my multi-token-pallet. So now I'm thinking about how to propperly implement that. So for exmple every pool of liquidity will be some struct in double storrage map, in my multi-token-pallet tokens are associated with AccountId but how can I transfer them to dex pallet, does it have AccountId? Maybe I can create fake signed accounts to deal with multi-token-pallet? Can somone give some ideas how to implement that?
Comment: Checkout Acala implementation, But just look at pallet asset and you can derive any logic with that pallet
Comment: Please review the kinds of questions which are supported on this forum: https://substrate.stackexchange.com/help/dont-ask
Answer: Adam Wierzbicki has created just this. It is based on Uniswap V1. You can find his pallet here: https://github.com/Wiezzel/substrate-dex.
The pallet is not production ready, but is very useful for learning.

How to create dex pallet?


SetUnhandledExceptionFilter and default exception handler when crash report dialog is disabled

Question: I have a situation where an application which uses SetUnhandledExceptionFilter (via google breakpad ) to set up a custom exception handler is hanging after the breakpad exception handler has caught a crash, saved a minidump file and returned EXCEPTION_EXECUTE_HANDLER from the SetUnhandledExceptionFilter function, to execute the standard exception handler.  The hang also happens if the filter function returns EXCEPTION_CONTINUE_SEARCH and the Windows Error Report dialog has been disabled via SetErrorMode().  If SetErrorMode() is not used to disable WER, the process closes once the user clicks the 'Close' button in the WER dialog.
The hang happens under Windows XP but not Windows 7.
The documentation for SetUnhandledExceptionFilter states that when 'EXECEPTION_EXECUTE_HANDLER' is returned, the associated exception handler will execute which "usually results in process termination," but I couldn't find documentation of the termination method used.
A stack trace shows that the hang is occurring in a global destructor for an object in a 3rd party DLL, called from the DllMain() prelude.
ntdll.dll!_KiFastSystemCallRet@0()
ntdll.dll!_NtWaitForSingleObject@12()  + 0xc bytes
kernel32.dll!_WaitForSingleObjectEx@12()  + 0x8b bytes
kernel32.dll!_WaitForSingleObject@8()  + 0x12 bytes
QtCore4.dll!QWaitCondition::wait(QMutex * mutex=0x03ad1b00, unsigned long time=4294967295)  Line 175 + 0x15 bytes   C++
QtCore4.dll!QThreadPoolPrivate::waitForDone()  Line 295 + 0x10 bytes    C++
QtCore4.dll!QThreadPool::~QThreadPool()  Line 429   C++
QtCore4.dll!QThreadPool::`vector deleting destructor'()  + 0x3d bytes   C++
QtCore4.dll!`theInstance'::`8'::`dynamic atexit destructor for 'cleanup''()  + 0x14 bytes   C++
QtCore4.dll!_CRT_INIT(void * hDllHandle=0x02b2b2d0, unsigned long dwReason=45265416, void * lpreserved=0x02b2b208)  Line 446    C
QtCore4.dll!__DllMainCRTStartup(void * hDllHandle=0x67000000, unsigned long dwReason=0, void * lpreserved=0x00000000)  Line 557 + 0x8 bytes C
QtCore4.dll!_DllMainCRTStartup(void * hDllHandle=0x67000000, unsigned long dwReason=0, void * lpreserved=0x00000001)  Line 507 + 0xe bytes  C
ntdll.dll!_LdrpCallInitRoutine@16()  + 0x14 bytes
ntdll.dll!_LdrShutdownProcess@0()  - 0xfe bytes
kernel32.dll!__ExitProcess@4()  + 0x42 bytes
kernel32.dll!7c81cb0e()
kernel32.dll!_BaseThreadStart@8()  + 0x2f479 bytes

I guess that the mutex was locked by a thread that has since been killed, but I can imagine other kinds of global ctors/dtors that might be executed in this way that would cause a similar problem.  My questions are:

Is it expected behavior that ExitProcess() is called in the event of a crash if WER is disabled via SetErrorMode()?
Does anyone know of a change in behavior of the default exception handlers between Windows XP and Windows 7 which might be relevant?
If ExitProcess() is expected behavior, what is the safest way to deal with this problem in third-party DLLs?  Calling TerminateProcess() from the breakpad exception handler fixes the problem but are there any caveats I should be aware of?
Comment: This is not untypical, use TerminateProcess.  Also better to have a guard process both call MiniDumpWriteDump and terminate the process.  Signal it with a named mutex.
Comment: I should add, the exact `SetErrorMode()` call I'm using is: `DWORD dwMode = SetErrorMode(SEM_NOGPFAULTERRORBOX);  SetErrorMode(dwMode | SEM_NOGPFAULTERRORBOX);`
Comment: And for the exact usage of `SetUnhandledExceptionFilter`, see the [google breakpad code](http://code.google.com/p/google-breakpad/source/browse/trunk/src/client/windows/handler/exception_handler.cc)
Comment: Some related reading I came across: [Mozilla bug](https://bugzilla.mozilla.org/show_bug.cgi?format=multiple&id=587406), [Native client bug](http://code.google.com/p/nativeclient/issues/detail?id=1545), [another Mozilla bug](http://robert.ocallahan.org/2010/01/three-lessons-from-pernicious-bug_30.html)
Answer: To answer "What has changed between XP and Windows 7," the following could be relevant: "The case of the disappearing OnLoad exception – user-mode callback exceptions in x64," http://blog.paulbetts.org/index.php/2010/07/20/the-case-of-the-disappearing-onload-exception-user-mode-callback-exceptions-in-x64/.


How is the Grover iteration realized in an actual quantum circuit?

Question: The number of the Grover's diffuser iterations is proportional to square root of the amount of input data, but how is this realized when building a real quantum circuit? Is it simply by concatenating the needed number of diffusers one after another (in which case I assume the circuit designer is required to know the needed number of iterations beforehand)?
Answer: A quantum logic circuit isn't a literal physical object, it's a plan for what operations to perform. It's like sheet music. The piano maker doesn't have to know how long the song is when making the piano, they instead provide reusable keys that the player will just keep hitting until the song is done.
In a superconducting qubit device, the "keys" are microwave pulses bounced off the superconducting circuit. Each pulse performs some fixed operation. You would run Grover's algorithm by decomposing its operations into those pulses and playing them until the song is done.
Comment: @Alternative7 No, you can have adaptive rules like "keep doing X until Y happens".
Comment: Following your piano metaphor, a person who builds a quantum circuit can then be said to be a music composer. If there is a number of iterations in a specific melody, he/she must know exactly how many times that melody repeats in order to complete the composition. Similarly, we as a circuit builder must know how many times the Grover diffuser will be run to configure the circuit. In other words, the number to iterate cannot be assigned by output from previous operations in the circuit. Have I got it right?
Comment: Ah, then I can let the algorithm/circuit set the number of repeat? That should make things a lot more convenient. Thank you!


How can focal length of the human eye lens change?

Question: Shortsightedness is due to reduced focal length and long sightedness is due to increased focal length.howcan the focal length of eye lens change overtime?
Comment: Would this be a better fit for https://biology.stackexchange.com? I don't see how it's related to physics.
Answer: The lens becomes stiffer as people age and people can't focus on objects close to the eye.  We focus our vision using muscles that pull on fibers to change the shape of the lens.  When the lens stiffens, it doesn't allow for good focusing of near objects.
https://www.mayoclinic.org/diseases-conditions/presbyopia/symptoms-causes/syc-20363328#:~:text=Presbyopia%20is%20caused%20by%20a,images%20appear%20out%20of%20focus.
Answer: Things like this are because of the curvature of the lens of your eye. If it's too curved the focal point of your eye will land somewhere in front of your retina causing myopia, or short-sightedness, and in the case of your lens being not curved enough, the focal point lands behind the retina and you have hyperopia or long sightedness.
How this happens over time (or if you are born with it) is a medical question and perhaps should be answered on one of the medical stack exchange sites.
Comment: I think you got "behind" / "in front of the retina" wrong. Otherwise a good answer.
Comment: @RalfKleberhoff You're right. Thanks Ralf.
Answer: You are going wrong

Shortsightedness is due to reduced focal length

Rather you wear

reduced focal length

or concave lens due to Short sightedness
So if your eye sight improves then the focal length of lens decreases
but it has no relation that change focal length will change your eye problem.
Comment: OP is not talking about glasses, but rather the focal length of the eye lens
Comment: then he should clearly mention human eye lens


Is it possible to make nanometers or angstrom a default unit to work with?

Question: I'm trying to make nanotube models and molecules in Blender but I'd prefer if I can work with the defaulted units as nm or angstroms.
Comment: The units are really relevant only for imperial system where the ratios are not equal. Use the Blender default units and pretend 1 unit is 1 nm.
Comment: @RobinBetts You should post that as an answer.
Comment: OK.... perhaps I should delete the comment?
Answer: See: https://docs.blender.org/manual/en/dev/data_system/scenes/properties.html?#units
That is the official explanation of units in blender and how to change them.
Answer: The short answer is no. (Blender allows you to set a Unit to 10μm, and will display 1μm in its numerical fields).
But apart from notational convenience, and possibly unambiguous communication with others when you share files, you would gain nothing; this is so far below anything in Blender that is scale-dependent: simulations of Newtonian physics, and the some aspects of the renderers' approximations to / imitations of the behavior of light.
This scale is smaller than the wavelengths of visible light. Since there is no possible visual representation of the world at this scale, only symbolization, @JaroslavJerrynoNovotny 's suggestion seems right. In Properties > Scene > Units, you set the units to 'None', noting for yourself and others that a Blender Unit is, say, 1nm.


How do you add the snap store to Synaptic?

Question: When I search in Synaptic for snaps that I know exist, they do not show up. How do you add the snap store to the repositories in Synatpic?
Comment: You can't and likely never will be able to as synaptic has  no snap support & it's development is pretty much nil
Answer: You can't, today, do this.
Somebody would have to extend synaptic to talk to snapd, either directly or using snapd-glib. While relatively straightforward (and nearly everything the snap tool does is by talking to snapd), it is quite a lot of work, especially if (as I imagine is the case) synaptic has not been thought from the start to talk to multiple packaging backends.
Given that synaptic has seen no work in two years, and that the project maintainer is a very active snapd developer, I suspect development on it has stopped.


ORA-01196: file 1 is inconsistent due to a failed media recovery session

Question: I have 3 database 1 primary and two standby, The storage of one the standby was abruptly removed without shutting down the database.
Few Days later the storage was returned, I restart the whole database server, and while trying to open my database instance(alter database open) I get the errors below:
ERROR at line 1:
ORA-10458: standby database requires recovery
ORA-01196: file 1 is inconsistent due to a failed media recovery session
ORA-01110: data file 1: '+DATA/drstby/datafile/system.306.918916853'
I tried to do "recover database;" and I get another error:
RMAN-03002: failure of recover command at 12/15/2017 17:11:50
RMAN-06054: media recovery requesting unknown archived log for thread 1 with sequence 62585 and starting SCN of 16625729961
Please what can I do.
Note:
the database in question has the following
FRA
ASM
RMAN BACKUPS.
THANKS
Answer: Looks like you need to restore archivelogs. If you have them backed up to tape, you may need to do something like:
<code>run
{
  allocate channel ch1 type 'sbt_tape' parms
  restore archivelog from scn 16625729961 until scn 999999999;
}
</code>
Replace the until scn 999999999 with an appropriate scn from your primary database.


How to convert 3D pose sequences to the bvh format?

Question: I am very new to animation and rendering softwares, so please let me know if I need to provide more information about this. I have a sequence of 3D positions of human joints (basically mocap data), representing different kinds of walking. I have managed to visualize the sequence using python, as I have shown in this video. Each data I have is a numpy array of size TxJx3, where T is the number of frames, J is the number of joints (21 in my case), and 3 represents the 3 co-ordinate values. So my question is, how can I convert these 3D positions into a BVH file, that I can load into blender? Or convert them to any other format so that I can load these data in blender?
Comment: Consult the code of the bvh importer / exporter.
Comment: Do you happen to know where I can find the code?
Comment: Is it this one? https://github.com/20tab/bvh-python
Comment: I went through the repo, but I didn't find a way to compute the 3D positions of the joints. For reference, I want to know how to recover the 'head' attribute of the PoseBone data structure in this link https://docs.blender.org/api/blender_python_api_2_77_1/bpy.types.PoseBone.html
Answer: OK, found the solution myself. Posting here in case anyone else finds this useful. In the BVH format, the following relationship holds between the joints:
$$pos_j = R_{P(j)}offset_j + pos_{P(j)}$$
where $pos_j$ indicates the 3D position of joint $j$, $P(j)$ returns the parent of joint $j$ in whatever DAG the positions are modeled in (generally the DAG starts at the root and points towards the end-effectors) $offset_j$ indicates the offset of joint $j$ relative to its parent $P(j)$ (aka the connecting limb), and $R_{P(j)}$ is the 3D rotation that determines how much should $offset_j$ be rotated from an initial pose (generally a T-pose). In the BVH format, for each parent $P(j)$, we need to store $R_{P(j)}^{-1}R_j$.
The main trouble I had then was working with joints that had multiple children, for example, the root joint, which has connections to both legs as well as the spine. I eventually came across this repo and digging through their function forward_kinematics inside skeleton.py, realized what to do. Basically, for joints with multiple children, I had to make copies with $offset=0$, and assign those as parents of the corresponding chains. Thus I made 3 copies of the root: one became the parent for the left leg chain, one for the right leg chain, and one for the spine. And similarly for the other parents with multiple children. And yes, the visualization works great!
Comment: How to calculate the rotation matrix?
Comment: Since I assumed the forward kinematics setup, the rotation matrix is part of my input, i.e., known apriori.


(Drupal 6.28) Blocks list suddenly doesn't show correctly

Question: I realized today that blocks list (admin/build/block page) doesn't correctly show  all the blocks implemented in the site.
Most of the blocks are set as disabled and others are set in the wrong region, while fortunately the site displays every block still perfectly.
I have a custom theme (Zen subtheme).
The only "strange" thing I did recently was the installation of a new module which involved the assignment of a new block to a region.
Is there any way to restore the correct blocks list?
Maybe if I don't make any change to blocks system (and save the new configuration), my site will always display correctly, but I would prefer to have the correct blocks list restored.
Yes, I could draw a manual scheme of all blocks in the site (pages, views, etc.) and use it in case of disasters, but if I could restore it with a module it would surely be better
Thanks
Comment: So `/admin/build/block` doesn't show correctly after enabling a new module. What happens if you disable and uninstall that module?
Comment: Thanks for replying. I uninstalled that module, but nothing changed in the displaying of blocks list.
Despite the incorrect displaying, I wonder: if all the blocks are still working in the site, is there some place where the actual blocks are stored with correct regions and correct settings?
Answer: That to me looks like an issue with theme update. Did you changed your theme recently. It yes, then try clearing your cache.
Comment: Thanks for replying. I also have read this issue can occur on updating a theme, but I did not update the theme.
All blocks are working. They are just not correctly listed.
This can be fine until I have to save the block system configuration again for a future block change.


IPN with Paypal recurring payments

Question: I'm trying to to set up a subscription system with recurring payments and IPN notifications but I was reading that there is no way of setting the notify_url from the API.
Answer: The CreateRecurringPaymentsProfile request doesn't by itself seem to do the actual recurring transactions,except for the "INITAMT" which you may set (any amount to be charged immediately upon profile creation, say a "set up fee" or something). If it does I think there should not be something like this Recurring Payments With Direct Payment for instance.
The DoDirectPayment request allows a RECURRING parameter and this requires the "recurring payment profile be already created" (using the CreateRecurringPaymentsProfile request) . DoDirectPayment also allows NOTIFYURL parameter .
So the IPN should be returning all that is applicable for recurring payments.
See IPN recurring payment variables - ones that are related to "recurring payment profile creation". You could identify this matching the "PROFILEREFERENCE" param you set with CreateRecurringPaymentsProfile against "rp_invoice_id" in the IPN .
IPN subscription variables - ones that are related to the first and following subscription  payments. You could identify this matching the "INVNUM" you set with DoDirectPayment against the "invoice" in the IPN .  It also returns "payment_date" and "payment_status" .
I haven't tested this though.
Answer: Unfortunately you are right. There is no way to send notify url with CreateRecurringPaymentsProfile. The only way to set it is to set it in IPN settings in paypal, but of course you won't be able to make different urls for different subscriptions.
Comment: Is there any way to test in the sandbox?
Comment: Yes, it is working the same way in sandbox. You just set IPN url in paypal IPN settings and it will work.

IPN with Paypal recurring payments


Difference between ImageSize and RasterSize when exporting to PNG

Question: I export 2D and 3D plots to PNG images using the export command. I can set ImageSize->{xsize,ysize} or RasterSize->{xsize,ysize} as an option for export. To me it looks like these commands have the same effect. Is there any practical difference between the two when exporting to PNG?
Comment: That is because PNG is a raster-based format recognising both raster-friendly ImageSize and RasterSize, whereas SVG is not but a vector-based format.
Comment: @RitaGeraghty Thanks! Just to be clear. You are saying it doesn't matter?
Comment: If you you're just looking to make sure the plot exports with high resolution, I usually skip both and use `ImageResolution -> 300` in my `Export` command.
Answer: You may first generate the rasterized image in Mma with specific RasterSize and ImageSize, and then export the already generated image as PNG or whatever.
When you do this, the RasterSize controls the image quality, while the ImageSize fixes its geometrical size. For example, let us vary the RasterSize by keeping the ImageSize constant:
 Column@Table[
  Rasterize[
   Plot[Sin[x], {x, 0, 2 \[Pi]},
    PlotLabel -> Row[{Style["RasterSize=", 14], Style[sz, 12]}]],
   ImageSize -> 200, RasterSize -> sz], {sz, {50, 100, 300, 500}}]

Let us now keep the RasterSize constant and vary the ImageSize:
Row@Table[
  Rasterize[
   Plot[Sin[x], {x, 0, 2 \[Pi]},
    PlotLabel -> Row[{Style["ImageSize=", 12], Style[sz, 12]}]],
   ImageSize -> sz, RasterSize -> 200], {sz, {100, 200, 300}}]

One concludes that each size of the image requires its own RasterSize to garantee the image quality.
Have fun!
Comment: Does this also control the image size in printer points and raster size in pixels for an exported png?
Comment: Or an exported pdf?
Comment: Many thanks Alexei, that's very helpful!
Comment: @David Keith By definition the option RasterSize determines pixel width and size.


Connecting oci8 to Apache & PHP

Question: We're building a Debian server to host a website which is built on PHP. It connects to an Oracle database. I need to connect the PHP to the Oracle database using oci8. However, some of the steps I've seen online require the use of oracle-xe-client package. I cannot find this package at the moment. Any help or advice about the process is appreciated!
Answer: Are you using these directions? Because they don't mention oracle-xe-client. Otherwise that package can be found here.
Otherwise, both PDO and ODBC have drivers for Oracle, albeit the ones for PDO are experimental.
Comment: Thank you. It does mention the Oracle client in the instructions. However, I really appreciate your help!

Connecting oci8 to Apache & PHP


How to recover iPhone SE when stuck in recovery mode without data loss?

Question: My wife's iPhone SE was overloaded with data and hung up while rebooting. Then after attempting to update to the newest firmware, I kept stuck in recovery mode. The problem is, that the iPhone is too full for recovery. Because my wife didn't make a backup, I want to avoid a flat install. Unfortunately every recovery tool I tried gave the same error and denied work in the middle of the process.
I discovered a windows tool called "no erase ipsw", that modifies ipsws up to version 10.3.3 in that way, that it makes the installer think that it does a complete new install, although it replaces only the system files and lets the user files untouched. Unfortunately is the resulting ipsw not accepted by iTunes, because it is "unsigned", let's say too old for Apples authentication process.
My only hope as far as I see now is to make iTunes think, it is a new ipsw.
Second possibility would be to modify a new / signed ipsw in that manner, that it becomes smaller than the original to save installer space, because the iPhone is over full.
Can anybody out there give a hint?
Answer: In iTunes hold down shift. A window should come up asking you to select your file. Try selecting it from there.


Object environment texture not showing

Question:
For some reason my environment texture isn't showing in my object. No matter what photo I put its always just a shade of white no matter how dark the photo I put in is. It never reflects the photo that's been input.
Comment: Could you please pack your image (FIle > External Data > Pack Resources) and share your file? https://blend-exchange.com/ or workupload and share the link
Answer: Not sure where your reflections should come from...
1.) If the Photo named "Empty" in your Outliner is the object that should be reflected, it never will reflect anything, because it is only an Empty.
These are (afaik) only used as reference images for modeling.
To get a "real" reflective Photo in your Scene, you need a textured Plane. There's a great addon inside Blender called "Import Images as Planes". You can just enable it under Edit > Preferences and see if this gives you reflections:

2.) If the "World" should be reflected, it seems your world shader is basically set up to do this:

So, just get rid of the mix shader and plug your jpg straight into the background input. The image mapping, however, will be another problem.

Object environment texture not showing


Can a home motion detector work through frosted or clear glass?

Question: I'm thinking of installing a motion detector inside an existing recessed fixture over my front door in order to illuminate the stairs and door lock at night. Home Depot sells a small sensor that can be installed between the socket and bulb but nobody has been able to tell me if the frosted glass cover will block the sensor.
Answer: IR (infra red) motion sensors are generally housed behind frosted plastic housings. Camera lenses also let infra red light pass through. Assuming yours is an infra red motion detector, frosted plastic or glass will not be a problem. The frosting is there to block your view of the lens and to "smear" the background contrast projected onto the IR sensor such that the IR blob's motion is better detected.
Comment: PIRs do work thru plastic, they generally do not work thru glass. Just type "does pir work through glass" into google.....
Comment: If, by "glass" you mean a typical window, you'd be correct. However, the OP will be hiding the sensor within a recessed fixture where the sensor will  be within millimeters of the glass. I tried an experiment with the motion sensor that was installed at my parent's small condo (in a corner of a living room, facing the front door). With the glass placed directly up against sensor, a red light briefly flashes when motion is sensed, just as if there were no glass in front of the sensor. The sensor fails to detect motion when the glass is more than a few inches away.
Comment: A viable question is how the heat generated by the light affects the sensor once it turns itself off. Presumably, it will take some time to cool back down to ambient temperatures, especially if it's isolated within  some recessed fixture.


Sobolev space $H_0^m$ and sobolev norm and seminorm

Question: I have problems understanding the definitions of $H_0^m(\Omega)$-spaces. What does the $0$ stand for? Does it mean that the functions are zero at $\partial\Omega$? Or does it mean that it is non-zero on all of $\Omega$?
Also the sum notation of sobolev seminorm and norm confuses me. The seminorm is defined as $ |v|_m = \sum_{\alpha=m} <D^\alpha v,D^\alpha v>_0 $ where as the norm $\|v\|_m = \sum_{\alpha\leq m} <D^\alpha v,D^\alpha v>_0$
Comment: Sometimes $H_0^m(\Omega)$ is defined to be the closure of compactly supported smooth functions, so $H_0^m(\Omega) = \overline{C^\infty_c(\Omega)}$, where the closure is taken with respect to the topology of $H^m(\Omega)$.
Answer: Yes, the subscript zero usually denotes that the functions are zero on the boundary. For the semi-norm, the summation is taken over all multi-indices $\alpha$ of order $m$ (also I think you are missing some square roots). This means you take all derivatives of order $m$, individually take their $L^2$ norms and then sum those. For example, if $m = 2$, and $v = v(x,y)$, then \begin{align*} \lvert v \rvert_{H^2}^2 &= \| v_{xx}\|_{2}^2 + \| v_{xy}\|_2^2 + \| v_{yx}\|_2^2 + \| v_{yy}\|_2^2 \\ &= \langle v_{xx}, v_{xx}\rangle + \langle v_{xy}, v_{xy}\rangle  + \langle v_{yx}, v_{yx}\rangle + \langle v_{yy}, v_{yy}\rangle \end{align*} where $\langle \cdot, \cdot \rangle$ denotes the $L^2$ inner product.
For the norm, you do the same thing, but do it with all derivatives up to the given order. In my example, we would have $$\lvert \lvert v \rvert \rvert_{H^2}^2 = \| v\|_{2}^2 + \| v_{x}\|_{2}^2+ \| v_{y}\|_{2}^2 + \| v_{xx}\|_{2}^2 + \| v_{xy}\|_2^2 + \| v_{yx}\|_2^2 + \| v_{yy}\|_2^2.$$ Does this help at all? If not, please explain what about the notation confuses you.
Comment: Thanks! This was precisely what I needed :)


Add flag for migration to Stack Overflow

Question: I have a suggestion for a feature i.e Adding a flag for migration to Stack Overflow.
Answer: We really don't migrate much.
In looking at the stats for the last 2 years, we have migrated 3 to Sound Design, 2 to Software Recommendations, 2 to SU, 1 to Signal Processing, and none to SO.
So I don't think it is a useful migration destination to have in our options list.
Remember you can flag for mod attention, and in the custom text field tell us where you think a post should be moved to.
Comment: And that answer means the mod looked at it and it did not need to be migrated
Comment: Yeah, I flagged for mod attention and it got declined beacuse "flags should only be used to make moderators aware of content that requires their intervention"


Is there a quick way to make this?

Question: I'm sure I can do that with array and some distortion, but how? I use Adobe Illustrator, and I've only tried the pen tool to draw each circle by hand so far, which shouldn't be the smartest thing to do. By the way, I need the exact same form.
Answer: A guess: You have it only as a low resolution bitmap and you need a vector version. In this case trace it and forget the idea to make an exact replica by applying some distortion transform to a less curved pattern of ellipses. There's too many unknown variables that should be guessed right. A mathematician might find the wanted transformation rule, but for most of us it's undecidable (=can be found only by an accident).
If you have it as vector and only want the transformation to make a replica starting from ellipses I cannot help.
Here's a snippet of the tracing result layered on your image and colored to blue.

This it the whole result(black is your image):

I'm sure it's as good as it's possible to get without any special software which tries to fit certain limited set of curve forms to your bitmap image.
I traced it in Inkscape because its default settings work well. In addition Inkscape's Node tool is extremely well designed to make fixes to paths. For example if you remove a node to get rid of an unwanted twist the curve stays as close as possible the original form, only smoother. In Illustrator removing a node generates a shortcut.
Answer: The original array may have had rotated ellipses (at the left had side). Possibly using a step blend to make each row. Maybe something like this

Then distorted with a 2 x 2 Envelope Mesh

This is fine if you are happy enough with something similar, however since you want the exact same form, I think it would be really difficult to use that method to recreate it exactly. Also redrawing and distorting circles manually is possible, but a lengthy process.
You could instead do an Image Trace. You can see the settings I used here.

The result of this depends on the quality of the original, which isn't great to be honest. However, if necessary, you could expand the trace and fix some of the curves manually if they got messed up.

Is there a quick way to make this?


Browser filters in SASS

Question: I'm working on an RoR app that uses SASS, I've been developing it in Chrome and it's come time to work on cross-browser compatibility.  For the most part everything looks fine but there's some small styling issues I'd like to fix.  Normal CSS filters haven't worked and I don't think that mixins are the solution, I don't need any variables in this situation and I haven't been able to find much on it.  Any pointers or advice?
Answer: You should use Modernizr to detect which features are available or not in the client's browser. It will append several classes to the  tag of your page, such as no-textshadowor boxshadow. Those are classes you can use in your css, even your mixin, like this :
@mixin box-shadow($shadow)
{
     @include box-shadow($shadow);
     .no-boxshadow &
     {
         @include box-shadow(none);
     }
}

UA sniffing in general, or detecting browser, is not anymore considered good practice as it is not future-proof. Modernizr detect features on the spot, and so even if there are a lot of browsers out there that don't support feature X or Y and that you don't know about, you can still work around them.
If it's just padding or magin issues, little style things like that, I recommand you use a css reset or better, a Normalizer to put every browser on the same level, erasing browser specific styles regarding elements.

Browser filters in SASS


What is an intuitive definition for "conjugate" in Group Theory?

Question: In Abstract Algebra, I learned about "conjugation" in the context of a group $H$ being a 'normal' subgroup of $G$ if the element $xhx^{-1}\in H$ for any $x\in G$. But this is not the first time I've seen the word 'conjugate'. The other times I've seen this are in pre-calculus, when trying to rationalize a denominator, or in the case where $(x+y)$ is the conjugate of $(x-y)$. Does the Group Theory version of conjugate have any link to the pre-calculus version (and other uses)?
Comment: This is a good question with the answer a bit "yes" and "no". The answer involves Galois Theory, where a "conjugate" in the "pre-calculus" situations is in fact an image of an element under the action of a group - and is generalised. But that is not the same thing as conjugation within the group.  I am sure someone will be able to give a fuller and clearer explanation, which is why this is a comment rather than an answer.
Comment: A conjugate element, in each of these contexts, is one that "acts like" the original. In your cases, the element $gxg^{-1}$ acts like $x$, and $a+\sqrt b$ acts like $a-\sqrt b$.
Answer: C. Falcon's answer refers to Galois theory, which is almost certainly the historically correct answer.  But the connection between the two ideas can be explained in slightly more elementary terms:
If $G$ is a group acting on a set $S$, and $s,t\in S$ lie in the same orbit, then the stabilizers of $s$ and $t$ are conjugate subgroups of $G$.  Specifically, if $t = hs$, then $gs = s \iff (hgh^{-1})t = t$.
Since we frequently study group elements in terms of their fixed points, it makes sense that the terminology for conjugation in a group gets sometimes mixed up with the terminology for orbits.
For example, we can check that $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ defines an action of the cyclic group of order $2$ on the field $\mathbb{Q}(\sqrt{2})$.  Then $a+b\sqrt{2}$ is conjugate to $a-b\sqrt{2}$ in the sense that they lie in the same orbit under this action.
More generally, Galois theory tells us that if $p(X)$ is any irreducible polynomial over a field $k$, and $k\subset K$ is a splitting field, then there is a group acting on $K$ that fixes $k$ and acts transitively on the roots of $p$.  In other words, the roots of $p$ are conjugate.
This is somewhat of an abuse of terminology, but it may explain the conceptual link between the two different uses of the word.
Answer: Arguably the easiest and the most intuitive way to understand conjugation is through linear algebra.
$T : V \to V$ be a linear transformation of the vector space $V$. Then given a basis $\Bbb B = \{e_1, \cdots, e_n\}$ for $V$, one can represent $T$ by a matrix $[T]_{\Bbb B} = [T(e_1), \cdots, T(e_n)]$, with each $T(e_i)$ expressed in terms of $\Bbb B$. That is, the columns of the matrix are the images of each element of the basis $\Bbb B$ under the transformation.
If $\Bbb B = \{v_1, \cdots, v_n\}$ and $\Bbb B' = \{w_1, \cdots, w_n\}$ are two bases of $V$, then we can try to see what happens if we start with the basis $\Bbb B$ and then switch to the basis $\Bbb B'$ by expressing $v_i$ as linear combination $\sum_k a_{ki} w_i$ of elements of $\Bbb B$. Similarly, we can try expressing elements $w_i$ of $\Bbb B'$ as linear combinations $\sum_k a_{ki}' v_i$ of elements of $\Bbb B$.
The matrix $P = [a_{ki}]$ is called the basechange matrix obtained from changing base from $\Bbb B$ to $\Bbb B'$. It can be easily verified that $[a'_{ki}]$ is simply $P^{-1}$.
The relevant fact is $[T]_{\Bbb B'} = P^{-1} [T]_{\Bbb B} P$. Thus, conjugation of matrices correspond to change of basis and conjugacy classes of matrices in $GL(V)$ correspond to basis-free linear operators.

This description of conjugacy can even be seen in general groups. $G$ be an arbitrary group which acts on a set $X$ by
$$G \times X \to X \\ (g, x) \mapsto gx$$
If $g$ acts on elements on $X$ by multiplication, $hgh^{-1}$ does the same job $g$ does, but on $hx$ instead of $x$, and then puts everything back at place. One can easily see this by letting $\text{Isom}(\Bbb R^3)$ act on $\Bbb R^3$ by isometries, say, and letting $g$ and $h$ to be elements of the isometry group (e.g., rotation/translation).
Thus, $hgh^{-1}$ is not really very different from $g$, but it just does the work of $g$ at $hx$ instead of $x$. This is more general in the sense that letting $G = GL(V)$ and $G$ act on the vector space $X = V$ gives me back the linear algebra picture.

To answer the main question, to my mind, Galois conjugates are not really related to conjugates as in groups (which has already been pointed out by several answers here). The real motivation for conjugation comes from Galois theory in a different way.
Namely, if $\text{Gal}(L/K)$ is the group of $K$-automorphisms of $L$, then given an intermediate extension $L/E/K$ (terminologies are explaned in C. Falcon's answer), $\text{Gal}(L/E)$ is naturally a subgroup of $\text{Gal}(L/K)$, as $E$-automorphisms of $L$ are automatically $K$-automorphisms of $L$ (pointwise fixing something larger automatically pointwise fixes the subset).
Now, if $L/E/K$ is a tower of Galois extensions, then $\text{Gal}(L/E)$ is normal in $\text{Gal}(L/K)$. This is where the notion of conjugacy comes to play, because Galois wanted to define group structure on the collection of all cosets.

If $G$ is a group, $N \leq G$ is a subgroup, it's easy to see where conjugacy  comes in trying to define group structure on $G/N$. If $g_1N, g_2N$ are two elements in $G/N$, then the most obvious way to try to define group structure is as $(g_1 N)\cdot(g_2 N) = (g_1g_2)N$. This can be made well-defined if and only if $gNg^{-1} = N$ (one direction is easy, for the other direction if $gN = g'N$, pick $h \in N$. $hN = eN$ implies $(hN) \cdot (gN) = gN$, which is - as we claim - the same as $(hN) \cdot (gN) = (hg)N$. Hence $(g^{-1}hg)N = N$, and $N$ is normal, as desired)

It's interesting as a remark that Galois deduced that the quotient $\text{Gal}(L/K)/\text{Gal}(L/E)$ is in fact isomorphic to $\text{Gal}(K/E)$, although that's irrelevant here.
Comment: @MrLolo Easiest mnemonic is to see what happens when $g$ does no job on $x$, i.e., $gx = x$. Then $hgh^{-1}$ also does no job on $hx$, i.e., $hgh^{-1} \cdot hx = hx$.
Comment: In your second paragraph, why does $hgh^{-1}$ do the same job as $g$ but on $hx$, and not $h^{-1}x $? Sorry if this is silly but I am often confused by the order of $h$ and $h^{-1}$ in similar transformations.
Answer: The notion of conjugacy was first developped by Cauchy for permutation groups. For permutation groups it's extremely simple to understand, and in fact since all groups are isomorphic to some permutation group, you can take this as a general understanding.
Say I've got $10$ different objects which I'm swapping about. I can consider two permutations: one in which I swap the first two, and another in which I swap the last two. But these permutations are really the same thing, right? They're the same permutation, just done on different objects - I perform the same "action" in both cases, if you see what I mean. For a more complex case, imagine if I take the first five objects and rotate their positions, then swap the first and fifth. This is "the same" as if I took the last five objects and performed the same action on them (the mere fact that you're able to understand what I mean by "perform the same action on them" without my having to explain it in detail is proof of this).
Two permutations are conjugate if they are "the same permutation done to different objects" in this sense. How can we make this more precise? Let's work out an example: the permutations $P = (1\ 2)$ and $Q = (3\ 4)$, on say the set of numbers between $1$ and $10$. The second one is just $P$, but done to $3$ and $4$ rather than $1$ and $2$. So consider the permutation $S$ which sends $4$ to $2$ and $3$ to $1$. We have:
$$P=S^{-1}QS$$
We first of all apply $S$ - this is basically just "renaming" the elements of our set. Then we apply $P$, and then we undo $S$ to get back to our original "naming scheme". It's admittedly a bit hard to explain this in writing, but hopefully with some thinking you should be able to see that the existence of an $S$ such that $P=S^{-1}QS$ really does capture the idea that $P$ and $Q$ are "the same thing done to different objects".
By the way, for finite permutation groups every permutation can be written as a product of disjoint cycles, and conjugacy is equivalent to having the same "structure" in one's cycle decomposition. Defining exactly what I mean by "structure" is one of those things that's hard to explain in words but easy to see with an example. The permutations $(3\ 4)(5\ 6\ 8)$ and $(2\ 3)(1\ 4\ 9)$ are clearly "the same", but done to different objects. The fact that both are the product of a 2-cycle and a disjoint 3-cycle suffices to show that they're conjugate.
Comment: +1000. I really enjoyed by your answer.
Answer: According to A. Jeanneret and D. Lines' book Invitation à l'algèbre the group conjugation arises from Galois' theory. Let me introduce you to the basic concepts.

Definition. Let $K$ be a field. If $L$ is a field and $\varphi:K\hookrightarrow L$ is a ring homomorphism $(L,\varphi)$ is said to be a field extension of $K$ and one writes $L/K$.

Remark. A ring homomorphism $\varphi$ from a field $K$ to a field $L$ is always injective. Indeed, let $x\in K$ such that $\varphi(x)=0_L$ and assume by contradiction $x\neq 0_K$, $x\in K^\times$. Since a ring homomorphism is unitary, one has: $$1_L=\varphi(xx^{-1})=\varphi(x)\varphi(x)^{-1}=0_L,$$ a contradiction, since a field is not the null ring. Hence, $x=0_K$ and $\varphi$ is injective.
Example. $\mathbb{R}/\mathbb{Q}$ is a field-extension for the natural inclusion.

Definition. Let $L/K$ be a field extension, $M$ is said to be an intermediate field if and only if $L/M$ and $M/K$ are field extensions.

Example. $\mathbb{R}$ is an intermediate field of $\mathbb{C}/\mathbb{Q}$.

Proposition-Definition. Let $L/K$ be a field extension and $M$ an intermediate field. If $\sigma\in\textrm{Aut}(L)$ (ring isomorphisms from $L$ to $L$), $\sigma(M)$ is an intermediate field called the conjugate of $M$.

Remark. The complex conjugation is a ring automorphism of $\mathbb{C}$. The field conjugation defined above is thus a generalization of the complex conjugation.

Proposition-Definition. Let $L/K$ be a field extension, $\textrm{Gal}(L/K):=\textrm{Aut}_{K}(L)$ (ring isomorphisms from $L$ to $L$ whose restrictions to $K$ is identity) is a group for the composition called the Galois' group of the extension $L/K$.

Proof. Let us show that $\textrm{Gal}(L/K)$ is a subgroup of $\textrm{Aut}(L)$.

$\textrm{id}_L\in\textrm{Gal}(L/K)$.
Let $\sigma,\tau\in\textrm{Gal}(L/K),\sigma\circ\tau\in\textrm{Aut}(L)$. Let $x\in K$, one has: $$\sigma\circ\tau(x)=\sigma(x)=x.$$
Hence, $\sigma\circ\tau\in\textrm{Gal}(L/K)$.
Let $\sigma\in\textrm{Gal}(L/K)$, $\sigma^{-1}\in\textrm{Aut}(L)$. Let $x\in K$, one has: $\sigma(x)=x$, hence one gets: $$x=\sigma^{-1}(x).$$ $\sigma^{-1}\in\textrm{Gal}(L/K)$.

Whence the result. $\Box$
The link with the group conjugation is contained in the:

Proposition. Let $L/K$ be a field extension, $M$ be an intermediate field and $\tau\in\textrm{Gal}(L/K)$, one has: $$\textrm{Gal}(L/\tau(M))=\tau\textrm{Gal}(L/M)\tau^{-1}.$$

Proof.

Let $\sigma\in\textrm{Gal}(L/K)$ and let $y\in\tau(M)$, there exists $x\in M$ such that $y=\tau(x)$, one has: $$\tau\circ\sigma\circ\tau^{-1}(y)=\tau\circ\sigma(x)=\tau(x)=y.$$
Hence, $\tau\circ\sigma\circ\tau^{-1}\in\textrm{Gal}(L/\tau(M))$ and one has: $$\tau\textrm{Gal}(L/K)\tau^{-1}\subseteq\textrm{Gal}(L/\tau(M)).$$
Let $\rho\in\textrm{Gal}(L/\tau(M))$ and let define $\sigma:=\tau^{-1}\circ\rho\circ\tau$. Let $x\in K$, one has: $$\sigma(x)=\tau^{-1}\circ\rho\circ\tau(x)=\tau^{-1}\circ\tau(x)=x.$$
Hence, $\sigma\in\textrm{Gal}(L/M)$ and since $\rho=\tau\circ\sigma\circ\tau^{-1}$, $\rho\in\tau\textrm{Gal}(L/M)\tau^{-1}$. Therefore: $$\textrm{Gal}(L/\tau(M))\subseteq\tau\textrm{Gal}(L/K)\tau^{-1}.$$

Whence the result. $\Box$
Remark. In other words, this proposition states that the Galois' group associate with a conjugate field of $M$ is a group conjugate of the Galois' group associate with $M$.
Comment: It's clear from your work, but perhaps you want to emphasize that your ring homomorphisms are always unitary. (Some people think of the zero map as a ring homomorphism.)
Answer: Conjugation usually relates to mappings on the algebraic structure called automorphisms.  These are bijective maps which preserve the operation of the structure.
In $\mathbb{C}$ there are two operations, addition and multiplication.  An automorphism of $\mathbb{C}$ is then a bijection from $\mathbb{C} \to \mathbb{C}$ which preserves both of these options.  Since $\mathrm{i}$ and $-\mathrm{i}$ are both solutions to $x^{2} +1 = 0$, it can be shown that the map $a+b\rm{i} \mapsto  a-b\rm{i}$ provides an automorphism of $\mathbb{C}$.  So if you have $z$ and $\bar{z}$ equivalent under this map, they are called conjugates. Notice that if you multiply two complex conjugates together you get $(a + b\rm{i})(a-b\rm{i}) = a^{2}+b^{2} \in \mathbb{R}$, this relates to the case in precalculus where you are rationalizing a denominator by multiplying $(a+b\sqrt{d})(a-b\sqrt{d}) \in \mathbb{Q}$; in this setting, there is also a connection to field automorphisms of $\mathbb{Q}(\sqrt{d})$.
In a group $G$ there is only one operation.  One common example of an automorphism of a group is conjugation by an element $g \in G$, this is a bijective map from $G \to G$ given by $x \mapsto g^{-1} x g$.  Two elements $h_{1}, h_{2} \in G$ are called conjugates if there exists some $g \in G$ such that $h_{2} = g^{-1} h_{1} g$.  (Note that there are sometimes other automorphisms of a group that are not equivalent to conjugation by an element. Group elements that are equivalent under these other types of automorphisms are usually not called conjugates).


How to set the number of mappers in new Hadoop api?

Question: I've switched to the new Hadoop api (mapreduce) from the old one (mapred). I can't set the number of mappers in the new api. I can use job.setNumReduceTasks() to set the number of reducers but there isn't any method for number of mappers. I also tried conf.setInt("mapred.map.tasks", numMapper) and conf.setInt("mapreduce.map.tasks", numMapper) but it is not working either.
Comment: I've seen this question come up a couple times and never have I seen a good reason to do this. Number of mappers = number of input splits sounds good to me.
Comment: I think the better and more appropriate question is "how do I change my input splits so that I have fewer?"
Answer: The number of mapper tasks is determined by the input split you have. Obviously, each part will be processed by 1 mapper. So essentially, your data determines the number of your mappers!
You can however use mapreduce.jobtracker.maxtasks.perjob to limit the parallelism (unfortunately, this affects both mappers and reducers!). But if you set this to 10, at most 10 mappers should run in parallel.
A more fine-grained control would be nice, but is an open ticket:

MAPREDUCE-5583: Ability to limit running map and reduce tasks
Comment: you mean there is no way to explicitly determine the number of mappers in the new api? as it is possible in the old api.
Comment: See the ticket I linked.
Answer: Starting from Hadoop 2.7, you can use mapreduce.job.running.map.limit and mapreduce.job.running.reduce.limit to control these at each job level.
Fixed by this JIRA ticket.
Answer: In YARN, one can set mapreduce.input.fileinputformat.split.minsize (its in megabytes) much higher than the blocksize of the files being read.  This will force more data through each mapper thereby reducing the number of mappers required.  However, some file formats may have their own minimum split size which takes priority over this setting.


Mean Value Theorem: Continuous or Defined?

Question: One of the requirements of the MVT is that the function has to be continuous at each point on the closed interval $a\leq x\leq b$. If we call the function $f(x)$, I am confused about how the function is continuous at points $(a, f(a))$ and $(b, f(b))$. The definition of a function being continuous at a point (let's call that point $(c, f(c)))$ is that $ \lim_{x\to c} f(x) = f(c).$ If we take the point $(a, f(a))$, in order for $f(x)$ to be continuous at $(a, f(a))$, $ \lim_{x\to a} f(x) = f(a)$ must hold true. However, it can't because the two sided limit doesn't exist. $x$ can approach $a$ from the right but not from the left, and $x$ can approach $b$ from the left but not from the right. So shouldn't "has to be continuous at each point" actually be "has to be defined at each point"?
Comment: Generally I think the mean value theorem is phrased as "continuous on the closed interval $[a, b]$" rather than "continuous at each point on the closed interval $[a, b]$" precisely for this reason: continuity on a closed interval is defined to mean two-sided continuity at the interior points and one-sided continuity at the endpoints.
Comment: Look up one-sided continuity.
Answer: The function $f:[a,b]\to R$ is continuous at $x=a$ if
$$\lim_{x\to a^+} f(x)=f(a)$$
Similarly $f:[a,b]\to R$ is continuous at $x=b$ if
$$\lim_{x\to b^-} f(x)=f(b)$$
The continuity at the interior points is two sided while the continuity at endpoints is one sided.
Answer: The other answers do a good job addressing the question from an analytic perspective.  Topologically, there is an analogous explanation (which exists because metric spaces like $\mathbb{R}$ are also topological spaces, a more general class of objects):

Definition: A function $f:X \to Y$ between topological spaces is continuous if, for every open set $U \subset Y$, its preimage $f^{-1}(U)$ is open in $X$.
Note: When $X$ and $Y$ are metric spaces, this definition of continuity agrees with the epsilon-delta definition from analysis.  The open sets in a metric space are unions of open "balls"; see here.

So user532874's concern can be expressed as follows:

Consider a purportedly continuous function $f:[a,b] \to \mathbb{R}$ and the open set $\Big(f(a) \! - \! \varepsilon, \ f(a) \! + \! \varepsilon\Big)$ for a small $\varepsilon >0$.  If $f$ is continuous, then the preimage of this set should be an open set as well.  But the preimage instead includes a half-open interval $[a,c)$ or even, if $f$ is constant, the entire closed interval $[a,b]$.  Ostensibly, this preimage is not open: in particular, there doesn't seem to be any "wiggle room" around $x=a$.  What gives?

And the answer is that, if $U$ is a subset of a topological space $X$ and we have a continuous function $g:U \to Y$, then the preimage of open sets in $Y$ will be open not with respect to $X$, but to $U$ which has been endowed with the subspace topology.  Under this topology, the open sets of $U$ are $\{V \cap U \ | \ \text{V is an open set w.r.t. X}\}$.
In our scenario, we have $X = \mathbb{R}$ and $U = [a,b]$.  Given the subspace topology, note that a half-open interval $[a,c)$ actually is "fully" open in $[a,b]$ because $[a,c) = [a,b] \cap (a \! - \delta, \ c)$ for any $\delta > 0$, and this interval $(a \! - \delta, \ c)$ is open in $\mathbb{R}$.  Likewise, $[a,b]$ is open with respect to itself since $[a,b] = [a,b] \cap (a \! - \! \delta, \ b \! + \! \delta)$.
Comment: I know this subspace topology appears arbitrary, but it's meaningfully chosen to be the coarsest topology so that the canonical injection $U \hookrightarrow X$ is a homeomorphism, and this has "desirable" implications; might add a section on this when I'm awake again.
Answer: Using the definition of limit,"approaching" $a$ means that for any real number $\delta >0$ (no matter how "tiny"), we consider the elements $x\in [a,b]$ such that $|a-x|\le\delta$, and the similarly for $b$. This implies that this elements as close to $a$ (or $b$) as they might be, must be in the domain of $f$ in the first place, and as a result the limit exists if it exists from the side we can approach $a$ (or $b$). (Note that $f$ is indeed defined at $a$ and $b$ because they are in the domain of $f$ by hypothesis)
Hope this helps!
Answer: We usually say that a function is continuous at an endpoint of its domain if it's continuous as one approaches the endpoint from all possible directions (which is just from within the interval here).
An example is the function $\sqrt x,$ defined on $[0,+\infty),$ which is continuous at $x=0.$ However, the function defined on the same domain by $0$ when $x=0$ and $1/x$ elsewise is discontinuous there.
Answer: For instance,  consider $f:[0,1]\to\Bbb R$ given by $f(x)=\begin{cases}x\,,x\in[0,1)\\2\,,x=1\end{cases}$.
Then $f$ wouldn't be continuous at $x=1$, and MVT wouldn't apply.
Answer: Let $S \subset \mathbb R$, and let $f:S \to \mathbb R$. To say that $f$ is continuous at a point $a \in S$ means that if $\epsilon > 0$, then there exists $\delta > 0$ such that if $x \in S$ and $|x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$.
In the above definition, $S$ can be any subset of $\mathbb R$ (not necessarily an interval), and $a$ can be any point in $S$ (including points that are on the boundary of $S$). So this definition still makes perfect sense in the case where $S$ is a closed interval and $a$ is one of the endpoints of $S$.


Password problems in ubuntu 13.04

Question: Ubuntu keeps asking me for my password to log in to my account when I enter the correct password, it asks for it again and will not accept it. Id does tell me if it's incorrect when I enter wrong password, though. Can anyone help please?
Answer: reboot the machine and while the grub loader menu appers choose it and edit it type e  then choose the kernel and press e to edit and just end of the line put 1 and press B to boot after getting into run level 1 change your password using passwd command

Password problems in ubuntu 13.04


Why am I getting this error with npm, and how do I solve it?

Question: I am trying to add a PayPal system into my html website. I am following this tutorial: https://www.youtube.com/watch?v=DNM9FdFrI1k&list=PL4eNesgScjZb3oktJK1786zI4nSET_tas&index=4
I am 9:10 into the video, and when I run npm run devStart as shown in the video, I get the following error:
PS C:\Users\dipla\OneDrive\Desktop\USB\Argo & coding\Examples\Web design\paypal-checkout> npm run devStart

> email@example.com devStart
> nodemon server.js

The system cannot find the path specified.
node:internal/modules/cjs/loader:936
  throw err;
  ^

Error: Cannot find module 'C:\Users\dipla\OneDrive\Desktop\USB\nodemon\bin\nodemon.js'
    at Function.Module._resolveFilename (node:internal/modules/cjs/loader:933:15)
    at Function.Module._load (node:internal/modules/cjs/loader:778:27)
    at Function.executeUserEntryPoint [as runMain] (node:internal/modules/run_main:81:12)
    at node:internal/main/run_main_module:17:47 {
  code: 'MODULE_NOT_FOUND',
  requireStack: []
}

I have node.js installed and I added its location to my environmental variables.
This is my first time using npm and Node.js
EDIT
I finally found what is causing the error, but I don't know how to fix it.
Here's what's happening:
When I run npm i express ejs dotenv @paypal/checkout-server-sdk it installs the node_modules folder in my project, it places the files where you would expect, which in my case is:
C:\Users\dipla\OneDrive\Desktop\USB\Argo & coding\Examples\Web design\paypal-checkout\node_modules

But the computer is looking for the \node_modules files in 'C:\Users\dipla\OneDrive\Desktop\USB\nodemon\bin\nodemon.js'
Normally you would see this error if you run the commands in the terminal from the wrong directory, however my terminal says
PS D:\USB\Argo & coding\Examples\Web design\paypal-checkout>

and when I run dir I get
 PS D:\USB\Argo & coding\Examples\Web design\paypal-checkout> dir

 Directory: D:\USB\Argo & coding\Examples\Web design\paypal-checkout

Mode                 LastWriteTime         Length Name
----                 -------------         ------ ----
d-----        2021-12-13  11:50 AM                node_modules
d-----        2021-12-09   6:57 PM                views
-a----        2021-12-13  11:50 AM         128194 package-lock.json
-a----        2021-12-13  11:50 AM            402 package.json
-a----        2021-12-13  11:25 AM            251 server.js

So, I tried moving the node_modules folder into the directory / my computer is looking, and it worked, however, this is not where I want to keep the node_modules files. It is a really bad location because when I make my website live, the mode_modules folder will not be inside my project folder, so it wont be included.
note:
I am running this project from my USB. If I create the project from scratch NOT on my USB (directly on my desktop), I get the following error:
PS C:\Users\evand\Desktop\paypal-checkout> npm run devStart

> firstname.lastname@example.com devStart
> nodemon server.js

[nodemon] 2.0.15
[nodemon] to restart at any time, enter `rs`
[nodemon] watching path(s): *.*
[nodemon] watching extensions: js,mjs,json
[nodemon] starting `node server.js`
node:internal/modules/cjs/loader:936
  throw err;
  ^

Error: Cannot find module 'node_modules/express'
Require stack:
- C:\Users\evand\Desktop\paypal-checkout\server.js
    at Function.Module._resolveFilename (node:internal/modules/cjs/loader:933:15)
    at Function.Module._load (node:internal/modules/cjs/loader:778:27)
    at Module.require (node:internal/modules/cjs/loader:1005:19)
    at require (node:internal/modules/cjs/helpers:102:18)
    at Object.<anonymous> (C:\Users\evand\Desktop\paypal-checkout\server.js:1:17)
    at Module._compile (node:internal/modules/cjs/loader:1101:14)
    at Object.Module._extensions..js (node:internal/modules/cjs/loader:1153:10)
    at Module.load (node:internal/modules/cjs/loader:981:32)
    at Function.Module._load (node:internal/modules/cjs/loader:822:12)
    at Function.executeUserEntryPoint [as runMain] (node:internal/modules/run_main:81:12) {
  code: 'MODULE_NOT_FOUND',
  requireStack: [ 'C:\\Users\\evand\\Desktop\\paypal-checkout\\server.js' ]
}
[nodemon] app crashed - waiting for file changes before starting...

So, my new question is: How do I make npm / my computer look for node_modules in a different directory?
Answer: I saw this solution, as you are using an external drive, this must be the problem
getting "Error: EINVAL: invalid argument, read" for "npm install --save-dev eslint --verbose"
"Since you are using drive letter I:, you might be on a portable drive. I had the same problem using an USB-Stick formatted on FAT32. I solved it reformatting to NTFS."
Comment: I made a copy of my USB, then reformatted it to NTFS, and the command "npm i express ejs dotenv @paypal/checkout-server-sdk" worked. But know I get an error with "npm run devStart". I am now 9:10 into the video. I also tried doing this directly on my desktop.
Comment: I edited my original post with more information
Comment: If there's another question that answers the question at hand, flag or vote to close as a duplicate. Link only answers are not particularly helpful nor do they add much value.
Comment: we still have not found any solution.
Comment: I have found what is causing the error, please see my original post as I made another edit to it.


Can you add emulsifying salts when making mozzarella?

Question: Pretty much what the title says, I want to improve the stability of melted mozzarella for a dish I want to make. Haven't made mozzarella before, but if I make it at home, is there a stage I could add sodium citrate (or sodium phosphate) like is done to make American cheese gooey like it is?
Answer: Well as such you can't add emulsifying salt in the preparation of natural mozzarella cheese. If you want to prepare processed mozzarella cheese then you can add emulsifying salts for better improvement of emulsification properties and protein solubilization, it helps in reducing free fat leakage. the stability of melted mozzarella cheese depends upon many factors associated with manufacturing steps i.e. cooking temperature, scalding time, the acidity of whey drainage etc... so you have to take care about each and every aspect of manufacturing step to make it more stable in their melted form.
Comment: The cheap American cheese brands got sloppy with this about a decade ago. Haven't bought it since. It's no good on pizza, and a disaster in Rigatoni. There at least I can substitute our fake American Parmesan.


Form visibility in a landing page

Question: On a landing page, is it better to make the form automatically visible once you get on the page or put a button and wait for the user to click it to make the form  visible ? what's best to get more conversion?
Answer: Depending on the content and (micro-/macro-) goals on your landing page but in most cases:

Less clicks -> better UX.
Answer: If you want to increase the conversion it should be visible in the first place. The fewer actions a user has to do, better conversion.
Form length and page speed are also important.
Answer: You should Plan the Path According to your Goal and the Product type.
But Always Remove Un-required Steps from the process to make it user-friendly.

Form visibility in a landing page


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Question: I'm trying to install Open CTI Connector to Salesforce (Professional license). The installation always fails giving me this: installing this package requires the following feature and its associated permissions apex classes. Is there any solution besides purchasing API access?
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Answer: I agree with JimRae on this one, purchasing API access won't solve this issue. The only option for pro would be to find an external cloud base solution that integrates only through a browser and not internally to Salesforce. Or you can look for an app exchange solution that would install and execute your apex solution. Or just upgrade to Enterprise Edition like what JimRae said.