Can a helicopter autorotate from a hover?

Question: If a hovering helicopter's engine fails, can the helicopter autorotate as it could if it were moving?
Comment: Vertical autorotation is not a [good idea](https://aviation.stackexchange.com/q/42969/21091)
Comment: Do you have a general altitude in mind as to when the engine fails during the hover?
Comment: @757toga no, I hadn't thought of that. I'm interested in theoretical possibility more than practicality, so assume there is enough altitude for any maneuvers to be made.
Answer: Look up the 'Dead Man's Curve' for a helicopter.  It gives the zones for safe flying (in event of engine failure) as a function of speed and altitude.  Hover is the left-side of the curve (where V=0).
Very low, a helicopter can safely hover and have an engine failure -- from low altitude, it won't pick up enough speed to have terrible consequences when it reaches the ground.
Up high, a helicopter has enough altitude to use that potential energy to safely autorotate.
There is a region of altitudes in-between where the helicopter can not safely hover in the event of an engine failure.
Answer: The helicopter will be able to maintain rotor speed using the same "windmill" effect.  However, unless the autorotation is initiated just slightly above the ground, the descent rate will become too large to be survivable (and control may be lost due to vortex ring state).
If the helicopter is moving too slow (KTAS < 50), but is above the H-V curve in altitude, the pilot would accelerate to a survivable autorotation speed where (1) the rate of descent will be smaller, (2) range will hopefully allow the pilot to find a suitable landing spot, and (3) forward speed will be used to boost lift and rotor speed in the flare, just before landing.
A pilot should be aware of many airspeeds related to this: (1) the minimum airspeed to use in autorotation, (2) the speed that will provide the minimum descent rate in autorotation, (3) the speed that will provide max range in autorotation and (4) the max speed in autorotation.
Answer: Yes it could, because the process of auto-rotation does not rely on the forward motion. Think of auto-rotation as gliding, but in a helical path.
A glider has to constantly descend¹ to maintain airspeed - when descending, a component of its weight counters the drag. This allows the plane to stay in motion, thereby maintaining the airflow needed by the wing to produce lift.
In a helicopter, this wing rotates while descending, and is called a 'rotor'. Other than this difference, the mechanics are identical to that in gliding. Since the blades follow a helical path, a component of helicopter's weight opposes the "drag" (or torque) of the rotor.
This allows the rotor to maintain its rotation speed, providing the blades the airflow they need to produce lift. Since blades don't rely on forward motion of the helicopter to produce this lift, auto-rotation is perfectly possible without forward motion.

¹The glider must constantly descend relative to the ambient air, not necessarily relative to the ground.
Comment: I think this answer misses the point. Forward speed helps immensely during the landing flare. As forward speed bleeds off, it adds energy to the rotor (just like an autogyro) and that energy can be used to arrest the vertical descent. While you can autorotate without forward speed, you will impact the ground without much flare.
Comment: @RainerP. You are right, but OP just asked if it was theoretically possible to have auto-rotation without forward speed; he didn't mention anything about landing.


I can't use my Ember attack with Fennekin

Question: So when I try to use my ember attack, it doesn't work, it says 0/25, and I understand I used it too much, but how do I raise it to 25/25 again?
Answer: What's happening is that the PP (Power Points) has run out on the move.
To fix that, you can:

Use an Ether/Leppa Berry or Max Ether on the move
Use an Elixir or Max Elixir on Fennekin
Go to a Pokemon Center and heal
Comment: With the last option being the easiest and cheapest!
Comment: @Mathias711. But only for Pokémon. We humans don't get free healthcare... A shame that is.
Comment: @Nolonar Depends where in the world you are :)
Answer: All moves have a limited amount of PP (Power Points), Ember has 25, which means it can be used 25 times before it needs to be recharged. Typically, the more powerful the move, the lower it's maximum PP. If a Pokemon runs out of PP for all of its moves, it will attack by struggling, which damages it's self as well as the opponent.
To regain PP, you have a few options: Ether or Elixer items. Ethers will restore the PP of a single move, while Elixers will restore PP to all of the selected Pokemon's moves. Bringing your Pokemon to a Pokecenter, or any other location that fully restores them will also work.
You can increase the maximum PP a move has by using PP Up items, or PP Max. However, these items are somewhat rare, and should be saved for more powerful moves that your Pokemon learns as you train it.
Answer: This is called PP or Power Points. They tell you how many times a move can be used. You must use a PP-restoring item such as an Elixir, which restores 10 PP to all moves. or Ether, which restores 10 PP for one move. There is also Max Ethers, which restore all of the PP of one move and Max Elixirs, which restore all PP to all moves.. You can buy these at your local Poke Mart. A Leppa Berry can also restore 10 PP. These can be found on Berry tree. You can also restore both your PP and HP at a Pokemon Center. To raise PP, a PP Up will raise the number of times you can move by one and a PP Max will raise it to the max.
Hope this helps!


Help me identify a science fiction story from the 60's or 70's. It begins with some post-human types having a "ball"

Question: It is then explained that the "ball" is about a mile in diameter.  I don't remember anything about the plot – just this opening gag line.  It was in an anthology I read as a kid. Possibly by a well-known author.  Of course the terms trans-human and post-human aren't used, but the concept is there.
Comment: _Dancers at the End of Time_ by Michael Moorcock immediately comes to mind.
Comment: @Spencer - They certainly had some epic-scale objects, but I don't remember a giant ball. Maybe in one of the later books of the series?
Comment: Hi, welcome to SF&F!  If you can come up with any other details - even if you can remember what the cover of the anthology looked like or any of the other stories in it - it would help a lot.  Check out the [suggestions](https://scifi.meta.stackexchange.com/a/9337/101407) to see if they can help you recall anything else to [edit] in.
Comment: Yes, I think you're on to something, Spencer.  Did Moorcock write short stories in that universe or am I thinking about a novel?
Comment: @Scramjet Yes, there were several.  I need to dig.
Comment: @Spencer Looking at descriptions of the End of Time, the story being about decadent immortals who like to throw parties sounds about right.
Answer: There is a line like this in Legends from the End of Time by Michael Moorcock.

My Lady Charlotina's Ball must have been at least a mile in circumference, set against the soft tones of a summer twilight, red-gold and transparent so that, as one approached, the guests who had already arrived could be seen standing upon the inner wall, clad in creations extravagant even at the End of Time.

The Ball is literally a ball:

The Ball itself was inclined to roll a little, but those inside it were undisturbed; their footing was firm, thanks to My Lady Charlotina's artistry. The Ball was entered by means of a number of sphincterish openings, placed more or less at random in its outer wall. At the very centre of the Ball, on a floating platform, sat an orchestra comprised of the choicest musicians, out of a myriad of ages and planets, from My Lady's great menagerie (she specialized, currently, in artists).

This is the fourth book in Moorcock's Dancers at the End of Time series, and that fits the description of post humans since the protagonists are immortal decadents with nothing better to do than mess around and indulge their fantasies.
Comment: _Legends_ is an anthology of previously published novellas.  This one is called "Pale-Roses" and was originally in _[New Worlds 7](http://www.isfdb.org/cgi-bin/pl.cgi?23797)_.  I read it in _OMNI_ but I can't find the listing.
Comment: Thanks @John Rennie!  I must have picked up the 4th book unknowingly.


Color Calibration in Ubuntu and Xerox 8570 printer

Question: The thumbnail size images show normal color of object. After double-clicking an image, appears off-color like it's shaded red. Printed photos are also off color. Ubuntu color setting states Xerox Phaser 8500DN is uncalibrated. I've been searching and searching and can't find any hints on how to fix this. My knowledge regarding Ubuntu (& PC's) is on the low side of moderate. I don't want to keep downloading stuff I know nothing about - any suggestions out there beside throwing my computer out the window.
Answer: Little theory. For each device, printer or monitor there is a ICC profile (ICC profile is a set of data that characterizes a color input or output device, or a color space (C) Wiki). If you want to find a match you can change settings of printer or monitor or both. As I understand you want to change only monitor settings. First of all try using different viewers, they can give different output on the same monitor. Second, check that problem is realy in output devices, not in pictures itself.
There is a usefull tool for managing ICC profiles, for sure it will help with monitor, check if it can help with printer. https://apps.ubuntu.com/cat/applications/natty/gnome-color-manager/


Emmc replace (inbuilt memory card mounter) with external memory card mounter

Question: Hi friends i want to increase inbuilt memory at infinity level by replacing Emmc holder (inbuilt memory card mounter) with external memory card mounter, can it be possible? If yes how? If NO Why??
Answer: There's more to just internal storage in eMMC - it holds a vast array of partitions. Here's an example I just got from my phone.

As you can see many of those "blocks" (mmcblk_p_) correspond to major functions of the phone. Even more, such as EFS (stores baseband info) and sbl* (bootloader) aren't even listed here because they're even more crucial, are loaded at boot time, and doesn't allow a regular user to tinker with.
Now imagine all those partitions all shifted to an external, removable source. With proprietary formats, is it possible in the first place? Even if yes, what if one removes the media in the middle of booting up?
And lastly, eMMCs (and recently-introduced UFS2s) are significantly faster than your average SD card. Running Android on a Class 4 SD card is actually a painful experience - if you don't believe it, burn Android-x86 onto an SD card and try booting from it.
Comment: But sir if we remove card in middle of booting system hanges  or stopped, just like a raspberry pie....ya it may hamper system but still I need to know how to replace it??
Comment: "with proprietary formats" - try and see if an SD card can have so many partitions of different and sometimes proprietary format.
Comment: Not understood, yes raspberry pie is proprietor formats but it's not restricting us to understand the concept of how?
Comment: @user190688 Out of my league, so I'll leave this up to someone else who's more knowledgeable. There are possible explanations but they're all just my speculations so I'd not bother confusing you with those.
Comment: To others: feel free to edit this if you have the privilege.


DMS vs Numeric format for latitude and longitude

Question: Is DMS format still widely used in industries these days? It seems to be way less suitable to operate comparing degrees with floating point.
Answer: Yes, it is still widely used. All navigational devices and GIS software offers DMS format for displaying, saving and loading data. And many consumer oriented platforms use it as the main or only coordinate format.
Beside practical things like the fact that a minute of latitude is one nautical mile. The main reason is that coordinates in DMS format are easily identified as coordinates by anyone. They are more "user friendly" in that sense.
There  are  many reasons to choose other formats or other  projections (like UTM/UPS), but it is a fact that DMS is still widely used.
Nevertheless, I'm sure all software make calculations with decimal degrees. DMS is used only for display and storing data in human-readable formats.
Comment: The minute-to-nautical-mile convertion is wrong. One minute of latitude is *approximately* one nautical mile **at the Equator** (~1800m). It is ~ 1300m around London latitude, and zero at the poles. It may have been a rough estimate centuries ago, but is entirely useless today, especially in a GIS context.
Comment: @RobertoRibeiro That's wrong. A minute in latitude is a nautical mile in the whole globe. What changes is a minute of LONGITUDE.
In nautical navigation (as well as in aeronautical navigation) despite the use of GPS they still use older methods. In the case of nautical navigation, they still use paper charts is common, and they measure distances, and set routes by setting theirs dividers to the rigth size in miles using the latitude grid of the map.
Comment: You are correct in the latitude/longitude difference, my mistake. However, the case still stands for the usage, or lack thereof, of DMS in GIS context. Especifically given that OP is talking about operating in one format or another, and not merely using it to print grids in a chart.
Answer: Not only is it still relatively common, but other forms of representation are also possible. Many GNSS (usually GPS) systems will report DD MM.mmmm (degrees decimal minutes). There are also packed and unpacked forms of DMS, DDMMSS.ssss or DD MM SS.ssss.
For computer processing, you're correct that just maintaining decimal degrees is easiest and only using other representation for display.


Good references for K-theory of modular curves?

Question: The title says it. I am looking for a good exposition on the K-theory of the curves $X_{i}(N)$, $Y_{i}(N)$, where $i\in\{0,1\}$.
I have some background in $K$-theory and also some background in modular curves.
Comment: Beilinson's "Higher regulators and values of L-functions", J. Sov. Math. 30 (1985) is supposed to be a good (general) reference here.
Comment: Maybe you should specify a little more explicitly what you are looking for. Beilinson's conjectures give a fairly general picture of what is expected of the K-theory (in particular of modular curves). Besides Beilinson's papers on higher regulators for modular curves, there are a couple of expository papers on Beilinson's conjectures by Nekovar, Deninger-Scholl, Schappacher-Scholl, etc. There is quite a follow-up literature on construction of special elements in K-groups, but essentially Beilinson's conjectures remain unproved even in the curve case.
Answer: I wouldn't recommend Beilinson's 1985 paper as a general reference -- it's terrifyingly compressed, developing an entire new subject in a single short paper, and crashes through the necessary material on modular curves in a couple of sentences.
A much gentler reference would be Flach's 1992 Inventiones paper "A finiteness theorem for the symmetric square of an elliptic curve", which gives a much fuller introduction to K-groups, Gersten complexes, regulator maps etc. You could also try looking at parts of my paper with Lei and Zerbes "Euler systems for Rankin--Selberg convolutions" for more on this.


Relay switching

Question: What will happen if we switch an electro mechanical relay more than the specified frequency? Eg if we switch a relay from On and then back to OFF before it's operate time and vice versa  , how relay would behave .
Comment: That would be a buzzer. You could adjust the amount of travel to change the volume. Many old style intercom sets use a buzzer like this.
Answer:
more than the specified frequency

That's by the very definition of "specified" ... not specified. It might switch, it might not switch. It might arc too much, it might break earlier, it might not.
Comment: Yeah, it might not have enough time to completely make or break the connection as it takes some time for the switch to physically move to the open or close position. If you really need a faster switch, use something electronic instead like a MOSFET
Comment: What type of load are you switching, it will have a bearing on the life of the relay.
Comment: can it possibly get contact weld permanently ? load is smps.  on pressing switch, smps turns on and off.
Comment: @Design4ec that's the thing: nobody can ever tell you. You're using it outside specifications.
Answer: When talking about "switching", one must distinguish between toggling electrical input vs. toggling the contact.
You can wiggle input voltage to a relay as fast as you want, but that doesn't mean that the contact will actually close or even move much.
For the contact to close quicker, you need to increase the drive voltage to counteract the coil inductance and to increase the magnetic field strength to overcome the inertia forces.
In most relays, you can thus close the NO contacts almost arbitrarily quickly by raising the initial coil voltage – until the wire insulation in the coil dielectrically breaks down, and/or until a ham radio operator living 2 miles away calls the FCC on you :)
I've driven some 12V rated reed relay coils from short 300-500VDC pulses, bringing the contact closure times into the 10-30us ballpark. This has limited uses – mostly just a silly exercise to demonstrate ultra-low-latency arithmetic on relays. You can make the relays do it about as fast as early microprocessors, but only once every few seconds :)
The downside really is that you can't do this to the poor relays very often. The contact opening times cannot be generally improved without constructing the relay differently – so the "recycle" frequency is still rather low. Close fast, open slow.
Relays have mechanical self-resonance frequencies at which their contacts close with more force than normally. This usually improves contact life vs same non-reactive load switched at a lower frequency.
The resonant frequency is higher than the maximum operate frequency specified in the datasheet. It's not very practical though, as due to drift/aging, the relay requires closed-loop control to stay on the resonance.
It also forces you to essentially have the relay act as a fixed duty cycle, fixed frequency PWM driver. Most relays are used in circuits where this mode of operation just doesn't arise and is of no use.
The effective slew rate of current at contact closure and contact opening has big effect on contact life – the faster the rise/fall time, the less wear on the contact. The faster and more forcefully the contact closes, the better. Thus, generally, it makes sense to close the relay as fast as possible without over stressing the armature – this is applicable to small armature relays and reed relays. Larger armature relays and contractors typically fail due to fatigue fractures of the armature if you try that.
Many relays have electrical time constants not much better than the mechanical ones, so faster switching requires higher coil voltages to improve contact life. This trades off thermal fatigue on the coil for contact life. Whether such trade off is warranted depends on the application.
And, finally, what any particular relay does in "off-datasheet" application is up to you to figure out. You have to develop a relay test setup not unlike the ones used by the relay manufacturer R&D and production test departments. There's lots of good experimental science here that integrates mechanical, electrical, electromagnetic and statistical knowledge. If you want to build a product around such off-label use, you have lots of work ahead to make sure you understand the relay behavior almost better than the manufacturer does.

Relay switching


I must be treated different if I don't have childrens with my husband?

Question: I have a muslim husband and me I am not. When he asked me for marriage I told him that I will not make him chidrens because I am much older than him and the nature say it's word. I had before children's with my previous husband. My question is: should I be treated different from the womans who have children's with their husband from the point of view of food, clothes or other rights of muslims wifes? Should I be forced to give 50% of the expenses in the house because I don't have children's?
Comment: Muhammad pbuh had many wives and many of them had not children, and muhammad Pbuh gave them equal right
Answer: No, every wife should be taken care of and treated kindly it doesn't matter whether she has kids with her husband or not. Some of prophet Muhammad's (pbuh) wives had kids from their previous marriages and he did his best on treating them all equally. Allah knows best.
Answer: In Islam, husbands are responsible for their wives and the lady of the house is not obligated to contribute to the earnings. Her responsibilities for which she will be questioned is did she keep herself chaste from others, did she look after the house well and did she bring up the children to be good Muslims.
In short, you should NOT be forced to give 50% of the expenses in the house because you are not a mother (yet).
But please note Islam lays a great emphasis on family and children are the bond that generally keeps a family united. Moreover, children are a source of 'Sawab e Jariya' for the parents.
Unless, you have a medical reason, I recommend you reanalyze your views on not having a child.


Copying multiple cells in Excel

Question: I am new to this but I am trying to copy multiple cells in an excel workbook and paste them into a separate tab of the same workbook.

Above is a sample of what my spreadsheet looks like, but my spreadsheet has over 800 lines of data.
I need the names to be copied and put into column A of Sheet2 and then the account numbers into column D of Sheet2.
I have tried this 2 different ways.

Using below code:
Sheets("Sheet1").Select
Range("A1,A3,A5,A7,A9").Select
Range("A10").Activate
Selection.Copy
Sheets("Sheet2").Select
Range("A1").Select
ActiveSheet.Paste
Sheets("Sheet1").Select
Range("A2,A4,A6,A8,A10").Select
Range("A10").Activate
Selection.Copy
Sheets("Sheet2").Select
Range("D1").Select
ActiveSheet.Paste

This gives me a Compile Error Syntax Error.
Code #2
Range("A2").Select
Selection.Cut
Range("D1").Select
ActiveSheet.Paste
Range("A4").Select
Selection.Cut
Range("D3").Select
ActiveSheet.Paste
...

This is keeping them in the same tab, instead of pasting them into a separate tab (I would just copy them over later). I repeat this for each customer. This one gives me a range error that basically says it's too large. Unfortunately, I can't recreate it because I deleted it.

Does anyone have a simpler way of doing this that won't cause an error?
Answer: Try this is assuming your data is consistently alternating (Name,acount).
Sub marine()
    Dim lr As Long, i As Long
    Dim sh1 As Worksheet, sh2 As Worksheet
    '/* declare the worksheets and use variables in the rest of the code */
    Set sh1 = Sheets("Sheet1"): Set sh2 = Sheets("Sheet2")

    With sh1
        lr = .Range("A" & .Rows.Count).End(xlUp).Row '/* get the last row in Sheet1 */
        For i = 1 To lr '/* loop to all rows identified */
            If i Mod 2 = 1 Then '/* check if odd or even, copy in A if odd */
                .Range("A" & i).Copy _
                sh2.Range("A" & sh2.Rows.Count).End(xlUp).Offset(1, 0)
            Else '/* copy in D otherwise */
                .Range("A" & i).Copy _
                sh2.Range("D" & sh2.Rows.Count).End(xlUp).Offset(1, 0)
            End If
        Next
    End With
End Sub

Above copies data from Sheet1 to Sheet2 but leaves the 1st row blank.
Also, it always copy data on the last row of each column in Sheet2 (A and D). So another approach would be:
Sub ject()
    Dim lr As Long, i As Long, lr2 As Long
    Dim sh1 As Worksheet, sh2 As Worksheet
    Dim rNames As Range, rAcct As Range
    Set sh1 = Sheets("Sheet1"): Set sh2 = Sheets("Sheet2")

    With sh1
        lr = .Range("A" & .Rows.Count).End(xlUp).Row
        For i = 1 To lr
            If i Mod 2 = 1 Then
                If rNames Is Nothing Then '/* get all the cells with names */
                    Set rNames = .Range("A" & i)
                Else
                    Set rNames = Union(rNames, .Range("A" & i))
                End If
            Else
                If rAcct Is Nothing Then '/* get all the cells with accounts */
                    Set rAcct = .Range("A" & i)
                Else
                    Set rAcct = Union(rAcct, .Range("A" & i))
                End If
            End If
        Next
    End With

    With sh2
        '/* get the last filled Names column in Sheet2 */
        lr2 = .Range("A" & .Rows.Count).End(xlUp).Row
        rNames.Copy .Range("A" & lr2) '/* execute 1 time copy */
        rAcct.Copy .Range("D" & lr2) '/* execute 1 time copy */
    End With
End Sub

Above code ensures that the correct account is adjacent to the correct name.
And you might gain execution performance too since one(1) time copy is executed. HTH.
P.S. As much as possible, avoid using Select.
Answer: Logic I implemented is to loop until last row in Sheet1 in step of 2. Loop variable indicates always row with name, the following row is account number, so it's easy in a loop to assign these values to particular columns on the other sheet. Also, I used another variable j, which indicates consecutive rows in Sheet2.
Solution:
Sub CopyData()

Dim sourceWs As Worksheet, targetWs As Worksheet, i As Long, lastRow As Long, j As Long
j = 1
Set sourceWs = Worksheets("Sheet1")
Set targetWs = Worksheets("Sheet2")
lastRow = sourceWs.Cells(sourceWs.Rows.Count, 1).End(xlUp).Row

For i = 1 To lastRow Step 2
    targetWs.Cells(j, 1) = sourceWs.Cells(i, 1)
    targetWs.Cells(j, 4) = sourceWs.Cells(i + 1, 1)
    j = j + 1
Next

End Sub

Copying multiple cells in Excel


Mathematica function intersection points with 3D grid

Question: I need to produce a 3-dimensional equispaced grid over a given function in a way, that I can calculate intersection points of the function with the grids edges.
So my first question is how to produce this grid in Mathematica. I don't really get if this is possible by modeling the grid by Polygons? For this I might need some help then, too. Or can it be done by using more simple graphics objects like done here?
Comment: What do you exactly mean by grid?  A grid of points?  Gridlines?  A grid of planes?  Can you elaborate?  It's not clear to me what the "intersection of a function with a grid" is.  Also, exactly what kind of function do you have?
Comment: In general, I have a function giving a surface of an object. Now, I want to get the points at which this surface intersects with edges of some kind of grid. Its like getting the intersection of a plane going through a cube. I hope this makes it more clear...
Answer: Do I understand it correctly that you are looking for the intersection of an implicitly defined surface with planes (planes that could make up a grid)?
Suppose we have this surface ...
j = 1.25;

ContourPlot3D[
 x^4 + y^4 + z^4 - (x^2 + y^2 + z^2) == -2/5,
  {x, -j, j}, {y, -j, j}, {z, -j, j},
 Mesh -> False, Axes -> False, PlotPoints -> 30]

... and we want to visualize the intersection with the plane $x + 2y - z$:
ContourPlot3D[x + 2 y - z == 0, {x, -j, j}, {y, -j, j}, {z, -j, j},
 Axes -> False]

The simplest way is to use custom MeshFunctions with ContourPlot3D:
gr =ContourPlot3D[x^4 + y^4 + z^4 - (x^2 + y^2 + z^2) == -2/5,
   {x, -j, j}, {y, -j, j}, {z, -j, j},
   Axes -> False, PlotPoints -> 30,
   MeshFunctions -> {Function[{x, y, z}, x + 2 y - z]},
   Mesh -> {{0.}}, MeshStyle -> Thick]

Or take the intersections with planes parallel to $yz$:
Update: You can extract the coordinates for the points making up the lines like this: Cases[Normal[gr], _Line, Infinity].
Comment: Thanks for your answer! But I do not only need to get it as a Meshfunction because I need to calculate some points at which the function is intersected by the plane. Like in the answer below. Maybe you have a idea?
Comment: @MathM I realise this is not ideal, but you can extract the mesh lines from the graphics expression.  Try `Graphics3D@Cases[Normal[gr], _Line, Infinity]`, where `gr` contains the intersection line.  It'll show you the lines only.  Now write `Cases[Normal[gr], _Line, Infinity]` only and you'll see all coordinates that make up the line.  Caveat: since the methods that compute the mesh lines are meant for visualization, they are not going to give very precise results.  `MaxRecursion` and `PlotPoints` can be used to increase precision a bit (at significant cost to computation time).
Answer: Here's an answer to my interpretation of the question:
Start with a three dimensional grid of lines:
lines = With[{j = 1.25, h = 0.25}, {
  Table[Line[{{x, y, -j}, {x, y, j}}], {x, -j, j, h}, {y, -j, j, h}],
  Table[Line[{{x, -j, z}, {x, j, z}}], {x, -j, j, h}, {z, -j, j, h}],
  Table[Line[{{-j, y, z}, {j, y, z}}], {y, -j, j, h}, {z, -j, j, h}]
}];

Turn each line into rule which gives a parameterisation of the line:
rules = Flatten[lines] /. Line[{{ax_, ay_, az_}, {bx_, by_, bz_}}] ->
  {x -> ax + t (bx - ax), y -> ay + t (by - ay), z -> az + (bz - az)};

Now solve t in whatever implicit equation you have, and extract the valid solution points:
points = Cases[{x, y, z} /. # /. NSolve[x^4 + y^4 + z^4 - (x^2 + y^2 + z^2) == -2/5 /. #,
 t, Reals] & /@ rules, {_?NumberQ, _?NumberQ, _?NumberQ}, Infinity];

And plot the results:
Graphics3D[{PointSize[0.015], Point[points], Opacity[0.25], lines}, Boxed -> False]

Answer: From your description, it seems that you need to find the points of intersection between a set of lines and your function, the lines being part of your "grid."
When you say intersection with the grid's edges, I am assuming that you mean points on the grid lines in the following sense: each grid unit makes cube of regular dimensions. You want to determine the intersection of the function with cube faces.
If this is the case, all you need to do is solve for intersection with sets of plane equations.
Comment: That is exactly what I like to do! So you mean using the Mathematica Solve-Command with plane equations. But I don't really know how do it with my function (which is a implicit defined surface like Szabolcs showed us on the top of this page).


Position of the wavefunction's collapse

Question: When a wave function is said to "collapse" to a single point during a measurement, is there uncertainty about the point's position or is it known infinitely precisely?
Comment: It is known precisely, although in practice one cannot construct experiments that measure to arbitrarily precision.
Answer: "Wavefunction collapse" doesn't mean collapsing to a single point. The wavefunction is always defined at all points in space.
A better understanding is that if you were to make a position measurement of your system, then the wavefunction "collapses" to a state with definite position. This wavefunction is $0$ everywhere except at the location where the particle was observed to be at.
Now, the above is an idealization. States with definite position are not physical. So yes, there will be some inherent "uncertainty", and the "collapsed wavefunction" will be a spike of finite width centered at the location the particle was observed to be at. As before, the wavefunction is still defined at all points in space.
More generally, measurement of any observable (not just position) causes the system's state vector to "collapse" to an eigenstate associated with the measured value of the observable. So once you go more general, "collapse" doesn't refer to any sort of physical (position) space at all.
Comment: So in actual physicality , the measured wave function will still be a combination of multiple eigenstates but heavily weighted at some point, am I correct?
Comment: @Sidarth Yes, that is correct

Position of the wavefunction's collapse


How does the appearance of the analemma vary with latitude

Question: How does the appearance of an analemma vary with latitude?
This what I take to be a definition of an analemma: If the position of the sun is plotted or photographed at noon clock time (say UTC) every day from a position facing due south (in the northern hemisphere) it forms a figure of eight shape as in the picture.
Photographs usually show it with the long axis at an angle. My question is, assuming that the angle is related to the latitude of the camera, how does this angle vary as the camera is moved from the equator to the pole, and why?
Answer: Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times.

Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu

Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu
The best way to understand the effect of latitude on the shape of the analemma is to consider the effect of earth's axial tilt on the shape of the analemma.

If the Earth's orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.
The 23.5-degree axial tilt of the Earth affects the Sun's apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.

You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.
Comment: Thanks @aeroalias, that is a very nice answer, and I have accepted it over my own because of the trouble taken. I realise now that it all depends on what you mean by the "shape". It will appear different in photographs because of the impossibility of reproducing a curved object on a flat surface. I believe that the shape measured in degrees of latitude relative to the equator and longitude either side of due south will be the same wherever it is viewed.See the Wikipedia article and in particular: https://en.wikipedia.org/wiki/Analemma#/media/File:Analemma_Earth.png
Comment: _If the Earth's orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped._ This is wrong. Under those conditions, the analemma would be a straight line. At all latitudes, not only the equator.
Answer: It is an effect caused by the inclination of the precession axis. This axis takes about 23,000 years to complete. If we could live so long we would see the anemalema spinning in the sky
Answer: I think that the answer is that the analemma would always be vertical, at right angles to the equator, if photographed at midday. The apparent slope in photographs is due to perspective, and being taken at a different time of day, not at noon.
At the equator the analemma will be directly overhead at the equinoxes, and off this by the angle of the ecliptic at the solstices.
At the poles the analemma will be centred on the horizon, the top half giving the long period without night in the summer, and the obscured half the long winters without daylight.


2009 macbook pro boots Ubuntu USB but not elementary OS USB?

Question: I Currently has Fedora 23 running on a Macbook Pro 2009.  Works great, but I miss elementary OS, which I use daily on my tower.  I used dd to create a bootable USB of Ubuntu 15.04 and when I restart the mac and hold down the option key, the Ubuntu USB is recognized and I can boot into Ubuntu from the USB.
I did the same dd command with the elementary OS freya .iso:
sudo dd if=elementaryos-stable-0.3.1-amd64.20150903.iso of=/dev/sdb bs=1M

after than finished, I rebooted and held down the option key and only see my Fedora drive pop up.  I then tried a different USB as well as an external USB HDD, all of which never show up when holding down the option key on reboot.
Also note, I tried these elementary USBs on a Windows box I have and all USBs showed up, and were able to boot correctly into elementary OS from the USBs.
So my question is, is there a way to get a macbook to boot from USB into elementary OS freya?
Also, does anyone know what Ubuntu's ISO has that allows it to be recognized on mac hardware that other Linux distros might not have?
Answer: You can use Rufus or Etcher for create a new bootable usb you can go Fedora and install Etcher and create a new bootable usb.


SELF HELP EVICTION OF ADVERSE POSSESSOR

Question: Can I Forcibly remove an adverse possessor without Notice or and Opportunity to be heard, before time bared by the statute of limitation in Florida.
Comment: Can you [edit] the question to provide a lot more detail please.  You should also reword the question to be a general question of law rather than a request for specific legal advice (which is off-topic).
Answer: Absolutely not. You have to use the legal system, whereby the sheriff is the one who uses force if it is necessary and ordered by the court. You can file an action at your local courthouse. If you want to do this self-help style, figure out how to file a petition, and figure out what you are petitioning the court to do. First off, of course, you need to figure out what you really want. For example, do you want a squatter to leave your apparently abandoned house; do you want a fence removed from your property; or do you just want to be sure that he can't claim possession of a chunk of your land in 3 more years (but the fence doesn't bother you)?
Since you're apparently talking about removing a person from your property, you might start by calling the police. If this is a former tenant as opposed to a stranger who broke in, don't bother (police don't get into civil matters until the court tells them to), just start the appropriate legal process. You might be filing an unlawful detainer action, but it would be a slower eviction if the person is a tenant.
Comment: If he knows about it and objects it can't be adverse possession.

SELF HELP EVICTION OF ADVERSE POSSESSOR


dataSource outlet

Question: I'm wondering how to use outlet, dataSource on a UITableView class. I'm stuck because I can't view any data on the table. Does anyone know how? I've looked at other websites, but their tips don't work. I tried to add the dataSource to the UITableView class, using the interface builder application
Answer: The dataSource property of a UITableView is a delegate what you assign to the tableView... Then the tableView calls that assigned class when it comes to collecting data.. In order to get data to display on the tableView you need to implement the UITableViewDataSource delegate methods...
A good place to start would be checking out the UITableView Class Reference:
http://developer.apple.com/library/ios/documentation/uikit/reference/UITableView_Class/Reference/Reference.html - UITableView Class Reference
http://developer.apple.com/library/ios/#documentation/UIKit/Reference/UITableViewDelegate_Protocol/Reference/Reference.html - UITableViewDelegate Protocol Class Reference
http://developer.apple.com/library/ios/#documentation/uikit/reference/UITableViewDataSource_Protocol/Reference/Reference.html - UITableViewDataSource Protocol Class Reference
Also, take a look at this sample code:
http://developer.apple.com/library/ios/ipad/#samplecode/TableViewSuite/Introduction/Intro.html
Answer: The UITableViewDataSource and UITableViewDelegate are usually set on the ViewController. If you are using a UITableView within a UIViewController you need to implement the delegate and datasource in the interface file (*.h)
@interface YourViewController : UIViewController <UITableViewDelegate, UITableViewDataSource>

Once you have the delegate declared you can implement 3 methods to make the tableview work
-(NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section
-(UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
- (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView

dataSource outlet


What is the effect of the size specifier of logrotate?

Question: When using the size specifier with logrotate, are the files rotated as soon as the size gets bigger than what is specified, or does it rotate when it is run from the cron and the size is bigger than what is specified?
Answer: It happens when it logrotate is run from cron (normally daily).
Answer: The "size" parameter is taken into consideration whenever you run logrotate. Whether or not you're running it through an automated system like cron/anacron, that is up to you; you can execute it manually without such automated systems just as easily. Anytime that logrotate is executed, automated or otherwise, the logs in question will not be rotated until they are equal or greater than the size you specified, even if they have already crossed the time threshold you've specified. From man's definition of the size parameter:

Log files are rotated only if they grow bigger then size bytes. If size is followed by k, the size is assumed to be in kilobytes. If the M is used, the size is in megabytes, and if G is used, the size is in gigabytes. So size 100,  size 100k, size 100M and size 100G are all valid.

https://linux.die.net/man/8/logrotate


Java: Generic Static Multidimensional Arrays

Question: If it is possible, how can I create a static multidimensional array in Java with different primitive datatypes per dimension?
By static, I mean the primitive array that is not dynamic like an ArrayList would be.
Comment: `Object[] arr = {new Integer[]{}, new String[]{}, new Double[]{}};`
Comment: @BheshGurung Yup, I figured it out. It was rather evident. ;)
Comment: Yes, I saw that I was typing too slow.
Answer: You can't.
A multidimensional array is, by definition, an array of arrays of arrays .... of something. So there's no way for any of those dimensions except the last to be anything other than an array. At least, by the traditional definition anyway. But if you mean something else by "multidimensional array", you'll need to tell us what that is.
As for "static", that is a heavily overloaded word in programming and every language that I can think of uses it to mean something slightly different. In Java, static means "belongs to a class, rather than to instances of that class." Again, if you mean something else by "static" here, you'll need to tell us what that is.
Edit: As originally posted, the question didn't include the word "primitive". That changes it a bit. Indeed, I agree that it would be nice from a convenience standpoint if Java allowed arrays to be indexed by char or even an enum rather than just int. But it doesn't.
Comment: By a static, I mean an array that is not dynamic like an ArrayList. I'll add that in my question.
Comment: Sorry about not including the word "primitive".
Comment: By `primitive` and `static` he means nothing but `unchangeable in size`, I guess.
Answer: After some testing, I have a simple solution:
Object [][] array = new Object [10][2];
array[0][0] = 2;
array[0][1] = false;

Comment: what would be best way to extract values from such array? In C# its Convert.ToInt32 etc, not sure about java
Comment: Combining different types in an array is bad practice and should be avoided. You need to know what element in the array is of which type, something, which the compiler can't guarantee, so this is circumventing the type-system aka begging for trouble, and you have to use an additional technique to memorize the types/positions. At position array[1][0], will you again have an int, and at (1,1) an boolean?
Answer: Dimensions in an Array are always from type int. Think about it!
int a = 4;
int b = 5;

Shoe shoe = new Shoe (Color.RED, 42, "Leather");
Hat hat = new Hat (17, Color.Black);

Foo foo = foos[a][b];
Zilch pop = bars[shoe][hat]; // no go

If you have a multidimensional array of Foos, the first dimension is an Foo, the second an array of Foos, the third an array of array of Foo. The only variable type is the one at the bottom.
Edit after update of question:
Arrays aren't called static or primitive. Their size is fixed on initialization, and what they have in common with primitives is, that they are a buildin, which is threated special in some cases. They are - in contrast to the so called primitive types, which aren't that primitive (they have, for example, operators, exclusively for their own, like * / -) but meanwhile they are objects, but not declared in the library.
Call them build in-types.
Using Bhesh Gurung's trick:
Object[] arr = {new Integer[]{}, new String[]{}, new Double[]{}};

is begging for trouble, and it is not made of different datatypes per dimension. Let's start with the dimensions:
// One-dimensional object:
JPanel [] panels = new JPanel [3];
// Two-dimensional object:
JPanel [][] panels = new JPanel [3][10];

You have JPanels on the bottom level, and an Array of JPanel on the next dimension. You can add more dimension, and will always get an additional (Array of ...) wrapped around.
You can not mix different datatypes in an Array like int and char, or JPanel and JFrame, or int and JButton. Only if you abstract over the difference, and use an JComponent for JPanel and JFrame as common parent, but this will not work for the build-in types int, char, boolean and so on, because they aren't objects.
But can't you use autoboxing, and use Integer instead of int, Character instead of char, and then use Object as common parent class? Yes, you could, but then you're not using the primitives any more, and you're begging for troubles.
Dan is talking about a different thing - using differnt types for indexing in the multi-dimensional array:
byte  b = 120;
short s = 1000;
String o [][] = new String[b][s];
b = 7;
s = 9;
o[b][s] = "foobar";
String foo = o[b][s];

There is no problem using bytes or shorts, but you can't restrict the size of an Array by declaring it as byte or short. In most cases the boundaries of a build-in integer type will not fit to a datatype (think 365 days per year), especially, since all types might get negative, so bounds-checking is necessary although and can't be restricted to compile time.
But now to the trouble:
We could declare the array as two-dimensional from the beginning:
Object[][] ar2 = {
    new Integer [] {4, 5, 6},
    new String [] {"me", "and", "you"},
    new Character [] {'x', 'y', 'z'}};

That works fine, and makes the inner arrays accessible immediately without casting. But it is only known for the compiler, that the elements are Object arrays - the underlying type is abstracted away, and so we can write:
ar2[1][1] = 17; // expected: String
ar2[2][0] = "double you"; // expected: Char

This will compile flawlessly, but you're shooting yourself in the foot and get a Runtime exception for free.
Here is the source as a whole:
public class ArrOfMixedArr
{
    public static void main (String args[])
    {
        Object[] arr = {
            new Integer [] {1, 2, 3},
            new String [] {"you", "and", "me"},
            new Character [] {'a', 'b', 'c'}};
        show (arr);

        byte b = 7;
        short s = 9;
        String o [][] = new String[200][1000];
        o[b][s] = "foobar";
        String foo = o[b][s];

        Object[][] ar2 = {
            new Integer [] {4, 5, 6},
            new String [] {"me", "and", "you"},
            new Character [] {'x', 'y', 'z'}};
        show (ar2);

        // exeptions:
        ar2[1][1] = 17; // expected: String
        ar2[2][0] = "double you"; // expected: Char
    }

    public static void show (Object[] arr)
    {
        for (Object o : arr)
        {
            if (o instanceof Object[])
                show ((Object[]) o);
            else
                System.out.print (o.toString () + "\t");
        }
        System.out.println ();
    }
}

Now what is the solution?
If your base-types arrays of (int, byte, char, String, JPanel, ...) are of equal length, then you have something like a hidden Object, a database-row. Use a class instead:
class Shoe {
    byte size;
    String manufactor;
    java.math.BigDecimal price;
    java.awt.Color color;
}

Shoe [] shoes = new Shoe [7];

If you don't have different types of the same size, they might be unrelated, and should not be put in a common container.
Answer: Well, you could define an array of an array of ... an array of Objects (nested with as many levels as dimensions) and at the bottom level fill each array with a different type ... and then, when you need to extract a value, cast it to the appropriate type. Too much work for what is worth, really. Java is no good for this kind of things, being a statically-typed language.
Maybe you should reconsider, why would you need such a data structure.
Answer: You can get the effect by using an object array:
final static Object tryit[][] = {
        {'a',4},
        {'b',7},
        {'c',8},
};
@Test
public void accessArray( ) {
    for (int i = 0; i < tryit.length ; i++) {
        char letter = (Character)tryit[i][0];
        int value = (Integer)tryit[i][1];
        System.out.println(letter + " has value " + value);
    }
}

The "@Test" is JUnit annnotation.
Note this approach would be subject to NullPointer and ClassCast exceptions at runtime if the wrong data is entered in the array.


How is display devices connected to the integrated Intel HD Graphics?

Question: Many new Intel processors have got integrated Intel HD Graphics so that there is no need to have a separate graphics card in a system with such a processor. How is a display device, such as a monitor, connected to the integrated Intel HD Graphics? Can it be connected to a display device on a system without a separate graphics card, such as a desktop computer without a graphics card?
Comment: There have been motherboards with such connectors for decades.
Comment: Er, through the standard VGA/ DVI port.
Comment: @Sathya But where is the VGA/DVI port if you have no separate graphics card? Is there motherboards with such ports?
Comment: Any motherboard supporting Intel's HD graphics will have a VGA/DVI port connected right to the motherboard.
Answer: The 'adapter' device is sort-of absorbed into the chipset of the system.
Look at this Intel H67 chipset diagram as an example.

The blue HD Graphics block is in the gray Intel Core processors block and there is a link to the PCIe Graphics block to its left.
The integrated graphics path goes down to the H67 block and out on its left to the Digital Display path into your monitor.
If you use a discrete graphics card, the processor communicates with it through the PCIe path.

This is also stated in the Intel article extract after the diagram.
Note that we are not referring to things like Nvidia Optimus here.

From Intel article Common Misconceptions of Intel® Processor Graphics,

In discrete graphics devices, the graphics subsystem is contained primarily on the PCI Express bus. However, with Intel X3000, the graphics subsystem sometimes uses the CPU in addition to its own GPU resources (vis-à-vis the Graphics and Memory Controller Hub (GMCH), system memory, and the graphics core integrated into the chipset). With 3x00 series graphics, the CPU is used for the first stage of 3D processing (geometry operations), while the chipset handles the rest of the 3D processing thereby placing less reliance on the CPU. In utilizing system memory for system and graphics usage, Intel X3000 balances both resources, yielding optimal price/performance for our customers.
Comment: Let's talk about Optimus now?
Comment: @mhambra, there is already a lot of talk about Optimus :-) http://superuser.com/questions/282734/how-to-disable-nvidia-optimus-on-a-dell-xps-15/284348#284348
Answer: On motherboards with Intel's H67 chipset and Intel's Z68 chipset the display device can be connected to a port in the motherboard such as a VGA, DVI, HDMI or DisplayPort.

I learned this via Do I need a GPU to use a built computer?:

Do h67 motherboards have this
  "integrated video chip"?
The board itself does not, but the
  Intel CPU's that are compatible with
  the socket 1155 chipset do. Every H67
  based board will have a VGA/DVI port
  that's driven by the Intel integrated
  graphics.

And via Intel Core i5-2500, Core i5-2400 and Core i5-2300 CPU Review. Page 4:

The differences between P67 and H67
  chipsets are quite logical. The first
  one is targeted for high-performance
  desktops, while the second one – for
  systems working with integrated
  graphics. So, H67 supports two
  independent monitor outs, but cannot
  split PCI Express x16 processor bus in
  two for a pair of graphics cards.

And via Intel Sandy Bridges H67, P67 and Z68 Differences Explained:

H67 – Onboard graphics are enabled,
  while overclocking is disabled
P67 – You need a graphics card for
  this chipset – onboard graphics is
  disabled, however overclocking is
  enabled here
Z68 – You guessed it…best of both
  worlds. Onboard graphics are enabled,
  as is overclocking


The meaning of "Yo, Pops"

Question: In the movie "Equalizer" (2014), the co-worker addressed Denzel Washington, "Yo, Pops, bet you didn't have to push dollies at your old job." and Denzel Washington replied, "Guilty as charged."
I just don't understand the meaning of "Yo, Pops" and why past tense "did" was used. Can anybody help, please.
Answer: "Yo, Pops" is a possibly insulting slang phrase that means "Hello, old man".
"Yo" is a slang term to get someone's attention.  It has connotations of either "friendly enough that it is OK to be rude", or "rude".
"Pops" is a slang term for "Pa" or "Father" or "old man".
The sentence is in the past tense because it is asking about something that happened in the past -- what Washington's character did in his previous job.
Comment: *Pops* is what the man said, so *pops* it is. What I mean is that *pops* as a singular noun for 'old man' is standard in some dialects of English. *Pop* is an informal term for 'father' used by most AmE speakers.
Comment: In this context, should it be "pop" instead?

The meaning of "Yo, Pops"


Is a module an inverse limit of finitely generated modules?

Question: Every module is the direct limit of finitely generated modules. Is it true that every module is the inverse limit of finitely generated modules?
Comment: Sorry, ignore my previous comment. I misread your question twice and ignored "finitely generated" so none of what I wrote makes any sense.
Answer: No. Consider $\mathbb{Q}$ or any other nonzero divisible group. Every homomorphism from $\mathbb{Q}$ into a finitely generated abelian group is zero: subgroups of finitely generated groups are finitely generated, quotients of divisible groups are divisible, and the only finitely generated divisible group is $0$.
More details: (using Wikipedia's notation)
Suppose towards a contradiction that $(X_i, f_{ij})$ is an inverse system of finitely generated abelian groups and $\pi_i \colon \mathbb Q \to X_i$ are the projections identifying $\mathbb{Q}$ as the inverse limit $\mathbb{Q} = \varprojlim\nolimits_i X_i$, in particular $f_{ij}\pi_j = \pi_i$. We know that $\pi_i = 0$ .
Consider the maps $\psi_i = \pi_i \colon \mathbb{Q} \to X_i$. Both $u = 1_{\mathbb{Q}}$ and $u' = 0$ are homomorphisms $\mathbb{Q} \to \mathbb{Q}$ such that $\psi_i = \pi_i u$ and $\psi_i = \pi_i u'$, hence the uniqueness statement in the universal property of the inverse limit implies $u = u'$, so $1_\mathbb{Q} = 0$. Nonsense.
(More generally, you can try to show that any map into $\mathbb{Q}$ would have to be zero.)
Comment: could you add more details? I don't understand why this is a counterexample


What would an orc do if it found the One Ring?

Question: If an orc found the One Ring, would it try to use it? Or were the orcs "programmed" to turn it over to Sauron?
Comment: "Oooh... a shiny. Mine!"
Answer: In his letter 246, JRRT spoke about the likely consequences should Frodo (or any other lesser being) decide to take control of the ring.
In short, the ring will drive anyone who claims it into a state of megalomaniacal madness. They'll become obsessed with trying to dominate the world and will ultimately seek to confront Sauron himself, at which point Sauron will utterly destroy them.
An orc that discovered the ring would suffer the same fate. Their arrogance, cruelty and lack of intelligence would surely make them prime fodder for trying to use the One Ring's power. They would instantly become a petty Lord of the Ring and a prime target for the Nazgul or Sauron himself.

He needed time, much time, before he could control the Ring or (which
  in such a case is the same) before it could control him; before his
  will and arrogance could grow to a stature in which he could dominate
  other major hostile wills.

and

But if he still preserved some sanity and partly understood the significance of it, so that he refused now to go with them to Barad-dûr, they would simply have waited. Until Sauron himself came. In any case a confrontation of Frodo and Sauron would soon have taken place, if the Ring was intact. Its result was inevitable. Frodo would have been utterly overthrown: crushed to dust, or preserved in torment as a gibbering slave.
Comment: @Darael, as has been mentioned elsewhere, if Gandalf had claimed the ring, he probably would have been able to defeat Sauron.  However, in the process, the ring would have corrupted him, turning him into something at least as bad.  Whether this counts as "claiming it safely" is a matter of opinion.
Comment: @MattGutting Possibly, given the sheer amount of power it took to *deal* with Morgoth, even later it would give him that little bit more oomph he would need to make a comeback.  But that's wild apocalyptic speculation.  Alternatively, Gandalf - being originally a being of the same order as Sauron who was sent back and further empowered by Eru himself - might have been able to claim it relatively safely (if it hadn't been far too late for that by the time Gandalf the White made his appearance).  Tom Bombadil, of course, could have claimed it safely, but to do so is not in his nature.
Comment: Thus it seems that the only being who could have possibly claimed the Ring without being "crushed to dust" might possibly have been Morgoth early in the history of Arda!
Comment: @mark - Yes. What Sauron fears is that a higher being will find the Ring and use its powers. Although they'll be corrupted by it, they won't be his servant.
Comment: @Mark safe for him.  Ish.  Not safe for anybody else.  Very good point.
Comment: I'm curious about the "lack of intelligence" description. In popular culture, and most fantasy settings, orcs are portrayed as dumb, but I haven't read anything like that on any tolkien book.
Comment: @flamma - That sounds like a separate question to me :-)
Answer: I don't believe there is any canon explanation for this. But I also believe that the Orcs are fairly weak willed and that the ring would eventually take over the Orc. Or it would move the Orc toward bringing the One Ring to Sauron.


Meaning of the verb "adorare" in a Christian context

Question: In the Christmas carol 'Adeste Fideles' there is a line 'Venite adoremus.' I'd like to hear an exploration of the meaning of this term. For instance, would it be appropriate to apply this word to anyone not God? From a Christian standpoint I mean. Given that God is One and he would have his worshippers give worship to him alone, would it be appropriate to apply this word to anyone other than God Himself? I guess I am asking if adoremus is the same thing as worship, or does it mean merely something like appreciation, Does it equate to 'adore' in English? We sometimes say we adore things and people other than God in English of course, but when using adore in it's most basic meaning it ought to really only ever apply to God for a real Christian, right? So what about 'adoremus'?
Comment: So your question is: in Christian Latin, is the term _adorō_ ever used for anything other than "worship"?
Answer: L&S on adorare: In ecclesiastical Latin, of the worship of the true God: adoravit Israel Deum, VULG. Gen. 47, 31: Dominum Deum tuum adorabis, IB. Matt. 4, 10: Deum adora, IB. Apoc. 22, 9; so of Christ: videntes eum adoraverunt, IB. Matt. 28, 17; adorent eum omnes angeli Dei
EDIT: Originally wrote that this comes from LSJ...
Comment: Thank you, this was very helpful.
Answer: In ecclesiastical Latin, adoratio is the outward expression of the reverence owed to the creator (called latria, from Greek λατρεία = service, worship), but also of the reverence due to certain outstanding creatures (like the Saints, the Virgin Mary, or Angels; called dulia from Greek δουλεία = bondage; or hyperdulia in case of the Virgin Mary).
See Summa Theologica II, II, 84, 1 for St. Thomas Aquinas' discussion of adoratio, which touches on all these points.
Nevertheless, I believe that there is a strong tendency to reserve the word for expressions of reverence for God only (including His only begotten Son, as in the song). There are instances – some cited by Aquinas – where adorare is used in the Bible for persons other than God. There are several such instance, e.g., in 3 Reg. 1.
See the entry for Adoration in the Catholic Encyclopedia for a rather thorough, well written and somewhat opinionated explantion of the matter (based on English usage, but (I think) nonetheless relevant.)
Comment: Very helpful. Thank you for the suggestions for further research.


What are the benefit of locking an already v-synced framerate?

Question: I've noticed some games run with less stuttering when I apply a 60fps lock, when already using (adaptive)v-sync on a 60Hz screen. Why is this the case if the frames are paced directly with my screen's refresh?
Answer: It depends on the game engine behind the game.
If you just run a v-sync, the engine probably don't know this. It will generate 60+ frames, but not all of them were directed to your display. Which means some of them just bounce.
If you define it in the game itself, the engine knows about it and can react. If the engine produces a world or a scene and probably can create 70 frames per second, it just stops at 60 frames. Now the engine can do something else for the 10 frames. It can probably load some textures which will be needed soon (in the next seconds). It can probably generate the models left and right from your viewport, which means they are just instantly ready if you spontaneously take a look to the left or right. Or it just can calm down and give your graphic card 10 frames time to rest and to cool (or not to heat up). Which will produce better performance in the following seconds.
Hopefully this will answer your question otherwise, just ask in the comment.
Comment: Thanks. I probably should've added that I am applying adaptive v-sync with the Nvidia control panel and the frame lock with Riva Tuner Statistics Server with v-sync disabled in game. I'm guessing in that scenario the engine knows nothing of either of those.
Comment: Well in this case I would guess that the engine would know about your Nvidia settings as they are often used for engine optimization.
Answer: Your comment about locking FPS in RTSS, disabling in-game v-sync and using adaptive v-sync in control panel changes the question quite considerably.
In this scenario, game can have less stuttering (and lag) because you are running it without v-sync, FPS will never go above your monitor refresh rate (here 60), so adaptive v-sync in drivers will never be utilized.
But this would mean you would see tearing because outputting frames from game and refresh rate of the monitor are not synced (even if it's 60FPS@60Hz). Do you see any tearing?


On TurboCharging, "Charging Slowly" message displayed on my phone Moto G4 Plus

Question: My Moto G4 Plus has the turbo Charging(30 min Charge 80%) feature.
 But when I connect to a Turbocharger, I get a display message on my
 lock screen as "Charging Slowly".  On connecting a general charger
 Phone charges as usual in 3 hrs. I am curious to know, with only the
 given knowledge, how can I predict whether the issue is in Mobile
 Phone Hardware or Software or the specific Turbo Charger.
Answer:
Moto phones are quite close to Nexus phones or pure android as commonly referred to, so the explanation below is valid
See Android phone "charging slowly": How to make it faster, for reasons on why charging slowly is displayed, in particular Andrew's excellent answer. To summarise, fast charging/ Turbo charging is achieved by increasing voltage or current or both, i.e. wattage ( being the product of current and voltage), as compared to conventional charging. Andrew's answer explains the values for which such notification appears ( Nexus phone on linked question uses a different type of which charging compared to your device, but that is a minor detail, since OS is flashing the notification based on wattage values and not on type of quick charging)
So, the most likely cause IMO, is that your Turbo charger is not working well or it is not receiving adequate voltage ( connect through a voltage stabiliser, avoiding extension boxes). You can verify otherwise by going to service centre and checking with the Turbo charger sample.

Do revert with your findings
Edit: I own Moto X Play and the charging cable is fused with the charger. If your charging cable can be detached from charger, try replacing with a good charging cable, to eliminate cable issue
Answer: I just hit the same issue. Check that you are using the microUSB cable that came with the Turbo Power charger.  I had accidentally switched around my USB cables with a generic one, and noticed that my phone was charging slowly, even after reseating both cable connectors.  When I replaced the generic cable with the original cable, it enabled TurboPower charging straight away.
Not sure if this is a case of the cheap cable having thinner power lines (i.e. higher resistance) or if the charger detects that this is the "right" cable. The original cable has a longer plastic lump on the microUSB end than the generic one, and it could be that there is a bit of circuitry in there that is needed for the phone to negotiate the faster charge with the charger, or that the phone checks whether the cable has a low enough resistance on the power lines to support higher current charging. If someone has access to the spec for Quick Charge 3.0 (a Qualcomm chipset feature which looks to be marketed as TurboPower by Motorola) feel free to revise this answer with the real deal.
FWIW, unless your mains connection is bad to the point of arcing, or you are in a severe brown-out, the power draw from the charger is not enough to see any problems from mains extension cords or power boards.  These are the output specs from the product page for Motorola's TurboPower 15 wall charger:
<code>Standard:   5V / 1.67A (8W)
TurboPower: 9V / 1.67A (15W)
TurboPower: 12V / 1.2A (15W)
</code>
So unless the voltage is very low due to some large load on the same power board, or a mains plug is making poor contact, it is unlikely that the charger will have problems drawing ~200mA from the mains.


Calling one asynchronous apex from another

Question: Can anyone provide me when we can call an asynchronous apex from another specially between when they are not of same type like schedulable from future or batch from schedulable,etc
Answer: The usual reason for this is that you need more resources or additional limitations than you can use in the current context for whatever reason. For example, a Scheduleable class might call a Batchable class so it can process hundreds of thousands of records every hour, or you might call a Queueable from a Batchable class to avoid MIXED_DML_OPERATION errors. You should avoid doing so when possible, but it's important to know the technique, since you may eventually need to do so. I think the most common use case is Scheduleable to Batchable (repeat a large job on schedule), but any allowed combination typically has some rare, exotic use, which is why such allowances exist.
Answer: When we are planning to update setup and non-setup objects at a same time or single transaction via batch we will face mixed dml operation error, so this case we use to call batch inside batch(Calling Asynchronous from another Asynchronous)
Please let me know if you need any more example regarding this.


Finding a piece to promote a pawn

Question: There is only one queen of each color is a chess set.  When a pawn gets promoted, say to a queen, where does that queen come from?
Comment: Related: https://chess.stackexchange.com/questions/33320/how-does-one-promote-a-third-queen-in-an-over-the-board-game
Answer: Most over the board sets come with 4 queens total (2 of each color) for exactly this situation. It's quite likely that at some point in your games, you'll need to have 2 queens on the board simultaneously, so they give you the main queen and a spare.
If you really need more than 4 queens, you can always "steal" another piece from your neighbors' game so that you have enough pieces to continue playing. Just make sure you give it back when one gets captured or they'll be mad with you!
Of course, it's possible to have 9 queens simultaneously, but it would be unnecessary for any chess set to give you this much redundancy on the off chance you decide to troll your opponents and promote every piece to a queen. Same thing with underpromoting so that you have 10 knights, 10 bishops, or 10 rooks...

Finding a piece to promote a pawn


Why are so different the temperatures of a compressed gas compared to a liquid/gas mixture of the same fluid and they are both at the same pressure?

Question: Why does C3H6 gas when compressed (in a reciprocating compressor) comes out with a pressure of 11 bar and 60 degrees celsius and when it is a mixture of C3H6 liquid/gas stored inside of a vessel reaches also the 11 bar but at a lower temperature of around 25 degrees celsius?
Comment: Cross-posted at Chemistry: https://chemistry.stackexchange.com/questions/125213/why-is-so-different-the-temperatures-of-a-compressed-gas-compared-to-a-liquid-ga
Answer: Compressing the gas will raise its temperature, whether it compresses it enough to turn to liquid or not.  If you leave it stored in a compressed state, it will cool to the temperature of its surroundings.
If you then release the gas from the pressurized container, it's temperature will be lower than ambient.
Putting many compressed cylinders in a row enabled air to be "liquified" by chemists such as Dewar over 100 years ago, hence the vacuum "Dewar flask" was invented to help prevent them from boiling away.  Vacuum creates an insulating layer that lessens heat transfer.


What does FIDU mean?

Question: I was wondering what does FIDU mean in the schematic below and what is it used for ?
Comment: "Fiducial" probably
Answer: In your image, FIDU is the symbol's name for fiducial. Fiducials are used by optical equipment (a pick and place machine) to visually see and locate the board's origin relative to the machine's origin. A small fiducial is probably a 1 mm diameter footprint. Typically, I would place 3 on the top-side of the PCB near the corners and 3 for the bottom-side. I suppose that's why there are 6 shown in your image. Sometimes, additional fiducials are added in close proximity to a large chip/ASIC to help perfectly place (minimize placement tolerances) the chip onto the PCB during PCB assembly.
Answer: "Fidu" is short for fiducial marker.
They are used on printed circuit boards (PCB) to assist in lining up the various layers.
Comment: Either for lining up the various layers, or for positional calibration for component placement machines.
Comment: At the place I work, we have to use them because the supplier of the boards wants them. Personally though, I've ordered boards from JLCPCB for small side projects, and have never included them. If you send without, they'll tell you if you need them or not. It's probably a bigger deal if they are loading the parts onto the board for you.
Comment: Are they really necessary ? I have never used them in the past.
Comment: Your PCB fabricator or assembly house will tell you if they are needed.  I know what fiducials are, but I've never had to use them.
Comment: It is clear for me now. Thank you guys your help.
Answer: If that is what I think it is, it is a Fiducial. They are used for the board manufacturer as a reference so that they know where everything is located on the board coordinate-wise.
If you want to learn more about them, there is plenty of information online if you look up 'Fiducial'.
Answer: I would guess it refers to Fiducial Marks, measuring points on PCBs for positioning.

What does FIDU mean?


Social Engine render form issue

Question: I'm adding some pages to default layout editor in SE4. What I do is just adding pages to core_pages table and editing core_content table to add a main container and a middle/right two columns layout. Everything works fine, but now I'm adding default Create Video page (videos_index_create) and I got some problems.
When I add this page via sql, I can obviously see and edit page layout by default layout editor. Actually, when I save changes these don't reflect to live page. If I go to the controller (Video/controllers/IndexController.php) and add
$this->_helper->content->setNoRender()

live page displays right sidebar and middle content, but with
$this->_helper->content->setEnabled()

it shows only the default video upload form.
So I edited create.tpl here
<?php if (($this->current_count >= $this->quota) && !empty($this->quota)):?>

  <?php echo $this->translate('You have already uploaded the maximum number of videos allowed.');?>
  <?php echo $this->translate('If you would like to upload a new video, please delete an old one first.', $this->url(array('action' => 'manage'), 'video_general'));?>

<?php else:?>
<?php echo $this->form->render($this); ?>
<?php endif; ?>

Changing the else deleting the echo statement and adding a simple echo 'foo';. Now, live page shows properly my layout (middle+right) with default content ("foo").
I figured thus that the issue is about this line in controller:
$this->view->form = $form = new Video_Form_Video();

but I can't go further… this custom form class (Video/Form/Video.php) seems nothing special, I can't really figure out why its rendering crushes default layout render.
Any social engine expert right here to help me? :)
Answer: If its a Zend_Form I am pretty sure you should be doing:
<?php echo $this->form;?>

In your view not call render yourself.
If it is NOT a Zend_Form please post the code for the form it probably has some css or whatnot breaking your layout.

Social Engine render form issue


What are weak squares in chess; how can they help to mate another player?

Question: I read this concept in a book but I didn't understand much about the concept. Can anybody explain this concept with an example?
Comment: Depends on what the game looks like. Can you add an example or more information?
Comment: https://lichess.org/practice Here's some examples of common weak squares/ combos exploited
Answer: This is a very common example.
<code>[FEN "6k1/5p1p/5BpQ/8/8/8/8/6K1 w - - 0 1"]
</code>
Do you see the weak dark squares around the king? Checkmate is possible because those squares are weak and can't be defended.
Comment: @OrangeDog It's completely obvious which king is being talked about.
Comment: This is a bit of a circular definition. The weak squares are the weak squares. They help mate a player because they are weak. You don't even specify which king you are talking about.
Answer: A weak square is a square of great great importance that is difficult to cover or get rid of.

Since the square is difficult to cover, it can become a great outpost for opponent's (your) pieces, usually a knight.
If you manage yo get the right piece on that square, it helps you in any kind of possible attacks. Whether the attack will result in checkmate or 'only' material advantage depends on the position.

It usually doesn't matter, whether there's a pawn on that square or not (as A. Nimzowitsch stated). You will have the same problems/advantages connected with it.
The only difference is that the pawn can be targeted for the sake of capturing it.

Recap:
There are 2 basic ways to take advantage of a weak square

Occupy it to get good outposts for your pieces and to get control over certain parts of the board.
Attack and capture the weak pawn

Note: Usually 1 weakness isn't enough to win the game, the opponent will be able to protect it. The principle of 2 weaknesses says that you have to find at least another one.


how is c useful in game development

Question: I know that C++ is the king in game development especially in the AAA market can anyone tell my how is C helpful I am a Computer Science student in my first year and are they any aspects of C i have to learn in depth?
Thanks
Answer: C is a subset of C++.
C is small and lean - a manageable subset of C++.
C++ on the other hand is huge and can be very confusing, with many questionable features and design decisions. C++ standard is gigantic.
There are quite a few notable industry figures who prefer using C.
Majority of professionals use a small and very conservative subset of C++.
My suggestion - learn how to use C thoroughly before jumping in with C++11.
There are too many features in C++ that look harmless on the surface, but once you start relying on them they blow in your face - exploding compile times, indecipherable template error messages - just to name a few.
Comment: This all seems like dubious advice to me. C is not a clean subset of C++. Whether or not learning C before C++ is advisable is also up in the air and highly depends on what you're going to use C++ for.
Comment: Compile times are NOT subjective. They are very measurable. C++ compiler is slower by an order of magnitude, that's very slow. That's why there's PIMPL idiom, unity builds and precompiled headers. Including and using *simple* and basic features of the standart library - things like STL containers will increase the compile times dramatically. When writing C++ code not only you might think about it's run-time performance, you inevitably end up thinking (and experiencing) it's compile performance.
Comment: Within the context of this question, calling C a subset of C++ seems very appropriate. C is the foundation of C++. As far as learning C++ goes, you basically end up learning the vast majority of C by doing it anyway. Whether you start full on with C++11 or C is up to preference. I suggest learning C in isolation first to get a better understanding of the language.
Comment: "C is a subset of C++" you're looking at it the wrong way. C++ is built on C. The original name given to it by Stroustrup was "C with classes".
Comment: "but once you start relying on them they blow in your face - exploding compile times, indecipherable template error messages" This is very subjective. You can't base a language recommendation on that.
Comment: wow this seems to be turning into a code war i just want simple advise

how is c useful in game development


Self modifying hash encryption algorithm

Question: I am looking for an algorithm to change it's keys every period and can still be deciphered.
Basically I am looking for a way to maintain a secured link between mobile clients and a stationary server in such a way that even if you intercepted the hash or even the authentication credentials themselves they would change on both sides every period.
Does it ring a bell to anyone?
Is there a better way to ensure that even if you will intercept the authentication credentials somehow it will only be valid to a certain request from a certain user?
Comment: Authentication cardinals? Surely you mean credentials?
Comment: It sounds like you've heard a few too many technothrillers. Just use SSL - it's already solved all the problems you're likely to encounter.
Answer: I suggest just using SSL.  It is designed to be reasonably resistant to man-in-the-middle attacks, which is what I assume is your concern.
Or maybe Kerberos.
Do not code encryption algorithms yourself.
Comment: +1 for the last line, the most important (and most often broken) tenet of cryptography
Answer: i'd recommend SSL instead of implementing some encryption algorithm yourself (it WILL be broken if the data you are trying to protect is important enough!). SSL is well tested. with SSL you can use certificates instead of logins/passwords. SSL prevents replay and man in the middle attacks (it uses a handshake at the beginning to make sure a new session key is used for every connection and that both parties are who they claim to be).
another interesting thing that comes to mind is RSA's SecurID. it provides a hardware key that changes every 60 seconds: http://www.rsa.com/node.aspx?id=1156
Comment: I'd say RSA's SecurID is exactly what I was looking for but what do you mean by "hardware key?"
Comment: if you follow the link above there's a pic of such a hardware key (the device that displays the current key). i just saw that they also offer software solutions (for PCs, mobile phones,..). i like the hardware keys though because they can't be stolen by hackers.
Answer: This is certainly possible -- just for example, the Japanese Purple cipher during the second world war did this. While such a cipher can certainly be difficult, it can also be broken (Purple was).
The reason is fairly simple: your sender and receiver have to generate new keys in synch with each other for the receiver to decipher the messages. That basically means you need something like a secure random number generator to generate new keys. Though that can be difficult to break (with long enough period and such) it's still a pretty normal encryption technique, and depends on having a secure random number generator. Once you have that, however, you're usually not gaining a lot over using the output from the generator more directly (e.g. as the key for a Vernam cipher).
Answer: I guess what you're talking about is finding some way of changing the cryptographic key used in you algorithm periodically, so that even if a key is discovered, then only the data encrypted with that key would be possible to decode?  If we avoid worrying about the start-up process, then one way to do it would be to encode part (but not all) of your subsequent keys in the set of data encrypted with one key, and as you switch keys, encrypt a different part of subsequent keys with the new key.
For example, say your keys are 8 elements wide (where an element may be a byte, or a 32-bit word, or whatever), and we'll label the keys you use to encrypt any given block of data as Kn, where 'n' is the block of data encrypted with that key.  We'll index elements of the key by saying Kn[0] for the 1st element, up to Kn[7] for the 8th.  We'll also call that block of data Dn.  Then, the plaintext Dn would include Kn+1[0], Kn+2[1], Kn+3[2], ..., Kn+8[7].  If you were able to decrypt Dn-7 .. Dn, then you'll have fully reconstructed Kn+1, so that you can then decrypt the next data block, and so on.  You need to get the plaintext for several blocks in sequence before you can reliably decrypt the rest of the data, though getting the plaintext for any given block will make attacks against the keys of the remainder easier.
Initial setup is a harder problem.  SSL would be a good way of distributing K0, K1[1..7], K2[2..7], ..., K7[7].
I'm not a professional cryptographer, so I'm not completely sure how secure this is.  This algorithm is offered to you AS IS, without warranty of any kind.


$\infty - \infty = 0$ ?

Question: I am given this sequence with square root. $a_n:=\sqrt{n+1000}-\sqrt{n}$. I have read that sequence converges to $0$, if $n \rightarrow \infty$. Then I said, well, it may be because $\sqrt{n}$ goes to $\infty$, and then $\infty - \infty = 0$. Am I right? If I am right, why am I right? I mean, how can something like $\infty - \infty = 0$ happen, since the first $\infty$ which comes from $\sqrt{n+1000}$ is definitely bigger than $\sqrt{n}.$?
Comment: Would you say that the limit of $b_n:=(n+1)-n$ is 0 for the same reason?
Comment: @doniyor The key thing is to remember that operations that we learn only hold for the integers, then the rationals, then the reals (and finally the complex numbers). Each time, when we extend it, we have to check that the definition makes sense (e.g. that we indeed have equivalence classes in fractions). In the case of trying to extend arithmetic to $\infty$, we have $\infty + * = \infty$, where $*$ denotes anything that we're adding to. Hence, $\infty - \infty$ could be any value, hence is undefined without knowing more. Same for $\frac {\infty}{\infty} = *$.
Comment: now i got, Charles. thanks a bunch
Comment: yeah, you are totally right Calvin. thanks
Answer: The argument $\infty-\infty=0$ will fail you if you consider $a_n=(n+1)-n$ where the limit is obviously $1$ or $n^2-n$ where the limit is $\infty$. Indeed if $a_n\to \infty$ and $b_n\to \infty$ you can say nothing for the covergence of $a_n-b_n$ (unless you have additional clues).
In our case, although $\sqrt{n+a}$ is strictly bigger than $\sqrt{n}$, both diverge to $+\infty$ in exactly the same way. In fact, $$\sqrt{n+a}-\sqrt{n}=\frac{n+a-n}{\sqrt{n+a}+\sqrt{n}}\to a\frac{1}{\infty}=0$$
for any $a\in \mathbb{R}$. For the first equality I used the well known identity:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$$
Comment: I know I will need this answer for my own class.+1000
Comment: wow, okay, thanks in tons. i had bad feeling for infinity all the time
Answer: No, $\infty-\infty$ is indeterminate.
This is what you can do:
$$
\sqrt{n+1000}-\sqrt{n}=\frac{(\sqrt{n+1000}-\sqrt{n})(\sqrt{n+1000}+\sqrt{n})}{\sqrt{n+1000}+\sqrt{n}}=\frac{n+1000-n}{\sqrt{n+1000}+\sqrt{n}}.
$$
Now you surely can see why this converges to $0$.
Comment: thanks julien, now i got the clue
Comment: @doniyor, julien wasn't just giving you a clue. What you have is an indeterminate form, and the general scheme of things is to turn to l'hopital's rule for such problems.
Comment: julien showed how to come to that $0$-limit. and tought me that my thought is completely wrong
Comment: Ok just wanted you to know if something similar came up.
Answer: But this is not really $\infty-\infty$. This is the limit of differences between two unbounded sequences. This is the key point here.
The difference is approaching zero, because $1{,}000$ is very small compared to $\sqrt n$. At some point, $\sqrt n$ is almost the size of $\sqrt{n+1000}$. For example for $n=10^{1000}$ the difference gets very small, that is $a_{10^{1000}}$ is a very small number.
Answer: You have to look at the sequence in particular.  In your case, a little algebra clears things up:
$$ \sqrt{n+1000} - \sqrt{n} = \frac{1000}{\sqrt{n+1000} + \sqrt{n}} \approx \frac{1000}{2 \sqrt{n}} $$
which obviously goes to zero as $n \rightarrow \infty$.
Answer: $$\sqrt{n+100}-\sqrt n=\frac{100}{\sqrt{n+100}+\sqrt n}\xrightarrow [n\to\infty]{}0$$
But you're not right, since for example
$$\sqrt n-\sqrt\frac{n}{2}=\frac{\frac{n}{2}}{\sqrt n+\sqrt\frac{n}{2}}\xrightarrow [n\to\infty]{}\infty$$
In fact, "a difference $\,\infty-\infty\,$ in limits theory can be anything
Answer: Note that
$$\sqrt{n}\lt \sqrt{n+1000}\lt \sqrt{n}+\frac{500}{\sqrt{n}}.$$
This is because
$$\left(\sqrt{n}+\frac{500}{\sqrt{n}}\right)^2=n+1000+\frac{500^2}{n}\gt n+1000.$$
If follows that
$$0\lt \sqrt{n+1000}-\sqrt{n}\lt \frac{500}{\sqrt{n}}.$$
Now by Squeezing we can conclude that
$$\lim_{n\to\infty}(\sqrt{n+1000}-\sqrt{n})=0.$$
Answer: I think you need to revise your definition of limit of a sequence.
$\sqrt{n+1000}$ comes arbitrarily close to $\sqrt{n}$. More precisely, for any arbitrarily small $\epsilon > 0$ there exists an integer $N_\epsilon$ such that,
\begin{equation}
\sqrt{n+1000}-\sqrt{n} < \epsilon \;\; \forall \; n>N_\epsilon
\end{equation}

$\infty - \infty = 0$ ?


Halacha B'Iyun Learning Help

Question: Where can I get hold of a semicha program syllabus or a guide to help me learn halacha b'iyun? I am not looking to learn for semicha just yet but want something to help be focus my halacha seder as just stam learning through shulchan aruch/rambam/mishna brura is getting a bit stale.  Ideally looking for a syllabus to help give breadth and depth e.g. the Israeli Rabbanut semicha.
Comment: Clever username +1
Comment: [Welcome to MiYodeya](https://judaism.meta.stackexchange.com/questions/3886/a-beginner-s-guide-to-my-how-is-this-site-different-from-other-judaism-sites) Shmuel
Answer: Are you aware of a program called Tzurba M'Rabanan? It is a halakha b'iyun program to learn most of Shulchan Aruch over 4 years. They publish brochures (or a complete set) - each brochure contains material for approximately 8 weeks of learning (with one in-depth topic per week).
Each week covers a topic and brings 30-40 sources, from gemara, rishonim, aharonim to elucidate the topic in depth together with summary notes from R Ben Tzion Elgazi who prepared the material.
This is studied weekly in over 300 places all over Israel and is now expanding to a program for youth and one for women.
There is now an English translation as well, see here for details.
Answer: R. Yitzchak Berkowitz' Jerusalem Kollel has source sheets available online for a range of halachic topics. You can find them here (in the column headed Level III).
They contain references to the relevant sources (starting from the gemara and rishonim and leading on to shulchan aruch and nosei keilim), as well as brief explanations and summaries.
I've personally used some of them before (on Issur v'Heter and Niddah) and found that they provide a good overview of the topic with a significant amount of depth.
Answer: The Dirshu Mishnah Berurah is good - it has a compilation of the differnt machloksen by all the halachos and that sort of thing. They also have a good program for learning it.
https://dafhalacha.com/

Halacha B'Iyun Learning Help


Searching content by taxonomy term

Question: I would like to search content depending on select menu. Something like the search box here: http://www.doyleia.com/anazitisi-ergasias
How can I do that with drupal 8?
Comment: Please notice that requests for implementing a feature that essentially are _I want to do like in this site [link to a site]._ are off-topic for us. That is more true if the site is not running Drupal; even if the site would be running Drupal, there would not be much difference. We should try to understand which modules the site is using, and with which settings.
Answer: You can achieve that with filters in Views.
Just create a view, add filters and make them "exposed", so that visitors can choose filters themselves. Filters can be taxonomy terms, list options or open-ended fields (e.g. text field) where users can write what they are searching for.
See the Chosen module for providing a more user-friendly interface for <select> elements that accept multiple choices.

Searching content by taxonomy term


Approximate roots using the binomial series.

Question: as the title of the question says I need to approximate roots using the binomial series, I am new to this subject and I have had a hard time understanding how to do it. I have been researching and I have found that the binomial series is: $(1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ...$ At this point is my first doubt, can I use this binomial series to approximate roots of index two, three, four and so on? For example, if I want to find an approximation for $\sqrt{67}$ what I would do is: $(1 + 66)^\frac{1}{2} = 1 + \frac{0.5}{1}66 + \frac{0.5(0.5-1)}{1*2}66^2 + ...$ However, even with the second term of the series (Result = $34$) I am far away from the real value ($8.185352772$). Could you help me and if possible give me some examples of how to approximate roots of index two, three, four, five? In a further example, if I want to find an approximation for $\sqrt[3]{264}$ I would do something like: $(1 + 263)^\frac{1}{3} = 1 + \frac{0.3333333333}{1}263 + \frac{0.3333333333(0.3333333333-1)}{1*2}263^2 + ...$
Comment: The series for $(1+x)^n$ only converges when $|x| < 1$.
Comment: You could use that $\sqrt{67} = \sqrt{100 \times .67} = 10 \sqrt{.67} = 10\sqrt{1-.33}$ and now apply the binomial series.
Comment: Yes. However, be aware that this is not the best way to approximate roots.
Comment: @TobyMak Thanks for the clarification, what could I do in that case?
Comment: @jjagmath For the expression $\sqrt[3]{3}$ could I use something like $\sqrt[3]{1000*0.003}$?
Answer: The series for $(1+x)^n$ only converges when $|x| < 1$.
So you need to find the nearest integer root. For $\sqrt{67}$, since $8^2 = 64$, you can write $\sqrt{67} = 8 \sqrt{67/64} = 8 \sqrt{1 + 3/64}$. Likewise, $\sqrt[3]{264} = 6 \sqrt[3]{264/216} = 6 \sqrt[3] {1+2/9}$.
Comment: I understand, what would happen in the case of $\sqrt[3]{3}$?
Comment: I have found something interesting, I want to calculate the following fourth root: $\sqrt[4]{1250}$, the nearest integer root is: $5^4$=$625$, So $5*\sqrt[4]{1250/625}$, $\frac{1250}{625}-1=1$ , However, the binomial series converges at $|x|<1$; How should I proceed in this case?
Comment: Then you have $5 \sqrt[4] {2}$ and now take a factor of $1.2^4$ out, since $\sqrt[4] {2} = \sqrt{\sqrt{2}} \approx \sqrt{1.4} \approx 1.2$, as $1+2x \approx (1+x)^2$ when $x$ is small.
Answer:
With respect to comments from OP we look at approximation using the binomial series expansion in order to calculate
\begin{align*}
\color{blue}{\sqrt[3]{3}}&\color{blue}{=1.442\,249\,570\ldots}\tag{1}\\
\color{blue}{\sqrt[4]{1\,250}}&\color{blue}{=5.946\,035\,575}\tag{2}\ldots
\end{align*}

Approximation of $\sqrt[3]{3}$:
Expansion of $\sqrt[3]{1+x}$ gives
\begin{align*}
\left(1+x\right)^{\frac{1}{3}}&=\sum_{n=0}^\infty \binom{\frac{1}{3}}{n}x^n\qquad\qquad\qquad\qquad |x|<1\\
&=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3+\cdots\tag{3}
\end{align*}
In order to calculate $\sqrt[3]{3}$ we are looking for an appropriate scaling factor $y$ and make an Ansatz
\begin{align*}
\sqrt[3]{1+x}=\sqrt[3]{3y^3}=y\sqrt[3]{3}\tag{4}
\end{align*}
We obtain from (4)
\begin{align*}
1+x&=3y^3\\
x&=-1+3y^3
\end{align*}
We know the expansion is valid for $|x|<1$, so we are looking for $y^3$ near $\frac{1}{3}$ which makes $x$ close to $0$. Since we want to do the approximation (at least in principle) manually we are looking for a rational value
\begin{align*}
y&=\frac{a}{b}\\
x&=-1+3y^3=-1+3\frac{a^3}{b^3}
\end{align*}
with small numerator and denominator. A table lookup of small values of $(n,n^3)$ gives
\begin{align*}
\begin{array}{c|rrrrrrrrrr}
n&1&2&3&4&5&6&7&8&9&10\\
n^3&1&8&27&64&125&216&\color{blue}{343}&512&729&\color{blue}{1\,000}
\end{array}
\end{align*}
We observe
\begin{align*}
\frac{343}{1\,000}=0.343\approx \frac{1}{3}
\end{align*}
is close to $\frac{1}{3}$ and we choose
\begin{align*}
\color{blue}{y=\frac{7}{10}}
\end{align*}
as scaling factor.

We obtain from (3) and (4) with $y=\frac{7}{10}$ and $x=-1+3y^3=-1+3\frac{7^3}{10^3}=0.029$:
\begin{align*}
\frac{1}{y}\sqrt[3]{1+x}&\approx\frac{10}{7}\left(1+\frac{1}{3}\cdot0.029
-\frac{1}{9}\cdot 0.029^2+\frac{5}{81}\cdot 0.029^3\right)\\
&\,\,\color{blue}{=1.442\,249}61\ldots
\end{align*}
which is according to (1) accurate to $6$ decimal places. Note that lineare approximation
\begin{align*}
\frac{1}{y}\sqrt[3]{1+x}&\approx\frac{10}{7}\left(1+\frac{1}{3}\cdot0.029\right)\\
&\,\,\color{blue}{=1.442}\,38\ldots
\end{align*}
is already accurate to $3$ decimal places.

Approximation of $\sqrt[4]{1\,250}$:
We can go on in the same way as before. Expansion of $\sqrt[4]{1+x}$ gives
\begin{align*}
\left(1+x\right)^{\frac{1}{4}}&=\sum_{n=0}^\infty \binom{\frac{1}{4}}{n}x^n\qquad\qquad\qquad\qquad |x|<1\\
&=1+\frac{1}{4}x-\frac{3}{32}x^2+\frac{7}{128}x^3+\cdots\tag{5}
\end{align*}
In order to calculate $\sqrt[4]{1\,250}$ we are looking for an appropriate scaling factor $y$ and make an Ansatz
\begin{align*}
\sqrt[4]{1+x}=\sqrt[4]{4y^4}=y\sqrt[4]{4}\tag{6}
\end{align*}
We obtain from (6)
\begin{align*}
1+x&=1\,250y^4\\
x&=-1+1\,250y^4
\end{align*}
We know the expansion is valid for $|x|<1$, so we are looking for $y^4$ near $\frac{1}{1\,250}$ which makes $x$ close to $0$. We are looking for a rational value
\begin{align*}
y&=\frac{a}{b}\\
x&=-1+1\,250y^4=-1+1\,250\frac{a^4}{b^4}
\end{align*}
with small numerator and denominator. A table lookup of small values of $(n,n^4)$ gives
\begin{align*}
\begin{array}{c|rrrrrrrrrr}
n&1&2&3&4&5&6&7&8&9&10\\
n^4&\color{blue}{1}&16&81&256&625&\color{blue}{1\,296}&2\,401&4\,096&6\,561&10\,000
\end{array}
\end{align*}
We observe
\begin{align*}
\frac{1}{1\,296}=0.343\approx \frac{1}{1\,250}
\end{align*}
is close to $\frac{1}{1\,250}$ and we choose
\begin{align*}
\color{blue}{y=\frac{1}{6}}
\end{align*}
as scaling factor.

We obtain from (3) and (4) with $y=\frac{1}{6}$ and $x=-1+1\,250y^4=-1+1\,250\frac{1^4}{6^4}=-\frac{23}{648}$:
\begin{align*}
\frac{1}{y}\sqrt[4]{1+x}&\approx 6\left(1-\frac{1}{4}\cdot \frac{23}{648}
-\frac{3}{32}\cdot \left(\frac{23}{648}\right)^2+\frac{7}{128}\cdot \left(\frac{23}{648}\right)^3\right)\\
&\,\,\color{blue}{=5.946\,065}\,28\ldots
\end{align*}
which is according to (2) accurate to $6$ decimal places. Note that linear approximation
\begin{align*}
\frac{1}{y}\sqrt[4]{1+x}&\approx 6\left(1-\frac{1}{4}\cdot \frac{23}{648}\right)\\
&\,\,\color{blue}{=5.946}\,75\ldots
\end{align*}
is already accurate to $3$ decimal places.