How do you find the center of a circle with a pencil and a book?

Question: Given a circle on a paper, and a pencil and a book. Can you find the center of the circle with the pencil and the book?
Comment: Open the book, maybe the answer to your question is written somewhere..!
Comment: offer the book to someone with a measuring device. give them the pen to mark the spot...
Comment: can't you just enclose a square around the circle, by using the corner of the book, and then the intersection of the diagonals of the square would be the centre of the circle?
Comment: @wim This is also a good solution!
Comment: This reminds me when I didn't got a "correct" in an answer I provided, when I should use 3 microprocessor instructions to have something done and I could do that with just 2 instructions... some teachers can be boring, sometimes... *You don't really need to use the book*.
Comment: Math answers should be completely general. Most of those submitted will not work if the circle is much larger than the book. But the book edge can be used to extend the normal to the chord.
Answer: Let the corner of the book just touch the circle's edge. You can draw two perpendicular lines along book's edge which intersect with the circle at two different points. Connect the two points and it's a diameter.
Repeat this and get another diameter. The intersect of these two diameters is the center of the circle.
Comment: See [Thales' Theorem](http://en.wikipedia.org/wiki/Thales%27_theorem).
Comment: @zdd You should not be so quick on the trigger with your Approved answer click.  This discourages others.
Comment: very good! any way else?
Comment: Could some one explain how those two points are the ends of diameter?
Comment: You assume that book corner are 90 degres. Fair enough. Elegant solution!
Comment: Thank you very much for your kindly reminder, I will keep it for a while next time.
Comment: Assuming a rectangular piece of paper, you could do this without the pencil or the book.
Comment: The picture made this make sense now. +1.
Comment: yes, was abt to post it http://en.wikipedia.org/wiki/Thales%27_theorem thanks!
Comment: I wrote another answer fully explaining why it works.
Comment: Even if the book corner wasn't a good right angle (which they usually aren't, since they tend to be rounded, especially for a hardcover), you can always use one of the pages.
Comment: @SenthilKumar Angle in a semicircle is 90 degrees
Answer: Not as simple as the accepted answer, but you have another solution:
Use the book to draw two parallel lines that cross the circle at different distances from the center to obatain 4 intersection points in the circle. Draw 4 lines connecting these points. The two points of intersection from these 4 lines define a diameter. Repeat the process with two parallels with different direction to obtain another diameter. Finally the intersection of two diameters returns the circle center.
Comment: The picture makes it make a million times more sense. This one gets a +1.
Comment: How do you use the book to draw two parallel lines?
Comment: @RahulNarain, hold the book so it's spine is on the circle, not it's face. But it would work even if the red lines are on opposite sides of the circle.
Comment: @RahulNarain The corner of book can be used to draw two parallel lines :)
Comment: I added an image to explain. The green line is a diameter because is perpendicular to the parallel lines (in red) and is crossing them at equal distance from both outer segments (in blue).
Comment: This solution is far better than the accepted one, since it does not depends on perception ("put the corner of the book..." how do you know that you are *really* drawing a diameter?). +1 for this answer!
Comment: The two points of intersection from these two lines define a diameter. Why?
Comment: And by the way, it works dosen't matter the size of the book or the circle.
Comment: Excellent, this one didn't use the corner of the book, I like it!
Comment: If @AlbeyAmakiir's idea of using the thickness of the spine is Luis's intended solution, this entails an awkward action and depends on the circle being big enough. Alternatively, use the book to draw two perpendiculars ABC and AD, BC being a chord. Then shift the book along AD and draw EFG parallel to ABC where FG is a chord and ED is on AD.
Answer: If the circle is a foldable material (like paper), you could fold it in half twice, the intersection of the folds will indicate the center.
Comment: How to rescue this: Suppose the circle is printed on stiff, opaque boxboard. Tear a page out of the book (let's assume the pages are reasonably thin and translucent). Using the pencil, trace the circle onto the page. You don't even have to trace the whole circle, just mark off four diametrically opposed arcs long enough for good alignment. Then do the folding trick with this copy. Make a hole in the center of the copy, line up with the original circle and then mark the center through the hole.
Comment: I stand corrected, it would require the circle to be an object by itself, not something printed on a larger body.
Comment: +1 you can see through most paper if you hold it up to the light.
Comment: Combine with the Thales solution: Fold one corner of the paper onto the circle perimeter and mark off the two edges with the pencil. Repeat with another corner of the paper. Proceed as above by intersecting the diameters. *Question:* Is there any situation of a circle on a rectangular paper, where you cannot use the (folded) paper edges themselves as ruler for this construction?
Comment: Yes you can fold it, but how can you make sure you are folding along the diameters?
Comment: Because you fold in half and a circle is symmetrical, both sides will overlap exactly.
Comment: How can you fold in half? the paper is not transparent, and you can not see the backside.
Comment: That's not the case, in that case, the pencil and the book was uselessness.
Comment: I can't vote this down because I don't have the reputation, so... virtual -1.
Comment: I have used this technique many times in the past. Regardless of what other people say, it has worked fine. So +1 from me :)
Answer: As pointed out, most of these answers will not work if the circle is much larger than the book. Instead, find someone with the proper size compass and straightedge, T-square, or other drafting tools, and threaten to hit them with the book if they do not find the center of the circle. See BAROMETER
Meanwhile, what is actually at stake here is Poncelet-Steiner
Comment: +1 for pulling out Poncelet-Steiner to explain, if nothing else, why the book having a corner is critical.
Comment: @Will Jagy, thanks for the link, but I think even the book can not cover the circle, it still works, since we can extend the line with the edge of the book. the key point is then the book has the corner as right-angle.
Comment: Everyone's a critic.
Comment: +1 for the BAROMETER link. And regarding size of book and pencil and circle: the OP provided the pictures of everything, and it's possible to observe that the circle is smaller than the book and the pencil...
Comment: No, Poncelet-Steiner is not the solution since it requires a known center. The center in this question is unknown, so Thales' Theorem is more likely.
Answer: Assuming the paper contains nothing but the shape of the circle:
<code> 1. Put eraser end of pencil on book. Or put pencil flush against side of book.
 2. Rest circle on tip of point.
 3. Circle will rest on the center.
</code>
Comment: That seems not very practical, but it's for sure something creative! +1
Comment: Hey, some people make a living doing this. http://www.youtube.com/watch?v=IRkZN27Hp_k
Answer: The only book I have is usually under my pillow. It has become skewed over the years - no 90 degrees joy. As a consequence I failed to have success applying the (elegant) description of Patrick Li. Moreover, my book is too small anyway to connect diametrically opposing points on the circle.
Therefore I had to revert to a more tedious approach.
Pick two points on the circle, close enough for my book's reach, connect them.
Align one of the book's edges with that line, the bookcorner at one of the marked points, draw a line along the adjacent edge of the book - which is non-perpendicular to the first line. Flip the book and draw "the other" non-perpendicular line through the same point. Repeat at the other point. Connect the two intersections that the four lines make, and extend this line - by shifting one bookedge along - to complete the diameter. If the intersections are too wide apart for my book, retry with the original points closer together. Repeat all of this with two other points on the circle, to get two diameters crossing at the midpoint.
If, by some miracle, my book has straightened out again (but still is tiny compared to the circle), I will quickly find out: the two "non-perpendicular" lines at one of the initial mark-points overlap. Then, I continue the line(s) until I cross the circle at the other side; likewise at the other initial point, and I wind up with a long and narrow rectangle. Repeating that from some other location on the circle, roughly at 90 degrees along the circle from the first setup, I get two rectangles and their intersection is a small parallellogram somewhere in the middle. It's diagonals cross at the center of the circle.
Answer: Draw a right triangle inside the circle, making all vertices touch the circle.
Transform it into a rectangle.
Draw a line joining the opposite corners and you will have obtained the center of the circle
Edit:
The idea is very simple: 1. Put a corner of the book touching the circle from the inside in any place.  2. Use the edges of the book for drawing the catheti of the right triangle until they intersect the circle.  3. Draw the hypotenuse using those intersections.  4. Put a corner of the book in any of those intersections overlapping one edge with the cathetus.  5. Use the other edge of the book to draw a line until it intersects the circle.  6. Draw a line joining this new intersection with the intersection opposed to the hypotenuse (right angle point). 7. The point in which this line crosses (intersects) the hypotenuse is the center of the circle.
Edit:

In fact, any rectangle inside the circle with every corner touching the circle will do the trick
Comment: This would work, but you should specify how you intend to do each step using only the tools available to you.
Comment: Thank you Rick. I edited my post following your suggestion
Answer: The pencil nor the book are necessary.

Given a circle on a paper

You do the following

Fold the paper so only a perfect semicircle is seen through the paper.
Unfold the paper
Rotate it
Repeat Steps 1 and 2
The intersection of the creases will be the center of the circle.
Comment: See also [this](http://math.stackexchange.com/a/206018/139) answer.
Answer: Another way to solve the problem is to pick 3 distinct points on circle say A, B, C. Then connect AB and BC. Halve A-B and B-C to obtain D and E. Draw lines through D and E perpendicular to AB and BC. The intersection of these lines will be the centre of the circle.
Comment: If you can't accurately determine the center of a circle, how can you accurately determine the midpoint of a segment?
Comment: @user113215 by taking off a page of the book, cutting it to match the size of the line AB, folding the piece of page that matched the size, putting it over the AB line again to mark what was the middle, and doing that again with the BC segment...
Answer: 1>
   Place your sheet of paper on top of the circle so that the corner just touches the
   circle's edge. Hold the paper in place, and use a pencil to make a small mark at the
   exact point where the two edges that meet at the corner cross the perimeter of the
   circle.
2>
  Using a ruler or straightedge, draw a straight line from one mark to the other.
3>
  Place the paper on the circle again with a different orientation. Make the tick marks
  again and connect these marks with a second pencil line using the ruler or straightedge.
  The center of the circle is the point at which this line crosses the first line you drew.
Answer: Using Thale's theorem and assuming that the corner of the book is 90°:

let corner of book touch the circle.
draw two lines along both edges of book touching the circle at the corner.
mark intersection of those lines with circle
draw line between those two intersctions: it's a diameter of the circle
repeat those 4 steps above for another point on the circle touching the corcer of the book
intersection of both diameters is the center
Comment: This appears to be the same as [Patrick Li's answer](http://math.stackexchange.com/a/205956). Did you have anything else to add?
Comment: @robjohn: Ups, you are right. I didn't check the other answers carefully enough. At least I mentioned in the answer the central theorem (Thale's) and the required property of the book (90° corner).
Answer: Hang the circle by its top edge on a wall close to a door or corner, draw a line straight down using the book's edge parallel to the door/corner edge. Rotate the circle f.ex 90 deg., draw a new line and you have the center and radius. This technique can also find the center of more complicated shapes. Gravity is the keyword here.
Comment: This answer is nonsense. What it finds is the center of gravity of the object onto which the circle is printed. It assumes that the circle is actually a disc object of uniform density, whose center of gravity corresponds to its geometric centre. The question clearly states that the circle is *on* paper, not *of* paper. Even if it's a cutout, paper is likely going to have uneven thickness so that the center of gravity doesn't correspond to the geometric center.
Comment: This solution uses additional tools, namely the ability to draw a line in a prescribed direction and the ability to find an apex point of the circle, neither of which is really in the (idealized) problem as specified.
Comment: @StevenStadnicki No just gravity as a reference.
Comment: I don't understand this answer at all
Comment: @Kaz :) ok, but you expect to draw very precise lines using the book and the pencil? The book edge will be a very straight line, without imperfections? The tip of the pencil is sharp enought or has some width associated ? Etc...
Answer: Let D be the center. Let A and B are points on the circle circumference. Let C be another point in the circle circumference.
Then angle ADB is twice angle ACB. I won't provide the proof. Actually it depends on whether C is between A and B or outside it. The theorem is still true but you also need to compute the angle ADB appropriately.
Now, a diameter is a point where the angle of ADB is 180 degree. That means we need to find a point where angle ACB is half of it namely 90 degree.
How?
That book has a 90 degree corner right? Put the corner in the circle circumference. Let's call it C. Now, take the intersection between the edges of the book to the edges of the circle (excluding C). Let's call them A and B. Now, if D is the center of the circle, we know that angle ADB is 180 degree.
We do not know what ADB is. However, 180 degree is simply a straight line. So if we draw a line between A and B, then D must be there.
If D is not on that straight line, then angle ADB won't be 180 degree. That's because the triangle ADB won't be "trivial". That means angle ABD and BAD won't be 0. By sum of angles in triangles, non zero angles of 2 of the angles in a triangle imply that ADB is not 180 degrees. Hence, we can conclude that point D is in a point AB. That means point AB is essentially a "diameter". It's a line that goes to the center of the circle.
Now repeat the process and draw a bunch of diameters. Of those diameters will intersect at some point.
That point is D, which is your center of the circle.
Answer: keep the circle on the book at one corner,so as the two sides of the book become tangents @ 90degrees to the circle. the join the two corners of the by a straight line which will pass thro' the center of the circle. now rotate the circle by few degrees closer to 90 degrees and draw another line again as done before.the intersection point of the two drawn lines is the center of the circle.
 another method.
1) fold the circle to make exactly half(semi)circle.
2) again fold the semi circle to make an exact quarter circle.
3) open and see the two folds meet @ one point, that is the center of the circle.
 both methods can get the center of the circle.
c v ramadoss. chennai 17
Comment: The second way was not true, you have no way to fold the circle exactly half, if you can do that, it means you already get the diameter.
Answer: Use the fact that the book is a straight edge to draw a straight line that touches the circle at one point.  Then use the fact that it has a right angle to draw a line perpendicular to that line at that point (where it intersects the circle).  This will be a diameter.  Repeat this at another point on the circle (not antipodal) to get another diameter.  They will intersect at the center.
Follows from the basic fact that any tangent line to a circle is perpendicular to the diameter that goes through the same point (and also from the fact that a standard book is rectangular in shape).
Answer: A silly answer, im joining to math.stackexchange.com only to answer this,  but it may help!

open the book
put the circle in the middle
close the book and see if both parts of the circle are matching (like a mirror)
pencil is not needed

valid?
Comment: Cute, but how do you put the circle in the middle (not to mention that you'd have to do this twice, to get two diameters)?
Answer: Using just the pencil

Mark a point A on the circle
Put the beggining of the straight edge of the pencil over the point A
Rotate the pencil, keeping the pencil's tip over the point A and the point of your finger on the pencil, accompaining the circle, so that you're measuring the distance from the point A to the point that the pencil touchs the circle
Immediatlly before this measure starts to decrease, you've found the circle's diameter.
Draw a line connecting point A and this point found on step 5. You've draw the circle's diameter.
Choose another point B. Do the same.
Where both line intersect, you have the center.

and as bonus:

enjoy that you have completed the task and read the book.


Disabling accessibility for a table column

Question: I have a normal NSTableView that displays a list of tracks. I have dedicated a table column to displaying an icon that indicates which track is currently playing.
I'm working on adding fuller VoiceOver support and I don't like how when each row in the table is selected the first thing said is "image". I would like to disable accessibility for that specific table column.
I know I can do this by subclassing either NSTableView and/or NSTableColumn to return YES from accessibilityIsIgnored.  Is there a way to do this without subclassing?
Answer: No. There is no such thing as "setAccessibilityIsIgnored" so you need to use a subclass that answers YES when asked -accessibilityIsIgnored.


How to ensure all nodes on CAN bus have received the message

Question: I know that that the ACK bit will be driven by the receiver of the CAN message. I'm wondering what would the situation be if you have multiple nodes listening to the same message. How would you confirm if all the nodes have received the message, since the ACK bit will only need to be driven by one of the nodes.
Comment: Talk to them to confirm if it's important.
Comment: See CAN Spec page 24 "7 FAULT CONFINEMENT"
Answer: If by "received" you mean "the CAN controller accepted it and passed it on to the software" - this is provided for by the workings of the protocol.
If any receiver (or indeed the transmitter) notices an error in the message on the bus (for example some framing issue, a bit stuff error, or the CRC doesn't match etc.) then it will create an error frame, and no-one will accept the message - and it will then be retransmitted automatically by the transmitter who sent it. So, even if there was some bizarre noise affecting only one small part of the bus, if even one receiver spots an error, everyone will throw the message away. Note that this even occurs if the node spotting the error is not configured to accept the particular ID - this filtering happens only after the message is accepted by the lower level hardware.
The exception to this would be if the node that notices the error has gone into an Error Passive state - due to seeing too many errors.  In this case, its error frame would include a "Passive Error Flag"(which is only made of recessive bits, so no-one sees it!). In this case, the other nodes (which did not see the error) would not be notified of this.
This is done to stop nodes with bad connections (for example) stopping the whole bus working.  It is a corner case, when something else has already gone wrong, but is worth considering if your system is critical and you wish to consider the case of multiple independent failures.
Answer: This is handled by the higher level protocol layer. Usually CAN is regarded to be so robust that you don't need to worry about individual messages getting lost. It also utilizes collision avoidance. Network supervising nodes rather tend to check if every node on the bus that should be there is present, and leave it at that.
Common strategies is to have each node send out a "boot-up" message when it is up and running, or to have each node report in at regular intervals (sometimes called "heartbeat"). In some cases it might make sense to implement a manner of handshaking, by having the receiver respond with another message. Some protocols also utilize the RTR bits of the CAN frame for this purpose: to send out an empty request frame for a certain type of data, after which the receiver responds with a data frame with the same identifier as the RTR frame.


Is it possible to prove ETH balance without revealing the holding address?

Question: I want to prove that I own an amount of ETH without revealing the address which holds it. I don't know if this is possible at all, whether with zk-snarks technology or any other way. I've found nothing on this and I'm not knowledgeable enough to figure out by myself.  Does anyone can shed some light on this? Thanks!
Answer: It should be possible with a ZKP.

without revealing the address

That's the "zero-knowledge" part.
The ZKP part will be challenging in itself but I foresee another problem.

prove that I own an amount

An account balance will tend to leak meta-data. A precise balance will narrow the field to addresses that have exactly the same balance which would often be exactly one. Oh oh ... it is probably merely a matter of searching for the account with the balance that matches the "secret". So, you will need to obfuscate that further. For example, "balance of at least x". This will further complicate the ZKP you would need to construct.
Not trivial. I might summarize the challenge by saying that Ethereum is optimized for transparency so this sort of thing is awkward.
Hope it helps.
Comment: awkward? I dont agree, there is valid reasoning behind that.
Answer: Here is a rough outline of 2 different methods.
#1 Provisions
The following paper presents a protocol along with security proofs (see the proof of assets section): [CCS15] Provisions: Privacy-preserving proofs of solvency for Bitcoin exchanges
The protocol involves Pedersen commitments (common in the crypo[currency|graphy] space) and a custom zero-knowledge proof. The idea is that you (the prover) hide your account inside a set of other accounts and through a zero-knowledge proof the verifier can be convinced that you own an account in the set with a balance within some range.
Here is a rough outline of what the prover (you) does:

Construct a set of Ethereum accounts, within which must reside the account that you own and would like to prove has a certain balance
Produce Pedersen commitments to a) your account balance and b) to zero for each of the other accounts in the set; publish these commitments on chain
Produce Pedersen commitments to a) the private key underlying your account and b) to zero for each of the other accounts in the set; also publish these on chain
Calculate a handful of additional variables related to the zero-knowledge part of the protocol and publish these on chain (this part of the protocol ensures the Pedersen commitments were produced correctly)

Here is a rough outline of what the verifier does:

Checks that the balances for the accounts given by the prover match those on chain
Verifies in zero-knowledge that the Pedersen commitments were constructed correctly
Adds together all the Pedersen commitments related to the balances, which gives a Pedersen commitment to the prover's account balance
The final Pedersen commitment can be used to do a range proof (show the account balance is between 2 values) but the details if this are not in the paper

Note the paper uses Bitcoin as the example but since Ethereum uses the same elliptic curve the protocol is the same. Note also that the accounts' addresses are unfortunately not possible to use, one must use the public keys of the accounts which you can get from the tx signatures for those accounts.
#2 zk-SNARK
The idea here is similar to the one above except that it uses a zk-SNARK to do all the work.
What the prover does:

Construct a set of Ethereum accounts, within which must reside the account that you own and would like to prove has a certain balance
Run the zk-SNARK
Post the resulting proof string + public inputs for the snark on chain

The zk-SNARK:

Public inputs: set of accounts (addresses, not public keys), set of ETH balances for the accounts, min balance for range proof, max balance for range proof
Private inputs: prover's private key, index in the set where the prover's address is
Computation:

construct the public address for the private key (unfortunately this is not a good computation to do inside a snark because it uses the keccak hash)
verify that the constructed address matches the one in the set at the given index
verify that the balance for that address is between the min and max amounts

What the verifier does:

Checks that the balances for the accounts given by the prover match those on chain
Verifies the zk-SNARK proof


Why is foreach iterating with a const reference?

Question: I try to do the following:
QList<QString> a;
foreach(QString& s, a)
{
    s += "s";
}

Which looks like it should be legitimate but I end up with an error complaining that it cannot convert from 'const QString' to 'QString &'.
Why is the Qt foreach iterating with a const reference?
Answer: As explained on the Qt Generic Containers Documentation:

Qt automatically takes a copy of the container when it enters a foreach loop. If you modify the container as you are iterating, that won't affect the loop. (If you don't modify the container, the copy still takes place, but thanks to implicit sharing copying a container is very fast.) Similarly, declaring the variable to be a non-const reference, in order to modify the current item in the list will not work either.

It makes a copy because you might want to remove an item from the list or add items while you are looping for example.  The downside is that your use case will not work. You will have to iterate over the list instead:
for (QList<QString>::iterator i = a.begin(); i != a.end(); ++i) {
  (*i) += "s";
}

A little more typing, but not too much more.
Comment: Minor nitpick: I prefer:
    for(QList::iterator i = a.begin(),end=a.end(); i != end; ...)
Answer: or you can use
QList<QString> a;
BOOST_FOREACH(QString& s, a)
{
   s += "s";
}

Comment: Umm, ok... Sure, you *could* use a different library!  Hopefully, you'd have a really good reason to be using Boost and Qt at the same time though, which extends beyond resolving this minor issue!
Answer: With C++11, Qt now encourages this standard for syntax instead of Qt foreach :
QList<QString> a;
for(auto& s : a)
{
    s += "s";
}

Answer: I believe Qt's foreach takes a temporary copy of the original collection before iterating over it, therefore it wouldn't make any sense to have a non-const reference as modifying the temporary copy it would have no effect.
Answer: Maybe for your case:
namespace bl = boost::lambda;
std::for_each(a.begin(),a.end(),bl::_1 += "s");


How do I archive and restore selected tables using SQL Server?

Question: How do I archive and restore selected tables using SQL Server?
Comment: If you mean real archiving then it depends on the application. If you want to keep the size of the database to a minimum only containing "live" data, you can use partitioning as one method or a separate database. But there can be a lot of work in creating an archive/un-archive solution. Some vendors e.g. Microsoft have some tools for AX Dynamics to handle purging and archiving.
Answer: You have to use a 3rd party tool such as Red Gate SQL backup pro to restore. Or restore as a different DB and copy across.
Selective restore makes sense, occasionally, but selective backup usually wouldn't. You-d have to roll your own selective backup with SMO for example
Answer: You could use the import/export feature in SSMS. You could export out the table to another DB, flat file, or other available formats. If you save your export package, you can rerun it every time you need to export your data. You should have integration services running though.
The same can be done for importing data back into the table. You can choose to append or overwrite the data. A bit of tweaking will have to be done to get the file formats the way you want it.
Answer: The only way to do this without 3rd party tools is by selectively backing up and restoring different filegroups, which requires some deliberate setup within the database.
In our environment, we use the "Object Level Recovery" feature of Quest Litespeed.
The workaround, as @gbn mentioned, is to restore a copy of the database with a different name, then copy the table/rows out that you need.
Answer: If it is a one time activity: Use import/export feature as already suggested. That feature uses SSIS on the background therefore you can tweak it.
If these backup and restore is continuous activity not a one time activity. Consider a database: some tables are not really important as others. For example invoice tables vs log tables (not transactional log files but monitoring tables). You would like to backup invoice tables more regularly than log tables. Log tables consume a lot of space but they are not as important as invoice data.
Consider using file group backups instead of full backups. Put all really important tables in one file group and other tables in another file group.
related links

http://msdn.microsoft.com/en-us/library/ms189906.aspx


Why does insufficient data lead to random errors?

Question: I've read up on the causes of random error and noticed insufficient data to be one of the causes of random error. However,  I'm confused as to why insufficient data could possibly lead to random errors. Could someone provide some context for why this might be?
Comment: Please define "random error" and "insufficient data". By the first, you probably mean "statistical error" (as opposed to "systematic"), but the latter can be many things.
Comment: Random Error - error in which there's equal likelihood of data value being too high or too low.
Comment: Off topic , recommended to move
Answer: More datapoints (if everything is done correctly) lead to more trustworthy results. With less datapoints, the possible (statistical) error grows, until at one point, when you only have as many datapoints as you have parameters, it looks like you have no error at all. (That is of course stupid, you should always have derived the error range of the individual data points.)
Now what is "insufficient"? Is that when the statistical error grows too large for your feeling? Or when you only have two points to draw a straight line through? Or when you only have two points to draw a parabola? That's your prerogative.
(note that I have also above used the word "error" in the same sloppy way your books seem to do. The "errors" are in your data, and they have a systematic component (hopefully small) and a distribution. The derived values (e.g. the slope of a linear regression) have a confidence interval.)
Comment: When you discussed the idea that too few data points are likely to exacerbate the size of random error, I think that's what my textbook was getting at when it mentioned insufficient data. Thanks!
Answer: When making a series of measurements we implicitly assume that there is a an 'error', or better, an uncertainty associated with the measurement and that this uncertainty is random and not so influenced by that of any of the others, i.e. the errors and random and independent of one another.
In virtually every case, no matter what the noise on a measurement is, if  a sufficiently large number are made then the histogram of the measurements will follow a gaussian or normal distribution, which is a 'bell shaped' curve. The peak of the curve is then taken as the average or mean value of the measurement. Depending upon how wide the bell shaped curve is allows us to put an error limit on the measurement. (The theorem supporting this is called the Central Limit Theorem)
However, we cannot always make a huge number of measurements and so if we make only a few then they could fall anywhere on this distribution; they could be exactly on the mean value or way off this value, one just does not know. In this case we have to use make estimate of the true value by averaging the measurements  and by assuming that the noise is random and gaussian (normally) distributed we can estimate the 'error', called the standard deviation.
Usually the error is stated as a mean value plus or minus the standard deviation ($\pm 1 \sigma$) for example $23.4 \pm 0.3$ which means that $\approx 68 $ % of the time the measurement falls within the limits $23.7$ and $23.1 $ but by random chance has a $32 $ % of being outside these.  This is not very convincing so it is better to try to obtain $\pm 2 \sigma$ which is $\approx 94 $ % chance of the measurement value being within the range, and by random chance $6$% of being outside this.  Of course it is up to you to decide what is appropriate.
An additional point to bear in mind is that the signal to noise ratio is reduced as $\sqrt n$ for n measurements. Thus to improve the signal to noise ratio of a measurement can involve a lot of extra work.
So to answer your question directly; the errors are there all the time, insufficient data does not make them greater or smaller but instead does not average them away to a smaller value.


Proving an Entropy inequality

Question: I wish to show that $H(X | Y, Z) \leq H(X | Y )$ for random variables $X,Y,Z$, where $H$ is the Shannon entropy. I know how to show $H(X | Y) \leq H(X)$ by using the Chain Rule on $H(X,Y)\leq H(X)+H(Y).$  I want to do something similar with $H(X,Y,Z)\leq H(X)+H(Y)+H(Z)$, but I'm stuck.
Comment: Well, that same argument applied to the second inequality gives \begin{align} H(X,Y,Z)&=H(X,Z|Y)+H(Y)\\&\leq H(X)+H(Y)+H(Z)\\\implies H(X,Z|Y)&\leq H(X)+H(Z)\end{align} But then I would think you can apply the chain rule once more to get $H(X,Z|Y)=H(X|Y,Z)+H(Z)$ which implies the desired result.
Comment: It might be helpful to include the work you did to show the simpler inequality.
Comment: We have $H(Y)+H(X|Y)\leq H(X)+H(Y)$, giving the result.
Answer: The inequality in the "conditional case" follows the exact same steps as in the "unconditional case" since the same equations and inequalities hold, however, with all quantities involved conditioned on $Z$. In particular, it holds
$$
H(X,Y|Z) = H(X|Y,Z)+H(Y|Z) \text{ (chain rule)}
$$
and
$$ H(X,Y|Z)\leq H(X|Z)+H(Y|Z) \text{ (conditional independence maximizes joint entropy)}
$$
You may want to verify these formulas using the same mechanics as in the unconditional case, however, with the conditional version of the joint distribution of $X$ and $Y$ (that is, given $Z$) in place of the unconditional version.

Proving an Entropy inequality


Do 4-part harmony rules also apply to piano arrangements?

Question: Does rules in 4-part harmony also apply to piano arrangements ?
Like for example no doubling the 3rd note of the chord, or moving in parallel 5, etc.
Answer: They don't fully apply, no. But it depends on what you're trying to do.
Remember that the voice-leading rules for four-part harmony are to mimic eighteenth-century chorale procedures. Piano arrangements are by default not in the eighteenth-century style, so by that logic there's little need to follow all of the voice-leading rules.
With that said, a few rules do still persist today, even in popular music. More than anything, I think the rule of resolving the chordal seventh down by step is the most-followed rule from the eighteenth century.
Forbidden parallels, however, happen all the time in music of the 20th century (and even earlier).
So depending on what you're arranging, you may find yourself breaking these rules all over the place. But that's okay, because if the piece you're arranging isn't in the eighteenth-century style, why arrange it like it is? (Unless you're trying to be hip with a different style arrangement!)
Lastly, I just wanted to note that doubling chordal thirds happens all the time. It's a rule that's been passed down, and it might be to prevent people from doubling the leading tone (which is the third of the V chord). But doubled thirds in tonic chords happen all the time, and they're a necessity (!) in some vi chords at a deceptive cadence.
Answer: The concept behind 4-part harmony excercises is to teach you the basics of voice leading. It is in reality completely unlike any real music you may find. All it really aims to do is teach you the basics. In actuall fact, real four part voice harmony does things like voices go over and under different each other, which is a real no-go for your excercises.
Also you may find octaves doubled in piano music that looks like the doubling which your are told to avoid but this is not a doubling of the harmonies perse but more just two of the same notes played to increase the sonority of a passage, those types of doubling are very piano-like.
Answer: In a piano arrangement, unlike in 4-part vocal writing, you don't normally have different voices. In a four-part Bach chorus, for example, or in a hymn, all the voices will be singing the same words and you'll have melodies and counter melodies going on.
On the other hand, most piano arrangements are one melody line being embellished and emphasized through creative use of harmony, style, and technique. You don't simply have multiple parts moving in similar rhythmic patterns like you do in vocal music. You may have counter melodies and such going on, but it is a very different style from vocal writing, for the simple reason that the piano is a totally different instrument than the voice.
For this reason, I'd say the rules aren't as applicable to piano arrangements.


Why does only half of my outlet work?

Question: I just moved into a new place. I plugged my vacuum into an electrical outlet, my vacuum won't turn on. So I tried the other side, my vacuum turns on. I then plug in a lamp to the side that won't work for my vacuum, lamp turns on.
Is the outlet defective?
Comment: Leave the lamp plugged in.  Are there any light switches or dimmers anywhere close enough to *conceivably want to control a lamp in this location?*  Try them and see if the lamp changes.
Comment: The spring clips are probably bad/old in one part of the receptacle, it happens. With the lamp, the amount of current it needs is so low that it works good enough. But with the large motor on the vacuum, the high current draw causes the spring clips to heat up instantly, pulling them away and making the voltage drop so low that something in the vacuum is shutting down to protect itself. Bottom line, that receptacle needs to be changed, that's a potential fire hazard.
Comment: Do you have any meters? Can you verify if that outlet has a low voltage problem or there is no voltage at all? Also see Harpers and Jack comments.
Comment: and try both plugs on that outlet.
Comment: I opened the cover, don't know why the owner has the outlet completely plastered to the wall, even if I wanted to replace it, couldn't.  I did try both plugs with my vacuum cleaner, one side will work the other side won't work for the vacuum, but will work for a lamp.
Answer: Maybe it is wired that way on purpose. Two receptacles in our bedroom are wired like that. One side is normal,always on. The other side is operated by a wall switch so you can turn on a plugged-in lamp with the wall switch and have a radio always powered by the other side.
Comment: If a standard switch is involved, how would the vacuum not work and the lamp would work at the same outlet? No information is given about any switch action between testing those 2 devices at the same outlet.
Answer: Definitely change it.it may not even be,or ever was originally connected to a stud. find the closest stud to that mess,cut that mess out then patch the sheetrock after you change the outlet and insert it into the proper box that's nailed to a stud.  Around here you could have it repaired for around $100. Done yourself about $15 to $20.
Comment: Where did you get the impression that the junction box is loose? Regardless, this doesn't address the question that was posed.

Why does only half of my outlet work?


Is there a GUI in xorg for changing the video driver?

Question: I was under the impression there was a GUI app for changing the Xorg video driver. If so, where is it? I don't see it in 10.04.
Comment: There's almost certainly a better question to be asked lurking under this one.  Why do you want to change the Xorg video driver? :)
Answer: If you're looking to switch between the proprietary and Free driver for your video card (e.g. if you have an ATI or NVidia card), try jockey.
On the default Ubuntu desktop this is in:
System - Administration - Hardware Drivers
You can enable and disable hardware drivers there.
Answer: If you mean a GUI for changing display-related properties, then you can find that at System->Preferences->Monitors.
If you are referring to a GUI tool for editing xorg.conf, I am not aware of one - you're better off editing the file directly (details here).
Comment: `System->Preferences->Monitors` Will only work when it's using xorg drivers that it supports xrandr, etc. I also agree though that there isn't (to my knowledge) a GUI for modifying the xorg configuration file. Though that sounds like it might be a helpful tool.
Answer: Perhaps you're looking for dpkg-reconfigure. It's terminal-based GUI for reconfiguring software. You should be able to change your xorg driver settings with sudo dpkg-reconfigure xserver-xorg.
Comment: This won't do anything interesting on a modern Ubuntu system (from 9.10 onwards, I think).  The packages no longer ship a default X configuration - it's all autodetected.
Comment: +1, although not a proper GUI, it's a nice interface to configure your xorg via a few keystrokes. Also you can use it in the virtual terminal if you break things up :)


What kind of objects can be reclaimed?

Question: What if you're purchasing a house and the previous owner wishes to reclaim some of the objects in your house? Wouldn't all the objects now be owned by the current owner, and if not, what kind of objects can be reclaimed?
Comment: "purchasing" or "purchased"?
Answer: When selling a residential property in the UK, you normally fill out a form called the TA10 Fittings and Contents Form, which outlines everything in the property included in the sale. This isn't a legal requirement, but your conveyancing solicitor will normally recommend it.
Anything else left in the property after the completion of the sale is still, legally, a possession of the previous owner and the new owner is obliged to inform them of these possessions in writing.
So, you do have to return it (and most people in such situations do return the possessions without question). However, if the presence of those possessions means you can't be reasonably expected to move into the property, then you can claim compensation from the previous owner for alternative accommodation (i.e. a hotel) until the situation is rectified. Or if the previous owner refuses to collect the possessions, you can likewise claim compensation for the costs of disposing of it.
Alternatively, if they do want the possessions back, but take their time getting it, you can charge them fair storage costs.
Comment: You can short-circuit the detailed list if both parties agree. I once bought a house from someone who was moving overseas at short notice to start a new job. We just added a clause in the contract that the sale consisted of the property and all the contents that were were not removed before the completion date.
Comment: @alephzero I've clarified that the form is not a legal requirement.

What kind of objects can be reclaimed?


how do you solve $y''+2y'-3y=0$?

Question: I want to solve this equation:
$y''+2y'-3y=0$
I did this:
$y' = z$
$y'' = z\dfrac{dz}{dy}$
$z\dfrac{dz}{dy}+2z-3y=0$
$zdz+2zdy-3ydy=0$
$zdz=(3y-2z)dy$
$z=3y-2z$
$z=y$
$y=y'=y''$
???
now, I'm pretty sure I did something wrong. could you please correct.
Comment: How did you write $y'' = z dz/dy$ ?
Comment: What method(s) have you been taught for how to do this?  For example, do you know what characteristic polynomials are?
Comment: $y''=\dfrac{dy'}{dx}=\dfrac{dz}{dx}=\dfrac{dz}{dy}.\dfrac{dy}{dx}=z\dfrac{dz}{dy}$
Comment: in this equation, since we don't have x we would consider $y'=z$ and solve the equation like a first order differential equation.
Comment: Your error is from the 5th to the 6th line after "I did this", where you cancel $dz$ off the left and $dy$ off the right.
Answer: HINT $\rm\ \ \ 0\  =\  y'' + 2\ y' - 3\ y\ =\  (D^2 + 2\ D -3)\ y\ =\ (D+3)\ (D-1)\ y, \quad D = d/dx $
i.e. factor the differential operator as you would any polynomial.
Comment: @Bill: I am sure if the OP understood what this meant he or she would have no problem with this question.
Comment: @Bill: +1 I think it is a little humorous that an answer which shouldn't make any sense was chose by the person who asked it as the best answer. =]
Comment: @Qiaochu: My experience with similar questions on sci.math and elsewhere over the decades shows that hints like that often suffice. Of course you are welcome to give your own answer if you like.
Comment: @night - That notation is not at all uncommon.  I've encountered it in 3 texts for undergrad study, and it's one of the most useful techniques discussed in the differential equations "Problem Solvers" book -- with plenty of examples given, albeit more awkwardly typed.
Comment: @night Operator-theoretic methods in the theory of differential and difference equations are not only very *natural* but also very *powerful* and *beautiful*. I cannot imagine why anyone would find them "weird" or "awkward". Moreover, in my experience, they are much more easily comprehended by students than are alternative approaches.
Comment: @Tyler: As I said, my experience shows that students pick up this idea quickly once they see an example. Most likely that's what happened here. I'm surprised that anyone thinks there is anything deep here. It is quite trivial - accessible to a bright high-school student.
Comment: @Bill: +1... For creativity. A little weird approach to the problem, but it make full sense. Just a bit awkard to treat the differential as a polynomial for this low level math.
Comment: @Bill: Sorry earlier about making the comment on the Differential Operator. I was reviewing a text recently and now I see some usefulness of knowing it.
Answer: So we see that we have a Linear Second Order Differential Equation. There are several variety of
methods to solve these type of equations which some are listed, but I think this method is the most
straight-forward one and does not require to much trickery or cleverness, its a pretty
step-by-step process.
HINT:$~~$ We know for these type of equations this must hold true:
$$A(x)\cdot y''+B(x)\cdot y'+(x)\cdot y = D(x)$$
: Dividing by $A(x)$ gives:
$$y''+p(x)\cdot y'+q(x)\cdot y = f(x)$$
$\underline {Our~~Case:}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ Homogeneous $2^{nd}$ order differential
$$y''+p(x)\cdot y'+q(x)\cdot y = 0$$
So for our case here we have the following: $$y''+2y'-3y = 0$$
I claim that this is not too bad of a general solution to obtain:
I am assuming you learned something about the characteristic equation thus far.
So for starts, try using $\underline{y = e^{rx}}$ for our problem and try to work it out. Gonna also assume you
know what the general solution form would like that:
$$y(x) = c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+...+c_ne^{r_{n}x}$$
Let me know how that works out.
P.S. NOTE.... $p(x)$ and $q(x)$ here is any function of the variable $x$ times the derivative of the function $y$ $(y')$ and the function $y$ itself. This means that $p(x)$ and $q(x)$ can also just be constants, which is what we have in our case, in case the notation is a bit confusing.
So now solving the equation for understanding:
Solution:  Let $y=e^{rx}$, then plug this into the differential equation. This would lead us to taking derivatives of $'y'$, but for now we will just plug it in to see what it will look like. Doing so leads us to the following:
$$y''+2y'-3y = 0$$
$$(e^{rx})''+2(e^{rx})'-3(e^{rx})=0$$
So now we see that we must take derivatives of each of the exponential terms that has primes connected to them. So doing this leads us to:
$$y(x)=e^{rx}$$
$$y'(x)=re^{rx}$$
$$y''(x)=r^{2}e^{rx}$$
So now substituting the derivatives of $y$ into the original ODE we can see that we get the following:
$$r^{2}e^{rx}+2(re^{rx})-3(e^{rx})=0$$
Now let's ask ourselves, what do we see in common in this new equation with the derivatives of our initial guess $y=e^{rx}$ to the homogeneous ODE? Well if you see it or don't see it let me point something out. All of the terms has our original guess $e^{rx}$ in them, so what we can do with that term is what we can do with any algebraic expression and that is factor it out of the equation as a common term as follows:
$$e^{rx}\left(r^{2}+2r-3\right)=0$$
Now what all of this work has led us up to is something called the Characteristic Equation or better known as Characteristic Polynomial, also can be referred to as the Auxiliary Equation which is what you see inside of the parenthesis as:
$$r^{2}+2r-3$$
To further simplify the equation we can divide out entirely by $e^{rx}$ which would just leave us with our Characteristic Polynomial being:
$$r^{2}+2r-3$$
NOTE:  All of the work we did to arrive at our Characteristic Polynomial can be saved in the future whenever we have constant coefficient ODE's, because the initial guess will always be divided out at the end. So here is the trick.
1.) Observe the order of the ODE (namely, highest derivative in the equation).
2.) Start off with the highest derivative and work your way down to the lowest term, (namely, being constant terms).
3.) Every term with a prime with have a $r^{m}$ corresponding to it, with (m = the amount of primes on that specific term) in the ODE, namely: $$y'=r^{1},~y''=r^{2},~y'''=r^{3},~ y^{(4)}=r^{4},~y^{(5)}=r^{5},~\ldots,~y^{(n)}=r^{n}.$$
4.) Any constants just multiply by the $'r'$ or sometimes used $'\lambda'$ terms.
5.) For terms not involving any primes, such as $'y'$, they will become just one. Because if we think back to the beginning, we used as our initial guess $y=e^{rx}$, so once we factored out $e^{rx}$ we will just have a one in the place where $e^{rx}$ once used to sit inside of the equation.
So for example, lets take a look at this ODE: $y^{(4)}+2y''+y=0$
The Characteristic Polynomial we seek will just be: $$r^{4}+2r^{2}+1=0$$
I know seeing is believing!, so I have some few examples for you to try out of yourself (not to fully solve) but to just see if you can carry out getting the Characteristic Polynomial by using the short method that I listed above:
Examples:
$i$ )  $$s''-4s'+4s=0$$
$ii$ )  $$i''+60i'+500i=0$$
$iii$ )  $$y'''-3y''-y'+3y=0$$
$iv$ )  $$x^{(8)}-5x^{(4)}+4x'=0$$
$v$ )  $$v''-6v'+13v=0$$
So not to lose focus of what our goal here was, and that was to solve $y''+2y'-3y = 0$. So using all of our knowledge gained thus far, we can apply it now to our problem. So by doing so we can conclude that our Characteristic Polynomial will be as stated previously: $$r^{2}+2r-3=0$$
From here we want to take the following measures in this order:
1) Factor is possible.
2) If factoring is not possible, use quadractic formula. If you forgot let me remind you:
$$r = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.$$
3) If both of above if not possible (usually for cubic or higher powers) then using long division/synthetic division and usually helpful by graphing the equation first and finding at least one zero and use that factor and take the original polynomial and try long division/synthetic division. So for example if you found $2$ as a zero of the polynomial, then you would use $(r-2)$ to try and reduce the polynomial down to a quadratic if possible or to something that can be factored.
But for our problem luckily, it is factor-able as so:
$$(r-1)(r+3)=0$$
With real-distinct roots being:
$$\text{ Characteristic Roots are: }~ r=1,~r=-3$$
So with this known, we could put together our general solution to the ODE, which was stated above, ( namely being,$~~~~~~~~ y(x) = c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+...+c_ne^{r_{n}x}~$).
So our general solution will be a linear combination of our homogeneous solutions of our  characteristic roots $r_{1} \text{ and }r_{2}$.
$~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~~~~$ $y(x)=C_{1}e^{x}+C_{2}e^{-3x}$.
No we have found our solution to the ODE, but this only a general solution because we still have undetermined coefficients in the solution (namely, $C_{1} \text{ and } C_{2}$). In order to find these, we will need to be provided with initial conditions which will then make the problem become an IVP (Initial Value Problem). So wrapping this all together, this is what we would do to solve any such $2^{nd}$ Order Linear Constant-Coefficients Homogeneous Ordinary Differential Equations, and the same applies for Higher Order.
There are more cases beyond this one of real distinct-roots. We will go over now what those other cases are.
The other type of situation that you can run into is when you have real non-distinct roots (i.e., having multiplicity  of two or greater). Now for the word multiplicity, all that means is simply the following: If you get some numbers that appear more than once (meaning that they are exactly the same).
$\underline{Example~1:}$   If we had this characteristic polynomial: $~r^{2}+6r+9=0$,
our roots to this polynomial will be the following: $r=-3,~r=-3$. This is a prime example of what multiplicity means. Our roots would be $r=-3$ (with multiplicity 2) because the same exact root appeared twice, this will hender having to multiply the linear combination of solution by an appropriate amount of powers of $'x'$ or $'t'$ whichever is your independent variable in the ODE. Let's take a look at a few more examples.
$\underline{Example~2:}$ If we had this characteristic polynomial: $~5r^{4}+3r^{3}=0$, our roots to this particular polynomial will be the following: $r=0,~r=0,~r=0,~r=-\dfrac{3}{5}$
This is yet another great example of when we would make the conclusion that we have yet again repeated roots being $r=0$ (now with multiplicity 3). Once again, you would have to multiply the independent homogeneous solutions by an appropriate power of your independent variable, which in our case we are using $'x'$. Let's take a look at one more example for exercise to get ourselves more comfortable with the idea. Also, by all means, please do take the time if you can and verify these characteristic roots to the polynomials to see if you get the same results and take it one step further and see if you can combine the homogeneous solutions into a linear combination of solutions to form the general solution correctly. So now, moving to our next example.
$\underline{Example~3:}$ If we had this characteristic polynomial: $~9r^{3}+12r^{2}+4r=0$, our roots to this particular polynomial will be the following: $r=0,~r=-\dfrac{2}{3},~r=-\dfrac{2}{3}$. This is once again yet another great example of when we would make the conclusion that we have found repeated roots in our polynomial being $r=-\dfrac{2}{3}$ (with multiplicity 2). Once again, you would have to multiply the independent homogeneous solutions by an appropriate power of your independent variable in the differential equation. So this pretty much covers the case of what will take place if we were to encounter such a polynomial that had non-distinct roots.
I hope this helps out. Let me know if there is something you need to be clarified a bit further. Thanks.
Good-Luck
Comment: Perhaps you could add the case when the (two) characteristics roots are equal.
Comment: @Americo: Sure thing I will. Thanks for that reminder. I guess I could go over all the cases: Such as complex roots and repeated roots. What do you suggest?
Comment: perhaps just the case of repeated real roots.
Comment: @Americo: Yea you are right, I was thinking that would be too much for one to swallow in one thread. :)
Answer: For this kind of differential equations Laplace transform is handy:
$$
{\cal L}[y''] + 2 {\cal L}[y'] -3{\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2{\cal L}[y] -sy(0) - y'(0)) +2(s{\cal L}[y] -y(0)) -3{\cal L}[y] = 0  \ .
$$
Now put $F(s) = {\cal L}[y]$ and $a= y(0), b= y'(0)$.
$$
(s^2F(s) -as - b) +2(sF(s) -a) -3F(s) = 0 \ \Longleftrightarrow \ (s^2 +2s -3)F(s) -as -2a -b = 0 \ .
$$
We solve this equation for $F(s)$:
$$
F(s) = \dfrac{as + 2a +b}{s^2 +2s -3} = \dfrac{as + 2a +b}{(s-1)(s+3)} = \dfrac{A}{s-1} + \dfrac{B}{s+3} \ ,
$$
where $A$ and $B$ are constants depending on $a$ and $b$, which I let you the pleasure to compute  :-)  and we get
$$
y(t) = A{\cal L}^{-1}\left[ \dfrac{1}{s-1}\right] + B{\cal L}^{-1}\left[ \dfrac{1}{s+3} \right] = A e^t + B e^{-3t} \ .
$$
Answer: $zdz=(3y−2z)dy$ does not imply $z=(3y−2z)$
See this
http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients
Answer: You can also write it in matrix form:
$u=(y',y)$, $u'=\big(\matrix{-2 & 3 \\ \hphantom- 1 & 0}\big) u$. Find the eigenvalues and eigenvectors, turn it into a diagonal system whose solution is simple. Go back to the original coordinates.
Answer: Start by assuming a solution of the form
$y(t)=e^{ct}$
Hence $y'(t)=ce^{ct}$ and $y''(t)=c^2e^{ct}$
Substitute into the original equation. You will find a quadratic for $c$. Solve it, and note that both solutions work, and as the ODE is linear and homogeneous, the sum of solutions also works.
Answer: Start by finding the characteristic polynomial. In this case it is $x^2+2x-3$. Find its zeros; they are $x=-3, 1$. The general solution to your differential equation is $ce^{x_it}$, where $x_i$ is a root to your characteristic polynomial. So we have two linearly independent solutions; $y=e^{-3t}, y=e^{1t}$. Because this is a linear differential equation, we can take any linear combo of those two solutions, and that will also be a solution. So the general solution for $y$ is $$y(t)=c_1e^{-3t}+c_2e^{1t}$$, where $c_1$ and $c_2$ are determined by your initial conditions ($y(0)$ and $y'(0)$).
Answer: Heh.  Figured I'd try something different for this.  Try this substitution:
$z=y'+3y$
$z'=y''+3y'$
$y''+2y'-3y=y''+3y'-y'-3y=z'-z=0$
$e^{-x}z'-e^{-x}z=0$
$(e^{-x}z)'=0$
$z=ae^x$
$y'+3y=ae^x$
$e^{3x}y'+3e^{3x}y=ae^{4x}$
$(e^{3x}y)'=ae^{4x}$
$e^{3x}y=be^{4x}+c$, $b=\frac a4$
$y=be^x+ce^{-3x}$
Answer: you can solve in in simpler way, convert it into auxillary form:
                          $ m^2+2m-3=0 $
factorise it,
                          $ m^2+3m-m-3=0 $
$ m(m+3)-1(m+3)=0 $
$(m+3)(m-1)=0$
$ m=1,m=-3 $
since, both roots are real and different so we can write its solution as:
$y=c_1e^x+c_2{{e^-}^3}^x$
Comment: I am not sure this adds anything to the existing answers.

how do you solve $y''+2y'-3y=0$?


What information does an IP address provide

Question: For example, if I use a web app that saves my IP address, what can be inferred from it?
What can be deduced from an IP address (and to what degree of certainty)?
Is it always possible to determine a user's geolocation given an IP address?
Answer: An IP Address is a network identification number that allows one address on the internet to communicate to another address. Given that one place can send data all over the internet to your location and back, it is logical to understand that if software can do it, so can other software.
There are 2 versions of IP Addresses. IPv4 and IPv6 addresses. Both work differently from the other, so I'll be talking about IPv4 (xxx.xxx.xxx.xxx) addresses.
On the internet, A corporation called IANA registers ranges of IP Addresses valid for different countries. Per country, each ISP is given a small set of IP Addresses that they are allowed to hand out. Because there are more people than available IP Addresses, IP Addresses are not assigned to a person or domain unless an additional fee is paid for it.
Lets assume this is not the case. If a user disconnects from the internet, or their DHCP lease expires, the ISP may give out a new IP Address. Usualy an ISP will only do so for disconnected users where it needs to assign an old used IP to a new user because it ran out. If that first person then connects to the internet again, and his/her IP was handed out, they'll get a new IP Address.
Whenever you query a whois on an IP Address, you always will see which company is responsible for handing out that IP Address. This is either the ISP or webhosting.
Because the whois information also lists the name + address details, it can be figured out upto a reasonable detailed GEO location to where the IP is residing.
If someone uses a major ISP in a country, the WHOIS information will tell anyone which country that person lives in. Some sites however seem to have a database that seem to be able to go up to city/street level about pinpointing where an IP Address originates from. Although these sites are not always accurate, there are ways to find someone's location.
But because someone's ISP can always be found by the IP Address, and the whois information always lists an email address to send abuse complaints to, it is always possible to find and punish people who are abusive on the internet in one way or the other.
Whenever you get an internet connection, you always have to agree with the terms, whicn state that no abusive activities are to take place. If complaints get to the ISP, they can suspend or even terminate the internet connection.
Answer: An IP address is often uniquely tied to you by your ISP. This means that if your ISP co-operates with whomever has the IP address, they can determine your identity and therefore what services you have been using on the Internet.
However you can never guarantee that the IP address you're connected to is actually tied to the identity you think it is. It is almost always possible that someone else has access to the IP, either because there's an insecure wireless access point there or because it's an open proxy/VPN service.
Regarding geo-location Often the fact that an ISP 'owns' or has access to a block of IP addresses means that it is quite well established who gets assigned an IP from this range. Hence it can be simple to identify an approximate location. However this comes down to how that ISP hands out IP addresses. Some will not pay any attention to geographic locations and therefore the precision of geographic location from the IP address can be very low.
Answer: An IP address is often a volatile thing - you may have one now, and a different one in five minutes. Further, it may only get to your "front door". If you have a wireless router at home, you may have several computers behind this wall. The IP address doesn't tell anyone which computer is connected.
The best information about the location of an IP address comes from the switch / router closest to your house. Your ISP is responsible for that "last mile" of your connection - and will know exactly what router (billing address) a particular IP address is associated with.
Importantly though, this might only be the entry point (firewall) to another network. My (large) employes uses a proxy server in California. When I connect to the internet while at work, you might think I was in CA but that is not so.
Incidentally a "trace route" can be used to find the intermediate points (routers) that get the signal from one place to another. These intermediate points often have names that tell you something about the routers' position - without needing to look things up.
Answer: When not using so-called IPv6 Privacy Addressing, IPv6 addresses might include your computer's unique MAC address. That means that the very same MAC address will be presented to the server, no matter to which IPv6 network (home, school, office, internet café) your computer is connected. A super cookie, if you will.
Luckily, nowadays IPv6 Privacy Addressing is the default setting for Windows and OS X. I don't know about Linux, and I don't know what the setting is for old systems that were upgraded to recent versions of Windows or OS X.
Details (and how to prevent it) in How to avoid exposing my MAC address when using IPv6?


Once made, why would anybody need access to their own Horcruxes?

Question: I understand why other people might want to get their hands on someones Horcrux, but once the bad guy/gal has split his/her soul and a piece of it has been interned into whatever vessel that was chosen to be the Horcrux, why was there a need for the person that created it to be able to retrieve it?
Comment: I don't think that there is a canon answer, but presumably, there may be other powerful dark magic that can be done using the Horcruxes.
Answer: As far as we know, you don't physically need the Horcrux object in your possession to resurrect yourself (at-least not with the method which Voldemort and Wormtail used in Goblet of Fire).
Note: don't confuse the Diary as a physical means of re-incarnation; as I understand it, it was the 'memory' part which was doing the resurrecting (using Ginny's life-force) and not the Horcrux part1.
Why Make Them Accessible?
Let's look at it from Voldemort's point-of-view (since his is the only one we know of - Herpo's details are quite obscure and won't really help us here).
How do I know they are ok?
Voldemort needed to check-up on his Horcruxes once he found out that the trio was after them. I believe this is the primary reason to have them accessible.
Kill two birds with one stone
Some of his Horcruxes had more purpose than just tethering his soul to this world, showing his cunning by not wasting an opportunity to make use of these hard-gotten devices:

Diary: Had his memories instilled in it which was designed to entrap and ensnare any unsuspecting readers.
Locket: Had a powerful curse which would cause the possessor's mind to get corrupted, designed to perhaps cause dissension among his enemies (maybe even confuse someone enough to walk straight to Voldemort?)
Nagini: Very powerful snake, able to communicate with and do Voldemort's bidding, mobile so V could keep her close at hand.
Peverell/Gaunt Ring: Extremely powerful curse placed upon it; touch it and you die fiend!
Quirell: Questionable (Quirell was a quasi/temporary Horcrux), but similar to Nagini, he was used to perform Voldemort's bidding.

I see you're on my Horcruxes' tail, let me just put it over here now
Another reason to be able to access your Horcruxes is to be able to move them if need be. Lets say you catch wind of someone being on your Horxruxes' tail; if you cannot access them, then you have no chance of grabbing them and moving them to another place.
Just Tell Me The Answer Already
In summary; It doesn't seem like it is essential to access your Horcruxes after you have made them, although you may see fit to check-up on them, or change their location if you see that someone is on their tail.

1. Correct me if i'm wrong; but that is what I understand from this: https://scifi.stackexchange.com/a/27492/21267
Answer: As I mentioned in my answer to this very similar question a couple minutes ago, it seems to me that a Horcrux that is accessible can be used for resurrection (like Tom Riddle's diary). As such, it is not normally retrieved by the wizard who created it, but by one of their followers (as most of Voldemort's horcruxes were).
Comment: Tom Riddle's diary wasn't used for resurrection, though. That would have been a totally separate entity, there would have still been the disembodied Voldemort around. Also very few, if any, of Voldemort's followers knew he had Horcruxes; Lucius Malfoy had no idea what the diary was.
Answer: They would need to know where it is in case they have to use it. Beside that small fact they never had to know where the Horcrux was. Also, the Horcruxes are capable of taking care of themselves to a small degree, but they also would have to make sure they have the same amount of Horcruxes and that someone didn't destroy them.
Reasons for knowing the location of your Horcruxes.

The object is important to you (Nagini). A lot of Voldemorts plan include Nagini he needed her and liked her companionship.
You want to keep tabs on the item in case someone killed one. Voldemort did this at the end of the Deathly Hallows. He searched for them and saw that they were being humted, if he had of kept tabs on them earlier then the problem would not have presented itself.
You for some reason have to use it. Possibly to reincarnate yourself, or there is a spell that you want to do with it.But I don't know how canon this is.
Someone is hunting them and you want to relocate them. Or you just want to relaocate them, in general.

Voldemort (as a example) made his Horcruxes out of rare, expensive, magical items. His Horcruxes were his wealth, backup plan and pride( no one else made 7). I think those are good reasons to know the locations. I mean everybody knows the locations of there favorite items. So in Voldemort's case I would say he knew, becasue it was vanity and he wanted to check on them. And because he had a case of paranoia.
But for all service purposes, since he left his Horcruxes alone it does not appear that they needed addtional maintenance beside being created. So that is not a reason to access them.
However, if you are asking if it is a common practice that people keep a tab on there Horcruxes then I would say possibly no and yes.
Herpo the Foul was the other holder of the Horcrux and he had 1. That means that he could have held the Horcrux as a ring on his finger or cast it into a chasm that he had put a extension charm on. But either way not much is known regarding his preferences. But we do know he died, so he obviously didn't take good enough care of his horcruxes.


Proof that a well-ordering is $\omega$-saturated only if it is finite

Question: I wanna prove that an infinite well-ordering is not $\omega$-saturated. How can I do this? Does the proof involve Lowenheim-Skolem theorem?
Comment: The title and body of your question say different things.
Comment: Yep, that's better.
Comment: I think I corrected the title.
Comment: I mean, the type of arbitrarily long decreasing chain is an obvious giveaway.
Comment: Do you know the theorem that $\omega$-saturated models realize types in $\omega$-many free variables?
Comment: No, I haven't covered that.
Answer: A useful fact is that an $\aleph_0$-saturated structure realizes any (partial) type in $\omega$-many variables with parameters from a finite set. Ie, if $\mathfrak{M}$ is an $\aleph_0$-saturated $\mathcal{L}$-structure, $A\subseteq M$ is a finite subset, and $\Sigma(v_i)_{i\in\omega}$ is a finitely satisfiable set of $\mathcal{L}_A$ formulas in $\omega$-many free variables, then $\Sigma(v_i)_{i\in\omega}$ is realized in $\mathfrak{M}$. See eg here for a proof. (The analogous result holds for all infinite cardinals.)
This fact allows us to show your result. Let $(M,\leqslant)$ be an $\aleph_0$-saturated well-order, and consider the collection of formulas $\Sigma(v_i)_{i\in\omega}=\{v_i>v_j\}_{i<j\in\omega}$; in other words, $\Sigma(v_i)_{i\in\omega}$ expresses that the $v_i$ form a strictly descending chain. Since well orders are total orders, if $M$ is infinite then $\Sigma(v_i)_{i\in\omega}$ is finitely satisfiable, and hence by $\aleph_0$-saturation is satisfied by elements $m_0>m_1>\dots$ of $M$. But now the set $\{m_i\}_{i\in\omega}$ is a subset of $M$ without a least element, contradicting that $<$ is a well-order.
Answer: The class of well-orderings has a very nice property: it's exactly the class of structures which $(i)$ satisfy a particular first-order theory (the theory of linear orders) and $(ii)$ do not have a certain countable "bad configuration." Re: $(ii)$, the relevant "bad configuration" is:

 An infinite descending sequence.

The "saturation slogan" is, "all possible configrations exist." Of course it's more complicated than that - how much saturation you have determines the size of the conigurations the slogan applies to - but the basic idea is sound. This suggests the following line of attack: show that in an infinite $\omega$-saturated linear order, the countable "bad configuration" above must occur. This will be a corollary of a more general fact:

 Suppose $\mathcal{M}$ is $\omega$-saturated and $p$ is a finitely satisfiable $\omega$-type over $\mathcal{M}$. Then $p$ is realized in $M$.
Answer: As indicated in Noah's and Atticus's answers, and my comment, it's probably best to understand this fact as an application of the theorem that $\omega$-saturated models realize all consistent types in $\omega$-many free variables, so an $\omega$-saturated infinite linear order realizes the type of a countable descending sequence. But in this answer, I'll try to explain how you could answer the question without using the more general theorem.
Let $(M;<)$ be an infinite $\omega$-saturated linear order. We would like to show that $M$ is not a well-order. To do that, we would like to build an infinite descending chain in $M$. Let's try to do this by induction. We start by picking an element $a_0$. Then, having chosen $a_n$, we pick some $a_{n+1}<a_n$.
There's an obvious problem with this strategy: What ensures that there is some $a_{n+1}<a_n$? That is, if we accidentally pick $a_n$ to be the least element of the order, we get stuck. Saturation is no help: if $a_n$ is the least element of $M$, then $M\models \forall x\, \lnot (x < a_n)$, so the formula $x<a_n$ is just inconsistent.
Ok, so instead let's pick our sequence by induction, making sure to never pick the least element. But now what ensures that there is some $a_{n+1}<a_n$ which is not the least element? If $a_n$ has only one element below it in the order, we again get stuck. Clearly, we're in trouble if we ever pick an element $a_n$ with only finitely many elements below it in the order.
New idea: Make sure that each element of the sequence has infinitely many elements below it. So we start by picking an element $a_0$ with infinitely many elements below it. Then, having chosen $a_n$ with infinitely many elements below it, we pick some $a_{n+1}<a_n$ such that $a_{n+1}$ has infinitely many elements below it. But what ensures that there is some $a_{n+1}<a_n$ with infinitely many elements below it? Now saturation is comes into play: This condition amounts to realizing a certain type. Details after the break.

For each natural number $n>0$, let $\varphi_n(x)$ be the formula $$\exists y_1\dots\exists y_n\,(y_1<y_2<\dots<y_n < x).$$ Let $\Phi(x)$ be the partial type $\{\varphi_n(x)\mid n\in \mathbb{N}\}.$ This partial type says "$x$ has infinitely many elements below it".
We construct a descending sequence $(a_n)_{n\in \mathbb{N}}$ from $M$ by induction, ensuring that each $a_n$ realizes $\Phi(x)$.
In the base case, we show that $\Phi(x)$ is consistent by compactness: Given any finite subset $\{\varphi_n\mid 1\leq n\leq N\}$, we can pick any $N+1$ elements of $M$, listed in increasing order $b_1<\dots<b_{N+1}$ (here we use the fact that $M$ is infinite). Then taking $x = b_{N+1}$, the elements $b_i$ serve as witnesses for the variables $y_i$ in the formulas $\varphi_n$ with $n\leq N$. Since $\Phi(x)$ is consistent and $M$ is $\omega$-saturated, we can find $a_0\in M$ realizing $\Phi(x)$.
Now suppose by induction that we have picked $a_n$ realizing $\Phi(x)$. We show that the partial type $\Phi(x)\cup \{x<a_n\}$ is consistent by compactness: Given any finite subset $\{\varphi_n\mid 1\leq n\leq N\}$, recall that $a_n$ satisfies $\varphi_{N+1}$. So we can find $b_1<\dots<b_{N+1}<a_n$ in $M$. Then taking $x = b_{N+1}$, we have $x < a_n$, and the elements $b_i$ serve as witnesses for the variables $y_i$ in the formulas $\varphi_n$ with $n\leq N$. Since $\Phi(x)\cup \{x<a_n\}$ is consistent and $M$ is $\omega$-saturated, we can find $a_{n+1}\in M$ realizing $\Phi(x)\cup \{x<a_n\}$.


PHP Determine If Monday or Thursday is next from date provided

Question: Is PHP able to determine if a Monday or Thursday comes next and what the date ('YYYY-MM-DD') will be if I provide it with a date to start from?
Answer: Calculate the date of the next Monday and Thursday using a given date:
$start = new DateTimeImmutable('2023-01-08');
$monday = $start->modify("next monday");
$thursday = $start->modify("next thursday");

Using diff with an if statement display the result
if ( ($start->diff($monday)->days < $start->diff($thursday)->days ) ) {
    echo "monday";
} else {
    echo "thursday";
}

Answer: Next monday or thursday is the smaller of the two. If DateTime is used instead of DateTimeImmutable, copies (clone) from $start must be used.
$start = new DateTime('2022-11-19');
$nextMondayOrThursday = Min(
  (clone $start)->modify("next monday"),
  (clone $start)->modify("next thursday")
);

echo $nextMondayOrThursday->format('l, d F Y');
//Monday, 21 November 2022

Try self: https://3v4l.org/Y5sE5
A solution that specifies the short form of the days of the week in an array offers more flexibility. In this way, 3 or more days of the week can also be specified.
$start = date_create('2022-11-20');
$next = ['Mon','Thu','Fri'];

while(!in_array($start->modify('+1 Day')->format('D'),$next));

echo $start->format('l, d F Y');

It is recommended to ensure that the while loop does not run indefinitely by validating $next.
Future times such as next Monday or next Thursday can also be conveniently encoded as a string using cron syntax. For Monday or Thursday 00:00 the string is:
'0 0 * * 1,4'

The days of the week are written as digits from 0 for Sunday to 6 for Saturday. A big advantage is that these strings can easily be stored anywhere like in a database. There are a number of solutions to implement it in PHP (e.g. cron-expression). Just an example using the Jspit\Dt class:
use Jspit\Dt;

$cron = '0 0 * * 1,4';
$date = Dt::create('2022-11-21')->NextCron($cron);
//"2022-11-24 00:00:00.000000"

Comment: I added notes about the cronjob keyword to my answer.
Comment: Thank you for helping me future proof this.  This is part of the cron job for creating new editions of the word game I'll be marketing to serve the disabled, elderly and those living with a chronic illness.


Measuring small amounts of lift in a wind tunnel

Question: I am doing a science project on how paper airplanes generate lift, and I need to know how to  measure the amount of lift generated by the plane.
Comment: For the record, I think off-topic by way of engineering is meant to exclude "what is the most cost-effective way to manufacture this?" questions, not "how do I perform a physical measurement?" questions.
Comment: And the question is how to measure the lift?
Comment: @ACuriousMind Yes.
Answer: When you throw a paper airplane and it flies in a more or less straight line, its lift is equal to its weight, because it is traveling in a straight line.
So all you have to do is weigh the airplane.
The more tricky problem is to know its velocity and it's angle of attack.
Lift goes as velocity squared, for a given wing area and angle of attack.
So, for example, if you reduce wing area by a factor of four, you will increase velocity by a factor of two, to fly in a straight line, at a particular angle of attack.
The angle of attack is controlled by the elevators at the tail of the plane.
If you angle them up, the plane will be at a higher angle of attack, so it will fly slower to get the same lift.
What you need to do is 1) measure the weight of the plane, and then 2) determine the relationship between wing area and velocity to achieve straight-line flight, for a given setting of the elevators.
Read this site.
It's great reading and will really educate you about flying.


Beam on top and beam at the bottom on a music sheet

Question: I am new to sheet music. I am trying to recreate in Guitar Pro, corresponding to guitar strumming, the music sheet notation of the picture below. The timing is 4/4 and tempo is 120. I am unable to create or understand the meaning of a top and bottom beam together.
Comment: This notation is not great. I hope Aaron explained the meaning of the beams. The rhythm notation is not typical. But also the tabulature doesn't match the score, so it's a bit confusing as a whole. I'm surprising Guitar Pro produced that, since it should make correct tabulature automatically...
Comment: @user1079505 I've never used Guitar Pro, but I've seen lots of situations that suggest that unless you *really* know what you're doing, how notation works and how to fix those issues, it really doesn't care that much about these *fundamental* aspects.
Answer: Beam meaning
The top beam, attached to all of the Ds, indicates that those notes constitute one "voice". The notes (chords) attached to the bottom beams are additional "voices".
Imagine instead of guitar, this were a work for choir with soloist. The top "voice" would be sung by the soloist, and the bottom "voices" would be sung by the different parts of the choir.
All the vertically-aligned notes are still played simultaneously. The way the separate beams affect the playing of the passage is that the "lead singer" voice — the Ds — should be played a bit louder than the "backup singer" voices.
Guitar Pro

You can access and enable the Multivoice feature in Guitar Pro 7 from the Edition Palet located on the left of your window, or even access it from the Edit > Voices menu. (Guitar Pro 7.5 support page)

How it should look when done properly
(Note: The notation is correct, but I didn't set things up to make the tablature match the OP version.)

This second version is also acceptable and, IMO, easier to read:
Comment: I prefer your version 1, except that it would be very easy to imagine that the 2nd and 3rd D's are tied, when they are not actually tied.


What is the command to know my PulseAudio version?

Question: What is the command to know my PulseAudio version?
Answer: pulseaudio --version

This should give you an output something like this:
pulseaudio 0.9.21-63-gd3efa-dirty

To get the version of the Debian package you have installed, use:
dpkg -l pulseaudio

which will give output like this:
||/ Name             Version          Description
+++-================-================-================================================
ii  pulseaudio       1:0.9.22~0.9.21+ PulseAudio sound server

Comment: Do you know how to upgrade to version 0.9.22?
Comment: I don't but someone else might - you should ask this as another question.


Can I install the all the software included with Ubuntu Studio on Kubuntu without losing the KDE environment?

Question: I am planning on installing Ubuntu on my new laptop, but I prefer the KDE Plasma 5 DE and I am likely going to use a lot of creation software like that which is included in Ubuntu Studio. I don't know if this is the right place to ask this but is there a way to install all the software from Ubuntu Studio in Kubuntu without switching my desktop environment? I just want the software.
Thanks in advance and I apologize if this isn't a good place to ask this!
Comment: Which applications do you want installed?   My experience with that combination was not that good.  Can be unstable.
Comment: I want to install at least Blender, OpenShot, FFMPEG, DVDStyler, GIMP, Ardour, and Audacity. Would it be easier to just install them individually?
Comment: GIMP, Ardour and Audacity will work just fine.  FFMPEG is part of Kubuntu-restricted-extras.  Don't know about Blender, OpenShot or DVDStyler. Yes, install individually.  Then check for any comaptibility issues after reboot.
Answer: This has been answered in a similar question here and on the Ubuntu Studio wiki (linked below), but since we want KDE rather than Gnome or XFCE (Ubuntu Studio's default DE) we'll do things a little differently. As of writing this, I'm running  Kubuntu 18.04 Bionic Beaver.

First Install Kubuntu, if it's not already on your system.
Open a terminal to update Kubuntu, install Ubuntu Studio packages and low latency kernel:
sudo apt-get update && sudo apt-get install ubuntustudio-audio ubuntustudio-audio-plugins ubuntustudio-graphics ubuntustudio-video linux-lowlatency

You may also want to add ubuntustudio-menus

Follow the configuration modifications section here to finish setting up Ubuntu Studio. note: most of the steps listed here are unnessecary for newer versions of ubuntu, however users should still be added to audio/video groups with  sudo adduser <username> video and  sudo adduser <username> audio

If you already have Ubuntu Studio installed, you can install KDE Plasma over Ubuntu Studio, although I do not recommend this as I've run into  problems like conflicting dependencies or things vanishing after upgrades. It just seems Unnecessary to have multiple desktop environments installed.


What's this brown growth on my yucca?

Question: There's a brown growth at the base of one of the leaves of my Yucca plant. Should I remove it? What is it?
Answer: This is an adventitious root.  Yucca's will grow roots out of the stem which can act to brace the main plant which can get quite heavy.
No action is necessary on your part.

What’s this brown growth on my yucca?


How to typset the Meitei Script in LaTeX and Overleaf?

Question: I am trying to write a language script which is represented by the UTF-16 code format: the Meitei script, occupying U+ABC0 to U+ABFF in the Unicode block. All i could find in the internet is related to UTF-8 or some special characters. How can I write the language script in LaTeX or Overleaf? Thanks.
Comment: Would that be the Meitei script or Meetei Mayek? You're in luck, Noto Sans Meetei Mayek is available on Overleaf, and so you can use XeLaTeX. I'll follow up with a quick answer and example; it would be great if you can edit your question to mention that you're specifically looking to typeset Meetei Mayek.
Comment: (Disclaimer: I'm a support personnel at Overleaf.) Overleaf converts all plain text files to UTF-8. Which language are you writing, by the way? You can contact firstname.lastname@example.com if you have a specific project there to debug.
Comment: @LianTzeLim The one I am trying to write has the code points range: U+ABC0 to U+ABFF in the Unicode block.
Comment: @LianTzeLim yes it is. Thanks a lot. If possible, please give an example of writing any unicode characters (UTF-8 or UTF-16) using the the code-points or hex code.
Answer: Overleaf converts plain text files to UTF-8. In this case the question is about typesetting the Meitei script (Meetei Mayek): this can be done using XeLaTeX, the fontspec package, and a suitable font e.g. Google Noto Sans Meetei Mayek.
(You would also need to have this font installed on your computer to view the Meitei glyphs in the example below and in the Overleaf editor panel.)
A small example:
\documentclass{article}
\usepackage{fontspec}

\begin{document}

This is some normal English text in Latin script. The next paragraph is Meitei:

{\fontspec{Noto Sans Meetei Mayek}
ꯀ ꯁ ꯂ ꯃ ꯄ ꯅ
\symbol{"ABC6}
}

Here we're back to English.
\end{document}

This example code needs to be compiled with XeLaTeX or LuaLaTeX. On Overleaf, you can click on the Overleaf menu icon above the file list panel, and then set "Compiler" to XeLaTeX or LuaLaTeX.
See this answer for more examples of entering characters by hexadecimal, decimal or octal code points.
Comment: Thanks for the answer. I get the symbol part and its working. But I couldn't write the Noto Sans font. I treid installing the font in my system, and tried typing some characters. Its typing only Roman characters.
Comment: Do you mean you want to type the Meitei characters from your keyboard? You may need a specific input method to do that, which is beyond the scope of LaTeX, tex.sx and Overleaf... I copied the characters from Wikipedia so I didn't do any typing.
Comment: Hi Lim, I have been trying to run the above code in overleaf. Till now, I have been compling with pdfLatex compiler. But when I changed to XeLatex compiler, it keeps on showing timeout. And when I changed to LuaLatex compiler, all the refrences cite, table and figures are showing error. And the symbols are also not showing at all. I did't get timeout in LuaLatex. Could youplease guide me on how to fix this?
Comment: You can contact email@example.com to let us know your project's URL, so that we can have a look at the project directly.


Community Moderator Elections: Can we get a subscription notification when each phase starts?

Question: I am new to the concept of these moderator elections. Is there any way I can subscribe to all all flavors of Stack Exchange websites for moderator elections? There is a badge caucus— "Visited an election during any phase of an active election". That's how I learned there elections exist.
Comment: Don't we already get notifications for the nomination and election phases?
Comment: For linking purposes: [Re-adding Election Notices to System Messages](http://meta.stackoverflow.com/posts/comments/434269)
Answer: When you are eligible to participate in an election phase, you get a notification in your global inbox on the site, no sign-up required.
Elections are also announced on the site's meta, in the community bulletin board on the site, and (sometimes, but we're trying to avoid it) in a system message banner.
There is no way right now to be notified of elections on any site, and I'm not sure what the point would be. Moderator elections are largely an internal matter on the site. They sometimes attract spectators from other sites, but for the most part, they're pretty self-contained.
You can read the gory details about the election setup and process here and see past (and current, though right now there aren't any) elections across the network here.
Comment: Note that you can also subscribe to email notifications for your global inbox, if you want more intrusive notifications of such things.
Answer: I think the best you can do is watch for posts tagged election on every meta site; you'll get other random posts about elections, and won't get one notification per phase, but you will hear about every election since they get mentioned on meta. There's a filter here you can watch (which is how I found this post in the first place)


How do I attach vertical blinds to drywall with a half inch space then a concrete wall behind it. They are heavy

Question: How do I hang vertical patio door blinds on drywall with a half in space behind it then a concrete wall.  I cannot access the concrete directly to manipulate certain screws etc.
Answer: Here's what I would do.
Make a standoff spacer to bridge the space between the concrete and drywall surface.
Make the spacer out of a piece of small PVC tube, or better, aluminum tube, say 3/8ths or 1/2" diameter at most, cut to match the distance from the concrete to the outside drywall surface.
Drill through the drywall into the concrete with a masonary drill to accept a Tapcon screw. Make sure the drilled hole is deep enough (maybe half an inch longer than the screw's reach) and work the bit in an out to clean as much dust as possible from the hole.
Enlarge the hole in the drywall to accept the spacer. Use something for a depth gauge that you can push in against the concrete and mark the drywall surface, pull it out and measure the distance, and cut the spacer to that length. Install the Tapcon through the bracket and spacer into the concrete.
Answer: One solution would be to get some 2" Tapcons from your home store and the appropriate drill bit. Drill your mounting holes 2.25" deep and screw the mounting brackets in. The Tapcons will go through the drywall, the half inch space and have an inch of concrete to secure the brackets/blinds.
Comment: One problem with this is if you tighten the tapcon too much you'll crack the drywall. When tightening, if you see the drywall starting to move, stop tightening.
Answer: This may not be ideal, or workable for some, but I've done this in many windows in a block home: Attach a 1x4 to the top inside of the window, using adhesive caulk and two tapcon screws. This gives me a wood surface that's much easier to screw into for hanging blinds. I caulked the slight gap between board and plaster wall and painted the board to match the wall.


Very strange rectangles in some of my photos. Canon 5D Mark IV. Just started yesterday. What is it?

Question:
Only 5 of 180 photos did this kind of display.
Comment: Please test with a different SD card and see if the problem goes away.
Comment: What is your example image? A screenshot? From what kind of device? How was the image saved on your memory card (raw or JPEG)? How was it transferred to the display device?
Comment: What is your example image, a screenshot? How was the image saved on your memory card (raw or JPEG)? How was it transferred to the display device (in terms of hardware connections)?
Comment: I did a "comp" between the file on the SD card and the file system. They are identical. I use Lightroom to import the photos. When I first open the file from the Lightroom grid view, it looks OK. But within a few seconds of processing, the rectangle appears. I see the same thing happen with Canon's "Digital Photo Professional" program.
Answer: You didn't say, but it seems you are saving raw files. The affected files are being corrupted in some way. With only five out of 180 photos affected, the best guess is that the memory card itself is the culprit.It is remotely possible that the files are being corrupted by the camera's data pipeline. It's also possible that a hardware issue between the memory card and your computer is to blame. If you are using a card reader, the reader or the connecting cable could have an intermittent fault. If the same hardware connection is being used to transfer images from the card to your internal file system, it is remotely possible the images are intact on the card and being corrupted in the transfer as well as when the system attempts to read them.
What you are seeing when you first open each image is a jpeg "preview" image that is processed by the camera and attached to the raw image file when it is saved on your card. You are seeing the rectangles when Lr or DPP replace the jpeg preview with their own processing of the raw image data.
Your next step should be to try a different memory card and see if the issue continues to present itself or not.
Comment: Yes, I will kill off my old 32GB memory card. I remember when it was HUGE.


Where can I find a complete list of Windows STOP error codes?

Question: Its all in the title, really. I need a list of STOP (blue screen) error codes so that when I see one, I can look it up. You'd think this would be documented somewhere, but I'll be darned if I can find it. Anyone got a link?
Answer: This site might be of help to you: Troubleshooting Windows STOP Messages
And an even better site with a lot of information is: Bug Check Code Reference (Windows Debuggers)
Answer: Not very official documentation, but quite a useful collection of stop codes nonetheless:
List of Blue Screen Error Codes (STOP Codes)
Answer: James Eschelman stopped updating the STOP codes at Aumha.org in 2007. In 2009 I took up the task of keeping an updated list here:
http://www.carrona.org/bsodindx.html

Updated through W10 (10240) - 480+ STOP codes

Working page for W10 (10586) at:  http://www.carrona.org/bsodindxNEW.html

Added another 48ish STOP codes