Thoughts precede neuron firing or vise-verse?

Question: I am at the verge of submitting a paper on 'scientific explanation of reality' to an international philosophy body. First submitted draft was returned by the evaluation expert, asking me to re-submit it, after re-checking the validity of my  assertion that 'science (biology) believes that neuron firing always precedes thought'. Can any one provide answer on biology's clear stand on above question ?
Comment: how do you define "thought" biologically?
Comment: see the "materialism" philosophy debate. biologists are more agnostic on this, a position not really required for biological work. maybe a better fit on [philosophy.se] or [cogsci.se]. there is possibly some data from neurobiology specifically brain imaging studies that does impinge on the question.
Comment: i'd be surprised if you could find a neuro-scientist who didn't go at least this far.
Comment: I would suggest doing reading on corollary discharge to answer part of your question.
Comment: this should be on the neuroscience stack. Voting to close until you migrate it.
Comment: @John — the question is too old to migrate, and the original poster hasn't been seen in the last seven years.
Answer: I think the only way to interpret these kinds of questions biologically is to assume that "thought" means "the perception of a thought occurring." The perception of a thought, like all perception, arises from neuronal firing.
The interesting biological (rather than philosophical) questions arise when we look at the relationship between the neural firing that gives rise to the perception of intent to act and neural firing that carries out the "intended" actions.
This paper is relevant. The authors are trying to argue that the initiation of a motor action comes before the perception of intent for that action. I'm not a cognitive neuroscience person, though, so I'm not very familiar with their techniques and personally have a hard time determining if they draw valid conclusions.
Comment: I didn't read the paper that you linked, but the reaction time literature supports that assertion as well.  See some of the papers in the early 2000s by JF Kalaska.  The work is done in primates, so it's difficult to gauge "perception" of the movement.  I'm sure there's been human work since then, I just haven't looked for it.
Answer: You've stumbled on an old and unanswered question that is still debated--- bordering on the issue of Free Will. The reason it is still in the realm of philosophy is because current experiments don't allow direct testing of which neurons give rise to specific thoughts. This area of consciousness is fraught with difficulties and we just can't get at an answer.
Now for a more philosophical answer.
All consciousness is generated by the brain so every thought must come from neuronal firing necessarily. The difficulty lies in the fact that thoughts themselves cause neurons to fire and one is never thoughtless so we are reduced to the chicken and the egg problem if we continue down this path. However, assuming we come from a thought free state- for example deep coma-- the first thoughts we have are indeed generated by neuronal firing so your assertion is correct.
Answer: Original thought is not perception. Perception is response to the afferent, reacting to something. Original thought is efferent not reactionary but inventive. This is the conundrum, chicken or the egg. Dennett would support that the egg creates the chicken in order to create another egg.  The egg that creates a rooster is well on the way to independent inventive thought (free will).
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).


Do multi-state of entangled particles exists?

Question: I haven't taken any course in Quantum mechanics. But I felt "Quantum Entanglement" quite interesting. I recently read some articles on it. But I am not sure if I understood anything or satisfied with the theory. What I learnt recently is a particle is in superposition state unless observed. For example talking about spin of an electron, it spins in both counter clockwise and clockwise direction until observed. So when two electrons are entangled and we observe one of the particles then it suddenly be in one of the states by which we can predict about other entangled particle. What's confusing me is how can a particle be in multi state when unobserved. Could it not be that they are already in predetermined state? Could it not be that superposition state might not exist at all and it is just a conception made by people? I am thinking it in a more classical point of view. Could anyone explain me in layman term?
Answer: What you're describing is known as a Local Hidden Variables theory, and those are ruled out by violations of Bell inequalities as observed in a wide array of experiments. In short, the kind of model you're talking about imposes some bounds on the amount of certain types of correlation that can be observed between results of independent measurements of the two particles, and the experiments show that the observed correlations exceed those bounds.
For more details, see the multitude of previous questions about Local Hidden Variables and Bell inequalities on this site.


Do 200cc and mirror cup stars/trophies stack?

Question: I know that trophies and stars on 150cc also count for the lower grand prix cups as well (i.e., they stack). Does 200cc and mirror mode work the same way? If I get three stars on a mirror mode cup, does it count as three stars for the same 150cc cup? What about for 200cc down to 150cc or mirror?
Answer: According to Gamefaqs, Mirror mode is independent from each other category.
But 200cc should stack with the lower 50cc, 100cc and 150cc cups, as it's just an upgrade of the same tracks.
Anyway you can try it by yourself. (I don't own the game, so I can't verify this.)
Comment: Note: in Mario Kart 8 Deluxe, I got a trophy in 200cc that did not stack to the lower cc ratings.


Altium: create a component that expands to multiple footprints

Question: I am using a component that requires two headers to connect to it. I currently have a single footprint for this component that simply includes two copies of the connecting header. The problem I'm running into now is that our company is moving to streamline our process for creating files for our pick and place machine. Currently, I use the "Generate pick and place files" option from Altium to create the files, and then manually edit it to expand the single component into the two headers, but my bosses want to remove any manual steps from the process. Is there a way to get this single component to expand into two placements in Altium?
Comment: Just use two instances of the same component.  Bundling them inside a new component complicates practical matters like bom and assembly.  Placement consistency across boards can be achieved using a snippet.  If you want to see it as a single component in your schematic with special numbering, you could place them in a subsheet, but honestly I wouldn't bother.
Answer: I ended up keeping the system in place as I have it, and created a script to parse and correct the generated Pick and Place files. Here's my reasoning:
Consistent pin <-> pad mapping
There are 480 pins on the referenced part. Mapping those pins to the corresponding connector pads was a lot of work and messing up a single one of those might ruin an entire PCB run. Keeping it all in a single library part guarantees that the mapping is correct for whoever uses it.
apalopohapa mentioned that I could place the two connector parts in a subsheet with the correct mapping to expose the pins. Expanding this idea to support consistency company-wide, we could instead create a device sheet out of the part. This method would also keep the correct mapping, but introduces some annoyances that I would rather not deal with:

Every user would have to explicitly add the device sheet directory to their Altium preferences in order to use the component.
Designers would just have to know to look for the component as a device sheet instead of looking in the usual libraries.
My experience using Device sheets has been somewhat of a pain. For example, if a component in a device sheet is pulled from a specific library, Altium requires the designer to look for and add the library to the project before anything can be exported to the PCB.

Consistent component spacing
Getting the spacing right between the connectors is critical. Martin mentioned that I could use a spare mechanical layer to call out the distance between the parts. This would work fine if it only had to happen once. But, this component is already being used in two separate products, and will likely be used again. Keeping the part as a single footprint guarantees that we only have to get it right one time.
apalopohapa also mentioned that a snippet could be used to guarantee spacing. This would also guarantee that we only have to get it right one time, but again introduces a few problems:

For company-wide deployment, every designer would have to explicitly add the snippet directory to their Altium preferences in order to use it.
Using a pcb snippet also appears to add several extra steps:

Remove the component designator on the existing component(s) (eg 'U5' -> 'U?')
If the component has already been imported in the pcb, delete it
Place the snippet
Modify project links so that the snippet is linked to the proper component(s)
Pray that whoever made the snippet used component designators that won't conflict with something you already have.
Push the 'changes' from the pcb to the schematic to update the designators in the schematic.

Ability to logically divide part across the schematic
Each connector is 240 pins, so representing the component in the schematic as two connectors would take up an entire page of the schematics and would rely on external NetLabels to make connections to parts on other pages.
I've seen this done before (sometimes it is necessary), but this practice has always annoyed me. To figure out what is connected where, I have to continually flip back and forth between pages. It makes the schematic much less readable and maintainable.
With the device entered in Altium as a single component, I can use Altium's part feature to logically group the pins together. For example, all of the power and ground pins can be grouped together, and the sub part placed on the schematic sheet that contains all of my regulators, etc.
Answer: While trying to make a fuse holder clip (Keystone 3518P), I'm having a similar problem because one fuse clip is made of two of these Keystone parts (and then you need a fuse!). I want to define the spacing, silkscreen, and 3D body once as a footprint library part as you've done.

What I ended up doing was creating three parts:

Fuse holder

Has a schematic symbol (with 2 functioning pins)
Has a footprint
Does not have a BOM line item
Type: Standard (No BOM)

Fuse endcap x 2

Has a schematic symbol (no pins)
Does not have a footprint
Has a BOM line item
Type: Mechanical

Fuse (the actual fuse cartridge)

Has a schematic symbol (no pins)
Does not have a footprint (although you could make a silk-only FP for the rating!)
Has a BOM line item
Type: Mechanical

This way all the components are represented in the BOM output, and you get a decent representation in the schematic for what's going on.
The only dangling issue for you is the pick-and-place origins, but you might be able to kludge this by creating a footprint for your two connectors that is just a crosshair that you have to put in the right place after you place your module component. This is not quite as dire because the assembly house will be looking at and tweaking those anyway.
Answer: I've just come across this thread.
I work in Altium's R&D department and am the Product Manager for a feature we're calling a Composite Component, which is designed to deal with use cases such as these.
This is the current thread on the subject:
https://forum.live.altium.com/#posts/236081
I've been working on a series of mockups and posting them in the thread.  The most current version is in progress at the moment and hasn't yet been posted.
Anyone is also welcome to get in touch with me directly at email@example.com to discuss any aspects of the feature.  I'm always very happy to hear from customers, and find out more about their requirements.
Comment: Was this feature ever implemented?  I am evaluating CAD packages and would be interested in such a feature.
Answer: The easiest way to accomplish this is to define the components correctly in the libraries. Each connector is a single component, hence in your schematic there should be two seperate connector components. The rule is, one component, one footprint. One component cannot use multiple footprints simultaneously.
It would be best to define the connector in the library and associate the footprint of one of the connectors only, then place two connectors on the schematic and thus two seperate connectors on the PCB. This will produce pick and place files with both the connectors. The mechanical spacing of the headers may be aligned on a masked mechanical layer from which Gerber files are not produced. This will ensure that the connector placement remains the same in the transision from a single footprint to one of two seperate components.
Comment: One of the primary reasons I put both connectors into a single footprint was to ensure proper spacing. You mentioned that "mechanical spacing of the headers may be aligned on a masked mechanical layer..." This is new to me; how is it done?
Comment: Altium provides 32 mechanical layers onto which you can place design/assembly datum. Your original post indicates Altium as one of the tags so this is directly applicable to that package. Set up a mechanical layer and name it whatever you like, for instance "Connector origins". Draw a cross on the center reference or pin 1 reference of that footprint on the new mechanical layer. Then update the PCB with the 2 new seperate connectors and re-align their centers/references on the crosshairs on your mechanical layer.
Answer: Regarding Daniels comment on the fuse holder I have exactly the same situation. I didn't want to cause an issue with the pick and place files so I did it the following way.
Fuse holder x 2
Schematic symbol
Footprint
Type: Standard
Fuse itself
Schematic symbol
Footprint (For silkscreen and 3D body)
Type: Standard
N.B Listed in design variants setup as DNF so it is not listed in PicknPlace file
In both the schematic and PCB I group these three parts together as a union so they move around together.
Cheers :)


Tension at topmost point of a vertical circle

Question: I read that for a body to move in a vertical circle, the minimum tension needed is 0 at the topmost point. Can someone explain to me what it means to have positive or negative tension on top. I'm not able to visualise it.
Answer: This is not a universal principle; to be able to speak of tension, you must have an object that attaches the moving body to some other object. Take a yo-yo as the moving body, which is attached by a string to your finger. When you spin it to a looping, and you don't have enough tension in the string, it will fall down before it reaches the top, because the string loses its tension, in other words, it collapses. You can't have negative tension in a string, but you can have it in a spoke: Intuitively, it just means that if you replaced the spoke with a string, it would collapse. If you have exactly 0 tension exactly at the top, then it's just enough for the yo-yo to keep doing the looping.
Answer: First you need to think what is tension. Tension is a strain that is created in the rope (or string) when a force tries to elongate the rope. It means if no force to elongate, then no tension.
Now taking the example you asked about. If there is a tension in the string at topmost point  then it will revolve because that tension is the centripetal force which causes it to make the circular motion.
What if the tension is $0$ at the topmost point? No problem. The linear speed (which acts perpendicular to the centripetal force i.e. tension in this case) will keep it in motion.

What is positive tension? It's the tension in the string that comes into play when, say, you try to elongate the rope.
What is negative tension? It's when you try to compress the rope. But wait, you can't do it because if you do it then it will bend but won't compress. So the tension in the string will be zero. You may get it in some numerical problems because it tells how much force we have applied till the tension became zero from positive.
Comment: Thank you. This was what I was looking for. Say while solving a question, I happen to get tension negative at the topmost point does that mean the string became slack before reaching the top most point?
Comment: @user93868 exactly...


Why doesn't the accretion of ice, cause the ice to melt?

Question: I was reading about the accretion of ice which describes the rate at which the ice layer above a lake thickens. It said that in order for the water just below the ice to solidify,heat had to be conducted by the ice from the ice-water interface to the ice-
air interface. Now this amount of heat is equal to the latent heat of fusion(as the water is changing into ice). My question is why won't the ice itself melt while it conducts heat equal to the latent heat of fusion?
Answer: Heat flux isn't temperature. As modeled by Fourier's law, for example, 10 cm of ice at -10°C and -5°C at its top and bottom will conduct approximately 30 watts per square meter—30 joules per second over that area—while staying frozen all the while.
Comment: The latent heat to solidify a millimeter of water into ice does nothing to 20 centimeters of already formed ice below the freezing point.
Comment: But in this case, the heat we are supplying is equal to the heat needed to melt it. And isn't it true that if you supply the latent heat of fusion, the object will melt?


Login failed for user 'my-PC\my'

Question: Why I have this error:
  Cannot open database "source" requested by the login. The login failed.
Login failed for user 'my-PC\my'  ?.
I use "windows" mode. From visual I can access data, but when I run it (asp project) I get an error.I get this eror when replaced database in project with modified version from Management Studio (I modified only stored procedure that's not invoking in project). I read some articles and in security option allowed 'my-PC\my' all operations.No effect.
EDIT:
here is my connection string:
<code>    "Server=(local);Database=source;Integrated Security=SSPI;"
</code>
Answer: Use connectionstring:
<code>"Server=.\SqlExpress;Database=<fill in your database name>;Integrated Security=SSPI"
</code>
If not working, make an SQL server account and use this connectionstring:
<code>"Server=.\SqlExpress;Database=<fill in your database name>;User ID=<fill in your username>;Password=<fill in your password>;"
</code>

Login failed for user 'my-PC\my'


Magento gets bought by Adobe

Question: Various news articles are circulating regarding the Magento-recently being owned by Adobe . What would be the apparent outcomes/consequences for the community/developer/freelancers and software firms , whose sole bread and butter is magento module development, themes, customization etc.
Comment: I am also really exited to know what will be the up coming changes !!
benifits/drawback don't know but what will be still question for us :)
Answer: Magento will remain same as now.
It will remain as an Open source platform as per Matt Asay who is the Head of Developer Ecosystem for the Adobe.
From below sources, you will get more information regarding Magento future with Adobe.

Check out the tweet of @MattAsay which showers hope.
Check out @Abobe official blog regarding acquiring of Magento.
Article from Mark Lavelle - CEO Magento.
Comment: I heard somewhere magento2 is not going to provide currency rate API free.is it right?
Comment: I don't hear this type of any news
Answer: Perhaps one of the things that might confuse some users (especially uninformed users) is that Magento has been rebranded to Adobe Commerce, making a name discrepancy with this site's name.
Magento Is Now Adobe Commerce

We are excited to announce that Adobe's acquisition of Magento is complete and Magento Commerce is now an Adobe company – part of Adobe Commerce.

Although, Adobe also still provides references to the original name in many places: Magento, now Adobe Commerce.

Magento gets bought by Adobe


Align width of dropdown menu to?

Question: In the past I ran often into the problem, that I did not know how to decide the width of a dropdown menu. I as a designer don't mind to just eyeball a good-looking width, but developers would love to see some system behind it.
In the given example (google current) the width is 192px. The dropdown is actually wider than necessary. In other google products I have seen less wide dropdown menu. So it can't be a minimum-width.
How do you guys set up width of objects like this?
I have the same problem with modals. Sometimes I see them super tiny and sometimes with lots of paddings.
Thanks!
Comment: Is your software localized (translated into multiple languages)?
Comment: Hey, yes, our software is available in 2 languages.
Answer: It's aesthetically pleasing when the left and right gutters of a menu are the same width. If the left gutter is 26px, then that should be the width from the widest menu label to the edge of the right-hand side as well.

You can eyeball your gutter widths to start; try using them with various labels to make sure they work with menus that might use long terms (especially in other languages). If you want to be more precise than eyeballing, you can also use your design program's grid system for sizing based on how tall each label item is (a 32-pixel label might look good with a 24-pixel side gutter, as an example).
Answer: If you look at the specs for the Material Design menu widget, you can see that they do define a set of different minimum widths. When to apply which minimum width depends on the context and usage of a given menu.
As Stacy mentioned, similarly-sized gutters improve the aesthetic appearance of the menu and its items. They also make it easier to "visually digest" the menu contents.
A minimum width ensure that the click target is large enough to make Mr. Fitts happy ;) even if the menu items' labels are very short.
How you define these dimensions for your own custom widgets is up to you, of course. If you do use standardized components such as those from Material design, however, stick with what their creators have defined. That results in more-consistent UX, and it makes developers' lives easier by minimizing customization overhead.

Align width of dropdown menu to?


Use only the left option key as meta key in Terminal.app

Question: There's an option in Terminal.app -> Preferences -> Settings -> Keyboard but it allows me only to set both option keys as meta keys.
iTerm.app has such an option but don't like that application.
Comment: Would a solution that changed the system key mapping be acceptable? You could remap the right option key to do whatever else you choose, system wide.
Comment: That wouldn't help since I need the option key functionality.
Comment: If the programmers of Terminal app added this functionality - it's hidden with a preference you'd have to set manually in the plist. You're probably out of luck and will have to choose another program (have you looked at both iTerm and iTerm 2) - but it's a great question and logical need for people that want a meta key. Emacs user, by chance?
Answer: If you aren't totally set on using Terminal, I believe that iTerm2 has a setting to just use one of the option keys as meta. If you are, is there a setting to maybe use the Esc key as meta? I know that it's more convenient to use the option key, but the Esc key might be easier to set.
Comment: "Meta key" specifically refers to a modifier key, like Control, Shift, Option and Command. The Esc key is never "the Meta key". The Meta key can either cause an ESC to be written preceding the character you type or it can set the high bit of the character byte; however, modern terminals usually use UTF-8 or other encodings that require all eight bits, so usually the ESC prefix is used. Terminal only supports sending the ESC prefix. So, typing ESC before another character is simply sending the meta escape sequence to the tty and doesn't require configuration (at least, not in Terminal).
Comment: Yeah, I'm using the Esc key as meta key currently (I think it's set as meta key per default?) and hate it. It's not only hard to reach, but its function is also slightly different. For example: In irssi if you want to switch between channels you can keep alt pressed while choosing your channel via the number keys. Using Esc you'd have to press it every single time before pressing a number key.
Comment: Yeah, I only mention Esc because I've had better luck getting it to work with a few of the command line programs I use.
Answer: I have built a solution myself and it doesn't require abandoning Terminal.app in favour of iTerm 2.
It is a simple status bar app that runs in the background and rewrites all left Alt + $KEY key events to two key events in rapid succession, Esc, then $KEY; however, it only does this if Terminal.app is in focus.
You can find the source for the app here. You'll need Xcode to build it.


What is responsible for ensuring all symbols are known/defined?

Question: Is it the C preprocessor, compiler, or linkage editor?
Answer: To tell you the truth, it is programmer.
Answer: The answer you are looking for is... the compiler it depends. Sometimes it's the compiler, sometimes it's the linker, and sometimes it doesn't happen until the program is loaded.

The preprocessor:

handles directives for source file inclusion (#include), macro definitions (#define), and conditional inclusion (#if).
  ...
  The language of preprocessor directives is agnostic to the grammar of C, so the C preprocessor can also be used independently to process other kinds of text files.

The linker:

takes one or more objects generated by a compiler and combines them into a single executable program.
  ...
Computer programs typically comprise several parts or modules; all
  these parts/modules need not be contained within a single object file,
  and in such case refer to each other by means of symbols. Typically,
  an object file can contain three kinds of symbols:

defined symbols, which allow it to be called by other modules,
undefined symbols, which call the other modules where these symbols are defined, and
local symbols, used internally within the object file to facilitate relocation.

When a program comprises multiple object files, the linker combines
  these files into a unified executable program, resolving the
  symbols as it goes along.

In environments which allow dynamic linking, it is possible that

executable code still contains undefined symbols, plus a list of objects or libraries that will provide definitions for these.
Comment: Surely the linker too, in the case of externally defined symbols.
Answer:
For symbols with internal linkage or no linkage: the compiler.
For symbols with external linkage: the linker, either the "traditional" one, or the runtime linker.

Note that the dynamic/runtime linker may choose to do its job lazily, resolving symbols only when they are used (e.g: when a function is called for the first time).
Answer: The programmer must make sure everything is defined somewhere. The programmer is RESPONSIBLE for doing so.
Various tools will complain along the way if they notice anything missing:
The compiler will notice certain things missing, and will error out if it can realize that something's not there.
The linker will error out if it can't fix up a reference that's not in a library somewhere.
At run time there is a loader that pulls the relevant shared libraries into the process's memory space. The loader is the last thing that gets a crack at fixing up symbols before the program gets to run any code, and it will throw errors if it can't find a shared library/dll, or if the interface for the library that was used at link-time doesn't match up correctly with the available library.
None of these tools is RESPONSIBLE for making sure everything is defined. They are just the things that will notice if things are NOT defined, and will be the ones throwing the error message.


What needs to be focused on when hardening a LAMP stack?

Question: I have a site running on a VPS.  I set up the LAMP stack and a few other programs personally.  This is only the second time I have set up a VPS and I am a bit worried about security.  Are there things that I should be aware of (regarding my Linux install and LAMP stack) which people could use to gain access to my VPS and steal/alter my site or shut it down?
Answer: Any external service is a security risk. You indicated you're running a LAMP stack, so I'm assuming you have at least the following services running:
<code> 1. SSHd
 2. Apache/Nginx
 3. PHP
 4. MySQL
</code>
MySQL by default is not externally facing, and you should not make it so unless you're involved in a clustering or replication environment (in which case there are alternatives to putting it on the wide web, like a VPN.)
Apache/Nginx + PHP cannot be made internally facing if you intend on providing content to people on the web, so these are going to be your main points of ingress.
SSHd itself has to be public facing (not always, but for the sake of simplicity in this case). Use of a Key pair for authentication and completely disabling password authentication is your best best for preventing people from obtaining access via this service.
Let's talk about webservices. Apache, Nginx, and PHP. These are the services that people will attempt to exploit in order to obtain access to your machine, so let's talk about securing them.
Firstly, you're going to want to isolate the services. If you've installed them yourself that means that ensuring that Apache and Nginx are running as non-root users.
Secondly, you're going to want to take a look at php-fpm or suphp, which allow PHP to operate as a user that is separate from your webserver even at a virtualhost or directory level.
This accomplishes two things, by taking root permissions away from Apache/Nginx, if someone gains control of that service they cannot impact anything but the operation of that specific service.
Changing the user of php (especially on a per-site basis) isolates the PHP module from interacting with other sites on the machine, and with the webserver itself, so if you have a client or website with exploitable code on it, your entire service, other sites, and your machine is not at risk.
Comment: I want to add. Make sure your stack is only a LAMP stack. If you install a bundle of software that is `LAMP`, it could have things you might not want in it. PHPMYADMIN, FTP, WEBDAV. XAMPP has these (even though XAMPP is supposed to be used as a development platform).
Comment: @Rell3oT trust no stock images! question everything! :)
Answer: Your application code.
Any server-side application code - be it PHP/Perl/Python (whatever the P in LAMP means to you) is your greatest risk. Whether you have written it yourself, or it's an open-source (or commercial) package, any coding flaws could lead to security vulnerabilities like SQL injection.
If you have written code yourself, you should follow the OWASP Top 10 at a minimum. You should also check your code, perhaps by scanning with Zap or Rips (which are free), or other tools, or hiring a pen testing company.
If you're using open-source or commercial software, you should check their history of vulnerabilities. If there are none at all that is a worry, it probably means the code has never been audited. If there is a small number, that's a good sign. If there are many you may want to avoid the project. Plugins can cause risks as well as core applications (a big problem with CMSs like Drupal).
Other people have mentioned infrastructure tweaks, but in general the defaults are pretty good these days. I would restrict SSH to key based login (instead of passwords) and ensure the application connects to the database using a low-privileged user.
Answer: What you are trying to achieve is known as hardening, i.e. making systems and services more secure. Obviously there is a lot to talk and think about. You probably want also to take a much broader perspective, because you probably want to harden your complete server, not only the few services you mentioned above.
Personally I think the NSA did a great job here. Take a look at their recommendations for various operating systems. Although they specifically speak about RHEL5 in case of Linux, most of the tips can be applied to any Linux distribtuion.
Comment: Both the links here point to wikipedia - did you mean to include a link to the NSA stuff?
Answer: Although it's a bit dated, the SANS check list is a good starting point (I'm about to go and look for the NSA list Karol mentioned).
But the big wins come from the simple things:

pick good passwords,
don't run stuff you don't need,
make sure you can identify when patches are available for the software you are using and install them promptly.


Changing the font style of line numbers in Microsoft Word 2007

Question: I added line numbers to my document in Microsoft Word 2007. How can I change the font style of these line numbers?
Answer: I just figured out how to do this.
Essentially, line numbers are assigned a style, just like almost any other special formating in Word.

Process

Go to the Home tab of the Ribbon after enabling line numbering.
Under Styles, click the Dialog Box Launcher arrow to open the Styles pane.
In the Styles pane, click Options.
Under "Select styles to show" in the Options window, select "All styles". Under "Select how list is sorted", select "Alphabetical"; this option makes it easier to locate the style in question.
Click OK to exit out of the Options window.
Back in the Styles pane, scroll down until you find the Style that is called "Line Number". Click the arrow next to it, hit Modify, and change formatting as necessary in the window that opens. Click OK once you're finished and then close the Styles pane.
Answer: I haven't found an option to modify the font of only the line numbers, but I've found they assume whatever font you set in the "Normal" style. So if you change the Normal style's font the line numbers will change, but it may upset other styles used in the document.
Comment: Thanks for your help, but I just found a way to get around that.


Broken Saddle Tee In Irrigation System

Question: I notice a broken line in my yard and when I dug the line up I have a white tee connector. Looking online it looks like a Poly Pipe Kwikseal Saddle Tee. The nut is missing, which I assume has caused the leak. I am wondering what the best fix would be. Should I replace it with the same tee or is there another way? Thanks!
Answer: I can see no one answered this and though I'm not a pro but after experiencing the same exact issue this week and doing internet research and asking staff at 3 different irrigation specialty supply stores, the common answer was: No, a new 100C Dawn Kwikseal saddle tee cannot be used to replace it because it won't have the same perfect seal as the original saddle tee install because of how the seal is done when the original hole in the poly pipe is made while the original saddle tee is clamped on.  They stated that eventually it will leak and promote more root growth from the nearby tree roots. Plus, those saddle tee's will have the same failure flaw in a few years if roots grow around it breaking the nylon bolt. They all recommend cutting the pipe at or around the perforation point and using an inserted T fitting with hose clamps on either end to provide a better water tight seal.  They also said you could try using the newer Kwik Tap poly saddle that has a pointed insert tube that goes into the existing hole but it may eventually leak as well.  bottom line is it's more work but the better longer lasting option is to cut the poly pipe and use a T coupler insert with clamps.

Broken Saddle Tee In Irrigation System


Set permissions based on metadata

Question: My client has a Document Center.  The whole document center make use of only one content type.  One of the metadata on the content type is Classification level (Classified, Confidential, Top Secret).  Everyone in the company can read all documents in the centre, except for those marked with metadata = 'Top Secret' documents.  How can restrict permission on those type documents?
Answer: I'm sorry but it doesn't work this way and there are no reliable workarounds. You might be able to use event handlers or workflows to manage security when documents are created or updated but those things can fail, and item level security is a maintenance and performance headache.
When dealing with security your best bet is to use a separately secured site, or at least a secured folder.
Comment: Thank you Derek.  In my mind I had that feeling that I'd might need folders, but just wanted to double check. :-)

Set permissions based on metadata


Linking SQL Azure and SQL Server

Question: In SQL Server you can link another database such as Oracle or Sybase and write queries that span both. Can you do the same thing with SQL Azure and SQL Server?
Answer: Well - this is a good question. Please see this link where I answered this question on the SQL Azure MSDN forum: http://social.msdn.microsoft.com/Forums/en/ssdsgetstarted/thread/5c8cbbb6-9027-4bf4-9239-e6c4cfc3426a
The short answer is that there is a way, but it is not supported. Here is the link that discusses how to make it work: http://sqlcat.com/sqlcat/b/msdnmirror/archive/2011/03/07/linked-servers-to-sql-azure.aspx
Here is the official support team statement regarding the workarount indicating that it is not supported: http://blogs.msdn.com/b/psssql/archive/2011/05/03/not-linkedservertosqlazure-where-provider-msdasql-and-dsn-sql-server-native-client.aspx

Linking SQL Azure and SQL Server


How to disable plugins on Foxit Reader?

Question: I'm using Foxit Reader to read PDFs on my Windows 8, and sometimes it crashes. According to the crash message, it crashed because of a specific plugin installed on it. When I go to Help -> About Foxit Plug-ins I can see the plugin's name, but there is no option to disable it there. I'd like to disable this plugin. How can I do it?
Answer: I had this problem too. Actually, the only solution that I found was to delete/move  *.fpi files (which are the files related to the plug-ins). Then close and re-open Foxit.
They can be found at --> "C:\Program Files (86)\Foxit Software\Foxit Reader\plugins"
Regards.

How to disable plugins on Foxit Reader?


Will Asarum europaeum add a ginger flavor?

Question: I was looking through a cooking catalog, and came across GINGER, EUROPEAN which made me wonder if this is like normal ginger that you cook with.
Comment: https://en.m.wikipedia.org/wiki/Asarum_europaeum implies it can be used as a spice but one of its references says it has been used in herbal medicine to induce vomiting. It's perfectly possible that both are true and the quantity makes the difference, but it suggests that some caution is needed.
Comment: Chris, if it's like American Wild Ginger, it's a matter of whether you eat the root pulp or not.
Answer: Short answer: kinda-sorta, in a weak way.
Long answer: European wild ginger is grown as a ground cover, not for culinary purposes.  It is closely related to American Wild Ginger, though, which can be used to cook with, but it has a weaker flavor than Chinese ginger, less pulp, and you shouldn't actually eat the roots.  I would presume European ginger to be a very similar experience, but only have experience with the American variety.


If a super massive black hole captures a very small black hole, would the smaller black hole be immune to spaghettification?

Question: I think to answer this question we would have to fully understand the nature of black holes. It seems to me that the smaller black hole could have a lower density than the larger one. In this case the smaller black hole might be effected by the tidal forces of the larger black hole. But if the densities of the two black holes were the same, wouldn't the smaller black hole be immune to spaghettification?
Comment: If we trust GR, then both black holes are already ring-singularities. What does it mean to spaghettify spaghetti?
Answer: The black hole event horizon is not a thing i.e. not a physical object. It is just a surface in spacetime from which light can never escape to infinity.
Also, if we take the Schwarzschild description of a (non-rotating) black hole then it is a point mass hidden away behind the event horizon. You can't spaghettify a point mass.
When two black holes merge, the surface in spacetime from which light can never escape evolves with time. It starts out as two separate spheres, evolves through a complex geometry as the black holes merge then ends up as a single (bigger) sphere. There is nothing in this process that can reasonably be described as spaghettification.
Comment: There is a serious question in here, but I don't think the OP is aware of it. How do two ring singularities merge? I am kind of curious myself now.
Comment: @CuriousOne How do they merge? Very quickly. ;-P


Do you need to explicitly include the primary key in a composite index?

Question: I recently read that when creating a composite index in MariaDB that the primary key is appended to the end of the index "under the covers". I access my data many times in order by these four columns: a, b, c, and id (id being the primary key). Can I omit id from the CREATE INDEX statement and still enjoy retrieving data via the index in order by all 4 columns? Or should I still explicitly include id in the CREATE INDEX definition?
Comment: Provide a sample - CREATE TABLE, including all indices, and a query. *Can I omit id from the CREATE INDEX statement and still enjoy retrieving data via the index in order by all 4 columns?* In general - no.
Comment: 3 columns in the index doesn't imply the remaining id key is in order in that index. Note its secondary composite indexes that have a PK reference appended as the PK is a clustered index, and secondary ones aren't. Note a primary key can be composite too. Creating a [sample fiddle](https://dbfiddle.uk/?rdbms=mariadb_10.6) with your query, and `explain {query}` will show index usage.
Comment: @danblack - I've never seen that happen, but that does not prove anything.  Have you looked at the code to see that it might actually scramble them?  I would expect creating a "temp" table of a,b,c,id, then sorting, then constructing the index BTree would be faster than sorting a,b,c, then tacking on id somehow, and finally constructing the BTree.
Answer: Yes you can definitely skip adding id to the CREATE INDEX
Let's say you already created that index with (a,b,c,id)
Here is how you can show that you can do with the id column:
Run pt-duplicate-key-checker against your MySQL Instance.
Let's say you ran this
CREATE DATABASE IF EXISTS mydb;
USE mydb
CREATE TABLE mytable
(
    id INT NOT NULL AUTO_INCREMENT,
    a CHAR(10),
    b VARCHAR(20),
    c TINYINT
    PRIMARY KEY (id)
    KEY abc_id_ndx (a,b,c,id)
);
INSERT INTO mytable (a,b,c) VALUES ('CHAR1','VARCHAR1',1);
INSERT INTO mytable (a,b,c) VALUES ('CHAR2','VARCHAR2',2);
INSERT INTO mytable (a,b,c) VALUES ('CHAR3','VARCHAR3',3);

When you run pt-duplicate-key-checker, it will report the SQL to change that index. In fact, the recommended SQL will be something like:
ALTER TABLE mytable DROP INDEX abc_id_ndx, ADD INDEX abc_id_ndx (a,b,c);

Comment: But, who's to say that pd-dkc's suggestion is not unnecessary?
Answer: MariaDB might do this but thinking more generically ...
Any secondary index has to have a "pointer" to the data record that it refers to.  If MariaDB uses the PK for that purpose, fine.  Other DBMSs do it differently (yes; I'm thinking of Oracle, here).
If you want to look something up by its [unique] Primary Key, then you don't need anything else to find it.  Just the Primary Key value will do and Primary Keys are always indexed.
If you want to look something up by any other value(s), then they need to be indexed.  If you regularly use multiple columns to do those lookups, then create a composite index on those columns (it's common for a database engine to "pick" only one index for each query).
So no; you don't need to include the Primary Key in any composite, secondary indexes.
Answer: I like to include the PK when I depend on id being part of the INDEX; eg,  to make it "covering".  This is to clue the human reader in that I need it.  (Suppose they convert to Oracle -- The efficiency of being "covering" would get lost.)
Bottom Line -- Add id if you need it.  Else, don't worry.


How to prove that $(a^m)^n=a^{mn}$ where $a,m,n$ are real numbers and a>0?

Question: I know how to prove the equality when $m$ is a rational number and $n$ is an integer, but do not know how to go about proving this for real numbers. On a semi-related note, I was trying to prove this when both $m$ and $n$ are rational, and found out that I have to prove that $(\frac{1}{z})^{\frac{1}{y}}$=$\frac{1}{z^{\frac{1}{y}}}$. Does this need to be proven or can I accept it as a definition?
Comment: Which definition of $a^b$ do you use for $a,b$ real?
Comment: Please, please, stick to the case $a>0$.
Comment: Does this answer your question? [$a^ma^n=a^{m+n}$ and $(a^m)^n=a^{mn}$ proof](https://math.stackexchange.com/questions/2495174/aman-amn-and-amn-amn-proof)
Comment: How about using Continuity?
Comment: @k.stm I have managed to find an alternate proof here https://math.stackexchange.com/questions/3727801/proving-exponent-law-for-real-numbers-using-the-supremum-definition-only I don't fully understand it though. Also, I have found proofs like yours before but I it is my fault for dismissing them. I didn't know that $a^b=exp(b•log(a))$ is a definition. I thought it used the fact that $(a^m)^n=a^{mn}$. I am sorry.
Comment: These is a fundamental property of powers. Example: $(2^3)^2=(2^3).(2^3)=2^{(3+3)}=2^{2.3}$. Regarding rational numbers, remember that $a^{1/2}= \sqrt{a}$. Since this looks like a HW question, we may not give you full answers on the site.
Comment: $$(a^m)^n = \exp (n \log a^m) = \exp (n \log (\exp (m \log a))) = \exp (nm \log a) = a^{nm}.$$
And by commutativity of the monoid $(ℝ,·,1)$, you have $a^{nm} = a^{mn}$.
Comment: @Basco It isn't a homework question. I just want to know the proof to have peace of mind and surprisingly, I can't find the proof for this anywhere on the internet.
Comment: @OrlinAurum You haven't searched properly. It also depends on your definition of $a^b$, as Hagen indirectly pointed out. You can find one proof three comments above this one.
Comment: This is also a decategorification of some more general principle. Not that it's what you're looking for, but it's related, see [Riehl]
Answer: The very first thing you need to do is ask yourself what the definitions are. Without proper definitions, you'll never have a complete proof. So, if $a>0$ and $m\in \Bbb{R}$, how are you even supposed to define $a^m$? This is not at all a trivial task.
For example, here's one possible approach to things:

First, define $\exp: \Bbb{R} \to \Bbb{R}$ by $\exp(x):= \sum_{n=0}^{\infty}\frac{x^n}{n!}$. You of course have to check that this series converges for every $x\in \Bbb{R}$.
Check basic properties of $\exp$, such as $\exp(0) = 1$ and for all $x,y \in \Bbb{R},$ $\exp(x+y) = \exp(x)\cdot \exp(y)$. Also, verify that $\exp:\Bbb{R} \to (0,\infty)$ is an invertible function.
Since $\exp:\Bbb{R} \to (0,\infty)$ is invertible, we can consider its inverse function, which we denote as $\log:(0,\infty) \to \Bbb{R}$. Then, verify all the basic properties of $\log$, such as for all $a,b>0$, $\log(ab) = \log(a) + \log(b)$.
Finally, given $a>0$ and $m\in \Bbb{R}$, we define $a^m := \exp(m \log(a))$.

From this point, it is a simple matter to use the various properties of the exponential and logarithmic functions: for any $a>0$ and $m,n \in \Bbb{R}$
\begin{align}
a^{m+n} &:= \exp((m+n)\log (a)) \\
&= \exp[m\log(a) + n \log (a)]\\
&= \exp[m \log(a)] \cdot \exp[n\log(a)] \\
&:= a^m \cdot a^n \tag{$*$}
\end{align}
Similarly,
\begin{align}
(a^m)^n &:= \exp[n \log(a^m)] \\
&:= \exp[n \log(\exp(m \log(a)))] \\
&= \exp[nm \log(a)] \tag{since $\log \circ \exp = \text{id}_{\Bbb{R}}$} \\
&:= a^{nm} \\
&= a^{mn}
\end{align}
where in the last line, we make use of commutativity of multiplication of real numbers.

Note that steps 1,2,3 are not at all trivial, and indeed there are entire chapters of calculus/analysis textbooks devoted to proving these facts carefully. So, while I only listed out various statements, if you want the proofs for the statements I made, you should take a look at any analysis textbook, for example, Rudin's Principles of Mathematical Analysis, or Spivak's Calculus (I recall Spivak motivating these things pretty nicely).
As for your other question, yes it is something which needs to be proven. This result can be easily deduced from two other facts.

For any $x\in \Bbb{R}$, $1^x = 1$. (proof: $1^x := \exp[x \log(1)] = \exp[0] = 1$)
For any $a,b > 0$ and $x\in \Bbb{R}$, $(ab)^x = a^x b^x$. The proof is a few lines, once you use the properties of $\exp$ and $\log$.

Now, if $z>0$, then for any $x\in \Bbb{R}$,
\begin{align}
z^x \cdot \left(\frac{1}{z}\right)^x &= \left(z\cdot \frac{1}{z}\right)^x = 1^x = 1
\end{align}
Hence, $\left(\frac{1}{z}\right)^x = \frac{1}{z^x}$. In particular, you can take $x=1/y$ to prove what you wanted.

Edit: Motivating the definition $a^x := \exp(x\log(a))$, for $a>0, x \in \Bbb{R}$.
The long story short: this definition is unique in certain sense, and is almost forced upon us once we impose a few regularity conditions.
Now, let me once again stress that you should be careful to distinguish between definitions, theorems and motivation. Different authors have different starting points, so Author 1 may have one set definitions and motivations, and hence different theorems, while author 2 may have a completely different set of definitions, and hence have different theorems, and motivation.
So, let's start with some motivating remarks. Fix a number $a>0$. Then, we usually start start by defining $a^1 = a$. Next, given a positive integer $m\in \Bbb{N}$, we define $a^m = \underbrace{a\cdots a}_{\text{$m$ times}}$ (If you want to be super formal, then ok, this is actually a recursive definition: $a^1:= 1$, and then for any integer $m\geq 2$, we recursively define $a^{m}:= a\cdot a^{m-1}$).
Now, at this point what we observe from the definition is that for any positive integers $m,n\in \Bbb{N}$, we have $a^{m+n} = a^m \cdot a^n$. The proof of this fact follows very easily by induction.
Next, we typically define $a^0 = 1$. Why do we do this? One answer is that it is a definition, so we can do whatever we want. Another answer, is that we are almost forced to do so. Why? notice that for any $m\in \Bbb{N}$, we have $a^m = a^{m+0}$, so if we want this to be equal to $a^m \cdot a^0$, then we had better define $a^0 = 1$.
Next, if $m>0$ is an integer, then we usually define $a^{-m} := \dfrac{1}{a^{m}}$. Once again, this is just a definition, so we can do whatever we want. The motivation for making this definition is that we have $1 =: a^0 = a^{-m+m}$ for any positive integer $m$. So, if we want the RHS to equal $a^{-m}\cdot a^m$, then we had better define $a^{-m}:= \frac{1}{a^m}$.
Similarly, if $m>0$, then we define $a^{1/m} = \sqrt[m]{a}$ (assuming you've somehow proven existence of $m^{th}$ roots of positive real numbers). Again, this is just a definition. But why do we do this? Because we have $a =: a^1 = a^{\frac{1}{m} + \dots +\frac{1}{m}}$, so if we want the RHS to equal $(a^{\frac{1}{m}})^m$, then of course, we had better define $a^{1/m}:= \sqrt[m]{a}$.
Finally, we define $a^{\frac{m}{n}}$, for $m,n \in \Bbb{Z}$ and $n >0$ as $a^{m/n} = (a^{1/n})^m$. Once again, this is just a definition, so we can do whatever we want, but the reason we do this is to ensure the equality $a^{m/n} = a^{1/n + \dots + 1/n} = (a^{1/n})^m$ is true.
Now, let's think slightly for what we have done. We started with a number $a>0$, and we defined $a^1 := a$, and we managed to define $a^x$ for every rational number $x$, simply by the requirement that the equation $a^{x+y} = a^x a^y$ hold true for all rational $x,y$. So, if you actually read through everything once again, what we have actually done is shown the following theorem:

Given $a>0$, there exists a unique function $F_a:\Bbb{Q} \to \Bbb{R}$ such that $F_a(1) = a$, and such that for all $x,y\in \Bbb{Q}$, $F_a(x+y) = F_a(x)\cdot F_a(y)$.
(Note that rather than writing $a^x$, I'm just writing $F_a(x)$, just to mimic the function notation more)

Our motivation has actually been to preserve the functional equation $F_a(x+y) = F_a(x)\cdot F_a(y)$ as much as possible. Now, we can ask whether we can extend the domain from $\Bbb{Q}$ to $\Bbb{R}$, while preserving the functional equation, and if such an extension is unique. If the answer is yes, then we just define $a^x := F_a(x)$ for all real numbers $x$, and then we are happy.  It turns out that if we impose a continuity requirement, then the answer is yes; i.e the following theorem is true:

Given $a>0$, there exists a unique continuous function $F_a:\Bbb{R} \to \Bbb{R}$ such that $F_a(1) = a$, and such that for all $x,y\in \Bbb{R}$, $F_a(x+y) = F_a(x)\cdot F_a(y)$.

Uniqueness is pretty easy (because $\Bbb{Q}$ is dense in $\Bbb{R}$ and $F_a$ is continuous). The tough part is showing the existence of such an extension.
Of course, if you already know about the $\exp$ function and its basic properties like 1,2,3, then you'll see that the function $F_a:\Bbb{R} \to \Bbb{R}$ defined by $F_a(x):= \exp(x \ln(a))$ has all the nice properties (i.e is continuous, it satisfies that functional equation, and $F_a(1) = a$). Because of this existence and uniqueness result, this is the only reasonable way to define $a^x \equiv F_a(x) := \exp(x \log(a))$; anything other than this would be pretty absurd.
The purpose of the rest of my answer is to try to motivate how anyone could even come up with the function $F_a(x) = \exp(x\ln(a))$; sure the existence and uniqueness result is very nice and powerful, but how could you try to come up with it by yourself? This certainly doesn't come from thin air (though at some points we have to take certain leaps of faith, and then check that everything works out nicely).

To do this, let's start with a slightly more restrictive requirement. Let's try to find a function $f:\Bbb{R} \to \Bbb{R}$ with the following properties:

for all $x,y\in\Bbb{R}$, $f(x+y) = f(x)\cdot f(y)$
$f$ is non-zero; i.e there exists $x_0\in \Bbb{R}$ such that $f(x_0) \neq 0$.
$f$ is differentiable at $0$.

The first two conditions seem reasonable, but the third one may seem a little strange, but let's just impose it for now (it's mainly there to try to motivate things and hopefully simlify the argument and to convince you that $x\mapsto \exp(x\ln(a))$ didn't come from thin air).
First, we shall deduce some elementary consequences of properties 1,2,3:

In (2), we assumed $f$ is non-zero at a single point. We'll now show that $f$ is no-where vanishing, and that $f(0)=1$. Proof: we have for any $x\in\Bbb{R}$, $f(x) \cdot f(x_0-x) = f(x_0) \neq 0$. Hence, $f(x) \neq 0$. In particular, $f(0) = f(0+0) = f(0)^2$. Since $f(0)\neq 0$, we can divide it on both sides to deduce $f(0) = 1$.

We also have for every $x \in \Bbb{R}$, $f(x)>0$. Proof:  We have
\begin{align}
f(x) = f(x/2 + x/2) = f(x/2)\cdot f(x/2) = f(x/2)^2 > 0,
\end{align}
where the last step is because $f(x/2) \neq 0$ (this is why in real analysis, we always impose the condition $a = f(1) > 0$).

$f$ is actually differentiable on $\Bbb{R}$ (not just at the origin). This is because for $t\neq 0$, we have
\begin{align}
\dfrac{f(x+t) - f(x)}{t} &= \dfrac{f(x)\cdot f(t) - f(x) \cdot f(0)}{t} = f(x) \cdot \dfrac{f(0+t) - f(0)}{t}
\end{align}
now, the limit as $t\to 0$ exists by hypothesis since $f'(0)$ exists. This shows that $f'(x)$ exists and $f'(x) = f'(0) \cdot f(x)$. As a result of this, it immediately follows that $f$ is infinitely differentiable.

Now, we consider two cases. Case ($1$) is that $f'(0) = 0$. Then, we have $f'(x) = 0$ for all $x$, and hence $f$ is a constant function, $f(x) = f(0) = 1$ for all $x$. This is clearly not very interesting. We want a non-constant function with all these properties. So, let's assume in addition that $f'(0) \neq 0$. With this, we have that $f'(x) = f'(0)\cdot f(x)$; this is a product of a non-zero number and a strictly positive number. So, this means the derivative $f'$ always has the same sign. So, $f$ is either strictly increasing or strictly decreasing. Next, notice that $f''(x) = [f'(0)]^2 f(x)$, is always strictly positive; this coupled with $f(x+y) = f(x)f(y)$ implies that $f$ is injective and has image equal to $(0,\infty)$. i.e $f:\Bbb{R} \to (0,\infty)$ is bijective.

Theorem 1.

Let $f:\Bbb{R} \to \Bbb{R}$ be a function such that:

for all $x,y\in \Bbb{R}$, $f(x+y) = f(x)f(y)$
$f$ is non-zero
$f$ is differentiable at the origin, with $f'(0) \neq 0$

Suppose $g:\Bbb{R} \to \Bbb{R}$ is a function which also satisfies all these properties. Then, there exists a number $c\in \Bbb{R}$ such that for all $x\in \Bbb{R}$, $g(x) = f(cx)$. In other words, such functions are uniquely determined by a constant $c$.
Conversely, for any non-zero $c\in \Bbb{R}$, the function $x\mapsto f(cx)$ satisfies the three properties above.

Proof
To prove this, we use a standard trick: notice that
\begin{align}
\dfrac{d}{dx}\dfrac{g(x)}{f(cx)} &= \dfrac{f(cx) g'(x) - g(x) cf'(cx)}{[f(cx)]^2} \\
&= \dfrac{f(cx) g'(0) g(x) - g(x) c f'(0) f(cx)}{[f(cx)]^2} \\
&= \dfrac{g'(0) - c f'(0)}{f(cx)} \cdot g(x)
\end{align}
Therefore, if we choose $c = \dfrac{f'(0)}{g'(0)}$, then the derivative of the function on the LHS is always zero. Therefore, it must be a constant. To evaluate the constant, plug in $x=0$, and you'll see the constant is $1$. Thus, $g(x) = f(cx)$, where $c= \frac{g'(0)}{f'(0)}$. This completes the proof of the forward direction. The converse is almost obvious
Remark
Notice also that from $g(x) = f(cx)$, by plugging in $x=1$, we get $g(1) = f(c)$, and hence $c = (f^{-1} \circ g)(1) = \frac{g'(0)}{f'(0)}$ (recall that we already stated that such functions are invertible from $\Bbb{R} \to (0,\infty)$). It is this relation $c = (f^{-1} \circ g)(1)$, which is the key to understanding where $x\mapsto \exp(x\ln(a))$ comes from. We're almost there.

Now, once again, just recall that we have been assuming the existence of a function $f$ with all these properties. We haven't proven the existence yet. Now, how do we go about trying to find such a function $f$? Well, recall that we have the fundamental differential equation $f'(x) = f'(0) f(x)$. From this, it follows that for every positive integer $n$, $f^{(n)}(0) = [f'(0)]^n$. We may WLOG suppose that $f'(0) = 1$ (other wise consider the function $x\mapsto f\left(\frac{x}{f'(0)}\right)$), then we get $f^{(n)}(0) = 1$. Finally, if we make the leap of faith that our function $f$ (which is initially assumed is only differentiable at $0$ with $f'(0) = 1$, and then proved it is $C^{\infty}$ on $\Bbb{R}$) is actually analytic on $\Bbb{R}$, then we know that the function $f$ must equal its Taylor series:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}
\end{align}
This is one of the many ways of how one might guess the form of the exponential function, $\exp$. So, we now take this as a definition: $\exp(x):= \sum_{n=0}^{\infty}\frac{x^n}{n!}$. Of course using basic power series techniques, we can show that $\exp$ is differentiable everywhere, and satisfies that functional equation with $\exp(0)=\exp'(0) = 1$.
So, now, back to our original problem. Given any $a>0$, we initially wanted to find a function $F_a:\Bbb{R} \to \Bbb{R}$ such that $F_a$ satisfies the functional equation, and $F_a(1) = a$, and such that $F_a$ is differentiable at $0$ with $F_a'(0) \neq 0$. Well, in this case, both $F_a$ and $\exp$ satisfy the hypothesis of theorem 1. Thus, there exists a constant $c \in \Bbb{R}$ such that for all $x\in \Bbb{R}$, $F_a(x) = \exp(cx)$. To evaluate the constant $c$, we just plug in $x=1$, to get $c = (\exp^{-1}\circ F_a)(1) := \log(a)$. Therefore we get $F_a(x) = \exp(x \log(a))$. This is why we come up with the definition $a^x := \exp(x\log(a))$.
Comment: Why/how do we define $a^m := \exp(m \log(a))$ ? I was under the impression that this somehow needed to use the fact that $(a^m)^n.
Comment: **Theorem:** For each $a > 0$ there exists a unique continuous function $E_a : \mathbb R \to \mathbb R$ such that the following hold: (1) $E_a(1)=a$; (2) for all $x,y \in \mathbb R$, $E_a(x+y) = E_a(x) \cdot E_a(y)$.
Comment: Existence is easy: $E_a(x) = e^{\ln(a) \cdot x}$. Uniqueness isn't hard either
Comment: @A-LevelStudent: This is quite an old post, but I suggest you look at the following [thread](https://math.stackexchange.com/questions/55068/can-you-raise-a-number-to-an-irrational-exponent/55078#55078). The truth is that it is actually quite difficult to define $a^x$ means when $x$ could be any real number. Clearly, $2^{1/3}$ does not refer to repeated multiplication, so for rational numbers we have to define $a^{p/q}$ as $(\sqrt[q]{a})^p$ so that the definition makes sense. That still leaves open the question: what does $2^{\sqrt{2}}$ mean? The answer is not as simple as you think.
Comment: @A-LevelStudent: That's okay. It's not helped by the fact that often when we are first introduced to the exponential function, it is not done in a very precise or rigorous way.
Comment: This is virtually what I wrote anout the other answer: The following proof that $\ln{a^b} =b \ln a$ is my area of difficulty with this response: $$\log_a {M^n} = x \implies a^x = M^n$$

$$\log_a M = y \implies a^y = M \implies a^x=M^n=(a^y)^n=a^{ny}$$                      $$x=ny=n \log_a {M}$$
 This proof is where you're comment and the answer come s from, but this proof uses the fact that $(a^y)^n=a^{ny}$!
Comment: @A-levelStudent nowhere in my answer did I ever define the logarithm to base $a$. By the way, what I defined in my answer as $\log$, you may instead want to call it $\ln$. Anyway, putting these small details aside, the proof for $\ln(a^b) = b \ln(a)$ using the definitions and theorems stated in steps 1,2,3,4 goes like so: $\ln[a^b] := \ln[\exp(b \ln(a))] = b \ln(a)$, because $\ln\circ \exp = \text{id}_{\Bbb{R}}$. No where in this proof have I even assumed that $(a^y)^n = a^{ny}$. All I did is use definitions which I have given.
Comment: @A-levelStudent if this is confusing, it's probably because (sounds ironic) you know too many facts, but you've probably never proven them all systematically. I know I had the same trouble when first learning these things. I had to pretend I new nothing about exponentials and logarithms and then start from scratch with definitions and prove every single thing.
Comment: @OrlinAurum There is nothing circular about this definition. I suggest you to temporarily forget everything you know about exponentials and logarithms and all their rules. Completely forget about the rules. Instead focus on how everything is defined. You need to convince yourself that everything has a precise definition (which is non-circular), for this one possible approach is to do things like I have written in steps 1,2,3,4.
Comment: @peek-a-boo I wasn't saying that it is circular (rather, I used to think proofs like this were circular, before reading your comment). I just wanted to know why $a^m$ is defined to be the same as $exp(mlog(a))$ Is it because it makes our lives easier?
Comment: For your first question about why/how we define $a^m := \exp(m\log(a))$, there's a simple answer to "why" : and that is "because I can, and because, why not?" Because if you follow steps 1,2,3 in that order then, everything is properly defined and nothing is circular in logic, so of course, I can do whatever I want with my definitions, so I can certainly define $a^m := \exp[m\log(a)]$. But, this is a very simple definition. A slightly better answer to this question is that it is "almost" the unique choice, if I impose a few other requirements (I'll try to edit and elaborate when I have time).
Comment: I'm sorry to keep on saying this, but I still believe you're using the questions as a proof for your answer. You say that $a^m=\exp(m\log(a))$. This is obviously derived using the log laws, ie that $\exp(\log(a^m))=\exp(m\log(a))$ (correct me if I'm wrong). However, the log law that allows you to write $a^m=\exp(m\log(a))$ is itself derived using the fact $(a^m)^n =a^{mn}$, as is demonstrated in my previous comment. To sum up: the only reason you know that $a^m=\exp(m\log(a))$ is because of the fact $(a^m)^n =a^{mn}$.
Comment: @A-levelStudent it is your very first line where you're going wrong. You need to forget everything you've learnt about log laws and exponential laws. Once again, if $a>0$ and $m\in \Bbb{R}$, then I am DEFINING $a^m := \exp(m\log(a))$. This is a definition, and I can define whatever I want. I suggest you re-read my answer carefully and follow the logic step by step and verify every single claim I make, in the order I have stated things. What I have done is defined $a^m:= \exp(m\log(a))$, and from this starting point, I am proving all other results like $(a^m)^n = a^{mn}$ and $a^{m+n}=a^ma^n$.
Comment: Ok. Then how did you come to your definition of $a^m=\exp(m\log(a))$? Surely you'd need to prove it?
Comment: @A-levelStudent how are you supposed to prove a definition? It is a definition, which means I am simply giving meaning to a certain symbol. I can define whatever I want as long as it is logically consistent and I define things entirely in terms of things I have previously defined. If you ask why this definition is reasonable/meaningful, then that's an entirely different question, which is worth looking into. The main reason is because if I define $a^x:= \exp(x\log(a))$, then it follows that $a^1 = a$, $a^{x+y} = a^x\cdot a^y$, and hence this agrees with the old one if we plug in integer $m>0$
Comment: @A-levelStudent It doesn't need to be proven as it is 'defined'. For example (and forget about fractions and irrationals and negatives and 0 here), $a^m$ is 'defined' to be equal to whatever you get if you multiply a, m times. Notice that you don't need to prove that $a^m=a•a•a•....a$ (m times). It is a 'definition''. But I think that there's more to this, as I guess $a^m:=exp(mloga)$ is chosen not out of thin air, but rather it is the most 'logical' definition. At least, that is what I am gathering from this. I do not know 'why' it is the most logical one though.
Comment: Neither do I understand why it is at all logical without the log laws.
Comment: @A-levelStudent I've tried to motivate the definition as far as I can. It is pretty much our desire to preserve a certain functional equation (along with some regularity assumptions) which motivates our definition. I apologize for the length of it; there's quite a bit to say (this is why in the very first paragraph of my answer I said very explicitly that going from rationals to reals is not at all a trivial task). Hopefully, you follow along each step carefully and keep track of what are the assumptions at each stage of my answer, and keep track of what has been proven at each stage.
Comment: To answer your question of "why" one defines $a^m$ this way, ...
Comment: @peek-a-boo that was a nice read. But shouldn't $c=\frac{g'(0)}{f'(0)}$ in order for the L.H.S. to be 0 ? Also, how do we WLOG suppose that f'(0)=1 ?
Comment: @OrlinAurum lol that's an embarassing algebraic mistake. Suppose $f'(0) \neq 0$, then consider the function $\phi(x) = f\left(\frac{x}{f'(0)}\right)$. Then, it is easy to verify that $\phi$ satisfies all the properties which $f$ does (we're applying the converse of Theorem 1 with $c=\frac{1}{f'(0)}$), AND in addition it has the nice property that $\phi'(0) = 1$. So when I say "WLOG suppose $f'(0)=1$", what I mean is that rather than considering the function $f$, we consider the slightly modified function $\phi$ (just that at the time I didn't feel like introducing too many new letters)
Comment: Also, I realized that in the second-last paragraph, it is actually not necessary to assume the function $f$ (or $\phi$ using the notation of my previous comment) is analytic. This can actually be shown directly by writing a finite Taylor expansion, with the Lagrange form of the remainder, and then showing that the remainder term goes to $0$ as $n\to \infty$, thereby proving that the Taylor series converges to the function for every $x\in \Bbb{R}$.
Comment: @peek-a-boo I've managed to show that $\phi$ has all the properties of f, and $\phi'(0)=1$. So, from here on, we can only work with $\phi$, and make no more conclusions for f, right? As in, we have to show that $\phi$, not f, is equal to it's Taylor series, right? Also, in theorem 1, how do we choose $c=\frac{g'(0)}{f'(0)}$? It's like saying "This theorem is true, but only in this specific case."
Comment: @OrlinAurum You can work with only $\phi$, but this is only a matter of simplicity, because it satisfies the simple ODE $\phi'=\phi$ (and $\phi(0)=1$). Even $f$ will equal its Taylor series... but we're just not that interested in $f$. If I ask you to solve $2c=1$, what would you do? Of course you'd say $c=1/2$. This equation has a unique solution depending on the LHS coefficient $2$ and the RHS coefficient $1$. Similarly, Theorem $1$ claims the existence of $c$ such that $g(x)=f(cx)$, so of course I have to find and choose a specific $c$ (depending on $f$ and $g$).
Comment: @peek-a-boo Hopefully this will be my last set of questions: Is there a standard way we came up with $\phi(x)=f(\frac{x}{f'(0)})$ , or was it an instance of intellect? Also, knowing that $\phi(x)=\phi'(x), \phi'(0)=1$, doesn't the fact that $\phi(x)=exp(x)$ follow in an easier way than the one you mentioned (i.e. writing a finite Taylor expansion and showing  that the Lagrange remainder goes to zero)? For example, we could think of $\phi(x)$ as $\sum_{n=0}^{\infty}a_n x^n$ And solve the differential equation from there. I'm only asking this because I'm not quite sure of the rigour of this.
Comment: @OrlinAurum No it's not a matter of intellect, you're just led to that answer by applying chain rule in reverse: if you know $f'(x) = f'(0) f(x)$, $f'(0)\neq 0$, $f(0)=1$, there is of course only one way to modify $f$ to get a function $\phi$ such that $\phi'(x) = \phi(x)$ and $\phi(0)=1$. In order to get $\phi(x)=\sum\frac{x^n}{n!}$, using Taylor's theorem (with remainder and show the remainder goes to $0$) is perfectly rigorous. At the time of writing my answer I hadn't thought of this yet, which is why I said "make the leap of faith $f$ (or $\phi$ in the new notation) is analytic blablabla"
Comment: @JoeLamond Thank you for your comment, I will look into that thread. My opinion has now changed (it's been 10 months); I'm now a bit embarrassed of my comments here :)
Answer: I am not yet allowed to comment, so I write here. For the case a>0 you can easily prove by use of logarithm. For negative a you need to use complex analysis to prove the same thing.
Use following $$\ln (a^m)^n=n\ln (a^m)=nm\ln a= \ln a^{mn}.$$
New edit for comments:
You have $y=a^x$ where $a>0$. By definition $x=\log_a y$.
Now instead in your problem statement you have $y=(a^m)^n$ so choose $b=a^m$ so that $y=b^n$ then by using the definition you get $n=\log_b y=\frac{\ln y}{\ln b}=\frac{\ln y}{\ln a^m}=\frac{\ln y}{m\ln a}=\frac{1}{m}\log_a y$ which gives you $mn=\log_a y$.
Choose $x=mn$ and use the definition again and you have proven the result.
Comment: Interesting answer, thanks for that! However, I think (correct me if I'm wrong), you may find that the proof for your answer, ie the log laws, is based on the question!
Comment: @A-levelStudent They come from the definition $a^b=\exp(b\ln(a))$.
Comment: @A-levelStudent What is your definition of $a^b$ ?
Comment: I do believe it's the same as yours, and the following proof that $\ln{a^b} =b \ln a$ is my area of difficulty with this response: $$\log_a {M^n} = x \implies a^x = M^n$$

$$\log_a M = y \implies a^y = M \implies a^x=M^n=(a^y)^n=a^{ny}$$                      $$x=ny=n \log_a {M}$$
 This proof is where you're comment and the answer come s from, but this proof uses the fact that $(a^y)^n=a^{ny}$!
Comment: See the substitutions I did in my new comment.
Comment: @A-levelStudent Well if you define $a^b=\exp(b\ln(a))$, then you are good to go because $$\ln(a^b)=\ln(\exp(b\ln(a)))=(\ln\circ\exp)(b\ln(a))=b\ln(a).$$ More work might be required if you use different definitions.
Answer: The rule is merely a specific case of the addition law of indices/powers. Consider:
$$(a^m)^n=a^m\times a^m \times a^m \times a^m \times a^m...\times a^m$$
where there are $n$ lots of $a^m$ multiplied together. We have the law that $$a^p \times a^q=a^{p+q}$$
Applying this to your question we have:
$$(a^m)^n=a^{m+m+m+m+m...+m}$$
where we have  $n$ lots of $m$. Now, $n$ lots of $m$ is equal to $n\times m$ which is equal to $mn$.
So, finally, we have that
$$(a^m)^n=a^{mn}$$
as required.
Edit: Since writing the above proof for positive integer exponents I have been shown a proof that $\ln {a^x}=x\ln a $ without using what we are trying to prove in the question itself, which allows us to use the laws of logs etc to answer the question, as previous answers have done:
Consider $$f(x)=\ln {x^n} -n\ln x\implies f'(x)=\frac{nx^{n-1}}{x^n}-\frac{n}{x}=\frac{n}{x}-\frac{n}{x}=0$$
This means that $f(x)$ must be equal to some constant, $c$, as only constants diiferentiate to $0$. Let's try to find $c$: We have
$$\ln {x^n} -n\ln x=c$$
Let $x=1$:
$$\ln1-n\ln1=c=0-0=0$$
So we have $c=0$, leaving us with $$\ln {x^n} =n\ln x$$
For the sake of completeness, I'll now go on to answer the question for all real exponents. Note that $\ln {x^n} =n\ln x$ is true for all $n\in\mathbb R$.
Let $x=a^m$:
$$e^{\ln {x^n}}=e^{\ln {(a^m)^n}}=(a^m)^n$$
But we also have
$$e^{\ln {x^n}}=e^{n\ln {x}}=e^{n\ln {a^m}}=e^{mn\ln {a}}=e^{\ln {a^{mn}}}=a^{mn}$$
So, at last, we have:
$$(a^m)^n=a^{mn}$$.
Thanks to peek-a-boo, among others, who attempted to make me understand this in more detail.
Comment: this reasoning only holds for positive integers $m,n$, but OP is asking for the case where $m,n$ are any real number
Comment: $m$ certainly doesn't need to be positive, or an integer. You have a point about $n$ though, I will think about it.
Comment: I agree that the rule $(a^m)^n = a^{mn}$ is true for all $a>0$ and any $m,n \in \Bbb{R}$, but your proposed proof only works in the case $m,n$ are positive integers. Part of the problem is in carefully defining what $a^m$ is even supposed to mean if $a>0$ and $m$ is a real number which is not a positive integer.
Comment: Why must $m$ be a positive integer for my proof?
Comment: Ask yourself the question how do you define $a^m$ if $m$ is not a positive integer. Once you answer this question, you'll see that your proof is incomplete and doesn't deal with these other cases


Emacs and Trac Integration

Question: Is there a way to integrate Trac and Emacs?
e.g. edit Wiki pages in org mode, edit and manage tickets from within emacs?
Answer: There is a Trac Wiki page editing mode for Emacs. It is supposed to ease editing Trac Wiki pages but also help with diffing, merging, highlighting and completion.
It is also mentioned at the EmacsWiki and in a blog post.
Answer: Personally for ticket management I use a dirty script in org-babel-python that would retrieve the list of tickets inline, into my org page:
#+begin_src python :results output
base_url = "http://your.trac.site.url"
ticket_list_url = "http://your.trac.site/query?..."
headers = {"Cookie": "trac_form_token=token-from-browser-session; trac_auth=your-auth-from-browser-session"}

import requests
tasks_html = requests.get(ticket_list_url, headers=headers)

from HTMLParser import HTMLParser

class MyHtmlParser(HTMLParser):
  in_link = False
  link_href = None

  def handle_starttag(self, tag, attrs):
    attrs = dict(attrs)
    if tag == "a" and attrs["href"].startswith("/ticket"):
      self.in_link = True
      self.link_href = attrs["href"]
    else:
      self.in_link = False

  def handle_data(self, data):
    data = data.strip()
    if self.in_link and data and not data.startswith("#"):
      ticket_num = self.link_href[8:]
      print "*** #" + ticket_num + " [[" + base_url + self.link_href + "][" + data + "]]"

MyHtmlParser().feed(tasks_html.text)

#+end_src

The script requires pip install requests plus org-babel with python enabled. Plus you need to set your *_url & set auth headers from your browser session.
Org-babel scripts can be run with C-c C-c
Just in case someone needs a dirty solution for listing tickets from trac in an org page..

Emacs and Trac Integration


From Tai O to downtown Hong Kong by ferry

Question: I plan to go to see the big Buddha by cable car and then take a bus/taxi to Tai O village. Is there an option to take a ferry back to the center of Hong Kong? If so, where can I find a schedule? Do I need to purchase tickets in advance?
Answer: The Tung Chung / Tuen Mun ferry mentioned in another answer only runs a few times a day, and doesn't bring you back to HK island...
The best way to go back to HK island by ferry is to take a bus from Tai O to Mui Wo, and then the ferry to Central.
Comment: It's bus number 1. Also, from the Big Buddha to Tai O is bus number 21.
Comment: Thanks. May I ask which bus goes from Tai O to Mui Wo and how often?
Comment: Can't remember the number offhand, but it's not like there's a million buses there. There's only a couple of lines. Ask/check the signs.
Answer: There is a ferry from Tai O that will take you back to Tung Chung or on to Tuen Mun. You can get on the MTR at the station in Tung Chung, from Tuen Mun I think there is a bus to the Tsuen Wan MTR station.
The ferry doesn't travel very often on weekdays; I think there's only one boat then and the trip from Tai O to Tuen Mun and back is almost an hour each way, so there is 2 hours between departures from Tai O. You pay when you get on the ferry. The public transit alternative is the bus from Tai O to Tung Chung, which leaves more frequently.


Linux distributions without selinux

Question: Are there linux distros without selinux portion  in kernel? Sorry if that's dumb question ;), but my distro (debian) has it and i see that it's in kernel on kernel.org so i was wondering if it's so popular that every distro has it or some delete this portion of kernel and use something else? If there are distros without this kernel part (and i mean totally deleted, not just disabled and waiting for being enabled on boot like in debian) could you give some examples?
Comment: well, maybe i want to use something else. what *you* have against someone controlling his own machine and disabling 'security' features which he found to not be as effective as their authors claim? there are some vulnerabilities which are easier to exploit with selinux enabled.
Comment: I can't speak for Phil, but in my opinion, any utility that breaks much, much more than it fixes is a candidate for avoidance.
Comment: for me it seems ubuntu ships the kernel with selinux

$ grep -i selinux /boot/config-2.6.24-23-generic
CONFIG_SECURITY_SELINUX=y
CONFIG_SECURITY_SELINUX_AVC_STATS=y
CONFIG_SECURITY_SELINUX_BOOTPARAM=y
CONFIG_SECURITY_SELINUX_BOOTPARAM_VALUE=0
CONFIG_SECURITY_SELINUX_CHECKREQPROT_VALUE=1
CONFIG_SECURITY_SELINUX_DEVELOP=y
CONFIG_SECURITY_SELINUX_DISABLE=y
# CONFIG_SECURITY_SELINUX_ENABLE_SECMARK_DEFAULT is not set
# CONFIG_SECURITY_SELINUX_POLICYDB_VERSION_MAX is not set
Comment: If it's not actually *enabled* by default, why do you care?
Comment: What do you have against selinux?
Answer: If you are concerned about which modules and features are compiled into your kernel, you should maintain your own kernel, compiled by you. Debian has a system named make_kpkg to facilitate this process. You can copy over a stable config from /boot to the new kernel source tree and load that into your custom build so you don't have to configure everything from scratch. Then you have complete freedom to manually scan through every single kernel configuration setting, including SELinux.
Answer: while it is in the vanilla kernel i presume every distro ships the system with that part enabled.
You can only disable it or recompile your own kernel without it.
<code>title SE-Linux Test System
        root (hd0,0)
        kernel /boot/vmlinuz-2.4.20-selinux-2003040709 ro root=/dev/hda1 nousb selinux=0
        #initrd /boot/initrd-2.4.20-selinux-2003040709.img
</code>
Comment: check you kernel config:

CONFIG_SECURITY_SELINUX=y

i presume this is the default for all teh major linux distros
Comment: that doesn't directly answer my question, but maybe recompiling kernel isn't that bad idea, +1
Answer: Ubuntu and Suse don't come with it in the default kernel, as they use app armor instead. I believe Gentoo uses GRsecurity, so doesn't come with it by default either
Comment: hardened Gentoo,  uses SElinux, but normal Gentoo doesn't.
Answer: The Wikipedia article on SE-Linux has a list of distros that use it.
Not being mentioned doesn't mean others don't, but I'd look with ones not on the list first :)
Distrowatch is a good spot to watch for updates on current and new distros.
Answer: Slackware Linux uses vanilla kernels.
Comment: selinux IS on vanilla kernels
Answer: SELinux ships with Debian, but it's disabled by default. There is a lot you need to do to enable it, so I wouldn't worry.  The only reason it's there is because a lot of companies that use Debian on their servers require SELinux -- so Debian offers it as a choice, but doesn't force it on any of it's users.
Comment: yes, but it's still there. just 'disabled' but it's there.
Comment: So why not compile a vanilla kernel from kernel.org without SELinux?

Linux distributions without selinux


Missing Features in Ubuntu 12.04

Question: I have recently upgraded to  Ubuntu 12.04, and 2 new features that I have seen advertised on various websites are not on my version:

Individual wallpapers on login screen
Privacy options in 'settings'

Is there a way of getting these features, or have they just been removed?
P.S. I upgraded to 12.04 from 11.10 via the upgrade tool.
Answer: 1. Privacy Options
To access the privacy options click on "Session Indicator" and select "System Settings".

In "System Settings" click on "Privacy".

If privacy options are not available, please try this:
To open the terminal press Ctrl+Alt+T, and run the following command:
sudo apt-get install activity-log-manager-control-cente
Then restart your system.
2. Personalised Login Screen
You'll get personalised login screen automatically when you'll change your desktop wallpaper (use something other than the default wallpaper). Your desktop wallpaper will automatically be used as your login screen background.
Comment: Hasn't happened for me since I upgraded to 12.04. Not to big of and issue for me but would like it to happen. I have had numerous backgrounds and only have had the default background at logins.
Comment: May be this will helps you: [What are the criteria for a user background to become login background on user selection?](http://askubuntu.com/questions/123721/what-are-the-criteria-for-a-user-background-to-become-login-background-on-user-s)
Comment: Thanks for the advice, but the privacy option doesn't even appear in system settings. Also, the wallpaper doesn't change automatically with the user.
Answer: Also, I never was a fan of upgrades. Sometimes they simply break your system, sometimes leave out features, and sometimes are just plain dumb. Next time try a clean install. Generally installing clean leaves you with a better boot time, better response, and guaranteed speed.
Comment: Yeah, I think your right. I'll try it tomorrow
Answer: I had this same issue with backgrounds not showing up on the login screen.  However, the problem only occurred when I tried to use files I transferred from a back-up external harddrive.  It turns out it is simply a permissions issue.  If you copied files from another location, they may only be user/group readable (but not other/world readable).  Since the global login screen doesn't have your personal permissions, it cannot access the background images unless they are world readable (don't make them world writable though).
So find the background(s) you want to use, and then change their permissions as so:
chmod o-r [some file with bad permissions]
This will hopefully resolve your login screen issue.

Missing Features in Ubuntu 12.04


Can I connect the two routers back to back to use HSRP? Please suggest

Question: Current network is like - MSP network--> Cisco 4431--> ASA--> N7k--> LAN.
I'm adding one more router and one more ASA in the network, Do I really need to add L2 switch to connect routers LAN side and FW outside interface to achieve HSRP redundancy?(as I dont have L2 switch), if L2 is mandatory for HSRP as I went through some docs, how the switchport on L2 and Router ports and ASA ports will be configured? on L2 switch - normal switch port on a single VLAN ? and on Router and ASA normal L3 ports ?
Comment: I'm not clear how the new ASA and router are going to be connected.  Will you have a new Internet connection with the new router?  Will the new ASA be a failover device for  the current ASA?
Comment: HSRP is a redundancy for the hosts, not for the routers. HSRP messages are exchanged between the routers on the LAN(s) where the hosts are connected, and that is usually on a layer-2 switch.
Comment: Hi Ron Trunk : Yes,new ASA will be connected to exisiting ASA in HA, new router will be connecting to existing router in Pair too on a new connection (MSP connection)
Comment: Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
Answer: You'll need some kind of L2 switch for the HSRP to work for you in this setup. The switch will give your Routers/ASAs what they need to "share" their Layer 3 address (as Ron Maupin said above). I'm guessing you must have these connections currently setup to run directly from each device to the next with no switch fabric in place?
Keep in mind, while you'll have redundancy at the hardware level for the routers you won't have redundancy in your switch fabric for the ASAs/Routers if you only add a single L2 switch. If the switch goes down, your hardware redundancy doesn't do much for you. Consider adding two layer 2 switches at some point in parallel (A and B side if you will) Just something to think about, but I'm sure you have.
L2 Switch ports - configured with a single VLAN shared for both Routers and the Firewalls.
Ports on ASA/Routers - Same you would have normally configured them except with your Routers using HSRP, and your ASAs setup for HA like Active/Standby failover.
Answer: "Do I really need to add L2 switch"
No, you already have one - the Nexus.  Simply carve out a few ports into a VLAN for that segment.  If you are really concerned that VLAN's do not effectively segment traffic the 7k has Virtual Device Context functionality whereby you can carve out a 'new' L2 switch.
Comment: Unfortunately, the Nexus doesn't connect to the routers. The router currently connects directly to the ASA, and the question is about adding another router and ASA, and having HSRP on the routers for the ASAs.
Comment: @RonMaupin The guy is asking if he needs to add a new switch, he is not asking about whether to add another router or ASA.  Do you understand the implications of adding a new switch?  He would need to insert a switch between the ASA(s) and Router(s) / undo that existing direct connection.  My answer is correct and complete.


What do you call the attach-y things on the business end of adjustable shelf brackets?

Question: See this here?  I am trying to find something like it so that I can create wall riser extensions for my cubicle at work.  I'd cut some lengths of 3/16"-thick board, attach to them these brackets, and then mount them on the inserts (or whatever those are called).  The thing is that I don't know what the end pieces of these brackets are called.  If I knew, I could execute a proper Google search to find more of them, but I don't, so I can't.
Answer: A medium duty, standard single slot, blade bracket?

Source: beta.lvmannequins.com
Another option might be removing the plastic caps off the uprights and slide something down there; put 'legs' on the board and slip it in. Isn't that how some partitions with glass tops work?
Answer: I doubt there is a common name for that mount but first thing you will need to do is find out the manufacture of the cubicle system.  Once you have that, you can then start searching for brackets or mounts for that system.  I quick google search, and it came up with a couple of items like that under the name of shelf bracket and tackboard clip.
Answer: I just did a search for "slot mount bracket cubicle" when I found your picture. Looks like you & I have the same problem.
Based on the results I got, there are different manufacturers, and their slot dimensions are not universal. However, since you have an existing bracket to measure, you should be able to figure out which manufacturer yours is.
The results I received included lots of manufacturer names and dimensions. Use terms like 'fits steelcase 900', or 'fits 1" on center', or '3/4" with 1.25" gap', etc. to do a more deailed search.


how to setup wifi on windows 2000?

Question: I heard that the configuration of the WLAN(WIFI) is special to windows XP and doesn't exist in windows 2000, is it true?
Did you know a small program who make that? Excuse me for my poor english, I'm French...
Answer: Windows XP has a special service called Wireless Zero Configuration (in English at least) that handles the various duties of scanning for networks and associating with them. This service, and the related GUI elements were never back ported to Windows 2000.
That being said, every wireless card I have ever used that had a Windows 2000 driver also came with a utility that would manage the connections as well. Almost all wireless cards are manufactured through Broadcom, Atheros, Intel and a handful of other vendors - it is quite possible that you could get a generic driver/connection manager working with almost any given card. You will have trouble finding a generic utility as the API for working with the card was not standardized prior to Windows XP and each driver can (in theory) behave differently.
Answer: Actually you'll just need to install the manufacture's Wifi driver and client software in order to get wifi to work under Windows 2000. Before you do that though, you'll have to upgrade the Windows Installer using the Windows Installer 3.1 redistribution or else the wifi installer will fail to work most likely. Actually quite a few software depends on this later Windows Installer to work. Just to give you an idea, I just made this posting from Windows 2000 Pro using Firefox 37 using an Intel wifi card. If you want to use later browsers, you'll need to apply some service packs and other fixes both official and unofficial.
Answer: Five or so years ago I was wondering how I could connect my old Windows 98 and 2000 PCs to my wireless network.  New wireless adapters didn't support 98 or 2000.  Then I found out about DD-WRT and wireless bridges.  DD-WRT (and probably a bunch of other open-source router firmware nowadays) lets you configure a wireless router as a client bridge.  This means you connect your old PC to the router (operating in client bridge mode) via an ethernet cable.  Since its connected via ethernet, there's no additional drivers or software that you need to install on the PC.  Then the client bridge is configured with the wireless settings to connect to your actual wireless router.  This is the best solution if you're unable to find an old wireless adapter with software/drivers for the operating system.

how to setup wifi on windows 2000?


I need to convert the date from mm/dd/yyyy to dd/mm/yyyy in vbscript

Question: So far I have this:
FormatDateTime(negativeItemsRS("ItemDate"), 0)
Its displaying the date in the format of mm/dd/yyyy. I want to convert that to a dd/mm/yyyy
Please help not sure how to do this.
Thanks,
Answer: You have to set the locale id to one that uses the date format that you want. I don't remember which format used where, but either UK (2057) or US (1033) should work.
You haven't specified your environment. In ASP you could use the LCID property in the Language directive or in the Session or Response classes, depending on what scope you want for the setting:
<%@Language="VBScript" LCID="1033"%>

or
Session.LCID = 1033

or
Response.LCID = 1033

Answer: To change the date from MM/DD/YYY to DD/MM/YYYY in VB Script very simple steps can be used as given below
Let say "date1" (MM/DD/YY) =  03/06/2014
I want "date2" to be in DD/MM/YY  as 06/03/2014
d = Day(date1)
m = Month(date1)
y = Year(date1)
date2 = d & "/" & m & "/" & y

Will give the required result.
Answer: What you need is a FormatDate function. I used to do this the hard way with manual concatentation, but I discovered that there are a few .NET libraries that are accessible from COM and thus from ASP Classic. My version leverages the fact that I have a StringBuilder class which is a wrapper around the StringBuilder class in .NET.
'******************************************************************************
Public Function FormatDate( sFormat, dDateValue )
'PURPOSE: To format a date with any arbitrary format
'ARGS:
'   sFormat is the defined formats as used by the .NET Framework's System.DateTimeFormatInfo.
'       Note that this format is case-sensitive.
'CALLS:
'   1. System.Text.StringBuilder class in the .NET Framework.
'EXAMPLE CALL:
'   Dim sFormatedDate
'   sFormatedDate = FormatDate( "MM/dd/yy", "1/1/1900 12:00 AM" )
'   Or
'   sFormatedDate = FormatDate( "MM/dd/yyyy", "1/1/1900 12:00 AM" )
'DESIGN NOTE:
'   This function leverages the fact that System.Text.StringBuilder is COMVisible.
'   Thus, we can call its AppendFormat function from here.
'   You can find more information about the format string parameters allowed at
'   http://msdn.microsoft.com/en-us/library/system.globalization.datetimeformatinfo.aspx

    Dim oStringBuilder
    Dim sSbFormatString
    Dim dDate

    If Not IsDate( dDateValue ) Then
        FormatDate = vbNullString
        Exit Function
    End If

    On Error Resume Next

    dDate = CDate(dDateValue)
    If Err.number <> 0 Then
        FormatDate = vbNullString
        Exit Function
    End If

    'if an empty format string is passed, then simply return
    'the value using the default shortdate format.
    If Len(sFormat & vbNullString) = 0 Then
        sSbFormatString = "{0:d}"
    Else
        sSbFormatString = "{0:" & sFormat & "}"
    End If

    Set oStringBuilder = CreateObject("System.Text.StringBuilder")
    Call oStringBuilder.AppendFormat(sSbFormatString, dDate)
    FormatDate = oStringBuilder.ToString()
    Set oStringBuilder = Nothing
End Function
'**************************************************************************
' Use this class to concatenate strings in a much more
' efficient manner than simply concatenating a string
' (strVariable = strVariable & "your new string")
Class StringBuilder
'PURPOSE: this class is designed to allow for more efficient string
'   concatenation.
'DESIGN NOTES:
'       Originally, this class built an array and used Join in the ToString
'       method. However, I later discovered that the System.Text.StringBuilder
'       class in the .NET Framework is COMVisible. That means we can simply use
'       it and all of its efficiencies rather than having to deal with
'       VBScript and its limitations.
    Private oStringBuilder

    Private Sub Class_Initialize()
        Set oStringBuilder = CreateObject("System.Text.StringBuilder")
    End Sub

    Private Sub Class_Terminate(  )
        Set oStringBuilder = Nothing
    End Sub

    Public Sub InitializeCapacity(ByVal capacity)
        On Error Resume Next
        Dim iCapacity
        iCapacity = CInt(capacity)
        If Err.number <> 0 Then Exit Sub
        oStringBuilder.Capacity = iCapacity
    End Sub

    Public Sub Clear()
        Call Class_Initialize()
    End Sub

    Public Sub Append(ByVal strValue)
        Call AppendFormat("{0}", strValue)
    End Sub

    Public Sub AppendFormat(ByVal strFormatString, ByVal strValue)
        Call oStringBuilder.AppendFormat(strFormatString, (strValue & vbNullString))
    End Sub

    'Appends the string with a trailing CrLf
    Public Sub AppendLine(ByVal strValue)
        Call Append(strValue)
        Call Append(vbCrLf)
    End Sub

    Public Property Get Length()
        Length = oStringBuilder.Length
    End Property
    Public Property Let Length( iLength )
        On Error Resume Next
        oStringBuilder.Length = CInt(iLength)
    End Property

    'Concatenate the strings by simply joining your array
    'of strings and adding no separator between elements.
    Public Function ToString()
        ToString = oStringBuilder.ToString()
    End Function
End Class

So, with this class you could do something like:
FormatDate("dd/MM/yyyy", RS("DateField"))

Note that the string passed in is case-sensitive.
EDIT I see that at some point I amended my FormatDate function to eliminate the use of my VBScript StringBuilder class and instead just use the .NET class directly. I'll leave the VBScript StringBuilder class in there for reference in case anyone is interested. (I did swap the order of the two however to make the code that appears at the top more applicable to the problem).