Conditions for Matrix to be Product of Near-Identity Matrices

Question: For $\epsilon > 0$, let $M_{\epsilon}$ be the family of $n$ x $n$ real matrices A such that $||$A$ - $I$_n|| < \epsilon$, where $|| \cdot ||$ is the standard operator norm. If $\epsilon$ is chosen sufficiently small, then all finite products of members of $M_{\epsilon}$ have positive determinant (i.e., they are orientation-preserving). Is this the only requirement for an $n$ x $n$ matrix to be expressible as such a product?  If so, that would imply the result that any non-singular $n$ x $n$ matrix can be expressed as a product of $n$ x $n$ matrices that each change only one coordinate (as this is clearly the case for any matrix  in $M_{\epsilon}$), which is what I'm trying to prove.
Answer: Let $A \in \mathbb{R}^{n \times n}$
Now if $\|A- I_{n} \| < \epsilon $ then I can express $A$ as product of two nearly orthogonal matrices. Right.
An orthogonal matrix is $QQ^{T} = Q^{T}Q = I_{n} $ now...each column of $Q$ is unit normal. So if we build an orthogonal matrix and alter slightly we manipulate the bounds on $\epsilon$ Like the following..
n= 3;
A = rand(n,n);
[Q,R] =qr(A);
I  = eye(n);
err = norm(Q*Q' - I);

now this is zero...for instance...
epsilon = 5;

Q1 = epsilon*Q(:,3);
Q1 = [Q(:,1),Q(:,2),Q1];
err1 = norm(Q1*Q1' -I)

err1 =

   24.0000

From the matrix norms it slightly less than 25...like I expected. This comes from the matrix norm equality
$$ \|AB \| \leq \|A\| \|B\|$$
and
$$ \| c A\| \leq  |c| \| A \|$$
illustrating that this bounds it closer change epsilon to $1$
n= 3;
A = rand(n,n);
[Q,R] =qr(A);
I  = eye(n);
err = norm(Q*Q' - I);

epsilon = 1;

Q1 = epsilon*Q(:,3);
Q1 = [Q(:,1),Q(:,2),Q1];
err1 = norm(Q1*Q1' -I);

err1 =

   5.1650e-16

Q1 = epsilon*Q(3,3);
Q2 = Q;
Q2(3,3) = Q1;
err1 = norm(Q2*Q2' - I)

Note that since 1 doesn't modify anything it will be close to machine precision or our $\epsilon$
In retrospect that was kind of dumb. We're going to create a matrix from the outer product of two other and subtract it from $I_{n}$
$$A =I_{n} -  vu^{t} $$
let $vu^{T}_{ij} = 0 , vu^{T}_{i=j=1} = \epsilon ,  $
So you have a zero matrix we create we subtract off epsilon from the identity.
$$ \| A - I_{n} \|  = \epsilon $$
$$  \|  I_{n} - vu^{t} -I_{n}  \| = \| vu^{T} \| = \epsilon $$
we can demonstrate this like the following..
n=3;
I =eye(n);
Z = zeros(n);
epsilon = 1e-3;
Z(1,1) = epsilon;
A = I-Z;

error = norm(A-I);

error =

    0.0010

So you simply create an $\epsilon$ and make it smaller.
Comment: I'm afraid I don't see how I could use this to answer my question. If an orientation-preserving matrix **A** is really far away from the identity, how does its QR decomposition help me get it as a product of matrices close to the identity?
Comment: I don't understand your concern. The QR decomposition is gram schmidt. I produced a matrix $Q$ which is orthogonal. I've demonstrated that with both code and math.

Also there are two parts..one where I create a diagonal matrix like you state because what you're saying is confused. If you subtract $\epsilon $ from identity in the first spot then subtract that from $I_{n}$ you get $\epsilon$ for the norm.
Comment: your constraint is to make something called $\epsilon$ typically people it is very close it can be any number. further more determinant of orthogonal matrices is 1.
Answer: I got the following solution from my algebra professor:
If $H$ is the group of $n \times n$ real matrices with positive determinant, we have that $H$ forms a connected set in the standard topology on $\mathbb{R}^{n^2}$. This is complicated to prove, but it essentially involves showing that any matrix in $H$ can continuously be moved to the identity by adding multiples of some rows to other rows and scaling rows by positive constants (most easily by induction on $n$). So given any matrix $A$ in $H$, there's a path $C$ in $H$ from $I$ to $A$.
We then fix a small $\epsilon > 0$ and let $V$ be the ball around $I$ in $H$ of radius $\epsilon$. We can choose a smaller open ball $U$ around $I$ such that for any matrices $X, Y \in U$, $XY^{-1} \in V$, that is, the multiplicative distance between any two matrices in $U$ is in $V$.
We then note that every matrix $X$ along $C$ yields a multiplicative translate $XU$ of $U$ that is still an open set (because $X$ is just a change of coordinates), and by compactness finitely many of these will cover $C$, so we have a chain $U_1, ..., U_k$ of open sets along $C$ linking $I$ to $A$, such that, by openness and the connectedness of the path, each $U_i$ overlaps with $U_{i+1}$. We can thus take a chain of points (matrices) $X_1, ..., X_{k+1}$ where $X_1 = I$, $X_{k+1} = A$, and for $2 \leq i \leq k$, $X_i \in U_{i-1} \cap U_{i}$. For each $i$, we have that $X_{i+1}^{-1}X_i \in V$, by the definition of $U$, so we can move from $I$ to $A$ via a product of $k$ matrices in $V$.


Jerk in Power plants due to breakdown of transmission line

Question: I read in news paper "heavy jerk in power plant due to breakdown in transmission line" what does it mean why does it happen?
Comment: Maybe the repairman is overweight and unfriendly?
Comment: in engineering terms, the 3rd derivative with respect to time is jerk.  1st = velocity, 2nd = acceleration and 3rd = jerk.  I Don't know if this applies here.
Comment: Some more context would be helpful. Link to the article? Or is this one of those newspapers that's on real newsprint (people still read those?)
Comment: I'm guessing it's a play on words, on the one hand it could mean an overweight obnoxious person, on the other hand it could mean some kind of disruption of power or operations at the power plant. But the second meaning isn't clear in American English---it probably depends on some specifics of International English. What country was that newspaper from? Maybe someone with better knowledge of English usage in that country can clarify.
Answer: When a transmission line suddenly becomes unavailable (trips due to fault) this can cause sudden frequency change in the system due to imbalance of supply to demand.  Some types of plant are very sensitive to frequency and a change of more than a few hz in system frequency can cause significant plant damage. For this reason they are usually equipped with frequency protection and the plant will trip if the system frequency becomes too high or low. The 'jerk' referred to here could be the mechanical effect on the plant caused by the change in frequency.
Answer: In the USA, for a generation plant, jerk is the rate of change of ramp rate. Because the plant jerk rate is limited for mechanical generation plant (the ramp rate is fixed, or almost fixed), a transmission system must have automatic load shedding: even if the plant could ramp up the power fast enough to handle the change, it cannot change the ramp rate fast enough to handle fault conditions.
To prevent automatic load shedding, power suppliers are required to meet minimum standards for Plant Jerk Rate Capability.
So a heavy jerk would be a sudden and sharp change in load. However, the word Jerk is not normally used except for power plant specifications.


How to retrieve a US DS-160 confirmation page?

Question: I completed a DS160 online application form for the US embassy in Barbados.  I got a confirmation page, but it was not printed legibly. I do not have the application ID. Can I retrieve the confirmation page without it, possibly with the confirmation barcode number?
Answer: The first step would be to get the application id, which you can do through the State Department's Electronic Application Center tool to Retrieve an Application ID.
When you complete that, it should take you to the last page you completed. If that is not the confirmation page print function, then go to the Consular Electronic Application Center and select the Embassy/Consulate at which you applied (Barbados). Select Option C-Retrieve Application and, with your application ID number, you can view and print your confirmation page.


The set of natural number is semi group?

Question: Because first two properties of group are satisfied but I have read in local author book it is not semi-group.
Comment: What book is that? Either the author's wrong, or you're misreading it, or they're understanding "semigroup" as what is generally called "monoid" and they don't consider $0$ to be a natural number (monoids have to have identity elements). Btw I didn't downvote, but this needs more context to be answerable.
Comment: Okay Sir you are right author is wrong.Thank you so much
Answer: Normally you would need to specify with respect to which operation you are checking if the given set is a semigroup. For example, the set of natural numbers is a semigroup with respect to addition and so is with respect to multiplication, but, for example:

You cannot define subtraction in an ordinary way on $\mathbb N$ - this is sometimes called out as "$\mathbb N$ not being a semigroup with respect to subtraction" (although this is incorrect - it is just that subtraction is not defined on $\mathbb N$)
A lot of other binary operations can be defined on $\mathbb N$ but are not associative. Think of exponentiation: $x^{(y^z)}$ is generally not the same as $(x^y)^z$.

So my guess is that maybe the author of the book is thinking of one of the two cases above, or of something similar.
Comment: Notice that $\mathbb N$ with subtraction is also not a semigroup for another reason, which is that, even where defined, it isn't associative. So for example, $(\mathbb Z,-)$, with $-$ seen as binary, is not a semigroup, as $$a-(b-c)=a-b+c\neq a-b-c=(a-b)-c,$$ where the inequality is verified if $c\neq 0$.
Comment: Okay sir! I have clear my point.so much thank you giving me information
Answer: Definition: A semigroup is a set $S$ together with a binary operation $*$ that is, a function $* :S\times S\rightarrow S$ that satisfies the associative property:
For all $a,b,c\in S$, the equation  $(a * b) * c=a* (b * c)$ holds.

The natural number $\mathbb{N}$ with usual addition and multiplication is semigroup.
If we define binary operation on $\mathbb{N}$ as follows
\begin{align*}
*: \mathbb{N}\times \mathbb{N} \to \mathbb{N}\\
(n, m) \mapsto n^{m}
\end{align*}
then this binary operation is not associative, because
\begin{align*}
2*(3*4)=2^{3^{4}}\neq (2^{3})^{4}=(2*3)*4.
\end{align*}
So $(\mathbb{N}, *)$ is not Semigroup.


Why do I not have the option to vote to delete?

Question:
Why do I not have the option to vote to delete? I'm not fussed about deleting, I can vote to close it but, as I have 20k score, I thought I should have these extra options? https://superuser.com/help/privileges/trusted-user
Comment: that's fair enough, I was just wanting to make sure you were aware that it probably would disappear one way or another.
Comment: You don't really *need* to delete it, it'll get [roomba'ed](http://superuser.com/help/roomba) pretty soon anyway.  It's not so painfully awful that it absolutely must be gone from everyones sight *right now*, it's just bad.
Comment: I wouldn't delete it either, I'd stick with the voting to close and some one else can delete (other than spam)! That is't the point though. It's more about why it shows something which doesn't appear to be correct? @Mokubai
Answer: You don't. Or more specifically the question has not met the criteria for you to try to put it out of its misery.

Voting to delete questions with a score of -3 or lower immediately after they are closed

The question needs one more downvote and to be closed.
Comment: Its a question tho
Comment: I've updated my post... I don't *think* this is correct :S
Comment: That's where I got it wrong :S Thank you


How to print bulk document from SharePoint site

Question: How can we print bulk documents from SharePoint site through the browser? We have to print the documents through browser means at client side not at server side. The printer is not in the server network. User will select multiple projects (that carries various documents) and click on print button. We have to silently print all the documents in client's printer.
Answer: An out of the box approach is to open the document library in Windows Explorer (Actions -> Open with Windows Explorer) select the desired files and right click on them.  Then select the print option from the right click menu.
Answer: Print Multiple Documents from SharePoint
Simply click on the venerable 'Open with Explorer' option;
In explorer, you can select up to fifteen documents at once, and a print button remains visible. This cuts down the work of multi-document printing a lot!
Regards,
Peter


How do you invite friends to the Halloween 2018 Google Doodle?

Question: I have been playing Halloween 2018 Doodle for a long time but I have not found a way to share a link with friends.  How can I invite them?
Answer: After starting the doodle, there is a Host Game button in the bottom right:

After you click this button, it will ask if you want to create a private match.  Click yes, and you will be prompted with a screen with a link you can give to your friends:

On your friends end, it should look like this after they navigate to the link:

Notice how there are now two ghosts on the screen, and it says 2/8 players ready. When you are ready, the host presses the start game button and you and your friends should load in.  I tested this locally with two different tabs in Google Chrome and it seems to work.


What are the risks of creating a gnupg private key with no passphrase?

Question: On a multi-user system (linux), I created a private key with no passphrase. I can now encrypt a file and decrypt it locally without supplyng a passphrase. This is very convenient and appears to be safe. The encrypted file has read/write permissions only for the owner of the file (me). However I am interested in what risks I am open to by not protecting my private key with a passphrase, or how could another user on the system gain access to my encrypted file? I am not intending to exchange encrypted files with other correspondents, I only want to encrypt and decrypt my own files locally.
Comment: The root user (system administrator) could decrypt the data.
Comment: your question is 2 part A)How can another user gain access to your private key file or any file  B)What can they do with it.    Re 'A' Dragonlord gave a great answer, the root user can.  Re 'B' of course they can decrypt the files encrypted with your public key, if they have a copy of your private key file and can use it, and they can then use it straight away if you put no passphrase on it.
Answer: If they get a copy of your private key then they can decrypt files intended for you (files others encrypted with your public key).
So if they crack into your system somehow and get your private key, then you lose the security.
That's the reason for a passphrase, so even if they get the private key file they can't use it because they don't have the passphrase.
Comment: And they can sign messages with your key claiming you did it.


Disabling GPU Compositing for All Chrome and Electron Instances on a Mac

Question: I have been experiencing severe performance issues with all Chromium browsers and Electron apps on my Mac for years. I've filed a bug report with Chromium and have contributed my findings to other bug reports for similar issues but I need a workaround in the meantime.
I've discovered that running Chrome with the --disable-gpu-compositing command-line switch fixes this issue almost completely, and in a way that is better than disabling hardware acceleration altogether.
On Mac, however, the only way to run Chrome this way is to run it from the terminal as this option is not available in settings or in chrome:flags. This is not ideal as it means I have to maintain shell aliases for several apps (Chrome, Chrome Canary, Edge, VS Code and Slack) and each of them spawns a terminal window with console output.
Ideally, what I would like is a way to globally disable GPU compositing, but keep other aspects of hardware acceleration, for all Chromium and Electron apps on this computer. I think something like this should be achievable by modifying the GPU block list. Is it possible to modify it on a local basis somehow?
Answer: add --disable-gpu-compositing
just like
chromium --disable-gpu-compositing
or anyElectronapp --disable-gpu-compositing
just that simple


Why can't we smell our own bad breath?

Question: Why can't we smell our own bad breath. Is there any way we can do that?
Comment: keep your hand in front of your mouth and blow on it.
Comment: People can, and do, do this all the time. Some people have a decreased sense of smell, and may not taste/smell bad odors (e.g. people with sinus problems, smokers, etc.) but it is quite possible.
Answer: What you might be experiencing is olfactory fatigue, the temporary, normal inability to distinguish a particular odor after a prolonged exposure to it.
It's a form of neural adaptation, a change over time in the responsiveness of the sensory system to a constant stimulus. This adaptative mechanism might prevent the overloading of the nervous system.
However, some people (maybe most of them?) are able to smell their own breath. At least, I know I do. I can't point you to any statistics, but it might be an interesting research project, figuring out how many people can smell their own breath.

Why can't we smell our own bad breath?


Track phone data usage from PC

Question: I manage my companies cell phones on a shared account.  We allot a certain amount of data to each user each month.  They are contractually obligated to this amount, else they are responsible for overages.  I am looking for a way to be automatically alerted via email whenever any one user exceeds their allotted data.  It would preferably be a third party software approved by Verizon or an app that sends the email.  If anyone knows of anything useful, please let me know.
Answer: I would see if I could download the billing information in CSV.  These records could be imported into a database.
<code>Select sum(minutes)-400,* from phones where sum(minutes)>400 group by phone_number,month(date),year(date)
</code>
Where 400 is the allowable number of free minutes.
Your going to store the date and time in a single column so the month() function exacts the month from the date/time column.  The same thing for the year.
Then your need a table containing:

The account or phone number of each company phone
Call date (and time is nice)
Duration
You probably want the destination phone number just for record keeping purposes, but not really needed.

Anyone who uses less than the approved 400 minutes won't show up on the report because we asked for greater than 400 minutes.

Track phone data usage from PC


Failure in obtaining new access token after expiration of old token using WSO2 Identity Server

Question: On expiration of access token I have been facing a peculiar problem wherein sometimes my calls are failing with a 401 error(Access token expired). As you would expect I should get a new token using refresh_token grant through OAuth2TokenRelayFilter. This works until a certain point after which the token stops refreshing. And the weird thing is on inspection I found that the token that expired was not the token used just previously, it was the first token that was issued on program run.
Any help would be appreciated.
Answer: Access tokens have some expiry time. But you can configure the access tokens expiry time in service provider level.
Please refer to this document to configure access tokens and refresh tokens expiry time in service provider level https://docs.wso2.com/display/IS580/Configuring+OAuth2-OpenID+Connect+Single-Sign-On
The refresh token grant can be used when the current access token is expired or when a new access token is needed. The refresh token is renewed when the refresh grant is used to get an access token. In WSO2 Identity server, A new refresh token is issued for refresh grant with a new expiry time and the previous refresh token is made inactive and can no longer be used.
There is a RenewRefreshTokenForRefreshGrant configuration to disable this property in the identity.xml file. By default this RenewRefreshTokenForRefreshGrant config in enabled true.
Due to this for each refresh grant request, the previous refresh token becomes inactive and a new refresh token will be issued with a new access token. You can disable the config in identity.xml file as below if you don't want to renew refresh tokens per request. Due to this RenewRefreshTokenForRefreshGrant property, if you use a previously issued refresh token will become inactive and you can't use that refresh token to refresh to access tokens further.
<RenewRefreshTokenForRefreshGrant>false</RenewRefreshTokenForRefreshGrant>
Please refer to this document to refer further about refresh tokens https://docs.wso2.com/display/IS580/Refresh+Token+Grant
Comment: Hi thanks for the reply. I have actually implemented both of the above. My problem is a specific case. In the normal scenario after a access token expires a new one is generated using refresh token, This works for me a few times after which it stops working. It says access token expired and something odd that happens is instead of failing with the last token it fails with the first token that was issued in this run.
Comment: Access_token will expire after the expiry time. That is expected behaviour. Can you please elaborate this "It says access token expired and something odd that happens is instead of failing with the last token it fails with the first token that was issued in this run" ?
Comment: Lets say I have a access_token expiry time of 5 minutes. Also i should mention I am using a self contained access_token.So I get a access_token "A" and refresh token "B". After 5 minutes it uses refresh_token "B" to generate a new access_token "C" and refresh_token "D". After 5 minutes it uses refresh_token "D" to generate a new access_token "E" and refresh_token "F".It goes on like this, working perfectly, till after some time it stops working. And in the chrome browser I can see "Access token expired : A". Not the last token but the first one.Also there is no error on the console..


Playing on a number of tables at one go

Question: It is usually advised to play on more than 1 table at 1 go to make reasonable money on poker. On the other hand, playing on too many tables can result in lack of focus needed on each table. How many tables is it advisable to play on at 1 go, in order to have positive outcome? Is there a particular number of tables professional poker players play on at the same time?
Answer: Really there can't every be a fixed number or fixed answer for this. The simplest answer is what ever number you can play without having a negative effect on your game and your win rate.
Don't look to emulate some of the online wizards who grind 20+ tables, you will get lost and lose money. Start off playing one, open a second, if it goes fine and doesn't impact you, open a third, a fourth, etc, etc. The moment you find an extra table effects your game negatively, well then you've found your answer on the number of tables you can play.
The number of tables you can handle will never be static either. You'll find lower stakes you could play an extra few, while tougher levels will require fewer tables. Also the more you play and work on your game the more tables you'll likely be able to add too. So just keep that in mind.
I suggest finding the number of tables you can handle either with play money or at the micros.
Answer: It depends of the type of games you are playing and the familiarity you have with it as well. The difficulty of the decisions you have to make should also affect the number of tables you play. Also the time you have between each decisions will affect it. So at cash game in NLHE or tournaments or speed poker you will generally play a different number of tables. aAs you improve more and more decisions will be automatic and knowing what to do becomes almost like a reflex. Basically if you have time to do something else while you play (browsing forums or the internet in general, or anything else), it is generally time to open more tables.

Playing on a number of tables at one go


How can I achieve this print look for a Bookcover?

Question: i'm currently designing a book cover and try to find a possibility to produce a golden square like in the picture, with text on it. I think in the picture they used a sticker. How can I achieve this Print effect, with a hot foil stamping in gold and the text basic in black without finishing?
Comment: UV varnish is transparent. If you do a UV varnish over a gold coloured ink, it will look shiny. It won't look metallic though. Also I don't think overprinting gold foil would work at all. Speak to your printer before you try something experimental. Might not be possible, or end up being too costly.
Comment: Hi. Welcome to GDSE. I don't think that's an example of hot foil work, it's also not gold.  Possibly spot UV varnish?
Comment: @BillyKerr thank you for your answer. I have already looked into UV Varnish. So far I have only found examples in which clear Varnish was used over normal ink. How can I get this shiny look? (It doesn´t matter at this point wether its gold or silver, tough :)
Comment: Although your image might not show an example of hot foil work, I think a nice result could be achieved with golden hot foil. What do you mean by "without finishing"?
Answer: There are printable foils out there, but it requires a special ink usually.
However, if hot foil stamping is an option, you could have the printer foil stamp the gold block, then come back and use a black pigment foil to foil stamp the words.  Make sure you check with them that none of the text is too small for a die.


Printer keeps pausing during prints

Question: I'm having a rather strange issue in that the printer pauses randomly during prints. This was first noticed when printing a large and complicated model (from SD). Since then a 20 mm  calibration cube has been printed (from USB) with no issues (paused briefly after printing the skirt), it was used in conjunction with M92 G-code command to calibrate the stepper motors. Now i'm trying to print a temperature tower but it pauses multiple times even on the first layer (tried USB and SD). How do i ensure that the printer does not pause during prints?
Here's a link to a video of the printer pausing (trying to print a temperature tower).
Slicer details:

Ultimaker Cura 4.0
Also tried slic3r 1.3 32 bit
The temperature tower had a change temperature at Z injected into the G-code but none of the others did.

Printer details:

Anycubic S
Firmware 1.1.2
filament PLA+
firmware reprap/marlin
Comment: Is the printer equiped with a optical filament sensor? I had similar pauses on a different printer and it was caused by filament sensor sencing false alarms.
Comment: The printer does not have a optical filament sensor
Comment: Does your gcode have multiple M109 (wait for new temp) commands?
Comment: It does but the temperature on the printers screen does not change to indicate that it's trying to change temperatures
Comment: Could it be the z-hop bug (at speed=c, yes c the speed of light)? It was a ridiculous cura bug.
Answer: I had a similar problem some time ago. The problem in my case was, that in the slicer I had a minimum layer time, which was pausing the printer until that time had passed. I can't remember the name of the settings and I'm not at home right now, but the slicer can either print full speed and pause or it can slow down, to maintain a minimum layer time. I guess the idea is to give the plastic time to cool down.
Comment: Do you mean **Cooling Thresholds > Slow down if layer print time is below:** ?
Comment: Yes. That will slow down. I have also seen that it will print full speed and wait. There was also a setting to make the hotend hover above the object. The problem is, that if you slow down, you will have the hotend melt the already printed parts. A pyramid, for example, will get the tip melted, if the hotend slows down. In that case, it would be better to print at full speed and move away and wait
Comment: See https://ultimaker.com/en/resources/52674-cooling in the "Lift head" paragraph. Quote: "[...]The print head will wait until the minimum layer time has passed and will then move back to the print to continue[...]"
Comment: This setting should just be removed, imo. What happens is melted filament will leak out of the nozzle during the wait time, and when you resume it's gonna end up in a bad place. Plus, the now the nozzle is short filament that will be needed elsewhere. Instead, they should estimate the time to print the layer, and if it's less than the minimum layer time scale the speeds (except travel speeds) appropriately so meets that layer times. I know some slicers have options to already do this. But this "wait in one place" setting is just not helpful and should be removed.
Comment: Yeah. Scaling the time exists as well. Problem is, that if the head moves too slow, it will heat up the part that is already printed. I usually print other parts at the same time, if that seems to be a problem or I just add a tower. Somewhere I have an STL of a 10mm diameter disc with 1mm height and I just scale it to the same height as the object I'm printing. The time spend printing this cooling tower is usually enough

Printer keeps pausing during prints


How do I use a button to toggle between highlighted in non-highlighted images in a UIImageView object?

Question: I am trying to use one button to toggle between two images in a UIImageView object.  I've been searching around for a problem/answer close to my specific problem but have had no luck, so here goes...
<code>MainViewController.h
 IBOutlet UIButton *selectionButton;
 UIImageView *selectionStatus;
 @property(nonatomic, retain) UIButton    *selectionButton;
 @property(nonatomic, retain) UIImageView *selectionStatus;
 - (IBAction)selectionButtonPressed: (id)sender;

MainViewController.m
 @synthesize selectionButton;
 @synthesize selectionStatus;
 - (IBAction)selectionButtonPressed: (id)sender
 {
   if (selectionStatus.highlighted == YES)
      selectionStatus.highlighted = NO;
   else
      selectionStatus.highlighted = YES;
 }

 -(void)viewWillAppear:(BOOL)animated {
   // selection images
   selectionStatus = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"NotSelected.png"]
                                       highlightedImage:[UIImage imageNamed:@"Selected.png"]];
 }
</code>
When I run this in the simulator, the images don't appear in "SelectionStatus", so I think I'm missing something big.  Help would be very much appreciated.  Thanks!
I should have said more about the button.  selectionButton is a custom button that overlays the selectionStatus imageview as well as a label that displays the results of a calculation.  Tapping selectionButton selects the value in the label for inclusion in a calculation of an average of up to three selected label values.  The image displayed in selectionStatus is a checkmark that changes colors depending on the selection status.  I've found another approach that works, but doesn't seem as elegant:
<code>- (IBAction)selectionButtonPressed: (id)sender
{
    if (toggleValue == YES) {
        toggleValue = NO;
        UIImage *unselect = [UIImage imageNamed:@"NotSelected.png"];
        selectionStatus.image = unselect;
    }
    else {
        toggleValue = YES;
        UIImage *select = [UIImage imageNamed:@"Selected.png"];
        selectionStatus.image = select;
    }
}
</code>
This approach does not require the use of viewWillAppear, and initialization is done in viewDidLoad.  Thanks for your comments and suggestions.
Comment: What do you mean the images `dont` appear?..after `initWithImage:` does it set as `NoteSelected.png`?
Answer: In ib , set the button's background image property.I think your image is very big,in fact it is show on your button when you set your button's state into highlight.
Just try,I am not sure.


Capture console.log to a file?

Question: I'm running a Chrome kiosk and want to retain and aggregate logs. The Chrome docs say that you can use --enable-logging --v=1 and it will log the to the chrome_debug.log file. Doing this creates a log, but it's all system-level stuff within Chrome and not the site-specific logs. Is there a way to capture all my console.log/info/warn statements straight to a file? Or do I need to engineer something?
Answer: As long as you are in control of the code, or you can inject code, you can override console.log to collect the messages.
var aggrLog = '';
(function(){
    var oldLog = console.log;
    console.log = function (message) {
        aggrLog += Array.prototype.join.call(arguments, ' ') + '\n';
        oldLog.apply(console, arguments);
    };
})();

This will aggregate all the console.log in variable aggrLog.
credit: https://stackoverflow.com/questions/11403107/capturing-javascript-console-log
Modern browsers allows to save local data to file, you'll need a trigger to decide when to save. see more here:
https://stackoverflow.com/questions/21012580/is-it-possible-to-write-data-to-file-using-only-javascript

Capture console.log to a file?


Where is my dividend?

Question: I have been holding a stock "ASG" for almost one year. By the  ASG Dividend History, I should get some dividend.  I checked my account on TD Ameritrade, it only shows my share amount. But where can find the dividend amount?
Comment: Did you look at your detailed transaction history for the periods that include the dividend payout?
Comment: Did your number of shares change post-dividend? You might be setup as a DRIP.
Comment: oh, I found it under history and statement if I set view date range. But how to get the dividend?
Comment: History & Statements
Transactions
Select dividends
Time range
Done
Answer: Your dividend should show up in one of a few methods:
(1) Cash in your trading account
(2) A check mailed to you
(3) A deposit to a linked bank account
(4) As additional new shares in the stock, as the result of a DRIP setup.
Comment: Although the question was about a particular instrument paying cash dividend I should also add that companies sometimes pay dividends in either stock or stock options (usually as warrants in the UK) for internal corporate reasons so even if you don't have a DRIP set up a dividend could automatically increase your stock quantity.

Where is my dividend?


Buoyancy and non-horizontal water

Question: With horizontal water and downwards gravity, buoyancy pushes upwards. However, in which direction is it pushing (say a boat) when the gravity is downwards but the boat is on a non-horizontal ground with water flowing down ? Is the bouyancy still pushing in the opposite direction of the gravity or along the normal of the water plane ? (Regardless of drag forces).
Thank you.
Answer: Think of it this way. In a non-accelerating reference frame, gravity is the only acceleration that creates body forces that balance the pressure force of buoyancy - created by the so called 'displaced' volume  - the volume that is cut off  below water by the water plane.
But if the body of water that supports the vessel is in a moving, accelerating frame, that acceleration adds with gravity to create a new net acceleration - probably in a different direction. Same equilibrium applies for this new acceleration as for the first case, gravity by itself.
At static equilibrium the buoyant force vector is colinear with the body force vector that both act through the center of pressure (buoyancy) and center of mass, equal to one another but in opposite directions.
But if you roll or pitch the vessel, the forces are no longer colinear. This creates a moment that causes the ship to oscillate. If it's stable, it settles back to the equilibrium condition. If not - watch out.
Comment: Thanks for your help. I'm coding a video game that simulates buoyancy and drag, but not water flow (that would be overkill), so the water doesn't actually flow and push the boat, it only resists to the boat movements. So my questio was only about buoyancy alone, ignoring other forces. After some experimenting it seems like the buoyancy has to push along the normal of the water plane.
Comment: That's right and the water plane at steady state aligns itself with net acceleration
Answer: Interesting question.  I think this is the answer:  In the case where water is flowing downhill and the surface of the water is slanted (not perpendicular to gravity), the buoyancy force acting on the boat from the water would be perpendicular to the water surface, and not parallel with gravity.  This will result in a net force on the boat in the downhill direction, so the boat (and the water) will accelerate.  Eventually (assuming a finite depth of of the river or whatever is channeling the water) viscosity of the water will limit it's velocity so that it doesn't speed up forever; and the water will in turn drag on the boat, creating a balancing force that limits the boat's downhill velocity to some finite value.
Comment: Thanks for your answer. I've tried to see what it does with the buoyancy being in the opposite direction of the gravity, and as it could be expected, the boat wasn't going down along the water plane. So I guess you're right, it should be along the normal of the water plane.

Buoyancy and non-horizontal water


SVM parameter selection

Question: Are there better alternative methods to choose C and Gamma which yield better training performance?
Answer: Grid search is slow as it spends a lot of time investigating hyper-parameter settings that are no where near optimal.  A better solution is the Nelder-Mead simplex algorithm, which doesn't require calculation of gradient information and is straightforward to implement (there should be enough information on the Wikipedia page).  There may also be some java code in the Weka toolbox, however I work in MATLAB and haven't looked at Weka in any great detail.
SMO is an algorithm for finding the model parameters, rather than the hyper-parameters.
Comment: There is one here http://theoval.cmp.uea.ac.uk/matlab/#optim but if you already have the optimisation toolbox then fminsearch is also an implementation of the Nelder-Mead method IIRC.
Comment: Could you provide your matlab implementation?
Answer: The Nelder-Mead simplex method can involve as many function evaluations as a simple grid search. Usually the error surface is smooth enough close to the optimal parameter values that a coarse grid search followed by a finer one in a smaller region should suffice.
If you're interested in gradient based optimization of C and gamma, there are methods like optimizing the radius-margin bounds or optimizing the error rate on a validation set. The computation of the gradient of the objective function involves something like one SVM train but a simple gradient descent may involve only a few dozen iterations. (Look at http://olivier.chapelle.cc/ams/ for an article and a Matlab implementation.)
Comment: In my experience, nelder-mead is usually faster than grid search and gradient descent is only slightly faster becuase while it takes fewer iterations, the cost of computing the gradient is high.  So if you have an implementation that provides gradient descent then use it, but Nelder-Mead probably won't be far behind.  Of course as soon as you have more than two hyper-parameters to tune grid search immediately becomes much the slowest method.  It would be interesting to see an study of the comparative efficiencies of each method.
Comment: You're right that if the number of parameters is more than a couple, grid search is not viable. But the same is true of Nelder-Mead, because the size of the simplex is determined by the dimensionality.
Comment: only to the same extent as for gradient descent, adding an extra dimension to the problem only adds one extra point to the simplex, so like gradient descent it scales roughly linearly in the number of hyper-parameters.  I've used it with problems with 40+ hyper-parameters and it is only slightly slower than gradient descent (you tend to get over-fitting in model selection either way though with that many hyper-parameters).
Answer: Here is an entry in Alex Smola's blog related to your question
Here is a quote:

[...] pick, say 1000 pairs (x,x') at random from your dataset, compute the distance of all such pairs and take the median, the 0.1 and the 0.9 quantile. Now pick λ to be the inverse any of these three numbers. With a little bit of crossvalidation you will figure out which one of the three is best. In most cases you won't need to search any further.

SVM parameter selection


What is the difference between pip and conda?

Question:
I know pip is a package manager for python packages. However, I saw the installation on iPython's website use conda to install iPython.
Can I use pip to install iPython? Why should I use conda as another python package manager when I already have pip?
And what is the difference between pip and conda?
Comment: http://stackoverflow.com/questions/20994716/what-is-the-difference-between-pip-and-conda
Answer: conda is a package management tool for installing scientific and analytical computing packages, which may be written in Python or other programming languages. conda also creates a virtual environment, like python-virtualenv does. conda is the package manager of Anaconda. Anaconda is a free Python distribution provided by Continuum Analytics, which includes over 195 of the most popular Python packages for science, math, engineering and data analysis, almost all of which are also available in the default Ubuntu repositories. Anaconda is known for making a big mess out of the default paths for many Python modules in Linux. For these reasons I don't recommend Anaconda for new Linux users.
pip is a general purpose Python package installer. In addition, python-pip and python3-pip are in the default Ubuntu repositories. In most cases you would choose pip instead of conda if you want to install a Python package management application. To create a Python virtual environment without installing conda, you can install python-virtualenv from the Ubuntu Software Center.
pip and conda use different packaging formats so they do not operate interchangeably, but you can use both tools side by side.
Comment: Footnote: Yes, you can use pip to install the latest version of IPython. I've done it myself. It's not necessary to install conda in order to install the latest version of IPython.
Answer: Conda is a Python Package Manager.  One of the advantages of using Conda is it can create and manage multiple environments having different versions of Python installed.  (I don't know if pip can do that.)


Do I need transit visa for Switzerland while traveling to Tanzania as Bangladeshi (US permanent resident)?

Question: I will be traveling to Tanzania on 10 November 2019 and my transit for two hours in Zurich, Switzerland and will be back on 17th
Answer: Only citizens of a few countries need transit visa for airside Schengen transfer. Those are: Afghanistan, Bangladesh, Democratic Republic of the Congo
Eritrea, Ethiopia, Ghana, Iran, Iraq, Nigeria, Pakistan, Somalia, Sri Lanka, Syria
and Turkey, and only whey they do not have meet one of the requirements to get an exception of this rule.
Although as a citizen of Bangladesh you'd normally need an airport transit visa, since you are the permanent resident of US, you are exempt from this requirement. See
https://www.eda.admin.ch/countries/usa/en/home/visa/entry-ch/transit.html and https://www.sem.admin.ch/content/dam/data/sem/rechtsgrundlagen/weisungen/visa/bfm/bfm-anh01-liste2-e.pdf
Comment: Beware that each Schengen country has its own additions to the common Schengen list of nationalities that need airport transit visas. They don't all match Switzerland's shown here.


Does ellipsis button work on left or right of table row?

Question: The problem here is that the ellipsis button in this table is placed to the left. I inherited this pattern, but I am not sure if it's correct, practical, or normal. From other things I've seen, the convention of putting an ellipsis to the right is most common. Thoughts?
Comment: What does this button do?
Comment: This button is a dropdown menu that has multiple options such as Edit (or View), Delete, etc. Secondary actions. Users can double click on the row itself to view it, but this button is mostly used for accessibility requirements (keystrokes).
Answer: Terminology
Luke Wrebowski calls this menu choice the meatballs menu.

Note that your buttons don't look like ellipses (•••); They look more like square dots (▪▪▪). I would fix this to ensure that users who have seen this pattern before can immediately get it.
Placement
You should take a look at this question for some ideas on how to think of placement of controls. But the short answer is to place it where the user reading the row would expect to find it. I have a strong urge to see this menu aligned to the right as you do. Firstly, it would group the Action and Menu controls, thereby keeping data separate from controls. Secondly, it aligns with left-to-right reading, so users would get to the controls after they have read the data in the row.
As a counter-example, below is a data-table from the webapp for the Deezer music streaming service. Each row has four controls including the meatballs menu and they are positioned on the sides of the row as well as within the row. I use this UI daily and I think it works.
Comment: Andre so you believe that the meatball menu should be placed on the RIGHT of the row, correct? Such that the user is reading the row left to right and then making their decision.
Comment: @bobbyo23 yup. I mistakenly wrote left instead.
Answer: The ellipsis (or meatballs, like Andre said) should be to the right, and there's an explanation for this, it's not like some day someone said "let's use meatballs as menu icons", but it's a legacy UI from a known convention.
See the images below:

Did you notice the pattern? Just in case: any action that has a further action has an.... ellipsis. A real one, not an icon.
From Ellipsis

The ellipsis is used as an operator in some programming languages. The precise meaning varies by language, but it generally involves something dealing with multiple items

Thus, as any ellipsis, it's placed on the right of the associated element, because you simply can't display a set of associated child actions before you show the parent, it simply makes no sense!
And the same way, when you use the icon, it should be to the right, because it means there's an action associated to that element.
All the above being said, I'm not sure that placing of the icon is the biggest problem in this UI. Just think about this: what if you want to delete several items at the same time? Opening a menu, selecting Delete, confirming it and starting again with the next item is very bad UX, so I'd suggest to replace that icon with a checkbox and add a set of global actions whenever possible, then the ellipsis on the right of the item, just like in Andre's example.
And if the actions aren't more than a few, then repeat icons per row. For example, I think just an edit icon should suffice and it will be way more clear than an ellipsis, no matter where it's placed
Comment: Interesting, I've never made the connection between this icon and the sub-menu hints from desktop applications.
Comment: Thanks Devin. In our particular use case, users do not need the option to delete multiple selections. If that were the case, then I would agree that a checkbox would be necessary to solve that problem. I do like the idea of just an "edit" icon with a label is best to make it obvious what actions are available. Thank you!


Vanishing save files that both do and do not exist on hard drive for Mac, what is happening?

Question: I am using the latest version of Blender (2.90.1) on an iMac pro with Catalina 10.15.7. I'm working on a project and I save the file in a save folder i created under the file name "Relax.blend" all goes fairly well for a while and then one of the random times i figure i better save it there is a message in the blender info strip in red that says file type not supported, file saved with an "@" symbol. When i check my finder window for the save file there is only a file named "Relax.blend@" there for a save file. If i close and reopen blender and try and select the original "Relax.blend" save from the blender startup window it prompts with an unsupported file type and the option to select it again disappears (same goes for the drop down menu of opening recent files under the file menu once starting a new file; the option disappears) if I return to my finder window and attempt to remove the "@" and turn the extension into a .blend file it tells me there is already a file under this name. if i change the name to "Relax1.blend" it will then open properly again... then, sure enough, the same thing happens with the "Relax1.blend" file and again when i try "Relax2.blend". I have tried using search on my entire hard drive for any of the relax .blend files but they dont show up, and i can no longer name any files those names because they are already taken. so the computer both does and doesn't recognize the files existence; i cant name files the same name but also can't find the files to delete them. What the heck is going on!? Where are these files? at this rate im going to have 500 versions of the same file that i can't see using up space and names on my computer. does anyone have a solution or idea what is happening?
Comment: Is it just that file, or does it happen if you start a new (different) project and save that as well? What happens if you save it in a different folder?
Comment: i have tried saving it in a different location with no luck. I failed to mention i was using the sequence editor and using hundreds of stills; if that makes a difference. I have found out something new, but still no explanation why its happening. turns out the file folder containing all data for the .blend file has the .blend hidden somewhere within. when i delete the entire folder and its contents i am able to rename the .blend back to the original name. I cannot find the file anywhere within it and still can't reload it normally, but i can at least delete it.
Comment: Yeah, that's strange. Unfortunately, I don't have a mac, so I'm unable to test anything. I was hoping it had something to do with that file itself, but based on your comment, it almost seems like it might have more to do with the filesystem. How does macOS hide files - is it possible some settings related to the folder are causing the save files to be hidden or otherwise restricted somehow ? Does @ mean anything specific with regards to Catalina file extensions, or is this the first time you've seen anything like it? Any chance you created the folder in a read-only partition?
Answer: Solved, sort of. Blender version 3.0 seems to have resolved the issue. It no longer happens.


Two part tariff-income effects

Question: I understand that a monopolist optimally sets a two-part tariff by pricing at marginal cost and then extracting CS as a lump sum fee.  However, wouldn't the lump sum fee induce a decrease in consumer demand?  Is the typical assumption made in two-part tariff problems that there is no income effect?
Answer: In the typical textbook treatment of the two-part tariff model, income effect is ignored/assumed to be negligible.
Goldman, Leland, and Sibley (1984) provided detailed comparison of optimal nonlinear pricing strategies (of which two-part tariff is a special case) with and without income effect. Their conclusion is

When [income effects] are absent, the aim of optimal pricing is to raise a given amount of profit with the least possible distortion to individual's consumption plans. Therefore increments of consumption are priced according to the  Ramsey rule. When income effects are present, the optimal pricing problem is equivalent to the optimal taxation problem, and the goal is optimal income redistribution.
Answer: In typical textbook two-part tariff there won't be an income effect because either demand will be just assumed to be given by some function or even if there is a utility function it won't feature budget constraint (e.g. see examples of such simple problems in Belleflamme and Peitz, Industrial Organization: Markets and Strategies pp 227-234).
In such cases the tariff won't have any effect on demand itself because it is a sunk cost from the perspective of consumer. Such assumptions are appropriate for items that are not too expensive (e.g. razors, club entry etc.) but would not be appropriate for more expensive items.
However, once we add budget constraint then there will be income effects and demand will be affected by the fee. For an example of such model you can see Ng & Weisser (1974). This changes the problem somewhat but authors show that even in such cases two-part tariff is still often optimal.

Two part tariff-income effects


What is swapfile and swapspace?

Question: I have made a swapfile following How to add swap space on ubuntu 16.04. And I have 4GB swap memory available. But in my root directory I see 2 file with names swapfile and swapspace , both occupying 4.3GB of space. I don't understand why 2 files are there. I guess there should be a swapfile but what is this swapsapce then?
Answer: You have 2 files probably because you created 2 files.
You can have a swap partition(recommended) or a swap file, or a combination of swap partitions and swap files.
Here you can get some more info:

Swap Partition VS File For Performance
Ubuntu SwapFaq
Wikipedia Paging Article
All About Linux Swap Space
Comment: I have been through the links. And I didn't created 2 files, I just followed the instructions that were given in link i mentioned. Even if i check  `sudo swapon -s`, I see `Filename: /swapfile  Type: file  Size:4194300  Used:52736  Priority:-1`. I don't see 2 swapfiles or swapportion. Also, how can I be sure which one should i delete and which I should keep.
Comment: From your output above, you can delete the second file `swapspace`  because this file is unused.
Answer: Do you have a program called swapspace installed?
# dpkg -l swapspace
Desired=Unknown/Install/Remove/Purge/Hold
| Status=Not/Inst/Conf-files/Unpacked/halF-conf/Half-inst/trig-aWait/Trig-pend
|/ Err?=(none)/Reinst-required (Status,Err: uppercase=bad)
||/ Name                     Version           Architecture      Description
+++-========================-=================-=================-===============
ii  swapspace                1.10-4ubuntu3     amd64             dynamic swap space manager

I see from the man page swapspace(1) that the default location for a swap file made by swapspace is /var/lib/swapspace, but the -s option lets you put it anywhere.
As long as swapon -s only shows that /swapfile is in use then the file /swapspace is not.
And if you have swapspace installed, and don't intend to use it, it should be safe to remove it with apt-get remove:
# apt-get remove swapspace
Reading package lists... Done
Building dependency tree
Reading state information... Done
The following packages will be REMOVED
  swapspace
0 to upgrade, 0 to newly install, 1 to remove and 5 not to upgrade.
After this operation, 122 kB disk space will be freed.
Do you want to continue? [Y/n] y
(Reading database ... 139800 files and directories currently installed.)
Removing swapspace (1.10-4ubuntu3) ...
Processing triggers for man-db (2.7.5-1) ...

Comment: I just deleted the swapspace file manually, I changed it's permissions and used `rm -rf swapspace`. Hopefully,I did't removed it in wrong way. After restarting my memory was free by 4Gb. ANd now if I check `dpkg -l swapspace` , I see : `dpkg-query: no packages found matching swapspace`
Answer: Try running swapon -s as root.   This will show you what swap is being used, and whether its a partition on a disk or a swapfile on a filesystem.
A 4GB swap partition as primary, and a 1GB swap file in /mythtv1/ as last resort.
 backend:~# swapon -s
 Filename           Type            Size    Used    Priority
 /dev/dm-4          partition       4194300 178756  -1
 /myth1/swapfile    file            1048572 512     -2

You can have multiple swapfiles and swap partitions, with equal or different priorities.  That is, you may choose to swap to two swap files, one on each of two fast hard drives, and have a third swap file with a lower priority as a backstop on a slower drive.
Effectively swapfiles and swap partitions function the same, its a matter of convenience which you choose.
Swapfiles don't need backing up, but they should have permissions to stop anyone but root reading it.

What is swapfile and swapspace?


How to Convert Property Survey Monument Marker Coordinates to State Plane?

Question: I have a set of Connecticut monument locations (a monumented property survey using monuments designated CGS Mon. No. 1577-1580).  Their coordinates appear to me to be in meters (244536.84 N, 386409.67 E).  I need them in CT State Plane feet (correct values should be approx.746000 N and 786000 E).  Simple meters-to-feet misplaces them, as does creating them in UTM and re-projecting to state plane.  The original data specifies no coordinate system or datum.  Everything I can find online anticipates the markers will be in geographic coordinates, but they're clearly not.  Does anyone know what these generally start in or how to convert them? (offset is about 10 miles and not just shifted, if the markers ought to be along the road -- not sure if so)
Comment: I've been involved in some large-scale engineering projects where using the wrong definition of foot caused significant movement of points. Your comment "Simple meters-to-feet misplaces them" causes me to wonder if this may be the case. How much is the displacement? Are you using the conversion for "US Survey Foot" or "International Foot"? When dealing with SPC you will notice a difference.
Comment: Note for others: I used [US NGS programs](http://www.ngs.noaa.gov/TOOLS/spc.shtml) to look up the state plane zone, then to test converting the xy points to lat/lon.
Comment: I've been letting GIS make the conversion, so I don't know.  I've tried transforming CT state plane meters to CT state plane feet.  I just went and hand converted survey feet to feet (using conversion factor 1.0000019685039), but it only changed the location by about 2 feet, not ten miles as needed.  It was a good idea, though.
Comment: I'm at home today, with no ArcGIS access. Do these values make sense: 41 30 6.60076 N,  73 31 47.02927 W? If so, it's NAD 27 state plane zone, so already in US survey feet.
Comment: Ha!  Perfect!  @mkennedy, you got it!  I was too discouraged to try NAD 27 and I figured, since the survey was from 1995 it wouldn't be.  Wow!  Thank you!
Answer: Without knowing the projection, or datum of the monuments, the only way to transform them correctly would be through a great deal of trial, and error.
Given the latitude of the area you are in, the Northing is too small to be in UTM meters, or feet, unless the first digit of the coordinate has been removed for space saving.
Knowing where the monuments are on the earth, download imagery for that area.
At least you will know the datum, and projection of the imagery.
Manually plot the locations of the monuments over the imagery using the same datum, and projection as the imagery.
Now comes the fun part.
Transform the monument data from its current projection, to its desired projection, see if you are in the ball park.  If not, try again with another target datum and projection.
It may also be necessary to transform from say a Cylindrical Mercator projection (Virtual Earth, etc) to NAD83, then from NAD83 to something else, say NAD27 using NADCON transforms, otherwise it is possible that the transformation is using Molodenski, or Snyder formula, and you could have significant differences.
Another solution would be to use Corpscon 6.
Then work backwards, taking the approximate values of what the points should be in CT State Plane Feet, knowing the datum, and projection, transform them to NAD83 Meters, Feet, UTM, and so on seeing if you find something that is within reason. Look past the first digit to see if the rest of the coordinate values are close. (Surveyors have been known to localize, or drop digits to save time, and reduce keypunch errors when doing calculations.)
I wish you all the best with this. This type of detective work can be incredibly frustrating.
Answer: From mkennedy, the answer: "it's NAD 27 state plane zone, so already in US survey feet . . .  I used US NGS programs to look up the state plane zone, then to test converting the xy points to lat/lon."


How to delete only a specific version of a question?

Question: I want to delete a specific version of a question, but deleting the question itself is not an option. Do you provide any such functionality?
Comment: You can make any particular version the current version, which is what people will see if that's what you mean. That's mainly used to revert bad edits.
Answer: We do, but you'll need to contact us directly. Let us know:

The link to the question itself
In what revision something was posted that shouldn't have been
What that something is so we can look for it
That you're certain the current version of the question is acceptable

We may, depending on which revision it is, have to delete all but the current revision, because diffs kinda break when you take chunks out of the middle. But, the first step is to contact us.
This is (for now) a hard delete, so it's only available to employees. We hope to make this 'soft' so that we can extend the functionality to moderators, but that's proven to be a bit more difficult than anticipated.


GPU option is missing in Blender 2.69

Question: I am new learner of blender3D. I installed Ubuntu two days ago, and was struggling installing my nvidia and blender.Finally I managed but now I have a problem.
My Preferences > System does not show GPU option, (Once it worked yesterday but I ruined my installation trying to fix left mouse button issue).
What I want to achieve is run blender and get the utmost performance of my hardware on Linux Ubuntu
Details of PC : i7 processor, with Nvidia 640 graphic card.
Things I have done are:
Fresh install of Ubuntu and then installation of Nvidia driver
sudo apt-get install nvidia-current

Then I installed blender
sudo add-apt-repository ppa:irie/blender
sudo apt-get update
sudo apt-get install blender

So, now my blender is working, but in the system preference it shows NO CPU and when I go to render view, I don't see the environment texture. Only green and gray colored background. I am in doubt as to whether my Nvidia is properly installed, or cuda installation must be done.
Answer: I found the problem. Now my GPU Rendering is working with one exception. The problem was the Blender package in ppa. I installed directly from the Blender website and now it works, exception is that it's not fast enough. CPU is faster.

GPU option is missing in Blender 2.69


Popular methods for outlier detection (right skewed distribution)

Question: What are the popular methods for outlier detection in univariate data, which do not assume normal distribution?
Comment: Many, many relevant questions on outliers here: have you consulted several? You need to have a specific new question to have much chance of a detailed answer. In any case, I wouldn't put much faith in popularity of methods any more than I judge newspapers by their circulation. Dropping presumed outliers just because they appear inconvenient is, I guess, one of the most popular methods, but arguably the worst of all.
Comment: One possibility is http://stats.stackexchange.com/questions/78063/replacing-outliers-with-mean/78067#78067 In that thread, as may happen here, the answers were wider than the question.
Answer: Generally, you should avoid trimming outliers in an ad hoc fashion and instead use nonparametric or robust alternatives.  A recent review with Monte Carlo studies can be found in Bakker and Wicherts (2014).  At least in psychology journals, Z-score cut-offs were most popular.  Of course, I wouldn't recommend that; the simulation studies in the same article demonstrate that Z-score cut-offs can inflate Type I error rates.
Although the review is focused on independent samples t-tests, most of their recommendations will apply more broadly. They concluded with the following recommendations:

• Correct or delete erroneous values.
• Based on prior research, it is not recommended to use Z scores
  to identify outliers. We recommend methods that suffer less from
  masking like the IQR or the MAD-median rule instead.
• Decide on outlier handling before seeing the results of the
  main analyses, and if possible, preregister the study at, for example,
  the Open Science Framework (http://openscienceframework.org/).
• If preregistration is not possible, report the outcomes both
  with and without outliers or on the basis of alternative methods.
• Report transparently about how outliers were handled.
• Do not carelessly remove outliers as this increases the probability
  of finding a false positive, especially when using a threshold
  value of Z lower than 3 or when the data are skewed.
• Use methods that are less influenced by outliers like nonparametric
  or robust methods such as the Mann-Whitney-Wilcoxon
  test and the Yuen-Welch test, or researchers may choose to conduct
  bootstrapping (all without removing outliers).

References:
Bakker, M., & Wicherts, J. M. (2014). Outlier removal, sum scores, and the inflation of the type I error rate in independent samples t tests: The power of alternatives and recommendations. Psychological Methods, 19(3), 409-427.
Comment: This is good advice, although I would greatly widen the list of approaches worth considering. I note that it does not answer the question as it says precisely nothing about popularity. As I consider that a dubious criterion, I am happy to upvote.
Comment: @Nick Cox, good points.  I've edited the answer. It now makes clear that the cited article  focused mainly on outliers with independent samples t-tests (that's the reason they emphasize Mann-Whitney-Wilcoxon and Yuen-Welch tests.)
Comment: @NickCox thanks for the link, it is quite useful. I have records of n individuals behavior expressed in seconds. I want to apply a clustering algorithm to identify some behavioral groups, however I found that some individuals did other behavioral activities in the recorded time, and therefore the distribution is rightly skewed. I believe that this can affect my conclusions and the whole analysis. Therefore I'm just trying to find a way how to reduce influence of such records.
Comment: You have a more specific question then. So either edit this question or (better, I think) start a new thread. However, I would advise expanding this outline greatly.


Is this the correct equation for Gravitational Potential Energy in Relativity?

Question: If I understand correctly the equation for kinetic energy in relativity is $$ E_k= mc^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right) \;,$$ and the equation for escape velocity in General Relativity is the same as in Newtonian Physics so $v_e=\sqrt{\frac{2GM}{r}}$ and when something is moving at escape velocity the kinetic energy must be equal to the opposite of the Gravitational Potential Energy, so $U=-E_k$.  If I I substitute in $\sqrt{\frac{2GM}{r}}$ for the escape velocity, I get
$$U=-mc^2\left(\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}-1\right)\;.$$  So is this the correct equation for Gravitational Potential Energy in General Relativity?
Answer: This has some correct elements, but there are flaws in your reasoning. You can't get correct results in GR by plugging relativistic correction factors into Newtonian equations.

So is this the correct equation for Gravitational Potential Energy in General Relativity?

There isn't really "the" equation for gravitational energy in GR. It's not an instantaneous action at a distance theory like Newtonian gravity. What you've cooked up resembles the square root of the time-time component of the metric in the Schwarzschild spacetime. The metric is what plays the role of a potential in GR, although we can define a scalar potential in a static spacetime. The earth's field, for example, is not static, because the earth is rotating.
One way to see that your equation can't really be valid in GR is that it predicts $U=0$ when $m=0$, but we know that light rays do interact gravitationally.

the equation for escape velocity in General Relativity is the same as in Newtonian Physics so $v_e=\sqrt{\frac{2GM}{r}}$

This logic doesn't make sense. The derivation of this equation uses $K=(1/2)mv^2$, which is false in relativity.
Answer: In GR, in general, the definition of a gravitational potential energy is meaningless. GR is a geometric theory, so you have to find the analogy with the concept of "potential" in a geometric object. It turns out that the geometrical equivalent of the gravitational potential is the metric tensor.
The metric $g$ is a symmetric tensor of second rank which basically tells you how to "measure" space-time interval. Here the link if you want to deepen the subject: https://en.wikipedia.org/wiki/Metric_tensor.
In GR particles move along specific curves called geodesics, the equation of geodesics is:
$$\ddot x^\mu = -\Gamma_{\sigma\nu}^\mu\dot x^\sigma \dot x^\nu$$ where the gamma factors are called Christoffel symbols, here the link: https://en.wikipedia.org/wiki/Christoffel_symbols.
You see that there is an analogy of equation of motion in newtonian mechanics where $$m\ddot{\textbf{x}}=\textbf{F}$$
and thus you can think of Christoffel symbols as something regarding gravitational force.
Since for the Levi-Civita connection Christoffel symbols are completely defined by the metric and, in particular, by the derivatives of the metric in this way
$$\Gamma^\mu_{\sigma\nu}=\frac{1}{2}g^{\mu\tau}(\partial_{\sigma}{ g_{\nu\tau}} +\partial_{\nu}{ g_{\sigma\tau}} - \partial_{\tau}{ g_{\sigma\nu}})$$ then you can think of the metric as the gravitational potential (the force is the derivative of the potential).
This is just an analogy and of course it has nothing to do with rigorous definitions (except for the geometric part)...


Is there an MKS unit system in which electric and magnetic fields have the same measurement unit?

Question: I'd like to know if there's an MKS unit system in which electric and magnetic fields have the same measurement unit. I couldn't find anything like that on the internet. Does that system exist? Do people use it? What is it called?
Comment: You could invent a unit system that is to MKS what the Gaussian EM units are to CGS - i.e. use the exact same derivational procedure but with m, kg, and s replacing cm, g, and s respectively. It is not SI, of course, but it fits your definition. E.g. the "MKS Gaussian" units would have a unit of charge where that two charges 1 m apart produce 1 N of force instead of 1 cm apart making 1 dyn of force. For "MKS HLU", produce $\frac{1}{4\pi}$ N of force.
Comment: Not sure how many floors go with 14.2 m either.  Now if you give it to me in feet, I've got it since I know there are 8-10 ft per floor. Not that science should convert to feet, only that your argument both for "civilized" and against "natural" are somewhat nonsensical. If people used the units more, then they'd have a better sense for them (presumably at the expense of what they use now, whatever that may be). Grace Hopper, for example, used to handout nanosecond-length pieces of wire so that students would build exactly that intuition. There's a permanent display of one at the USNA.
Comment: @Brick You have a point. Forgive my childishness.
Comment: @The_Sympathizer I know that. I am asking if such a system has been already done/used by someone else, and if it has a name.
Comment: Two votes down? Why is that? Alright guys.. I must say I am a bit surprised. I'll edit and keep only the question per say. I won't kid anymore around here. I apologize.
Answer: It is called natural units. Very simply, all dimensional conversions are set equal to 1:
$$c=\hbar=\epsilon_0=\mu_0=1 , $$
and so forth. It is commonly used in particle physics.


I'd rather + past perfect/present perfect

Question: Help guys
I can't understant what's the difference between I'd rather present perfect and I'd rather past perfect.
For example:
I'd rather have spent this money on the holiday (money wasn't spent on the holiday) = we regret or smth...
AND I'd rather you hadn't rung me at work (suggests that he/she did this call then why I can't use present perfect?)
Answer: I think your example "I'd rather have spent this money on the holiday" is a shortened version of "I'd rather I'd have spent this money on the holiday", in which case the tense following "rather" is expressed by the "(woul)d" and what follows, "have spent this money", is non-finite, meaning there is no tense in "have spent this money".  It is logically a past tense, but a perfect form is the regular replacement in English for a past tense in a non-finite context.  Compare "It's not possible that she ate clams" = "She can't have eaten clams".
In your second example, "I'd rather you hadn't rung me at work", the morphological past expresses something contrary to fact -- a sort of subjunctive -- rather than a real past tense.  Compare "I'd rather you didn't ring me at work", where the sense of "did" is future.
Comment: I think I'd disagree with your first point. "I'd rather have spent this money on the holiday" makes perfect sense and I wouldn't say it's a shortening of anything. Although you could rephrase it as "I'd rather I had (I'd) spent this money on the holiday".
Answer: The construction of a sentence with "rather" depends on whether the subjects of the two clauses are the same or different. If the subjects are the same, the verb takes the present form, whereas if the subjects are different, or the subject is restated, then the simple past is used. For example, compare:

I'd rather leave.

with

I'd rather you left.

Now in the case of a predicate in the perfect tense/aspect, such as "have spent this money on the holiday," have is the verb to which this tense change will apply, so you could say

I'd rather have spent this money on the holiday.

or

I'd rather you had spent this money on the holiday.

Similarly, you could switch your second sentence around, so if you had rung the person you are addressing, you might say

I'd rather not have rung you at work.

Or you could restate the subject (I), and make it

I'd rather I hadn't rung you at work.


In command-line mode, how to avoid delete key erasing last character?

Question: On command-line mode, Delete key erases next character, but if cursor is at end of line, it acts as Backspace, erasing previous character. Can I disabled that?
Answer: Try this:
:cnoremap <expr>  strlen(getcmdline()) == getcmdpos() - 1 ? '' : "\"

This binds the delete key on the :-command line to running an expression and executing the keystrokes the expression returns.  In this case I'm using a ternary operator to return nothing if the length is the same as the cursor position, and returning the delete key otherwise.


Verb forms with "I"

Question: I am totally confused in the usage of pronoun "I". We use the verb "have" with "I" not "has" though it is singular. "I" is singular but again we don't use (s, es, ies) with the verb while making the sentence of present simple tense.
Comment: That's right. Yes it's odd. What's your question?
Answer: The inflection with -s is used only in the simple present and only with the third person singular, not with the first person (the person speaking) or second person (the person addressed).

{I / we / you / they }  {have / run / push / &c}
    {he / she / it}   { has / runs / pushes / &c}

One verb, BE, employs an inflection with -s (was) in the simple past for both first and third person singular; but BE is quite different from all other verbs.
A few verbs, the modals can/could, may/might, must, shall/should, will/would are defective: they have no  inflection with -s, even in the present third person singular. They also lack participles, and the 'plain' form is never used as an infinitive or imperative.
Comment: Thnx but I am confused with the pronoun "I".       Why we use has with he, she, it, singular name?? Answer is that the subject is singular, so my question is that the pronoun "I" is also singular, so why we use have with pronoun "I"?
Comment: Thnx I got it. :)
Answer: Thinking about 'singular' and 'plural' pronouns doesn't really help you here. In English, meaning (including 'singular' and 'plural') doesn't always match up with grammar (including verb agreement).
Thinking about 'singular' and 'plural' may help with regard to {she/he/it has} (singular) and {they have} (plural). It certainly doesn't help in thinking about {you have} (singular) and {you have} (plural) - 'you' takes the 'plural' verb whether I am talking to one person or more than one person.
And it doesn't help with 'I' (singular in meaning but plural in grammar (most of the time)) and 'we' (plural in meaning and plural in grammar).
Some words just work differently. StoneyB has mentioned verb BE as working differently from other verbs. In different ways, the pronouns 'I' and 'you' work differently from other pronouns, in particular in terms of verb agreement.

Verb forms with "I"


Can we copyright a product without revealing our identity?

Question: What I want to ask is if we can have product copyrighted without the public knowing our identity (only the government would know) or we can be traced back by our copyrighted product? Example like does GPL can be traced back to my identity? P/s:Sorry, guy with copyright condition like something like you can do anything to my work but leave the sentence on that when you done :'If you like it please support me to donate to me at xxx.xxx.xxx(many way to donate without identity (sorry because I'm wasting your time). Anyone ro im going to accept One of the answe
Comment: One practical issue is that a tax ID number must be made available to anyone who is paying royalties under a licensing agreement. This is easily solved with a company owning the copyright, but if more problematic if a mere pseudonym or anonymous ownership is used.
Comment: Ahh, Banksy bingo.  Guess how many answers before a reference.
Answer: Anonymous/pseudonymous works are still copyrighted.  See, for example, 17 U.S. Code § 302(c), which provides for the length of copyright for an anonymous or pseudonymous work - obviously, they wouldn't have such a clause if these works did not have copyright.  And you don't need to register a copyright to have copyright.
The copyright office has instructions on how to register an anonymous work.  I'm not quite sure how that works; presumably you could keep the author secret, but I'm not sure if you can keep your registration secret.
But even if you can somehow register while remaining anonymous, what if someone infringes your copyright?  What are you going to do - sue them?  You'll be listed as the plaintiff.  Courts usually won't allow anonymous plaintiffs in lawsuits unless there is a very good reason.
One thing you might try is creating a corporation to own the copyright.  If you could somehow arrange the work to be a work-for-hire for the corporation, then only the corporation would appear in the copyright records.  Then it's just a matter of making sure you aren't publicly linked with the corporation.  You could probably get an attorney to help you with that.
Comment: If you register a copyright as a work for hire, be aware that (in the US) the [duration of the copyright changes](https://en.wikipedia.org/wiki/Copyright_law_of_the_United_States#Duration_of_copyright) from life+70 years to 120 years from creation (or 95 years after publication, whichever is shorter). Things are similar in other countries.
Comment: @ErikF : Of course, for most people over 50, that will *extend* the period their heirs own the copyright, and there are only a handful of people who could be personally affected.
Comment: It just that I want to recognize as the person owning copyrighted of my stuff I don't planning to sue anyone I even will register it as general purpose license but I want to be the person owning it
Comment: And be private !
Comment: The government will know my identity but the public don't
Comment: I just have one condition is when I anyone modify or using it must have the word something like: "If you like it please donate money to support me to ..... thanks!" (there is many way to donate money without knowing the sender identity) .
Comment: @AlphaBetA in theory perhaps, but I'd love to know the method in which "donations" couldn't be traced back to you in any way shape or form?
Comment: It sound weird I hate Bitcoin but it one of the example
Comment: Im sorry to inform you, that even BitCoin can be traced back to you; depending on how badly they are trying to trace you, of course.   The Ethereum BlockChain is more secure, at the moment, but eventually even that will be able to be traced.
Comment: presumably an author's lawyer can register a copyright without revealing the author's identity.  the identity would remain known to the attorney and remains secret because of attorney-client privilege.
Answer: What you propose is already the default scenario. Any newly created work is copyrighted, even if there is no explicit author. Registration is not necessary to own a copyright.
The problem is that you can't do a lot with the copyrighted work in that case. You apparently want to share the work, and have conditions attached to its use. This requires a license. A license is a legal contract, and the parties in a legal contract must be identified.
(And as D M notes, what are you going to do if someone breaches the license of violates your copyright?)
Comment: Nothing I just want people to don't sue me and only have 1 condition is add the word I'm the author please donate to me to support me at xxx.xxx.xxx(many way to transactions without knowing identity)
Comment: Like does LGPL(like GPL) can be traced back ?
Comment: @AlphaBetA: From the [FSF](https://www.gnu.org/licenses/gpl-howto.html) : "If several people helped write the code, use all their names". Even with multiple authors, you cannot leave out some names.
Comment: I know only me and can use GPL can the pupublic traced back my identity from the license ? You didn't answer it .
Comment: @AlphaBetA: The point is, you have to list all authors who are granting the license (GPL = General Public License). It's trivial to "trace back" your name when that name must be explicitly stated !
Comment: I'm work in 1 person and that only me
Comment: Does that mean I can be easily traced back by my license?
Answer: In Canada, you don't have to register copyright while staying anonymous, but if you do this, copyright will only extend to the lesser of (publication+50 years) or (creation+75 years) (Copyright Act R.S.C. 1985, c. C-42, s. 6(1)-(2)). Registration is where you'll hit a bit of a problem: The registration form requires a valid address even if you use a pseudonym (so that people can contact you for licensing!) Post office box addresses are acceptable, so you could probably use that. All registered copyrights are searchable.
You will have to make sure that you can be contacted during this time, or else someone can make application to the Copyright Board under s. 77 to get a licence from them (as an "orphan work"), where the Board gets to make the conditions of licence.
How you want to licence your work is your choice: This is a separate matter from copyright registration.
Comment: you can't have a registering agent (an attorney who keeps your identity secret)?  their address would be the point of contact.
Answer: Hypothetical, leaving out a few minor things, but this covers the main points to accomplish the end goal:

I would form a private trust with a business attorney
I would have a 2nd business attorney form a business entity, perhaps a 401c or an S Corp or an LLC, depending on laws in the state the entity is formed in. The business or non profit would be setup as an entity for lawyer 2 in his name, and caused to be filed by only him, and of course he'd be the only name listed on the business entity state docs filled
I would purchase stake in the newly formed business entity, majority share holder, in the company with it being legally binding through a private deal, with a private contract between yourself and the lawyer; which only you and the lawyer retain copies of this agreement
Then I'd go back and have business lawyer 1, set up the private trust with the only listed member being the company name of the company owned and filed for formation of by your other lawyer #2
I'd then have the private trust listed in the registration of the copyright; and the tax ID associated with the business listed as well. Of course the tax ID being obtained by the owner and creator of the company, lawyer #2
Amy royalties or pursuit of copyright infringement can be filed on behalf of the private trust, and the money would come back to you, and your name would never appear, except in 2 private contracts between yourself and 2 separate attorneys
If the lawyers both cannot be retained on an ongoing basis, then I'd have them sign NDAs related to the business they created, but which you own stake in and gave pre-arranged signed contracts on how money made by this company is handled in terms of it all funneling back to you, minus attorney fees, and business associated fees and taxes, etc..

In summary:
The Copyright is now in private trusts name, with a business listed as private trusts sole beneficiary, then that business listed in the trust shows ownership and formation of by attorney; you have a legally binding contract that gives you ownership stake in the company through a private deal and private contract between lawyer 1 and yourself and lawyer 2 and yourself; all money going to trust which has one sole beneficiary, the business you have private and legally binding ownership in.
Not perfect, but close. hopefully you get the gist of this tact


Non Linearity used in LeNet 5

Question: I was looking at the original implementation of LeNet-5 and I noticed a disparity in different sources. Wikipedia suggests that the non linearity used is the same sigmoid in each layer, some blog posts use a combination of Tanh and sigmoid while Andrew NG said it used some crude non linearity which no one uses today without naming it. I looked at the original paper but it's like 50 pages long and the diagram does not mention the activation functions used explicitly. I searched a bit and the sigmoid function was there and mentioned in context of activations while the tanh function is taken as a squashing function. I'm not sure if that is the same or different as then it used other terms when referring to the sigmoid ones. Anyone knows what's up with this?
Answer: In the original paper there are some clarifying statements:

The four inputs to a unit in S2 are added, then multiplied by a trainable coefficient, and added to a trainable bias. The result is passed through a sigmoidal function. (p.7, col.1)

Here, sigmoidal function is generic.

As in classical neural networks, units in layers up to F6 compute.. This weighted sum .. is then passed through a sigmoid squashing function .. The squashing function is a scaled hyperbolic tangent. (p.8, col.1 - but see also Appendix A for details)

Here, "sigmoid squashing function" is used to indicate a scaled "tanh" (remember that tanh is a rescaled logistic sigmoid function).
Therefore, I think Wikipedia's suggestion to use the same "sigmoidal function" is correct. For the sake of precision, the tanh should be used.

Non Linearity used in LeNet 5


What did God Create Genesis 1:1 - Heavens or Heaven?

Question: In Gen 1:1, some translation say 'heavens' and others say 'heaven.' Which one is correct?
Comment: הַשָּׁמַ֖יִם (_hassamayim_) is the dual of an unused singular from an unused root meaning 'to be lofty'. [see Strong 8064](https://biblehub.com/hebrew/8064.htm).
Comment: @Awo Amo since the process is just beginning there is little information at the point of the first verse. My KJV says heaven ( singular ) as the creation continues in the following days more information is given. Since heaven or heavens are infinite your question is not easily answered based on the first verse only. You may have to be more specific on when.
Comment: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
Comment: Paul in 2 Cor 12:2 alluded to idea of multiple heavens(1st, 2nd and 3rd heaven).
And in other passages of scripture(2 Chr 6:18, Ps 115:16, Duet 10:14), same idea is also suggested. As I understand the creation account in Genesis 1:1-, the first verse is a grand summary of what's decscribed in subsequest verses and continuing in Ch.2.
Here's my question: Which heaven did God create in Gen 1:1 - the first or the second or all of them(however many there are)?

Hope this clarifies the question.
Comment: This is like asking what Clint Eastwood said to the helpless man lying on the ground before him, was it either "**Do you feel lucky, punk?** or was it  **Do you feel lucky punk?**  "
Answer: Actually, in Hebrew, nouns have three forms of number - singular, dual and plural.  English only has singular and plural. [Side note: classical Greek had singular, dual and plural but by the time of the NT Koine Greek, it only had singular and plural.]
The word in Hebrew for "heavens" in Gen 1:1, as in all places throughout the OT, is the dual form, הַשָּׁמַ֖יִם (shamayim), literally "two heavens".  [It never occurs in the singular form.]  This noun can refer to any of the following:

the atmosphere or air where birds fly, Gen 1:20, 2:19, Deut 4:17, etc.
the starry heavens, Gen 15:5, Deut 4:19, etc.
the abode of God, 1 Kings 8:30, Ps 2:4, etc.

Note that the Hebrews made no distinction between these three ideas as we do today.
[Side note: In the NT the equivalent word is οὐρανός (ouranos) which can occur in both singular (eg, Matt 5:18, 34, 6:10, 20, etc.), or plural (the most common, eg, Matt 3:2, 16, 17, 4:17, 5:3, 10, 12, 16, 19, 20, etc.  Indeed, the NT writers appear to make no difference between the singular and plural form of οὐρανός]
Comment: A rabbi I knew in my grad school days, commenting on the idea of "the third heaven" (mentioned by Paul), said that in second temple Judaism the tripartite heaven as you indicated above was standard.
Answer: In English, the word heaven can be pluralized, but in Hebrew, there is only one form, and this form ends with the morpheme that generally marks the plural in the Hebrew language. So the Hebrew word שָׁמַיִם does not have a singular form, it only exists in this form, which is why most translators translate "heavens". ׁׁׁׁׁ
Answer: It is heavens, plural. Except the original word was one which doesnt have a singular and plural version. Almost like how we use the word deer. There is only the one word by which we change its meaning thru context. That's the best example I could think of sorry.
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).


How to duplicate monitor in lxpanel?

Question: When I boot Raspberry pi 4b without a desktop (no matter buster or bullseye), the image is duplicated on two displays. If I boot the operating system with a desktop, then the displays show an extended picture. How can I load the Raspbian desktop on two displays with duplication? So that the desktop on the displays is shown the same.
Comment: Several options. e.g. use the display config utility and drag & drop 1 display on top of the other
Comment: @Dirk if you have time to write a full answer or provide a link to a tutorial on how to do this I would appreciate it.
Answer: Write in console or in script:
<code>xrandr --output HDMI-2 --same-as HDMI-1
</code>

How to duplicate monitor in lxpanel?


Moment of inertia of hollow body and solid body

Question: I read in my textbook about the various results of moment of inertia for different geometrical shapes like solid and hollow cylinder, sphere, disc and ring etc. Something general I noted is that $M.I$ for a hollow body is always greater than a solid body, say for a sphere
$$MI_{hollow}=\frac{2}{3}MR^2,$$
$$MI_{solid}=\frac{2}{5}MR^2.$$
So clearly $MI_{hollow}$ is greater.
And this works even for 2D shapes like ring and disc for which $MI_{ring}$ is twice $MI_{disc}$.
My question is is true for any irregular shape that the hollow will have more MI than the solid version of the body. And secondly, is there any proof to it?
Comment: Things will become far clearer if you actually derive the moments of inertia for, say, a half-dozen different geometries. It's just a matter of working through the integrals, if you've taken calculus. It would be easy to simply tell you that the moment of inertia is higher for mass farther from the revolution axis, but you'll really understand the concept by working through the calculations.
Comment: and, it's just the 2nd moment of the density distribution
Comment: You are making comparison using the same mass for two different shapes. Of course if your hollow object has more mass concentrated in the periphery it will have a larger moment of inertia compared to a solid shaped object of similar radius and same mass because of the physicial meaning of moment of inertia.
Comment: MMOI measures how far away from the axis of rotation is mass concentrated. For a hollow object the mass is obviously as far away as possible since the middle is hollow.
Answer: You can find the general approach to calculating the moment of inertia here: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#:~:text=For%20a%20point%20mass%2C%20the,a%20collection%20of%20point%20masses.
The moment of inertia is a measure of an objects resistance to angular acceleration. For an object of given mass the moment of inertia will be greatest when the mass is concentrated further away from the center of rotation. This is why the moment of inertia is greater for the hollow sphere than the solid sphere of the same mass.
Regarding any two irregular shapes, one hollow and one solid but having the same mass, the hollow shape will have the greater moment of inertia about the center of mass. This follows from the moment of inertia of a particle rotating about a point being $I=MR^2$. The moment of inertia varies linearly with mass but as the square of the distance away from the center of rotation.
Hope this helps.
Answer: It's important to carefully define the objects we are comparing, in particular how the mass is distributed in the hollow object. If the hollow object has a constant mass per unit area, the filled object has a constant mass per unit volume, and the two objects have the same total mass, then the hollow object doesn't necessarily have a higher moment of inertia. Here is a 2D counterexample.

The filled shape is a gear and a disk, both with radii $r$, connected by a very thin rectangle of length $L$. The gear has many narrow teeth on its periphery so that while the outer and inner radii that bound these teeth are both close to $r$, the perimeter of this shape is many times greater than that of the disk to the right, by a factor we will call $\alpha$. The rectangle is narrow enough that its mass is negligible. Since the gear teeth are short, both the gear and the disk have approximately the same mass $M/2$, where $M$ is the total mass of the shape. The axis of rotation is at the center of the gear. Using the parallel axis theorem,
$$I_\text{filled} \approx \frac{1}{2}\frac M2r^2+\frac M2\left[\frac12r^2+(L+2r)^2\right]= \frac{5}{2}Mr^2+\frac12 M(L^2 + 2Lr). $$
In the hollow shape, we can no longer neglect the mass of the rectangle. The mass per unit length of the shape is
$$\lambda = \frac{M}{\alpha2\pi r + 2\pi r + 2L}.$$
The hollow gear has all its mass distributed at a distance approximately $r$ away from the axis of rotation, so just like a circle it has $I_\text{gear} = m_\text{gear}r^2$. The total moment of inertia is
$$I_\text{hollow} \approx \lambda\left\{\alpha2\pi r^3 + 2\pi r\left[r^2 + (L + 2r)^2\right] + 2L\left[\frac{1}{12}L^2 + (r+L/2)^2\right]\right\}$$
$$= M\frac{\alpha2\pi r^3 + 2\pi r\left[r^2 + (L + 2r)^2\right] + 2L\left[\frac{1}{12}L^2 + (r+L/2)^2\right]}{\alpha2\pi r + 2\pi r + 2L}$$
We can make $\alpha$ arbitrarily large by increasing the number of teeth on the gear (while making the teeth narrower). As $\alpha \to\infty$, $I_\text{hollow}\to Mr^2<I_\text{filled}$.

If we are comparing moments of inertia about the centers of mass, we can modify the objects by putting another disk to the left of the gear so that the object is symmetric. Then the centers of mass of both the filled and hollow objects are still the center of the gear. The calculation proceeds the same way and the conclusion is still the same.