Is my RAMPS 1.4 board fried?

Question: I was adjusting the current on my Pololu stepper drivers when I accidentally fried two of my stepper drivers when I accidentally shorted the contacts with my multimeter probes. When I changed the fried drivers the board wasn't working properly. My Arduino Mega works fine alone when connected to my PC and executes the codes properly. However, as soon as I plug the RAMPS shield into my Arduino, the board disconnects from my computer. I cannot figure out the problem.
Comment: Hi and welcome to SE.3DP. Which version number of the RAMPS board do you have? Exactly which contacts did you short? A photo or diagram *might* help to describe the issue better. Please [edit] your question and add a photo/diagram, the contacts that were shorted and the version of the RAMPS board.
Answer: It is quite likely that something else (other than the stepper drivers) on the RAMPS board (and maybe even the Mega) has been damaged as well... The Mega board may still execute code correctly but maybe a random component has been damaged that you won't notice until it gets used.
It most probably would be possible to test both boards to locate the exact issue and faulty component, given enough time and patience.
The best thing to do, initially, is to look closely for any component that looks shot/burnt, or smells bad. It may sound odd but smelling a board can sometimes (but not always) locate the general area where a component has failed.
Another (more logical) approach would be to look at the schematic of the RAMPS to follow the circuits connected to the contacts that you shorted. Then check the components along that path. This method often is the most fruitful.
Assuming that you have a RAMPS 1.4 (you don't specify which version you have), there is the schematic:

However, even if you are lucky enough to locate the damaged component, do you have the skills to replace a SMD component?
Sadly, to be honest, it is probably cheaper, easier and quicker to just first buy a new RAMPS and substitute that in. The test to see if everything works OK. If the issue is still not resolved, then buy a new Mega and repeat the test.

Is my RAMPS 1.4 board fried?


what is the problem with this photo and how can i fix it?

Question:
We had a wrong camera setting on alot of pics. Is there any software to clean this up and make it clearer?
Comment: What problems do you think there is? What did you expect this to look like? What "wrong camera setting" did you have, and why do you think it was wrong?
Comment: It is not the camera settings. You need to understand thoose setings. :)
Comment: This photograph looks great to me. I bet you could print it at about 80cm × 1m and sell hundreds of copies to a hotel chain.
Comment: Was the camera set to some sort of 'art' setting?
Answer: Unfortunately, that photo is much too far gone for any software to be able to clean it up. One of the big advantages of digital over film is the ability to instantly review photos - get in the habit of using it.
Answer: Assuming you wanted a crisp picture of the flower, there are two basic problems:
Out of focus.  That really should be obvious.
Very grainy.
If this is from film, then you are blowing up a tiny area of the picture much bigger than the film resolution can support.  Don't do that.  You can't infinitely enlarge pictures without this kind of artifact.
If this is digital, then extreme high ISO was probably used.  Again, don't do that if you don't want this effect.  ISO isn't something you can magically increase without trading off with other things, like noise.
Comment: The "nikon-d3100" tag strongly implies digital :-)
Comment: Sadly I have seen a number of beginners leaving iso set to maximum setting to avoid blurry pictures in dim light. I assume they were told to do so in a dark indoor setting etc but not told to switch it back afterward.
Comment: The grain might come from trying to "sharpen" the image in software, with the maximum possible sharpening value. That of course did not improve the image, which was to much out of focus and therefore much to blurry to be improved by any software.


Using objects in Ajax calls PHP files

Question: I have instantiated a class in my index.php file. But then I use jQuery Ajax to call some PHP files, but they can't use my object that I created in the index.php file.
How can I make it work? Because I don´t want to create new objects, because the one I created holds all the property values I want to use.
Answer: Use the session to save the object for the next page load.
// Create a new object
$object = new stdClass();
$object->value = 'something';
$object->other_value = 'something else';

// Start the session
session_start();

// Save the object in the user's session
$_SESSION['object'] = $object;

Then in the next page that loads from AJAX
// Start the session saved from last time
session_start();

// Get the object out
$object = $_SESSION['object'];

// Prints "something"
print $object->value;

By using the PHP sessions you can save data across many pages for a certain user. For example, maybe each user has a shopping cart object that contains a list of items they want to buy. Since you are storing that data in THAT USERS session only - each user can have their own shopping cart object that is saved on each page!
Answer: mm, you should store in session, $_SESSION["someobj"] = $myobj;,  and ensure that when you call the Ajax PHP file this includes the class necessary files which defines the class of $myobj and any contained object in it.
Could you be more specific? I can try.
This is how I create an object then assign it to a session variable:
include(whateverfilethathastheclassorincludeit.php)
$theObject = new TheObjectClass();
//do something with the object or not
$_SESSION['myobject'] = $theObject;

This is how I access the object's members in my Ajax call PHP file:
include(whateverfilethathastheclassorincludeit.php)
$theObject = $_SESSION['myobject'];
//do something with the object

Comment: could you be more specific? how so i create an object then assign it to a session variable. how do i do to access the object's members in my ajaxcall php file?
Comment: Hm. I'm not sure whether I approve of the use of the `` tag for non-key-related issues :) but still +1.
Answer: Another option if you dont want to use sessions is to serialize your object and send it through a $_POST value in your AJAX call. Not the most elegant way to do it, but a good alternative if you don't want to use sessions.
See Object Serialization in the documentation for more informations.
Answer: You have not provided code, but what I guess is that you need to make your instantiated object global for other scripts to see it, example:
$myobject = new myobject();

Now I want to use this object elsewhere, probably under some function or class, or any place where it is not getting recognized, so I will make this global with the global keyword and it will be available there as well:
global $myobject;

Once you have the object, you can put it into the session and then utilize it in the Ajax script file.
Answer: If you don't want to move your object that is in your index.php, have your ajax make a request to index.php but add some extra parameters (post/get) that let your index.php know to process it as an ajax request and not return your normal web page html output.
Answer: As others have suggested, $_SESSION is the standard way to do it, in fact, that was one of the reasons, that sessions where invented to solve. Other options, i.e. serializing the object rely on the client side to hold the object and then return it untampered. Depending on the data in the object, it is not a good solution, as a) the object may include information that should not be available on the client side for security reasons and b) you will have to verify the object after receiving it.
That said, and if you still want to use the object on the client side, then JSON is an option for serializing object data, see JSON functions in PHP.
Answer: Based on most of the answers here, referring to storing the object in $_SESSION, is it more efficient to store only the individual properties that need to be accessed in AJAX as opposed to the whole object, or does it not matter?
E.g.
$_SESSION['object'] = $object;

vs
$_SESSION['property1'] = $object->property1;
$_SESSION['property2'] = $object->property2;

I know the OP is asking about accessing the entire object, but I guess my question pertains to if it's just a matter of only accessing certain properties of an object, and not needing to access methods of a class to alter the object once it's in AJAX.

Using objects in Ajax calls PHP files


Serialization objects in Java

Question: I want to send some objects through sockets from client to server. I can serialize they like object or convert to xml. Which of this methods take less memory?
Comment: Can the server deserialize Java objects?
Comment: Yes..I only want to choose best method to don't spend much memory
Comment: Are you wanting to protect from DoS too?
Answer: Serializing them will take A LOT less space. You can also try kryo to get an even better size for your serialized objects. It supports Deflate compression/decompression. Take note however that it's non-standard, so the other side of the socket must use the library as well to de-serialize.
Answer: Naturally serialization takes a lot less memory that converting to XML... think of all those <...> and </...> tags! Serialization takes care of all that with numbers, not ASCII characters.
Also, you can serialize to xml! http://x-stream.github.io/
Answer: Converting to XML takes up more space on the client and server than just sending them serialized, since you are basically copying the content into a new variable. Sending them serialized may not use the full capacity of a packet, but you can always just process the first packet and overwrite it with the next to save some space (At least that's how I'm currently doing it).
However, serializing it will probably make the transfer slower, since you have to send multiple packages. On the other hand, if you put everything into one XML, you might run into size restrictions on the packets
(I'm talking about DatagramSocket and DatagramPacket here, since these are the ones I use. I dont know how the situation is with other transfer methods).
Answer: XML vs Java Serialization, one may use more bandwidth, but the main memory used will be your objects. If you are worried about memory used, I would make your object structure more efficient (assuming it is a real issue)
You can stream XML and Java Objects as you serialize/deserialize which is why they shouldn't use much memory.
Obviously, if you build your serialized data before sending it, this will be inefficient.

Serialization objects in Java


Question about IDE features

Question: After looking at the comments and the votes, I believe that this question warrants a discussion.  As I don't want to preference my own thinking above others in the community, I'll put my thoughts in an answer below.
Comment: This question is being [discussed in chat](http://chat.stackexchange.com/rooms/60102)
Answer: I am perplexed about why this was closed in the first place, as I don't see how it can be a question about aesthetics at all.
Colors in a block-based language are an aspect of how organizing principles of code are communicated. A quick look at Snap! shows that these colors are not simply to be pretty, but rather to allow someone to easily be able to follow the flow of the code (yellow), or to permit a kid to easily spot why the pen is not writing properly (blue-green).
I haven't placed my vote yet, but I believe it should be re-opened (and I currently intend to cast my vote that way, unless some comment comes along here that I find persuasive).  Perhaps others can explain why they believe it should be closed.
Answer: After a group discussion, we decided that we were unclear about OPs intentions, as there were multiple possible readings of the question.  We discussed two possibilities:

OP was asking about color choices specifically
OP was asking about how to logically group commands by color in order to enhance student understanding.

In the case of (1), this question was deemed to be off-topic for cseducators, and a duplicate (or near duplicate) of this question from ux.SE.  We believe that the question would possibly be on-topic at ux.
In the case of (2), then the question is on-topic for cseducators.  It suffers from being unclear, however, and should be modified to make clear that it is about how to organize commands into color groups, at which time the question can be reopened.
Answer: I've been going back and forth on this question. I originally voted to close because I saw this less as a pedagogical discussion and more of a UX one. The design of an app for use in the classroom, while educational in the end, is initially about user experience; it just so happens that the users in this case are students. That fact doesn't necessarily make educators the experts. (I hope that makes sense: I'm trying to be nuanced.)
To explore this issue more, I went through UX.SE and found this question which has been viewed more than 10K times:
Should software for 8 to 14 year olds be colourful and "childish"? (and which links to this bit of relevant research)
Based on those answers, which are informed by a deep understanding of color psychology and cognitive development, I see this as an area where UX expertise, as opposed to CSEd expertise, provides the better answer.
Maybe that's one way for us to draw a dividing line as we decide what's in and what's out here: is this the community with the greatest expertise on SE to answer that question? In this instance, I would have to say no.
Answer: To me this seems fairly clear cut: Is it a question about teaching computer science? It is, but it indirectly, and in my opinion, although I do really think it's an interesting question, it's too far from direct teaching to be on topic here.
I believe this to be too off topic because although it's a question about language design for a language used for teaching. You probably wouldn't call a professional language designer an educator (although some language designers are educators also). This is a language design and UX question because the core question is "How can I design a language for xyz (language design)? Which colors (UX design) should I use?" The goal of the OP is not to teach, but to design a language for use by teachers.
If questions indirectly related to computer science are made on topic for this site, I believe that we'd also have to permit questions like "How can I choose a UI design for my remote desktop tool for classrooms?" which sounds off topic to me.
The language design question is really asking about creating a tool for teachers rather than teaching and as such is off topic.

Question about IDE features


Can you use the Seeming spell to appear invisible?

Question: The Seeming spell gives any number of creatures you can see within range a new appearance. Is it possible to use this spell to make the caster and their friends transparent?
Comment: Shocked to find we haven't had this asked before, but it doesn't...  [**seem**](https://rpg.stackexchange.com/search?q=is%3Aq+%5Bdnd-5e%5D+seeming) that we do.
Answer: No.
It's just a disguise.

This spell allows you to change the appearance of any number of creatures that you can see within range. You give each target you choose a new, illusory appearance. [...] The spell disguises physical appearance as well as clothing, armor, weapons, and equipment. You can make each creature seem 1 foot shorter or taller and appear thin, fat, or in between. You can't change a target's body type, so you must choose a form that has the same basic arrangement of limbs. Otherwise, the extent of the illusion is up to you.

Seeming is a disguise, not an open-ended illusion (as an aside, Mirage Arcane actually calls out "disguise" and "conceal" separately: "the spell doesn't disguise, conceal, or add creatures"). There are some clear boundaries on how much larger and smaller you can appear: you can be taller, but only 1' taller. You can be heavier, but only up to "fat."
It seems like the question here is mostly based in "otherwise, the extent of the illusion is up to you." 5e doesn't define "disguise" in a rules context here to necessarily exclude invisibility, but I would take the stance that "invisible" does not constitute "a new, illusory appearance." Beyond this, we get into quibbling about the intent and the "natural language" of 5e, so I think it's simpler to just consider the balance implications of the ruling and work backwards from there.
It would be massively imbalanced.
Invisibility at 2nd level can make one character invisible for an hour, but "the spell ends for a target that attacks or casts a spell." You can upcast it to 5th level to make 4 characters invisible for an hour.
Greater Invisibility at 4th level can make one character invisible for a minute, lasting through attacks and spellcasting.
As Seeming lacks the clause from Invisibility, it would behave as Greater Invisibility: losing invisibility on attack/cast is not an implicit aspect of being invisible; it is an explicit part of the spell Invisibility. This would yield a spell that, at 5th level, casts as Greater Invisibility with more targets, lasts 480x as long, and doesn't require concentration.
In a similar vein, this would make Disguise Self a self-only, 60x duration, no-concentration Greater Invisibility from a 1st-level slot.
Comment: In theory, what you _could_ do is create a disguise that mimics the background behind the target. That way, they would be effectively invisible, but only from a specific angle. (Of course, the better term for that would be "camouflage".)
Answer: No: the spell does not grant the invisible condition.
Being invisible has specific mechanics and it is a well-defined game term.
The Seeming spell does not grant the invisible condition in any way.

Beside the excellent analysis made by Shivers in their answer, one should note that by using this spell

You give each target you choose a new, illusory appearance.

Any meaning of "appearance" refers to the visible aspect of a person, and illusory stands for something that looks real but it isn't.
Answer: Invisibility, no; camouflage, maybe
As Shivers and Eddymage explain very aptly, "Invisible" is a very specific state a creature can be in. Seeming does not allow you to gain access to that state.
I would rule however that you could give yourself and your friends a brick-wall "texture", which you could then use to camouflage against a brick wall. How well you succeed in creating this textured appearance would be subject to an Arcana check, and how well everyone manages to line up with the wall behind them would be another check (probably Stealth). But it seems doable!
Answer: Transparent: no reason why not, but don't expect it to help with stealth
You can probably use Seeming to become ghost-like, if you're trying to haunt people. Hollywood ghosts are often transparent but clearly visible.

Invisible: no, but infinitesimally small might be possible

You can make each creature seem 1 foot shorter

What happens if the creature (say, a familiar) is less than a foot in all dimensions, can you reduce its height to zero? (Given the discussion of thin/fat build, the spell presumably gives you a width and height appropriate for a zero-height creature, so you are zero size in all dimensions)
What if a much-shorter-than-average halfling (average height, 3') is Reduced to less than 1.5', and Seems much less than six inches tall?
This is very clearly not the intention of the spell, so I recommend not trying this unless your GM appreciates rules abuse at the table.


How to combine a object to an mixamo character animation in blender

Question: I created an animation in blender using mixamo character. The animation consists of three animations combined. Walk, Run and Jump. After completing the mixamo animation I realized I forgot to add a backpack to the mixamo character, but when I add the backpack and click animation the backpack doesn't stay on the character. Is there a way to join the backpack to the mixamo character without messing up anything. I am using Blender 3.01. Below is the image.
Answer:
move your backpack where it should be

(sorry for taking this simple box - i am lazy)

select your box

shift select your armature

go to pose mode

now select the bone which should be the parent of your backpack - now this depends how you want it to look...but i chose this one here:

press CTRL-P -> Bone
Comment: Wow. Thank you so much. It worked. I saved so much time thanks to you. Bless you. :) @Chris
Comment: you are welcome


Isn't the parameter A denoting technological progress from Cobb-Douglas already included in alpha and beta?

Question: Does it not follow that if capital is better performing than labor it would be used more frequently? What exactly does parameter A actually mean, then?
Comment: In crude terms $A$ in $Y=AL^\alpha K^\beta$ captures impact of *quality* whereas $\alpha, \beta$ will capture impact of *quantity*. Two different production units with same $\alpha, \beta$ and same quantity of $L, K$ employed can have different output if, say, the workers in one factory are more skilled (or the machines/tools are more advanced) - i.e., higher $A$..
Comment: A being a general parameter of quality is then the average of the parameters of quality for both L and K? Still, do alpha & beta not capture the ideas sufficiently (since their sum is equal to 1, the addend that is more efficient will be greater than 0.5) @Dayne
Comment: $\alpha>\beta$ simply means that an extra unit of labor contributes more than an extra unit of capital. So in *relative* terms $L$ is better than $K$ but $A$ allows you to compare capture improvement within one (or both) factor(s).
Comment: Why is it so that α>β  means that an extra unit of labor contributes more than an extra unit of capital? Why would an economy use more of a factor that is less efficient?
Comment: because of the declining marginal product for both factors.
Answer: The word technology might have slightly different meaning depending on a field. For example, Varian in his Microeconomic analysis (3ed) calls the whole production function technology. He literally writes on a first page of his book:

The simplest and most way to describe technology of a firm is the production function

However, in macroeconomics it is often equated to the $A$ parameter (see Romer Advanced Macroeconomics chapter 1 where technology growth and growth of $A$ for Cobb-Douglass function is used interchangeably throughout the chapter and also elsewhere in the book).
In the broader (Varian Meaning) you can consider alpha and beta parameters to be part of technology or determined by available  technology (since whole production function is by definition in the Varian meaning determined by technology) but they are not technologies themselves per se (just their component). Rather they are output elasticities and in addition their sum tells you if there are increasing, decreasing or constant returns to scale, but technological progress does not necessarily mean that firms production will have different returns to scale or that output elasticity increases or changes in any way. Also, note these are virtually always considered to be constants.
In the macroeconomic sense they are not technology and the technology would be solely captured by $A$. In macro sense the $L^{\alpha}K^{\beta}$ would be some sort of 'vanilla' production which is then augmented by technology parameter $A$ (which could by the way also enter in a way that would augment only labor ($(AL)^{\alpha}K^{\beta}$) or only capital ($L^{\alpha}(AK)^{\beta}$) or also both ($AL^{\alpha}K^{\beta}$)).
So depending on which subfield you look at the terminology will differ, but I can't recall ever seeing in the literature someone claiming that alpha and beta are the technology, at best you can argue they are determined by technology. Also depending on which field you specialize in you will want to follow the preferred terminology in the literature.


Anticipated Delphi Switch

Question: When is the Tezos mainnet expected to switch to Delphi?
Is there anything that node owners should do in anticipation?
Answer: If everything goes as expected, the network should switch to proto 7 in ~18 days (Thu, 12 Nov 2020 10:47:01 GMT).
<code>Sun Oct 25 22:41:40 UTC 2020

$ tezos-client show voting period

Current period: "promotion_vote"
Blocks remaining until end of period: 25207
Current proposal: PsDELPH1Kxsxt8f9eWbxQeRxkjfbxoqM52jvs5Y5fBxWWh4ifpo
Ballots: { "yay": 6476, "nay": 0, "pass": 22592 }
Current participation 34.21%, necessary quorum 56.37%
Current in favor 6476, needed supermajority 5180

</code>
Agora Link: https://www.tezosagora.org/period/36
About the nodes, it should be fine since this proposal didn't include sapling.
Answer: If you upgrade your node to version 7.4 or 8.0~rc1 you won't need to download the new protocol from your peers so the activation will be simpler and this also lets you check before the activation that the protocol compiles.

Anticipated Delphi Switch


Why is there internal energy change in a bomb calorimeter?

Question: In a bomb calorimeter, the reaction occurs at a constant volume, hence we say that the heat absorbed by the water in the surroundings is equal to the change in internal energy for the reaction. But,  problems in textbooks involve the reaction inside the calorimeter to be occuring at constant temperature, but for an ideal gas, internal energy depends only on temperature, so if temperature remains constant, what actually causes the internal energy change?
Comment: Try to pay more attention to the question formulation. While the topic may be confusing to you, the question itself, including your own prior analysis, should be clear.
Answer: The total differential for the inner energy is:
$$ \mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V + \mu \mathrm{d} n$$
In a chemical reaction, the chemical amount $n$ for each species changes (and the chemical potential is not the same for each species). So you expect a change in inner energy (even ignoring the $T \mathrm{d} S$ term, which is discussed further here).
Another way of saying this is that you break bonds and make bonds (and they don't have the same strength, and you might have more or fewer bonds after the reaction), turning one ideal gas into another, and that changes the inner energy.
Take the simplest example:
$$\ce{2H(g) -> H2(g)}$$
The number of particles changes, you have bonds in the product and none in the reactants - surely the inner energy is allowed to change.
Comment: That formula just describes the translational energy of the gas. This is correct for mono-atomic ideal gases. Here, all your reactants and products have bonds, and most of them are not even in the gas state, so you are missing a lot of "types of energies" in your formula.
Comment: If I for example consider the combustion of 1mol liquid oxalic acid to give carbon dioxide gas and liquid water at 27°C in a bomb calorimeter, if I subtract the values of the absolute value of the internal energy of the products from reactants, by using the formula U=f/2nRT for finding absolute value of internal energy, I find that the internal energy change for the reaction is +ve, but it is exothermic reaction so, Q is-ve, how is this consistent with the 1st law of thermodynamics?
Comment: @Kartsen , ok I understood that, can you also tell why when finding the enthalpy change for a reaction occuring in a bomb calorimeter at constant temperature, in the formula delta H=deltaU+deltaPV,  in place of delta PV we use delta n(g)RT, where delta n(g) denotes difference in number of moles of gases after and before the reaction, this formula is only for ideal gases, but as in the reaction of oxalic acid is stated above, some of the products are liquid, but an ideal gas can never liquify, but still how we use the ideal gas formula for enthalpy calculations?
Comment: The melting point of anhydrous oxalic acid is about 190 deg C.
Comment: @Kartsen , is the formula U=f/2nRT not valid for diatomic or triatomic gases? I thought it was valid for all ideal gases, only the values of f will change accordingly as the gas is diatonic, monoatomic or triatomic
Comment: @AdityaMukherjee Please ask a new question about this. It will invite others to reply as well, and will encourage you to do a bit of your own research and to explain the quantities in the formula, and allow you to format it properly.
Comment: Ok, I have posted a new question, could you please also reply to the previous question I asked.


When do MySQL DBA need root access

Question: For a MySQL DBA, what operations require root access. I know that we need to be root for installation, starting and stopping services. With DBA's not having root access, is there a way to perform majority of the operations?
Answer: Basically the mysqld is started as mysql user (in the init script is something like su mysql). Therefore there is no other option beside having access to this user.
But you can setup your system to run mysqld as a normal user.

When do MySQL DBA need root access


Safety stock calculation

Question: I am trying to calculate a optimal safety stock based on time-series analysis of the sales and the previous lead times of the suppliers.
For that purpose, I am using the classical formula which is:

That is based on the average and standard deviation of the two components, which are the forecasted demand and the lead time.
The forecasting is made using the auto.arima function of the forecast package in R.
The problem I am facing, is that I have the sales per day, and the lead times are also days, but if I try to input that directly to the auto.arima, the forecast is bad because there are a lot of days where sales are 0, and also doesn't get the yearly seasionality right.
So what I do is aggregate the sales monthly and forecast monthly, that gives good forecasts, but then to estimate the standard deviation everything has to be in the same time unit right?
So I split the monthly forecast equally amongst its days, which gives for example 20 units a day, then compare to the actual data to get the forecast error, and that is for example many days at 0 and one day 100, which as a result means high standard deviation of the error, and this leads to high safety stock values.
I think I am doing something wrong here when I forecast monthly and then compare the error on a daily basis. It feels statiscally incorrect, however I am not sure it is.
The other option I was thinking is converting lead times to months and use the formula with all the quantities in a monthly basis (I think forecast error in months will be relatively smaller), do you think it is a better choice?
Any other options I am missing?
Answer: Sparse sales data (many days with zero sales) is a well known problem in safety stock optimization. This pattern is often referenced to as intermittent or erratic demand.
This is a great post on how to design a demand prediction model for intermittent demand and use it to optimize safety stock - https://blog.griddynamics.com/safety-stock-optimization-for-ship-from-store/

Safety stock calculation


Is there a cost to running the NewYork Stock Exchange? Does it affect the price of stocks sold on the exchange?

Question: Is there a cost to running the New York Stock Exchange? For example electricity costs, management costs,  rents on the building(s).  How do these costs (if any)  affect the price of stocks sold on the exchange?
Comment: The price of the many securities is affected by the ease with which they are bought or sold---their liquidity. The costs of running affect the transaction costs of trading assets. These transaction costs contribute to the liquidity of the assets, a measure of which could be the bid-ask spread. I'm not sure how much the exchange's costs contribute, though. They might be quite small.
Answer: Yes, there's a cost to running the New York Stock Exchange. There's the premises, utility bills, staff, systems, and so on.
Yes, this affects the price of stocks on the exchange. The exchange recoups its fees, ultimately, from the companies listed on it, pushing up their costs.
However, for pretty much all companies listed on it, the stock exchange fees are a tiny part of their overall costs. Which means that the impact on the stock prices is pretty much negligible.
Comment: Stock exchanges also earn income from selling market data, and from renting out spaces for colocated servers.
Answer: Stock exchanges do not own the stocks traded there. They are just like markets you buy your grocery from. Market stalls are rented out to traders and traders pay small fee/rent in return. Similarly, companies that list their stocks in exchanges (like NYSE and LSE) pay fees in return for having their stocks listed and traded. This has nothing to do with the prices of stocks that vary every second. The stock prices reflect the performance of the companies which issue them among other economic factors.
As to the cost of running exchanges, it really depends on the size of the exchange, you should get in touch with the NYSE for information about running it. Here you can find info about LSE registration and annual fees.
Comment: Company costs (such as paying fees) have nothing to do with the performance of the company?
Comment: Yes, the fee the companies pay can be 0.5%-1% of total trade volume. The largest fees companies pay are for investment banking services and this is not related to registeration or listing.
Comment: @denesp, company performance really depends on the market it operates, competition, factor costs and its ability to maintain its market share. I cannot imagine Tesco's performance being affected by the meagre fee it pays to keep its shares listed on LSE or Amazon/Apple on NYSE.


Novice-Question: is this a 68pol mini SCSI Sub D socket in this controller?

Question: I have this SCSI controller LSI20320IE SCSI-Controller and would like to know what SCSI cable is the right one. I want to connect a LTO2 with 68 pol female. Is that a 68 pol MINI SUB D SCSI socket ( I have read: "For Wide SCSI-2, the most common connector was the larger sibling of the MD50, known as the MD68, HD68, MiniD68, or HPDB68. This is about 1 7/8" (47mm) wide. IBM used the HDCN68 on some RS-6000 systems, and it seems likely that a few other manufacturers used other alternatives". Source: WIKI ) on the controller? Where to get a proper calbe for it? THANKS
Answer: According to this Product INFO PAGE LSI20320IE the external connector is a VHDCI SCSI connector. And the cable could be get here: Gekko-computer


Is using a stop gradient on a residual connection the same as not using the residual connection?

Question: Stop gradient operation prevents the gradients to be calculated for the proceeding graph. However, skip connection outputs are added to the sub-network being skipped over.
Comment: According to the question it would be $r(x) = sg(x) + f(x)$, where $sg$ is the stop gradient function.
Comment: Say $x$ is the identity, and $f(x)$ the main path. A residual block implements $r(x)=x+f(x)$ basically. I think you want to stop the gradient of $f(x)$, right?
Answer: The question is about placing a stop gradient ($sg$) operation on the identity path, $x$, of a residual block $r(x)$, that is: $r(x) = sg(x) + f(x)$, where $f(x)$ usually represents a couple of convolutions with batch-norm.
In general the stop gradient operation tells auto-diff to treat that expression as a constant, so when asking for the derivative of $sg(x)$ it should be the same as differentiating a constant value.
In practice, I think that $x$ or, $sg(x)$ (since it returns $x$ but without tracking its gradients), still accounts for the forward pass of your network but not in the backward pass since backprop would only consider $f(x)$ and not also $x$.
Comment: So would training be the same as not using a residual connection?
Comment: No, because with $sg(x)$ (indeed, I assume doing so after each residual block) backpropagation cannot anymore consider the impact of the identity (i.e. its derivative), $x$, which initially is the input (x=input) but afterwards is the output of the $i$-th residual layer, say $x=h_i$.


UI Navigation issue sharepoint 2010

Question: I'm not sure if anyone else has seen this issue but I've first noticed it after I applied the Feb 2013 CU for SharePoint 2010. I have configured a site with global navigation with links under each heading. When I hover the navigation, a white box appears behind the items. When I scroll up or down, the white box scrolls down with the page and covers other items, almost as if it's z-index is higher than any other item. I've only seen this issue with IE and have been unable to debug it using IE Toolbar since it only appears when I hover over an item. I've tried adding all the classes that are added via css when I hover over an item, but that does not make the whitebox appear and the nav appears correctly. Has anyone else seen this issue? If not, what would be the best way to go about debugging this?
Comment: We too noticed same issue. BODY #s4-workspace{ *+position:static; } using above CSS this worked fine for me for menu but page has scrolling issue? I am still searching for solution. Kindly let me know if you have fix for this issue. Bests
Kiran
Comment: Kiran, I haven't had much luck in finding a solution for the issue. I'm starting to suspect that this may be an issue with the CU update
Answer: The answer can be found here:
https://social.technet.microsoft.com/Forums/office/en-US/442c8543-e67b-4d37-9ded-578789c7635e/sharepoint-2010-navigation-ui-bug-with-feb-2013-cu
There's an iframe being added to the DOM dynamically. Just add this style to hide it:
<code>iframe[id*='TopNavigationMenuV4_FrameID_1'] {
   display:none;
}
</code>

UI Navigation issue sharepoint 2010


Will driving through heavy rain with bike on my car damage the bike?

Question: I'll be driving for 4 hours with my bike on the rack through heavy rain. Will this damage the bike?
Comment: Roof rack or rear mount?
Comment: Also what's the bike? The answer for this year's MTBs will be very different to those for a lovingly restored 70s racer.
Comment: I'd argue that if the rain is so heavy and the traveling speed is so high as to cause damage to the bike, a major concern is the road hazard this driver/vehicle represents.
Comment: Depending on how full your vehicle is, consider taking off both wheels and maybe the saddle, wrap the chain area in cling film (saran/glad wrap) and stash the frame across the rear seat.  Remember to take enough tools to reassemble at the far end though!   A bike frame might fit in your boot/trunk/storage too, like this.
Answer: The prevalent thought among many members of the Cycling Illuminati (trademark) is that the high vehicle speeds will drive rain into your bearings on a bike if it is mounted to the roof rack. Seems kinda plausible, but here is a GCN video about how hard it is to actually drive water past seals and into bearings with a pressure washer:

Should You Jet Wash Your Bike?

Spoiler: It took them a long time, with the pressure washer jet focused directly on the bearing itself at point blank for 3+ minutes to finally drive some water past the seals.

Unless you are traveling down the Autobahn at top speed with your bike mounted to a roof rack you hyper-car, you will probably be just fine.  Pressure washer water velocity is about 243 mph (391 km/hr). A Bugatti Veyron 16.4 Super Sport is the world's fastest road legal production car with a top speed of 431 km/hr (268 mph).  Technically, if you had a  special rack designed for your Veyron (wouldn't want to upset the handling at 400 km/hr), you might be able to generate enough velocity for this to be a concern. Even then, it is likely only the head set bearings (or maybe bottom bracket if you have external bearing version) that will take the brunt. Even then, these bearings are oriented in a plane perpendicular to the plane of motion (which will greatly reduce the velocity of the water that actually interacts with the bearing seal face).  Plus, don't forget that the front wheel usually rides inside the car and the rear hub is left on the bike, but will be running in disturbed air, which will further reduce the force of the rain water.
Comment: Looking at that video it appears that he was using external bearings with fairly good seals.  A standard bottom bracket on even a modestly expensive bike is either a loose bearing unit or a cartridge.  I have seen several loose bearing BBs which were rusted completely out -- no sign of grease -- and several cartridge setups where water had gotten to the BB housing and was essentially sloshing around the cartridge.  Plus, the guy in the video was not holding the wand nearly as close as most people would, if they didn't know any better.
Comment: @DanielRHicks - BSOs or bikes without grease will always be hyper sensitive to any water exposure.  At the end of the video the presenter held the wand against the BB for 3 minutes 25 seconds to finally get a scenario where there is water ingress. He also held it within an inch of the BB for a minute without any water ingress.  I don't see how it is  physically possible for "most people" to hold it any closer.
Answer: If you're driving on the freeway, then you might be going 75 mph (120km/h) with a headwind adding another 15mph (25kph) so a total of 90mph (145kph).
That's fast enough to drive rain into the bearings in the headset, which is oriented right into the wind. Rain can also get into the bowden cables for your derailleurs and brakes, which will hasten rusting and sticking.
You can get rack bike covers (condoms). The main purpose of these is to reduce the amount of bugs that collect on your handlebars and seat, but they'll also protect your headset -- as well as your brifters and brake/derailleur cables.

NB: Your bottom bracket and axles are oriented perpendicular to the wind when on a roof rack so they'll get less water driven into them.
Comment: Do you have empirical evidence that 145 kph wind (perpendicular to the bearing orientation) will drive water into modern sealed cartridge bearing?
Comment: The OP hasn't specified what model year and type of bike he or she has.
Comment: Hope it doesn't have those ill-fated [potassium](https://www.youtube.com/watch?v=Jy1DC6Euqj4) bearings.
Comment: Could be cup and cone for all we know.
Answer: It will be about equivalent to exposing it to maybe 4-10 times the rain standing still.  And the higher the speed the worse the effect will be.
The problem is that the forward motion will tend to drive rainwater into the bearings, and, to a lesser degree, into cables and frame members.
But 4 hours is probably not long enough to be especially concerning, if the bike is well-maintained to begin with, and if it has a good chance to dry out afterwards (vs, say, being immediately locked away in a damp shed).
I'm trying to think if there's any good reason to mount frontwards vs backwards, but can't think of any.  Sideways would be worse, in addition to the obvious wind resistance problem.
Comment: Yeah, I was thinking roof rack.  Sideways on a rear rack you have car breaking wind for the bike.  Would be better to be out of the rain entirely, of course, but probably better than any orientation on the roof.
Comment: @mattnz - You probably have a point that a bike on a rear rack will accumulate "dirtier" dirt than one on the roof, and hence, for two bikes that both look equally dirty, the one from the rear rack should be cleaned and lubed first.  But the main concern here is rain (and grime) being forced into the bearings via air pressure, and my gut feel is that the rear rack is better from this standpoint.
Comment: What about sideways on a rear-mounted rack? That would keep the worst of the horizontal rain/spray off at any reasonable speed
Comment: Sideways on the rear rack there's a lot of fine mist and swirling. The mist is laden with road grime and exhaust fumes and probably exhibits high pH. Over a long time, I would expect this to do more overall damage to a bike than clean rain water from the roof.
Answer: There are many people who commute or compete in inclement weather with out issues. just prepare for the conditions.  I would apply a coat of wax to the frame and any chromed parts. Look for a chain lube designed for wet conditions. Liberally apply the lube to the derailleur pivot points, exposed cables and the chain. Do not get lube on the brake surface of the rims or rotors. When you arrive at your destination wipe off any standing water and relube if you feel you need to. Avoid covering the bike with a tarp or plastic cover. The wind will cause the cover to flap and buff the paint off the frame.
Comment: On the back of my van I almost always use a cover. I've never had any paint damage from that. I do strap the cover down well, but you can't stop all flapping.


What's the name of this flower with red colored leaf?

Question: I found this stock photo from dreamstime.com listed as "Firecracker plant" or "Flor Cebacia".
Answer: Could be a single flower from the African Tulip tree, Spathodea campanulata, though it looks more like a photograph of a painting of the flower than the actual flower. The flowers on this tree grow in clusters, not individually as shown here, and there aren't any leaves or other clues as to ID to be 100% sure, so I'd have to say 95% certainty. Considered an invasive plant in some hotter regions of Australia, its a tropical plant which is highly sensitive to cool or cold temperatures. Usually evergreen, can be grown in other places such as California, where it might behave as a deciduous tree because of cooler temperatures.


How to customize admin edit page - custom type

Question: I create a custom post type to manager my custom form. How i can display my details data in admin dashboard. It may look like
https://managewp.com/wp-content/uploads/2013/05/WooCommerce-Overview-Orders-Edit.png
Answer: I believe, you are interested in Meta Boxes. In the picture, meta boxes are all over the place. You should take a look at

WordPress add_meta_box function
Example How To Add Meta Boxes To Edit Area
Meta Box Plugin to create metaboxes.


Pipeline Analog to Digital Converter

Question: I was studying for my Design of Analog Circuits course and I have two questions about this ADC. If someone can explain me why in the figure below in the Pipeline Analog to Digital Converter, why Vin is compared with Vref/2 instead of Vref and why after doing the summation, the result is multiplied by a gain of 2? What does that has to be with functionality of the ADC.
Comment: If you have to ask that, you need to go back to the very beginning and understand what an in general ADC does, more specifically how voltages map to binary codes.
Comment: If Vref is the full scale range, then above what voltage must the signal be for the MSB to be set?
Comment: If Vin > Vref/2, then MSB = 1. If Vin < Vref/2, then MSB = 0. Why is that? Why comparing with Vref/2?
Answer: The comparison with vref/2 is because that is the place-value of the MSB of the ADC output.
the next bit has a value of 1/4 vref, so the difference value is doubled before comparison with vref/2

Pipeline Analog to Digital Converter


tab bar application

Question: I have made a tabbar application which display images in detail view i have two views first and seconed  at the image view i have a button addtofavorities i want to click on addtofavorities and add the image name into seconed view.seconed view is a table view
Comment: Please edit **your** post, not other people's posts
Answer: If you want to save the image name to other view then, just take a string in other view import other class in first view add that value to the string of other class and send to other class, it will carry that value to your other view.
Answer: Please try to explain yourself a bit better. What do you want to do? Does your implementation crash?
Based on your (lack of) explanation, it's as simple as set a selector to your favourite button and change second view.
Comment: IMHO, this should be a comment, not an answer.

tab bar application


Forward and delete individual messages in Hotmail

Question: Can I setup my Hotmail to automatically forward a message from a single sender (not all messages coming into my Hotmail account) to another email account (like Yahoo or Gmail) and then have the message automatically removed in Hotmail?
Answer: While the new look of Outlook.com is awesome, some of these types of advanced features are simply not yet available.  Currently Outlook.com only supports "basic" Exchange rules and therefore only support one action.  Once a rule matches and executes, the remaining rules are ignored as well, so there is no way to bypass this limitation.
Obviously you can enable a forwarding rule, but you would have to delete the email from your inbox manually or employed client rules in the Outlook desktop app.
Here is a recent Reddit post from Microsoft indicating that they are aware of this and working to improve things:
http://www.reddit.com/r/IAmA/comments/1m926j/we_are_the_outlookcom_team_ask_us_anything/cc6yqhp


Table Section Header showing when returning from detailed view after in UISearchDisplay Controller

Question: the following set up I have:
- A TabBarControler
  - Inside a Tab a UINavigationController
    - Inside Here a UITableViewController
      - Inside here the possibility to search entries with a UISearchDisplayController
I may not post pictures which would explain everything better but SO won't allow it.

So I have aTable with sections and entries and a search bar.
Now I do a search and select one of the search results to enter a detailed view
After selecting one of the results and returning back to the search results

The Sections headers are there but I made sure in the appropriate section callbacks that there is returned a nil respective 0 value when the search table is active.
I am guessing that it is connected with the NavController but I have no clue how to send the section headers into the background again.
Any help appreciated,
Thanks,
Andreas
Answer: ... found it ... shame on me, it was a table reload call in the parent class in the ViewDidApear Method.


How exactly ethereum is a smart-contract platform

Question: I am new to Ehterum and Blockcain. I am using ehtereum smart contracts api web3.js for my front-end interaction. I am using solidity for writing contract using truffle. I want to automate the transaction on baisis of some condition such as time interval. I have posted this question on other forums such as gitter community and got answer that I can use Oraclize or Reality keys to achieve this? So I was wondering why It is called smart-contract if it is not smart enough to even execute an automated transaction? Can someone please explain that is it true that we cannot write such automated contract in ethereum or maybe I have wrong understanding of things?
Comment: Related: https://ethereum.stackexchange.com/questions/304/why-cant-a-contract-wake-up-and-run-itself
Answer: Saying smart contracts aren't smart because they can't schedule a function to run is like saying your smartphone isn't smart because it can't phone a person you don't know the number for.
Yes, it's true you can't schedule something out of the box, when we were "promised" contracts that are smart enough to check conditions and enforce certain actions without depending on humans. When you get into it you realize it is not as straightforward as you thought and there's a lot of coding involved and things we are used to in other languages do not exist in Solidity yet, you feel like you went back in time at least 20 years. You just have to understand the shortcomings and rethink the logic of your code.
Enough rant, yeah you can't schedule a function to run at a later time. You could use Oraclize or Ethereum Alarm Clock to do it. Or, as I said, rethink your logic so there's no need to schedule the function call.
Comment: thank you @pabloruiz55 for the answer. So I just wanted to confirm that this feature is not part of smart contracts currently or will it be added in future? Ans secondly if we use third party platforms such as Oraclize won't it make our dapp centralized again as we wil have to trust someone else again?
Comment: I am not defending Ethereum's shortcoming that a contract cannot schedule a function to run. I just want to add that I feel Ethereum does not really do away with the necessity to trust someone. If you want to use a smart contract written by someone else, you are trusting the person or organization that wrote and deployed it to write bug-free code that is not malware. That trust may not be justified. Parity multi-sig wallets had a bug that someone exploited to steal about $32M. The fix for that bug introduced another, which was accidentally triggered by an Ethereum "newbie" locking away $300M.
Comment: Continuing because previous comment hit max character limit .... If humans are involved in an activity, trust will be necessary. The difference is that an Ethereum app may make it possible for you to do business with many unknown people thousands of miles away. Now, you do not have to have trust in all of them but only in the company that made the app available. So a smaller trust "footprint" and the company has more motivation to be honest because it has a bigger stake in continuing to exist.


Set of Turing machines that halt after exactly 14 steps

Question: Let $M_i$ be the Turing machine with Gödel number $i$. Let
$$A = \{i \mid M_i \text{ with input \(x\) halts after exactly 14 steps}\}$$
Is the set $A$ recursive?
Comment: Welcome to [cs.SE]. What have you tried? Tell us more about what you know, and where you got stuck.
Comment: How do you think one can check whether $i\in A$, I mean whether $M_i$  halt after 14 steps on input $x$.  I suppose that $x$ is a fixed string.
Comment: Hello! We discourage posts that simply state a problem out of context, and expect the community to solve it. Assuming you tried to solve it yourself and got stuck, it may be helpful if you wrote your thoughts and what you could not figure out. It will definitely draw more answers to your post. Until then, the question will be voted to be closed / downvoted. You may also want to check out our [reference questions](http://meta.cs.stackexchange.com/q/599/), or use the search engine of this site to find similar questions that were already answered.
Comment: I opened a discussion on meta about this question: http://meta.cs.stackexchange.com/q/1048/755
Answer: I assume you mean $M_i$ is a TM which is determined by Gödel Number $i$ such that from $i$ one can always construct $M_i$ in finite time and that $A = \{i \mid M_i \text{ halts after exactly 14 steps on \(x\) as input, where $i$ is a G}\ddot{o}\text{del Number}\}$ where $x$ is constant.
$A$ is recursive if there exists a TM for $A$ that always halts.
Such a TM can can be constructed as follows:

If $i$ is a Gödel Number construct $M_i$, else $i \notin A$, halt.
Run $M_i$ on the constant $x$ for 14 steps, if it halts after exactly 14 steps then $i \in A$, else $i \notin A$, halt.

Both 1 & 2 always halt which means that the TM always halts. So $A$ is recursive.
Comment: You might consider not to encourage [undesirable posting behaviour](http://meta.cs.stackexchange.com/questions/594) in the future.  Up to you -- I respect whatever call you decide to make.  (Not intended as a criticism...)
Comment: @D.W., complete_idiot: note that claiming that this answer "encourage[s] undesirable posting behaviour" is a personal claim, not a site policy. Some people consider answering lazy questions a waste of time, but if you want to do it, that's your choice.
Comment: Note that you can use [math formatting](http://cs.stackexchange.com/editing-help#latex) here.


Integral of $\frac{\ln (x^2)}{x}$

Question: I'm not sure where to start with this problem. I have the answer, but do not know how to work it out. I always get confused when it comes to integrating natural logs. Do I use u substitution or integration by parts? I've tried working it out with both, but I'm not sure what to use as u. I tried $(\ln x)^2$ as u, but ended up with $\frac{(\ln x)^3}{3} + C$ when it should be $(\ln x)^2 + C$ (I apologize for improper format).
Comment: Do you mean $\ln(x^2)$ or $(\ln x)^2$?
Comment: @CBenni: If by "he" you mean Jaden M., then I disagree, because Jaden M. did not know what it meant, and did not know how the answer was obtained. If you meant what is intended in the problem, then I agree, as mentioned in my answer, and as Amr mentioned in a comment.
Comment: The question is unclear, but I assume it is (ln x)^2
Comment: @JadenM.: Your title seems to have a different meaning.  I don't think that assumption is safe.
Comment: I simply do not understand which one it is. It wasn't specified in the problem, so I have no idea whether to use (lnx)^2 or ln(x)^2.
Comment: @JadenM. The things in your comment mean the same thing to me.  I think the question is whether to use $(\ln(x))^2$ or $\ln(x^2)$.
Comment: I assume he meant $\ln(x^2)$, because the antiderivative of $\frac{\ln(x^2)}{x}$ is $(\ln x)^2$, which correlates with the "coorect" answer given by him.
Answer: Note that $\frac{1}{x}=\frac{d}{dx}\log(x)$, therefore:
$$\int \frac{(\log (x))^2}{x}dx=\frac{(\log(x))^3}{3}+C$$
In case you meant $\log(x^2)$, check Babak's answer.
Comment: Then you wanted to integrate $\log(x^2)$ and not $(\log(x))^2$
Comment: @Jaden M.: If you want to see whether the answer is correct, you can differentiate it.  The derivative of $\frac{1}{3}(\ln(x))^3$ is $\dfrac{(\ln(x))^2}{x}$, and the derivative of $(\ln(x))^2$ is $\dfrac{2\ln(x)}{x}$.
Comment: When I did the problem, that is the answer I got as well. However, the answer should be (lnx)^2 + C. I don't understand how.
Answer: If you have $\int\frac{\ln(x^2)}{x}dx$ then note that $\ln(a^b)=b\ln(a), a>0$ so you have $\int\frac{\ln(x^2)}{x}=\int\frac{2\ln(x)}{x}dx$ and this latter one can be evaluated by setting $\ln(x)=t$. For another one look at @Amr's answer.
Comment: Well done, again!+1
Answer: If you are trying to find an explicit formula for the antiderivative of a function given by an explicit formula, it is essential that you know what the explicitly given formula means.   While you could solve the problem both ways for additional practice, clearing up what the actual problem was in the first place is still needed.
When notation is ambiguous, there is not generally a correct mathematical way to figure out what the notation means.  It requires some way of finding out what conventions are used in the source of the problem.  If the problem is in a textbook, you often find explanations of the conventions earlier in the book, or at least enough context from earlier examples to determine the conventions.  If the problem came from a teacher, you can ask the teacher what is meant.  If the problem arises in the real world, you can look at the context of where it came from, and you may want to kindly ask the person who wrote it down to use less ambiguous notation in the future.
In calculus textbooks, it is common to denote powers of some functions by putting the exponent between the function name and the argument.  This allows less parentheses to be written, and thus laziness may be a key source of the popularity of the convention.  This is most commonly seen for trigonometric functions.  E.g. $\sin^4 x$ means $(\sin(x))^4$.  When this convention is used, appearance of something like $\sin x^4$ can usually be safely inferred to mean $\sin(x^4)$.  The reason it would be expected to not mean $(\sin(x))^4$ is that the latter would be written instead as $\sin^4 x$.
The same convention may be applied to logarithms, and thus $\ln^2 x$ could mean $(\ln(x))^2$ while $\ln x^2$ would mean $\ln(x^2)$, which turns out to be equivalent to $2\ln(|x|)$ for all real nonzero $x$.  The context of what you say the answer "should be" indicates that $\ln x^2=\ln(x^2)$ is likely to be the convention used in your problem, whereas $(\ln x)^2$ is used when the function is squared rather than the argument.  (So while I don't see evidence of your source using the "$\ln^2 x$" notation, they do seem to keep the convention that exponents after the argument without parentheses apply to the argument, not the function.)
Answer: Can be solved using substitution method:
Edit: $ln{(x^2)}$ can be written as $2ln{x}$. (Basic properties of logarithms) So the main equation now looks like: $\int\frac{2ln{x}}{x}dx$  Now, the substitution part:  Put $ln{x} = t$
differntiating both the sides:  $\tfrac{dln{x}}{dx} = \tfrac{dt}{dx}$  => $\frac{1}{x}dx = 1dt$
Now substitute $ln{x} = t$ and $\frac{dx}{x} = dt$ in the main expression
The expression so formed is: $\int 2tdt$  Now, it's easier to integrate this.  We get: $2\frac{t^{2}}{2}$ + C = $t^{2}$ + C  Now substitute $ln{x}$ back with $t$
We get $(ln{x})^2$ + C.
Comment: Thanks.  "In the question, it is $\ln(x^2)$."  That depends on what part of the question you are looking at.
Comment: How I would interpret "$\ln(x)^2$", it cannot be written as $2\ln x$. Unless you meant $\ln(x^2)$ (and $x>0$), but $\ln(x)^2$ would reasonably be interpreted to mean $(\ln x)^2$. That is, those parentheses serve the purpose of showing what is the input of the logarithm.  A major problem the OP had was interpreting what was even meant by the notation, so using $\ln(x)^2$ to mean $\ln(x^2)$ might not help. Otherwise, I agree with this as a solution to what is probably the intended problem, and it gives more detail on the outline of Babak's answer. (To me, $\ln(x)^2=(\ln x)^2\neq \ln(x^2)$.)
Comment: @JonasMeyer Ooops... Sorry. I mis-typed it. Since, in the question, it is $ln({x}^2)$
Comment: @JonasMeyer Oh! The question is in the "title": $\frac{ln(x^2)}{x}$. And the desired answer is $(ln{x})^2$.

Integral of $\frac{\ln (x^2)}{x}$


Short story about an astronaut that enters an alien space ship and discovers it is a sarcophagus

Question: Upon realizing it is a sarcophagus, he decides to sacrifice himself (I think by disconnecting his air) and exist in it for eternity along with the original creature. It may have been in an old Omni magazine, or some other periodical. I thought it may have been in the anthology "Dead Astronauts" but I did not find the story in there.
Comment: Sounds vaguely like something Lovecraft would write, but sounds much newer
Comment: duplicate of https://scifi.stackexchange.com/questions/38659/book-series-with-an-astronomer-exploring-a-golden-egg-shaped-spaceship-in-earth (not accepted but confirmed by OP comment)
Answer: Ben Bova's Voyagers (1981), the first book in a four-book series.  An alien ship is detected coming into the Solar System.  The only astronaut able to make physical contact with the ship does so as a one-way journey; he will die inside the alien ship.
Spoiler:

 The astronaut revives because the alien sarcophagus is covered in alien nanotech robots.  Nanobots infect the astronaut, bringing him back from the dead.  The astronaut is recovered by other astronauts and sent back to Earth.  The nanobots become the focus of the next two (three?) books, because the alien nanobots include the dead alien's consciousness, hence the titles The Alien Within (1986) and Star Brothers (1990).  The nanobots give regeneration, healing, different powers, but all at a price - the alien judges each person's life up to the point it is infected into, deeming them worthy or not of the nanotech gifts.  Several Earth mega-corporations try to strip the alien consciousness from the bots, to no avail.  The alien nanobots are trying to educate humans into accepting aliens.
Comment: Well spotted. Since the OP was looking for a short story that he read in *Omni*, perhaps you should add a link to the original short story (also titled ["Voyagers"](http://www.isfdb.org/cgi-bin/title.cgi?64115)) by Ben Bova in [the August 1981 *Omni*](http://www.isfdb.org/cgi-bin/pl.cgi?59796) which led to that series of books, and which can be read [here](http://archive.org/details/omni-magazine-1981-08).
Comment: This is the short story as I originally read it. I was not aware of the book series. Checking it as the correct answer. I'll let the moderator decide who gets the points.
Comment: Since I was answering according to the novel, and not the original magazine short story, I don't mind giving credit to user14111 for the correct answer.
Comment: I do want to add that it seems like it took comments from both jhpace1 and user14111 to solve this one. It appears that user14111 identified the short story after seeing the comment from jhpace1. Thanks again to both of you.


Why does the temperature of the working fluid have to be close to the temperature of the cold reservoir during a Carnot cycle?

Question: I understand why the temperature of the hot reservoir has to be minimally higher than the temperature of the hot working fluid during the isothermal expansion phase of the Carnot cycle (to limit new entropy being produced in the working fluid that we have to get rid of). But during the isothermal compression phase why do we need the cold reservoir to be only minimally cooler than the cold working fluid? The working fluid looses the same entropy independent of the temperature of the cold reservoir, so why do we need to minimize the new entropy created in the cold reservoir? What is wrong with just letting the cold reservoir get more and more entropy as long as the working fluid returns to the same state as before the cycle?
Comment: If there is resistance to heat transfer within the working fluid, the volume average temperature of the working fluid will differ from that of the hot and cold reservoirs.  So which temperature do we use, and, with an irreversible process and non-uniform temperature taking place within the working fluid,  what temperature do we use to calculate the work?
Answer: The work $W_0$ produced in the Carnot cycle is the result of transporting a certain amount of entropy, say, $S_0$ from a higher temperature $T_h$ to a lower temperature $T_{\ell}$. Here $W_0=S_0(T_h-T_{\ell})$ and this is done by isothermally absorbing  $S_0$ at $T_h$, then adiabatically expanding after which isothermally rejecting the same amount of entropy $S_0$ at a lower temperature $T_{\ell}$, followed by an adiabatic compression stage to close the cycle. If you increase the temperature at which the entropy $S_0$ is rejected to the environment or lower it at which it is absorbed or both, then less thermal work is done than otherwise would be available.
Answer: Energy transfer down a gradient (e.g., in temperature) generates entropy.
The Carnot cycle represents the (unachievable) ideal of zero entropy generation.
Therefore, the Carnot cycle can't incorporate any heat transfer across a finite temperature difference.


How do I connect a new remote to my ps4?

Question: Just bought a ps4 remote. How do I connect it to my ps4? I tried pushing the PlayStation button & connecting the usb cord. No luck.
Comment: Do other remotes work on that PS4? Have you tried that remote on other PS4s? Did you buy it new or used?
Comment: Remote that came w ps4 works, but brand-new remote not working.
Comment: Sounds like a dodgy remote. I'd return it.
Answer: The standard way of connecting the PS4 controller is to connect it through USB and pressing the PS Button. If that fails for you, you can try to manually connect it through bluetooth. To do that, you have to make sure the controller is off, and press the PS Button and the Share button at the same time. The light on the controller should start to blink. Next, you will have to go into the bluetooth settings of the PS4 and find the controller and pair with it.
I have had to manually connect it through bluetooth before when the USB method wasn't working.


How this $\int_0^{a}x^\text{erf(exp(-x))}dx$ behaves?

Question: Really i'm confused about behavior of this integral : $\int_0^{a}x^\text{erf(exp(-x))}dx$ using wolfram alpha for some values it's seems to me that integral satisfying : $I(a)=a-0.14...$ , for $a \geq 10$ , but i didn't succeed to get it closed form since it is divergent , Then I want to know more about it's behavior ?
Comment: Is the integrand strictly positive over the interval of integration? Does it tend to some non-zero value?
Answer: The leading behavior as $a\to\infty$ is easy. Just define
$$
I(a)=\int_0^{a}x^\text{erf(exp(-x))}dx\ ,
$$
and take the derivative
$$
I'(a)=e^{\log(a)\cdot\mathrm{erf}(e^{-a})}\ .
$$
One can show that
$$
\lim_{a\to\infty} I'(a)=1
$$
(which implies the behavior $I(a)\sim a$) by the substitution $e^{-a}=t$:
$$
\lim_{a\to\infty}e^{\log(a)\cdot\mathrm{erf}(e^{-a})}=\lim_{t\to 0}e^{\log(-\log t)\cdot\mathrm{erf}(t)}
$$
and using $\mathrm{erf}(t)\sim 2t/\sqrt{\pi}$ as $t\to 0$.
To compute the subleading contribution, we need to study
$$
J(a)=\int_0^{a}x^\text{erf(exp(-x))}dx-a=\int_0^{a}(x^\text{erf(exp(-x))}-1)dx\ ,
$$
which leads to a convergent integral as $a\to\infty$. Therefore, the sought asymptotics reads
$$
I(a)\sim a+\int_0^{\infty}(x^\text{erf(exp(-x))}-1)dx+o(1)\ ,\qquad\mbox{for }a\to\infty\ ,
$$
where the constant term
$$
\int_0^{\infty}(x^\text{erf(exp(-x))}-1)dx\approx -0.14113...
$$
(there is no hope to find a closed form for this integral, though - but it's fine, it's just a constant).


Internal forces in a system

Question: Newton's third law of motion says that the internal forces in a system cancel out each other --- what is the system here? what can be considered as a system? and why do the forces cancel out each other's effects?
Answer:
what is the system here? what can be considered as a system?

A system can be anything.
For example, if we consider the Earth and Moon as one system, then their mutual gravitational attractions are internal forces, which cancel themselves out. If we considered only the Moon as the system, then the gravitational force from the Earth is not internal anymore.

and why do the forces cancel out each other's effects?

Because, by Newton's 3rd law, whenever there is a force present then there is also always a reaction force of the same size but opposite. Naturally, these cancel each other out if they are both acting on a system.
Those two forces do not act on the same bodies, though. If you kick a bowling ball, then the ball feels your kicking force, while your foot feels the ball's reactions force. But if you consider both ball-and-you as one system, then both forces are included in the system. Each object feels a non-zero net force, making them accelerate (the ball starts moving, while your foot is slowed down). But the system as a whole doesn't feel any net force. It doesn't accelerate. The two parts that the system consists of (ball and you) are accelerating, but the centre-of-mass of this ball-and-you system doesn't accelerate at all.
Answer: Consider a system of two cars colliding. The system consists of two cars. That's what Newton means.
Internal forces are any forces due to bodies inside the system, hence the name. Although for each car, there is a net force due to the other car (as they just collided with each other and there is a contact force), for the system as a whole, nothing really acts on it.
If this is the case, momentum within that system is conserved.
Now if there is an external force (a force by anything that is outside the system acting on bodies in it), momentum conservation law still does not fail (as Bill pointed out in the comments) as the final momentum is simply adding over the impulse of an external force to the initial momentum:
$$p_{final} = p_{initial}+ \int Fdt$$
This is in fact the general expression for momentum conservation in a system. In cases where there is no external force, $F=0$ and we get our familiar $p_{initial}=p_{final}$ relation.
Comment: @AaronStevens ya, I have edited them.
Comment: No, the momentum conservation law is $\vec{p}_{final}=\vec{p}_{initial}+\int \vec{F} \mathrm{d}t$. If $\vec{F}$ is zero, the momentum is *constant*, but momentum is still conserved even if $\vec{F}$ is not zero. Conservation *includes* the possible transfer of momentum into or out of a system. Conservation is not the same as constancy. Constancy is a subset of conservation. Momentum conservation law *never* fails.
Answer: There  are many general misconceptions about Newton's 3rd law.Read the statement of 3rd law in simple terms:-
"If a particle A exerts a force F on a particle B,then particle B would exert a force --F on A".
We don't know why particle A would exert a force and in counter attack particle B would exert a equal and opposite force.It is a law and we have to follow it just as we follow definitions in mathematics.
So,this law never talks about systems,it talks about particles.
You can take anything as a system and a force which is applied by particle of the system on other particle is internal (for multi particle system).
It is a consequence of 3rd law of motion that internal forces    cancel out in the system.
A force applied by an agent other than the system is external force for the system.I am saying again that internal forces cancel each other in a multi particle system as a consequence of 3rd law and we have to follow it.
A research is being done in LHC for the proof of statement of 3rd law on gravitons.
Answer: Firstly Newtons 3rd law doesn't mention systems it just says that is object A exerts a force of object B then object B will exert an equal force but in oposite direction on object A.
This is practical physics and as you mention systems lets use the system that is the human body.
Instead of object A use the palm of your hand.
For object B use your forehead.
If you place the palm of your hand (object A) on your forehead (object B) you will ferl your forehead exert preasure on the palm of your hand that is equal to the preasure that the palm of your hand is exerting on your forehead. This is a practical example of Newtons 3rd law within the system of a human being.
However due to the Law of Energy Conservation which states energy can neither be created or destroyed the energy used by the forces relating to Newtons 3rd Law of Motion in the Human System are transformed into sound energy which is of course vibrations on a quantum level.
The theorem for this energy conversion is as follows
IF force energy > doh THEN quantum vibration will be (argh)

Internal forces in a system


What's causing this buzzing sound on the high e string?

Question: My Takamine acoustic has a tendency to buzz when I play the high e string fretted around frets 2 and 3.  I figure it's something to do with the neck and tension rod.  Are there any other parts of a standard acoustic guitar that can be adjusted to fix string buzz?  What's a safe procedure for adjusting the tension rod?  Is there anything else that could be causing the buzzing?
Comment: I lean toward your frets needing to be checked and maybe leveled and recrowned. If the saddle was too low you'd tend to hear the buzzing up and down the neck. The nut can't affect a fretted note because the finger pressing down negates the nut's effect. It's also possible the neck is a bit warped or twisted but that's pretty rare in comparison to frets that aren't seated right or not leveled.
Answer: Buzzing is almost always caused by a string vibrating against a fret. This could be due to a worn spot on the fret you are pressing on, which results in the string being lower at the point of fretting and higher, unworn frets being in the path of vibration. Finding out which fret is the culprit can be done with a straight edge, but I recommend that you take your guitar to a luthier for a look see. Those guys are professionally trained to spot issues like this, and nearly all consultations are free (at least with the one's I know).
Considering the truss rod, I wouldn't monkey with that. There's a lot of information on the web about how to adjust those, but there's a really good chance you could warp the neck if you don't know what you are doing.
Also, a mis-cut nut, action that is set too low, or playing with a heavy hand can cause these types of buzzing. If you are a hard strummer, consider dealing with the pain of higher action for less buzz. It's a trade off we all have to manage.
Comment: A mis-cut nut won't cause buzzing on a fretted note, only open notes. The saddle could cause the problem on fretted notes though.
Comment: "Buzzing is always caused by a string vibrating against a fret." No. There can be loose parts inside, or, like I had, a string buzzing against the bridge.
Answer: I had exactly this problem a while back, which I suspect was caused by the nut bedding down a bit. I sort of fixed it by changing my string gauge so I now have slightly heavier at the bottom end (56 instead of 53 if I remember rightly). The higher tension has pulled the neck up just enough to clear the buzz.
I should say that I did this to get more bass out of my acoustic (it's a narrow body) rather than to fix the buzz, but it sorted both.
Obviously, this is not a proper fix but an interesting little anecdote. (Maybe?)
Jduv is right, go and see a luthier or experienced guitar tech. Don't go fiddling with the truss rod!
Answer: Regarding the truss rod, lots of people say not to mess with it. However, if you read up on what it does and have a minor adjustment to make to correct high action or fret buzz by low action then a making a couple of half turns the correct way followed by tuning and checking the action is simple enough. Though i must admit the first couple of times i needed an adjustment i went to a guitar tech and watched what they did to build the confidence to tackle it on my own later on.
Answer: Like the above said, take it to a luthier if you can.  But if you want to mess with it, I'd suggest first checking where the string rests in the slot of the nut (usually made of bone).  If it's just that high E on the first couple of frets, try slipping a thin piece of paper, or something between the string and the slot, then tune it back up.  If the buzzing goes away, here's another trick.  Instead of replacing the nut, get a tiny bit of baking soda and super glue.  Mix it up, and fill in the slot on the nut, for the high E.  Then you will need a tiny file (just the width of the string, and slowly file the hardened baking soda/glue mix a little bit.  Reset the string, tune it, and if you like it, leave it.  If not, file in small increments, until the string is where you like it, and the buzz is gone.  The dude at the music store said the baking soda/crazy glue is harder than the bone.  Good luck!
Answer: On my vintage Goya acoustic, the high e string was buzzing like the drone string on a sitar; open and in all fretted positions. The fix was easy. I loosened the string, bent it some more before putting it down the bridge hole and made sure the knob at the end of the string was making good contact with the bridge plate. I held the bridge pin down while I tightened it and Voila! no more buzzing. The better the contact the knob at the end of the string makes with the bridge plate, the better the sound...period.


How does an air conditioner produce liquid

Question: I always notice that when an air condition is working and cooling the room it, is heating outside the room and a Liquid is dropping from the other side of the air conditioner. how does this happen and is it drinkable?, what is the name, is it water(H2O)?
Comment: I'm voting to close this question as off-topic because (a) the answer is easily found using an internet search engine; and (b) I'm not comfortable advising anyone on the consumption of water condensate from an air conditioning unit.
Comment: Hmmm... I've very comfortable advising not to drink the condensate. The condensate is not at all like distilled water.  Quite possibly lead solder on the coils plus heaven knows what sort of biological gunk (bacteria, mold, algae...) has grown from the air pulled through the coils.
Comment: I'm voting to close this question as off-topic because it is not about chemistry.
Comment: Just because the condensate is contaminated post-condensation doesn't mean it is isn't being distilled.
Answer: At the risk of answering a question that may be better suited in Physics.SE, yes, it's water. It's condensation from the evaporation coil, which is where the refrigerant expands in order to cool the air drawn through the unit. That is how an AC unit also dehumidifies the air.
Chemically speaking, distilled water is drinkable, but it isn't great to drink. It can sap electrolytes from your body over time. From a unit like that, however, it may be contaminated with oils, rust, dirt, etc.


With multi-line labels, how to show the 2nd line only for labels that need it?

Question: Mac OSX 10.6.8 QGIS 1.8.0 : I've two columns in the attribute table, name, and name2 for those labels that need to be 2 lines; multiline labelling is checked. But when I use "name"||'\n'||"name2" I only get labels when there is a printable entry for both name and name2. Can you spot what I'm missing to show the 2nd line only for labels that need it?
Answer: Thats possible only with QGIS Master at the moment:

"Strecke" || '-n' ||   CASE WHEN  length( "Kilometer"  ) > 0  THEN  "Kilometer"  ELSE ' ' END

You don't have the CASE statements in Lisboa.
Comment: Am I not correct that '' above is a null string? If so, I don't see why the case statement is needed when "Kilometer" is already null.
Comment: No, I used a single blank. A Null string seems to break the labelling.


Bava Metzia: Dual meaning of "Metzia"

Question: Has anyone ever commented on the fact that Bava Metzia, literally the "middle gate," can also mean the gate of "found things," and Bava Metzia is the one masechta that deals with finding lost objects?
Comment: are you asking whether there is discussion of the pun connecting the Aramaic word and the Hebrew word?
Answer: They sound alike in English but I'm afraid there's nothing more to it.
The tractate's name is מציעא,
which is Aramaic for  אמצע (middle). Mem-tzadi-'ayin. It goes between "Bava Kama" (first portion of civil law) and Bava Basra (last portion).
Lost objects occupy the first part of Bava Metzia, but are not its sole (or even main) topic.
"Finding" is mem-tzadi-aleph; just because most Ashkenazim pronounce 'ayin and aleph interchangeably doesn't mean the letters are linguistically identical.

Bava Metzia: Dual meaning of "Metzia"


How Short-time Fourier transform (STFT) is different from from filtering?

Question: If we want to know frequencies in a signal over time I know we can use Short-time Fourier transform (STFT). but I read in a paper "The spectrogram representation was obtained by first filtering the speech stimulus into 16 logarithmically spaced frequency bands between 250 and 8 kHz" so it seems they defined 16 bandpass filter and filter the data.
I want to know how different are these two methods?
Comment: you describe pretty well how differently they are: one is a fourier Transform, the other is a logarithmically spaced filter bank, so they yield fundamentally different bands
Answer: STFT is also bandpassing. The method the quote refers to is likely the Continuous Wavelet Transform. The fundamental difference is, STFT uses fixed-resolution kernels spaced linearly, while CWT uses varied-resolution kernels spaced logarithmically.
Example CWT filterbank in frequency domain (source; x axis from 0 to pi radians):

An STFT with Gaussian window would have same shaped filters, but with fixed width and peaks incremented linearly. A key difference is, CWT kernels are guaranteed to be zero-mean, STFT aren't, so STFT may leak in dc information which is undesired for bandpassing. Nice tutorial here; also see this answer.
If the quote doesn't refer to CWT and it's arbitrary bandpassing, I complement other answers in that STFT is a time-frequency representation, that enables powerful non-linear analysis methods via manipulating the joint 2D representation. E.g. synchrosqueezing for denoising or estimating instantaneous A.M. and F.M.
Answer: Using an STFT usually implies a set of FFTs or DFTs, each of which is identical to a bank of fixed length FIR filters (depending on any window(s) applied), that FIR filter length being the size of each STFT.
If you use a separate pre-filter, you can use an IIR or much longer FIR filter kernel, which may provide better filter characteristics (stop band attenuation, low frequency cut-off, Q factor, etc.) than can be accomplished within just the FFT length.
Answer: Both are basically convolution of a signal by one or several IIR or FIR filters (a bank of complex filters), that can be implemented in the Fourier domain. However, the "several filters" of the STFT are design together to possess certain "invertibility properties". They cannot be "any bunch of bandpass filters". For instance, they should (altogether) preserve the DC or zero-frequency component.
Short-Time Fourier transforms are inherently a signal representation (analysis only, onto a 2D time/frequency plan) that preserve all the original information content of the data. Filtering (linear ones) denotes operations that potentially modify the content of the data, by incorporating both analysis and synthesis back to the 1D initial time domain.
One advantage  of the STFT is that it allows you to perform analysis, detection, masking, etc. in a nonlinear or non-time-invariant fashion by inserting processing blocks between analysis and synthesis, allowing to synthetize a result which is different from traditional filtering.
Comment: _"nonlinear or non-time-invariant"_ I presume refers to manipulating the spectrogram, worth distinguishing from the representation itself which is LTI.
Comment: Yes, unless one tries tp have different subsampling operators on branches of the filter bank
Comment: Assuming uniform subsampling that's equivalently variable hop length, that makes sample-to-sample time scales different across rows, yes, though individual rows remain LTI. Unsure that disqualifies it as "LTI representation" but it does prevent 2D convolution over it.


Magento 2 pwa Module "Magento_UpwardConnector"

Question: Module "Magento_UpwardConnector"
Fatal error: Uncaught TypeError: file_get_contents() expects parameter 1 to be a valid path, bool given in /var/www/html/vendor/magento/upward/src/Resolver/File.php on line 94
Answer: Sorry, this is not an answer but I'm lacking the rep to comment.
I'm one of the author's of this plugin and will need additional information to help debug. Can you please link me to the documentation you were following to get to this step, and any additional context that led up to this stack trace?
Just a guess, but this appears that you've enabled the UpwardConnector module without configuring a path to an UPWARD file, which is required. To remedy that you would log into your backend and configure the path to that file in Stores > Configuration > General > Web > UPWARD PWA Configuration.


What is the reason for radioactivity?

Question: In any radioactive series we can observe that the half-lives of a daughter nucleus might be less than that of the parent nucleus. If the reason for radioactivity is gaining stability, why does a more stable nucleus (the one with higher half-life) turn into a less stable (the one with smaller half-life value) one?
Answer: As you say, the daughter nucleus may be less stable, therefore the reason for radioactivity cannot be to gain stability.
Instability simply means that there exists an allowed transition to a lower energy state. For a very unstable state that transition typically takes a short amount of time. For a completely stable state that transition takes an infinite amount of time. Therefore, once it gets to a stable state then it will not decay further, but the "reason" for the decays is the loss of energy rather than the gain in stability.
Stability is more of a stopping criterion than a driving force.
Answer: There are several reasons why that might be true. Here is one, and I invite the experts here to weigh in with more.
We know from experiment that there are several different fundamental physical processes that can take place during radioactive decay. They have different characteristic time scales that in turn depend on things like conservation laws, which allow some decay mechanisms while prohibiting others at each step in the overall decay process.
So an unstable nucleus may begin its path towards greater stability with one particular decay mechanism that is relatively slow, and then follow it with a second step that is faster, and a third which might be slower, and so on.
Comment: How does the decay process 'now' know that it would lead to stability after a large number of such decay processes?
Comment: this has to do with the number of protons versus neutrons in the nucleus. for all nuclei heavier than iron-56, there are not quite enough neutrons to overcome the electrostatic repulsion of the protons, and the nucleus will take any chance it gets to get rid of the excess protons. whole books have been written about how this works; it is a complicated business but one which is well-understood by now. have  look at "radioactive decay" on wikipedia for a start.

What is the reason for radioactivity?


Can I obtain Dutch citizenship if my father received it "by option" without a request?

Question: My father received a certificate from the embassy in 2003 titled "Confirmation of the acquisition of Dutch citizenship by option," stating that he has Dutch citizenship. However, he never applied for a Dutch passport; he died in 2007. I am now 19 years old and my sister is 13. Will either of us be able to obtain Dutch citizenship? Does he have to be alive to declare this for us?
Answer: It may depend on why your father was able to use the option procedure to acquire Dutch citizenship; at least some option procedures are retroactive.  More importantly, it depends on whether your father was Dutch when you and your sister were born.
For example, if your father was Dutch from birth, but lost that nationality because he was born abroad and lived abroad for more than 10 years after his 18th birthday, then his option procedure will have restored his Dutch citizenship retroactively ("met terugwerkende kracht").  In this case, it does not matter when you and your sister were born.
If your father's option procedure is not retroactive, then you and your sister can be Dutch only if you were born when he was Dutch.
In addition, he must have been married to your mother or else he must have acknowledged you formally as his children.
If these conditions were met, then you are already Dutch and all you need to do is apply for a passport.
The only declaration that might be needed would be the "acknowledgement" that you are his children, which isn't necessary if he was married to your mother at the time of your birth.
Comment: Hi @phoog . Yes he had dual nationality. Dutch citizenship and another. He lost it by living outside of the Netherlands for 10 years after his 21st birthday (under the 1985 law). He then made this option statement. This option certificate was indeed retroactive, which made him "deemed to have never lost dutch citizenship". More of it can be read here: [https://en.wikipedia.org/wiki/Dutch_nationality_law#From_1_January_1985_to_31_March_2003]. My sister and I have now both received our Dutch passports by applying with this certificate. My mother and father was also married at the time of birth.
Comment: @Brandon success! Thanks for coming back to let us know.
Comment: Thanks for your contribution. Your knowledge is very helpful and correct :).
Answer: It depends, it depends on where you live and have lived.  You also may have to renounce your existing citizenship.
Whether or not you are eligible for the "option" to become dutch depends on these criteria: https://ind.nl/en/dutch-citizenship/pages/option.aspx
As an adult it sounds like it's unlikely for you and perhaps possible for your sister as a minor.  I would reach out to your local dutch embassy or consulate for more information.
Comment: Since the certificate came from the Embassy, it appears to be one of the option procedures that does not require residence in the Netherlands, so it seems likely rather than unlikely that the OP is Dutch.
Comment: If the father was Dutch when the OP and her sister were born, and could pass on his nationality to them, then they are already Dutch.  It doesn't matter where they live, because they're both under the age of 28, so they could not have lost Dutch nationality because of residence outside the kingdom or the EU.  At least some option procedures are retroactive, including one that was available from 2003 to 2005 but is no longer, in which case the father will have been Dutch when the children were born.  That option procedure also does not require renunciation of other nationalities.


Vanishing gradient problem

Question: In a neural network, does gradient vanish during a great number epochs as well, rather that only vanishing through different layers?
Comment: What do you mean by "rather that only vanishing by different layers"?
Comment: Sorry, I mean "Vanishing through different layers" I expressed myself badly
Answer: Gradient vanishing means drastic decrease of gradients when backpropagating through many layers. This problem is also known for recurrent neural networks as they are mathematically equivalent to very deep networks.
That is true that gradients decrease during training. However, that is not gradient vanishing. That is the sign that the network has been trained, i.e. that is what you normally expect in the end of network training.

Vanishing gradient problem


4.5 year old will use diaper instead of toilet because "it's not a big deal"

Question: My 4.5 year old step son, uses the toilet during the day for number 1 and 2, but still wets the bed. He is consequently still sleeping in diapers. This allows for a small time in the morning where he is awake, and in a diaper. If he needs to use the toilet during this time, he will make no effort to go to the toilet and just go in his diaper (poo or pee). We try to be as kind and understanding with night time accidents, but it seems he is understanding this "no big deal" attitude to mean that using a diaper if he has it on is perfectly fine.
How can we encourage him to use the toilet rather than his diaper at all times? We already have "take your diaper off immediately when you wake up" kind of rules in place but they often fall to the wayside if he gets distracted.
Answer: I have to sympathize with your 4 year old.  It takes a lot of self-discipline, likely more than he has available, to go to the bathroom when there's a much more convenient option.
Have you considered trying to night-train him yet?  That would solve your morning problem, and most four year old children are capable of staying dry at night.
Here are a couple of resources on night training.
Best of luck!
https://betweencarpools.com/bedwetting-child-these-solutions-make-cleanup-way-easier/
https://betweencarpools.com/toilet-train-a-child-for-the-night/
Answer: Stop using diapers and get ready to do lots of laundry.
Honestly, there's not much reason for him not to use a diaper. You are right in that it should be no big deal, but if it's no big deal, then why change? The "change" and "big deal" can come from his own discomfort. There is nothing "dangerous" about a kid wetting his pants or the bed (if cleaned up in a timely matter). It's just messy and a general pain.
So take a three step approach:

Perpair for extra laundry. If that means more loads at home or more laundry trips, repair what you need for that. Make sure he is a part of the prep. "We're going to try to not need diapers anymore. I know there will be mistakes, so we're going to get ready for them."

Stop buying diapers.

Clean up the mess; make him a part of that, too. Ohhh, you wet the bed. No biggie. Let's get the sheets to the laundry.

And remember, we all went through it. Just take a deep breath, and remember to get mattress protectors or whatever you need. (Chances are, with him taking an active role, you are looking at 2-3 first-night accidents and then 1-2 accidents a week for a short term, then 1 a month or two or so for "quite a while".)
Comment: Yup, my 5yo still wets the bed if she doesn't go to the bathroom before bed. She's 6 in May. Some kids just take a while. We just have lots of mattress protectors and spare linen, and do lots of laundry.