character degree in Gap

Question: Let S be a finite set of integers, do I can check with Gap that this set be a set of character degrees of small group?
Comment: No one has mentioned yet: the order of the group must be a multiple of the LCM of the elements of S.
Answer: Dima is correct that the problem is much easier with multisets. Furthermore, taking direct products with Abelian groups, it is clear that for any given set of irreducible character degrees which actually occurs, there are infinitely many groups for which it occurs, if we allow multiplicities. I had a recollection that Isaacs and Passman proved a kind of converse of this: if a finite group $G$ has all its irreducible character degrees of degree less than $d,$ then $G$ has an Abelian normal subgroup $A$ whose index is bounded in terms of $d.$ I could not locate a reference at the time, though I did not have Isaacs' book to hand.
Since my original post, however, Yemon Choi kindly provided a reference in the comments below (see MR0167539). The wording of Isaacs and Passman's result is perhaps mildly ambiguos: they speak of an Abelian subgroup of index bounded in terms of $d$, omitting the word "normal", but of course if $G$ has an Abelian subgroup $A$ with $[G:A] \leq h,$ then the intersection of the $G$-conjugates of $A$ is normal, and has index at most $h!$.
   Hence, in general, if we know the irreducible character degrees of a finite group $H,$ there is an Abelian normal subgroup $A$ of $H$ such that $|H/A|$ is bounded in terms of the largest degree. So in small cases, it may be possible to use GAP to determine all possibilities for $H/A.$
Comment: I am out of the office right now so cannot view it, but I think the result of Isaacs and Passman that you refer to is in their paper
*Groups with representations of bounded degree*, Canad. J. Math. 16 (1964) 299--309. http://www.ams.org/mathscinet-getitem?mr=167539
Comment: Thanks, that figures- I put a date range in Mathscinet starting at 1965!
Answer: Given an arbitrary set $S$ of powers of some prime $p$, subject only to the condition that $1 \in S$, there necessarily exists a $p$-group $P$ such that $S$ is exactly the set of irreducible character degrees of $P$. In fact, it is always possible to find $P$ having nilpotence class $2$. This theorem appears in my paper in PAMS 96 (1986).
There is no analogous result for arbitrary sets of integers. For example, no group has irreducible character degree set $\{1,2,3,5\}$.
Answer: You most probably mean "the set of degrees of irreducible characters". Indeed, given a finite set of integers, it should not be hard to construct a finite group that has irreducible characters of that degrees (and many more other degrees).
Still, for "the set of degrees" the problem does not look easy, as degrees can repeat, and you don't have a good way to control the order of the group $G$. For "the multiset of degrees" $d_1$,...,$d_t$ things are much better, as then $G$ has order equal to $\sum_{j=1}^t d_j^2$, and so you can just look through GAP's database of small groups for such a group.

character degree in Gap


Are my aged hops ready?

Question: I've got about 3oz of cascade hops that spent 2 years in the freezer before I decided around January 2011 to take them out, put them in a paper bag, and let them sit at room temperature to age so I can brew a lambic.  Over the course of the last several months, anytime I opened the closet that I was storing them in I would get smacked in the face with a cheesy aroma.   However, a couple of weeks ago, I realized that I couldn't smell the cheese anymore, and today I finally opened the bag up, stirred my hand around in it, and stuck my nose in and inhaled, and no noticeable cheese aroma.   Is this a useful indicator that they are aged enough to put into a Lambic without causing off flavors or excess bitterness?
Answer: Try chewing a small leaf, or making hop tea with a couple of cones. This will help you gauge how much bitterness and aroma is left.

Are my aged hops ready?


Mach Kernel vs. XNU

Question: When I ran $ls on my root directory a file /mach_kernel was found.  I'm assuming this is the OSX kernel.  I always thought that OSX used the XNU hybrid kernel, so why is this file called "MACH"? I'm running Lion.
Answer: Based on what the Wikipedia page on XNU says, XNU is based on Mach. Since OS X started using the Mach kernel and now uses a derivative of it, they probably just kept the kernel's file name the same so they didn't have to tweak firmware during upgrades or with new machine releases.
Answer: The file is called /mach_kernel for historic reasons; XNU is, indeed a hybrid of Mach and a BSD layer (forget Wikipeida, there's http://www.amazon.com/Mac-OS-iOS-Internals-Apples/dp/1118057651 which seems more definitive).
In actuality, this kernel isn't the exact image which gets loaded - there is a pre linked kernel cache (with all the necessary kexts) in /System/Library/Caches/com.apple.kext.caches/Startup.
Comment: preloaded not prelinked

Mach Kernel vs. XNU


What is a good French language equivalent for the English expression, "like pulling teeth"?

Question: In English, when somebody is being very uncooperative in telling you information or performing task(s) you want them to, it is often said that getting them to cooperate is "like pulling teeth".  Another similar expression is "like getting blood out of a stone".  Is there a good French alternative for these expressions?
Answer: Tirer les vers du nez is only appropriate in sentences like "Prying information from X was like pulling teeth", since it is specifically about informations.
If you want to refer to such difficulties in general case, you'd better use être la croix et la bannière, e.g. "Getting him to cooperate is like pulling teeth" would be "C'est la croix et la bannière de le faire coopérer".
And in the case of a reticent person you can say he is une tête de mule.
Answer: In French, you would use tirer les vers du nez (literally: pull worms from his nose) when you try to get information from someone.
See also the entry for tirer les vers du nez on another site.
Funny sidefact: it's easier to do when one has un verre dans le nez (is drunk, literally has a glass in the nose). Notice that verre and vers are homophones.
As others rightly say, this doesn't apply to making one do something, or to otherwise difficult situations. In these cases, in addition of the already mentioned calvaire (casse-tête is nice too), you can also compare your task to an impossible one:

— Tu as réussi à faire manger sa soupe à Pierre ? (Did you succeed in having Pierre eat his soup?)
— Non, autant essayer de faire danser la java à la tour Eiffel ! (No, it would be easier to try to make the Eiffel tower dance!)
Comment: Comment degoûtant. :-)
Comment: -1 pour la spécificité : c'est intéressant, mais ça ne répond pas (m'est avis).
Answer: Quite similar to "la croix et la bannière", a very general idiom to say that something is extremely difficult:

Arriver à le faire coopérer, c'est un calvaire !
Comment: Et dans mon expérience, le *calvaire* n'oublie jamais d'en être un *vrai*. Mais ça serait **un vrai calvaire** que de traduire ce commentaire.
Answer: We could also see this from another point of view: When someone doesn't want to say something, we could say that he is "muet comme une tombe" (mute as a grave).
(You can also answer someone that your are "muet comme une tombe" when someone asks you if you can keep a secret)
Or you can simply tell him he's not very cooperative. "Tu n'es pas très coopératif" (or "vous n'êtes pas très coopératif").
Answer: A biblical analogy is: Il est plus facile pour un chameau de passer par le chas d'une aiguille — It is easier for a camel to go through the eye of a needle

What is a good French language equivalent for the English expression, “like pulling teeth”?


Movable Type on GoDaddy

Question: I have deluxe linux shared hosting with godaddy, and it supports perl. but i dont know how to setup and use perl scripts.
Also i have no folder named CGI or CGI-BIN or anything like that.
As movable type requires me to put my content in the CGI folder, i cant. creating it also doesn't work.
So can you guide me on how to install movable type on godaddy. Also i dont want to spend money and install it from the godaddy hosting connection.
And please remember as i'm on a shared hosting, i might not be able to edit my apache server configuration file.
Thanks.
Answer: Unfortunately, GoDaddy makes installing MT annoying, in several ways. There are some notes in the wiki(and a pointer to the installation instructions) on some things to expect.

Movable Type on GoDaddy


How does Quick Scoping or "Feather" Scoping work in Modern Warfare 3?

Question: I know some may cry "foul" or "cheating" but I'm really fascinated by the people who seem to be amazing marksman in the Modern Warfare series via a quick scope kill.
Can anyone detail out the mechanics of quick scoping for MW3?
I'm on xbox 360.
Answer: The key to QS is to make sure before you take your shot - you are completely stationary. Too many people who try quickscoping are still moving when they take their shot and this results in about a 1:4 shot-to-kill ratio ... it's frustrating as I used to fire while still moving and even though you are lined up and it is dead on them.  I was sometimes too keen to get that shot off. When you are still moving it seems as though the bullet passes straight through them, when in fact it actually hasn't gone where you intended it to go.
Secondly try to shoot from cover at first. Get into the habit of ducking and going prone, then standing - taking your shot and dropping as you shoot. This is known as a drop shot.
When you get good - you can start to drag scope, which is basically where you go to scope in and your crosshairs aren't lined up with the enemy - you have to VERY QUICKLY drag the scope right or left and time the shot and fire when crosshairs are over your opponent.
Once you have mastered the basics - you can look at 360's, stalls and no-scopes.
Remember, line up target, stop, shoot.  You want to consider a sensitivity of something between 6 and 10. I play on 9, but worked up from 5. 10 i find is just a bit too erratic for anything other than a run and gun.
Really you just have to practice, there is no shortcut to no-scoping.
Hope this helps
Comment: +1 for mentioning drag scoping. Wherever I see kill cams of people drag scoping it is utterly amazing to me that they were able to correct a shot by so much in less than a second
Answer: Basically, Sniper Rifle's are extremely inaccurate while not scoping, but will instantly become extremely accurate when you scope in.
This happens extremely quickly after holding in the scope button (not sure if it's instant or just really really quick), so people will click both scope and shoot at almost the exact same time, which will make you shoot very accurately at whatever you're aiming at, so instead of trying to "noscope" someone, they'll try to "quickscope" them instead.
Comment: Would you care to elaborate on the "how" part and leave the "why" part out?
Comment: @Bora "people will click both scope and shoot at almost the exact same time". That's all there is to it. Can't really elaborate on that.
Answer: I've not been able to do much quickscoping in MW3 yet, but I use to do it some in MW2. The basics are this:
First do some target practice. Find something not moving. Aim at it without scoping and then press the scope button(left trigger on xbox) and about a quarter of a second later shoot. Repeat until you're hitting your target. The basic thing you have to do is automatically know where the center of your TV screen is. Some people actually put a physical "dot" sticker in the center of their TV for this. I think that's taking it a bit too far though.
I haven't leveled up too far yet in MW3 so I'll include some relevant information on what the perks and proficiencies should be for quickscoping later.
Answer: This may sound like babble, too lazy to reread it but I tried to get everything in 1 go. This is how I taught my friends to quickscope but I was helping them step by step so it'll be different. I use ps3 btw.
By looking into the scope, your crosshairs get smaller until it's almost perfect due to the sniper scope. Quickscopers utilize this by shooting their gun as soon as the crosshairs reach it's most accurate point as a)the shot will be on target b)it doesnt give the enemy time to kill you as if you were trying to hardscope them.
To learn to quickscope, get in a game by yourself. Tap the aim button repeatedly and watch your crosshairs get smaller and smaller. Try to anticipate when it's the smallest, thats usually right before you can see inside your sniper scope. When it's at the smallest, shoot. As you get better, you'll be able to learn how to walk and do that until it becomes natural.
If you are a good learner and you can pick up how to do it the hard way, I recommend playing black ops and quickscoping there. I practice quickscoping on black ops in combat training by setting the computer to veteran. Black ops quickscoping is focused on the drag-scope.
MW3 is the easiest to quickscope, then mw2 (but that has a ton of aimbots), then blackops is the hardest. You can only hit someone in black ops by drag scoping so I dont recommend doing that first.


Segregation of Managed Metadata between site collections

Question: Is it possible to create a particular set of metadata and use it in a site collection, and create a different set of metadata terms to be used in a separate site collection without them seeing the entire list of terms from both site collections?  is this the purpose of groups in the managed metadata service?
Answer: The term set group is intended as the security boundary for managed metadata. But usually all users can see (read) all the groups; the security applies to permissions to modify and manage the metadata.
If you want to make certain termsets hidden from certain users, for example if you have a term set that is confidential, you will have to create a separate Managed Metadata service application and control access to it by restricting read access to the service application connection. This is managed in Central Administration by selecting the service application and clicking on Permissions in the ribbon. This will be easy to manage if your two site collections are hosted in separate web applications.
It is also possible to manage user access by partitioning the managed metadata service if you are running your web application in partitioned mode. This is more relevant to a hosting situation.
This subject is covered in more detail on TechNet.


How can I cause a browser to download a new copy of a web page when it thinks that its cached copy is still valid?

Question: Sometimes when I make small changes to a website, I notice that certain browsers are very stubborn about downloading the page again.  Both F5 and CTRL-R are worthless and the only way to fix it is to go through the options and manually tell it to clear the cache.  I've used PHP before to send a cache-flush header, but is there a simpler or better way?
Answer: It depends on what's cached. If it's the main page then the only way to prevent caching is using the no cache HTTP header.
However keep in mind that JS and CSS files are also usually cached. One trick to invalidate them when they change used by many websites - including this - is appending a dummy number after the '?'. Everything after the '?' is ignored by the web server, but since the URL changes the browser always re-downloads it.
An example taken from this very page: http://sstatic.net/webmasters/all.css?v=ac775b7bd951. The "ac775b7bd951" comes from mercurial, the versioning system used. You can do something similar with svn (example: all.css?revision=123).
Answer: F5 is refresh using the cache.  Ctrl-F5 will ignore the cache.  I'm not certain, but it sounds like Ctrl-R is equivalent to F5 rather than Ctrl-F5?
Comment: Yep that's right. Although I believe in Opera Ctrl+F5 reloads ALL the pages.


Separation of solids from liquids using magnetic waves

Question: Using magnetic forces, I want to separate solids from liquids in a solution as a centrifuge would do.  Is there a way to hit a volume of liquid and get it to separate liquids and solids with heavy pulses of magnetic waves?
Comment: ""with heavy pulses of magnetic waves"" Could You explain those waves? Such waves might earn You a Nobel Prize.
Answer: For magnetism to work for seperation, the primary thing needed is ht e magnetic properties be different for the constituents i.e either of them should be a combination of paramagnetic, diamagnetic or ferromagnetic material, which may vary in magnitude.
Now, paramagnetism and diamagnetism are awfully weak to work in commercial situation. A trivia: Human body is diamagnetic and can be magnetically levitated by enormous magnetic field, a frog has been levitated by using 5-6 T.
So, the only viable situation arises with ferromagnetic material, which is used presently as such for separation of iron ore.
So coming back to your problem, the liquid and solid you want to separate needs to be having different magnetic property and probably with sufficient string magnetic field, you can separate out them.
As a matter of fact, in centrifuge separation, the inertial mass is used for separation, which makes up the centrifugal force.


Wireframe iOS design apps for iPad?

Question: Is there an app for iPad to assist in wireframe design for iPad and iPhone apps?
I don't mean actual development, just placing standard widgets and the things you usually do with photoshop and associated kits.
Answer: I've adored Briefs app since running across it a few years back. Nothing comes close to the power and design of this app. It for a long time was probably too powerful (sadly) to ever ship on the iTunes store - but the app has finally launched on the Mac and iOS App stores.

http://giveabrief.com

The trial Mac version is free as is the iOS app, Briefscase, which run the briefs. Heavy users of the app will want to buy the actual app on the Mac App Store.
Answer: Blueprint is, in my opinion, the most powerful and most useful, and comes the closest to the promise of Briefs. You get customizable standard controls, it's easy to work with, and most importantly it allows you to set up actions associated with taps, swipes, and rotation. The key is that not only can you work out a complex app interface, but you can then distribute the completed mockup so that other people can play with it using the free Blueprint Viewer app. Super, super handy.
The only downside of Blueprint, imo, is that you can't easily create mockups that don't look like a finished app. When working with outside clients or with semi-clueless management, if you present something that looks finished they'll sometimes confuse it with the idea that the app actually is close to being finished. Rougher wireframes are also useful for getting people to focus on the functionality and not how it looks. Still, a great app, the best imo.
Now to the also-rans:
Interface HD holds a lot of promise, especially the idea that it can generate some basic Xcode template stuff, but it's very unfinished and doesn't feel very polished. Definitely more promise than followthrough at this point.
UI Sketcher is designed for rough sketches and can be used for a variety of projects, not just iOS design. Unfortunately that lack of specialization means that it's not actually very good at iOS app mockups. It's so crude, really, that I've never been even slightly satisfied with a result. Maybe if I used a stylus.
iMockups and SketchyPad do have more iOS-focused controls and options, but both fall prey to the UI Sketcher problem of being designed for other things, too. Additionally neither feels sufficiently complete or polished, and everything is harder to do than I'd like. I've used iMockups to create something I actually presented to clients, but felt the need to constantly apologize for the ways the wireframes weren't quite like I intended the app to be. SketchyPad exports to Balsamiq Mockups format, so if you're already a big Balsamiq fan then it might be a good match for you. Both are definitely a step up from the looseness of UI Sketcher (or any other sketching app), but just aren't good enough to actually use for real work.
I've not used Mocking Pad, but it appears to be nearly identical to iMockups.
Answer: Briefs and Briefscase might work for you, though the design app is a Mac app.
This Google search on ios prototyping on iPad also leads to a number of solutions.
Comment: Might you want to expand on your search? (Perhaps explaining which results are most promising and why) Posting links to google results are generally not well received. Briefs is a great app so we certainly agree on that.
Answer: Interface HD does this.
You could also use Keynote with the excellent Keynotopia templates.
Answer: WireframeSketcher comes with dedicated stencils for iPad applications. You can use links for simple interactivity. You can even test your prototype on iPad by exporting it to HTML that you can put online via Dropbox.
Answer: I have found AppCooker to be robust enough to create a mockup with 40 > screens. You can specify your own tappable areas and links between the screens so that those mockup screens come to life.
However I'm really interested in finding alternatives too: AppCooker is great. Don't get me wrong. However I really wish that it had the ability to handle different layers. It requires a lot of cloning and violates the DRY principle: if you want to change an element that's already been copied into 5 different screens, well, you gotta make the change 5 times! Not fun.

Wireframe iOS design apps for iPad?


Possible to "retire" a board?

Question: Is it possible to retire or deactivate a board for projects that are over? I'd like to be able to move projects that are completed, canceled, or on hiatus out of the main organization board and into a hidden or greyed out area without losing any of the people associated with it or having to invite people to multiple organizations.
How are other people dealing with finished project boards?
Answer: Yes, using Options -> Close Board...
You will be able to re-open it on Boards -> View Closed Boards

Possible to "retire" a board?


Where is the User Profile Service in Sharepoint Foundation?

Question: I have installed Windows Server 2008 R2 & SharePoint Foundation 2010 in my machine. But I cannot find 'Forefront Identity Manager' in the Administrative Tools-> Services section. Due to that, I cannot activate my 'User profile Service' in SharePoint Foundation 2010.
Is there a resolution for this?
Answer: There is no User Profile Service with SharePoint Foundation.
Answer: You should NEVER start, stop, or modify any SharePoint services through the Services dialog; these should always be managed through Central Administration -> Services on Server. During the initial provisioning of the User Profile Service, the FIM services will be installed and provisioned. Build the User Profile Services Service Application, Start the User Profile Service on the applicable server, then start the User Profile Synchronization Service on the applicable server (this will provision the FIM services). Make sure you double-check all of the permissions, and service account configurations to make sure everything is correct before starting the UPS Sync service. If your service fails to start, or gets "stuck" on starting, 99.9% of the time you have permissions wrong.
Edit: As Mike pointed out, there is no UPS in SharePoint Foundation. I overlooked in your post that your environment was Foundation.


What should I use to mount a camera to a wall, with the option to swivel and tilt?

Question: We need a solution to mount a camera to a wall. We'd like an option that swivels and tilts, which this option does. This option uses a vise to mount, which means we need to install something into the wall to attach the vise to. Does anyone have ideas? Or another solution that is effective for our goal?
Answer: There's a wall mount system available at Adorama that sounds exactly like what you're looking for.
Also, depending of the surface and the use you are planning on, a suction mount might be a good option too, since it wouldn't require any holes in the wall.
Answer: First thought that came to my mind: Umbra Conceal Wall Book Shelf.
Those things are pretty cool to just see a pile of books floating on a wall. They should be able to more than handle the weight of a small camera on that device.
Answer: There are a lot of cheep wall mounts designed for security cameras, generally they are small, cheep and can swivel and tilt.
Here are two example products in this category: A simpler mount ($6), A more complex system ($8)
They should be able to support a compact camera, consumer video camera and maybe even a small DSLR without any problem (but I wouldn't try to mount a larger DSLR or a camera with an heavy tele lens on them).
They do require drilling


Fermion Field of Standard Model

Question: Why fermion field is treated as anti-commuting and boson field as truly classical in standard model?
Answer: First of all, the Standard Model doesn't treat bosonic fields as classical. They're quantum mechanical i.e. non-classical, they're just not anticommuting or Grassmann-odd. Second, a consistent theory just requires the relationship between spin and statistics, see e.g. the

http://en.wikipedia.org/wiki/Spin-statistics_theorem

Combining integer spin with Fermi statistics leads to ghost or energy or the norm that isn't positively definite, and vice versa (half-integer spin with Bose statistics). It was proved by Pauli.
However, your very question didn't actually talk about the integer vs half-integer spin at all. It was talking about the relationship between fermions and anticommuting fields. This is almost a tautology. A fermion is a particle whose wave function for many particles is antisymmetric, $\psi(x_1,x_2)=-\psi(x_2,x_1)$ etc., so the fields that create these particles must be anticommuting, $a^\dagger(x_1) a^\dagger(x_2)=-a^\dagger(x_2)a^\dagger(x_1)$. The multiparticle state in QFT is written as
$$ |\text{2 fermions}\rangle = \int d^3 x_1 d^3 x_2\, \psi(x_1,x_2) a^\dagger(x_1)a^\dagger(x_2)|0\rangle $$
Because the wave function $\psi$ is antisymmetric, only the antisymmetric combination $a^\dagger(x_1)a^\dagger(x_2) - a^\dagger(x_2)a^\dagger(x_1)$ contributes to the state, and in fact, only this combination is nonzero. The sum – the anticommutator – vanishes. That's why the antisymmetry of $\psi$ is "automatic": if there were a non-antisymmetric part of $\psi$, it would vanish in the integral above because the product of the creation operators is antisymmetric.
The same for bosons and "commuting", without the minus sign.
The answer to your "why" question is that your statement is really a tautology, pretty much a definition of bosons and fermions, up to the possibly confusing comments about "one antisymmetry" implying the "other antisymmetry" above. Of course, you could also ask why one uses commuting or anti-commuting fields to describe particles at all. Well, Nature just works in this way. Quantum fields naturally reduce to multi-body quantum mechanics with the automatic symmetry or antisymmetry – and they may give rise to automatically Lorentz-invariant theories, too (something that would be hard in the "non-relativistically styled" multiparticle quantum mechanics).
Answer: If you think anti-commuting field is too arbitrary and do not like anti-commuting field, you may ask do we really need to use anti-commuting field
to describe fermions? Can a theory with only bosons have fermionic excitations
emerging at low energies? The answer is yes! So we do not need to
use  anti-commuting fields to describe fermions, and this is true in any dimensions.
Comment: I thought this was only true in 2D?
Comment: A theory with only bosons can have fermionic excitations emerging at low energies.  This is true in any dimensions.
Comment: @Xiao-GangWen I would be interested to read more about only boson-theory. Do you have any reading recommendations or the technical term for this theory?
Comment: @JacobH: See http://arxiv.org/abs/1210.1281 and  http://arxiv.org/abs/cond-mat/0302460

Fermion Field of Standard Model


Passing Emissions Test: Can't figure out how to fix code P0705 on a 1997 Nissan D21 / Hardbody / Pickup

Question: So hello. I have had a lot of luck with my other vehicles on forums and a recently became the (hopefully-soon-to-be)-proud owner of a 1997 Nissan D21/Hardbody/Pickup XE. It has 156k miles on it and is completely stock. 2.4l 5spd 4WD. It runs like a dream, gets great gas mileage, little rust on frame, none on body and bed. All in all, a great truck.
I wanted this specific model truck after reading extensive reviews on their reliability (some say "to a fault"!) and longevity and the fact it is such a badass utilitarian vehicle. I went to go see about 12 before I found the one I currently own. Now, because NH is notorious for it's salt in the winter and D21s are notorious for their frame rust, I had to go to Maine to get it. Maine, where I find out they will slap a sticker on a watermelon. Once I get to NH, I realized, for the first time, just how nasty computerized emissions tests can be.
So far, I have replaced 2 air sensors, a catalytic converter, a speed sensor, fixed an EVAP system leak. Now I am at the one I am stuck on. Where the first list of repairs took about 2 weeks, this one has taken about two months and is still not fixed no matter what we do. Now I am stuck at my mechanic's (unless I pay $450 to have it towed to the closest Nissan dealer) unless I am trying to do the drive cycle and even that is technically illegal. If I do drive, cops around here have a boner for people flying the "illegal vehicle" flag and I will most certainly be arrested and the truck impounded (I already have a ticket AND I was pulled over 6 times while under a 60 day emissions sticker).
The code is P0705: Transmission Range Sensory Malfunction (PRNDL Input)
I replaced the neutral switch (which the dealer said is the range sensor) but lo and behold, the "service engine soon" light came on again after about 21 miles.
Weird things:
-Before the neutral switch was installed, I had to drive over 53 mph for more than a mile AFTER at least 17 miles since cleared codes before the check engine light flipped on. Otherwise, I could drive the car all day under 53 and it would never trip the light, but it would also never "be ready" according to the state's inspection computer.
-Before the NS, the car sucked at higher speeds. We're talking accelerating from 60 mph to 70 mph took over a minute with pedal to the metal.
-On the latest drive cycle test after I had the neutral switch installed, the car was running better than it ever had while I owned it. Better acceleration and lower RPMs at high speeds. The accelerator pedal was much more responsive, especially in 5th gear.
-After the neutral switch was installed, I could drive for as long as I wanted without tripping the check engine light (would still register not ready) until I stopped the engine, restarted it, then drove above 53 mph for more than a mile.
-Checked blown fuses, but all are OK.
Regardless, the tranny does not buck or anything. As I said, the truck runs like a dream.
I feel like my mechanic's are getting annoyed with the truck and their motivation has ceased. As much as I am going to change my mechanic as a result, I am committed to these one's to fix the truck unless I want to risk getting arrested.
Any help would be greatly appreciated, especially things I can do/check in the parking lot of my mechanic's shop.
Thanks for any assistance. I know I am new but please help! It's getting too cold to ride the motorcycle to work!
Comment: Did you ever get this resolved, Filip? [Electronic Versions](http://www.nicoclub.com/FSM/hardbody/) of the Service Manuals are available online, and there are a couple of nice online communities specifically for the D21 Hardbody at [Nissan Forum](http://nissanforum.org/nissan-forum/forumdisplay.php?f=20) and [Infamous Nissan](http://www.infamousnissan.com/forum/forumdisplay.php?f=21).
Answer: If the sensor was replaced and you still have problems, then it's either the sensor, the sensor's circuit, or the ECU. Since it's a new sensor, I would check the circuit. If the mechanic it is at cannot do this, then your best bet is to head to a Nissan dealer.
The current mechanic will likely need the factory service manual for this model to properly check the circuit, which some mechanics do carry depending on how specialized they are.


How can I save as Excel files (.xls) in Calligra sheets?

Question: Would someone tell me if it is possible to save as Microsoft Excel (<code>.xls</code>) in Calligra sheets?
Answer: Using calligra you can only open ms documents, but you cannot save them in .xls , you always will have to save them in .odf(Open document format). Sorry to be the bearer of bad news but that's one of the important functionalities that is missing in calligra.
You can use this url to know more about it : http://maketecheasier.com/is-calligra-a-great-alternative-to-libreoffice/2012/06/18
"One thing though, calligra doesn't support saving to Ms-Word .doc and .docx format. It only supports the Open Document Format (ODF). You can open, view and edit .doc and .docx file, but you can only save to .odf format."
Comment: [A related bug report with some more information (*funding is needed*)](https://bugs.kde.org/show_bug.cgi?id=298115)
Answer: Unfortunately, Calligra does not (yet) have a filter to create/save to Microsoft formats. The only workarounds are to save in the Open Document Format (ODS for a spreadsheet) or as a Google doc.
Excel 2007 or later will open ODS files, but you may need to download a "reader" file to make that work. Otherwise, you would have to go through LibreOffice/OpenOffice to get an XLS or XLSX file for Excel to open.


How useful is carrying out a heuristic evaluation in this circumstance?

Question: Working on a website redesign. We have documentation from workshops that give us insight into the pain points of the existing site from the clients and their business requirements. How useful would conducting a heuristic evaluation on the existing site be if we are planning to redesign the whole site? I was thinking initially the learnings from the existing site would help me design the new site but I'm questioning how valuable doing a heuristic is?
Answer: In this case it would probably depend on how deep you want to go into the existing issues with the website.
Sometimes clients and businesses' requirements are not enough, as they might refer to some particular problems, but not to the underlying structure.
Having said this, I would probably invest my time/resources on the new version instead. You could do some tests before the relaunch, making sure everything is easy to understand and interact with.
Answer: I would restate the question slightly: do you learn enough about what improvements to make, as well as what you're currently doing well and should continue doing, such that it's worth the opportunity cost of doing it?
I would say that it depends on what you already know from these workshops and whether you feel like you have some open questions or an insufficient understanding of the strengths and weaknesses of the website.  One quick way to answer this is to do a very fast heuristic evaluation: in 20 minutes, go through the existing site and see what you learn.  If you don't come across anything that you didn't already know from the existing documentation, you can probably stop; if you uncover new issues that weren't already known, then you should continue.  A 20-minute investment is small enough that it probably won't have a negative impact on the time that you would spend working on the new version.


Hemp socks and wicking properties

Question: I am an avid snowboarder and I have noticed that socks made of merino wool are incredibly good at wicking away moisture while keeping my feet warm. However, I was wondering about other textiles that can compare. In particular, I was curious about hemp. I am aware that hemp is very durable, but I am uncertain of its wicking properties.
Would hemp be a good alternative to merino wool and are there other textiles that can work?
Comment: Ever used hemp rope?  Rough as sandpaper is being generous.  I can't imagine wearing socks made out of it, no matter how much it wicked.
Comment: Note that hemp is illegal in some jurisdictions.
Answer: One problem with hemp compared to wool is how it conducts heat when it is wet. Wet hemp conducts heat very well when wet which means your feet could get freezing cold. Wet wool is a poor heat conductor so even if wet they will not be as cold.
Answer: I can't recommend Hempsocks, except you use them as second socks to reduce friction. They're very durable but not comfortable to wear.
A good alternative to Merinowool is Bamboo, it's also very smooth and dries very quick.
So in short, hemp is good to protect your socks or reduce friction because it's a very cheap and durable material, doesn't cost much - lasts a life. It's just uncomfortable to wear directly on your skin
Answer:
and are there other textiles that can work?

One option is of course polypropylene, it is hydrophobic (unlike most other materials used in clothing) and has a lower thermal conductivity than wool - this does not mean better insulation, but it is a safe bet it can perform fairly well over natural fibres and is fairly cheap.
You may see these as liners to prevent saturated socks from touching your feet by providing a hydrophobic buffer (although film of sweat does stay on the fibres even if not absorbed, such as with nylon blends) or as a bulkier option such as a boot sock.
I would recommend looking to this material as a possible alternative to wool, in terms of performance in cold environment over natural fibres. Nearly all natural fibres such as cotton, hemp, cellulose (bamboo/rayon etc.) perform substandard in the cold and very poorly in the wet. Some silk blends can retain insulation while wet but may end up costing more.

Hemp socks and wicking properties


Have Sterling Engines Been Used In Outer Space?

Question: I was reading about Stirling Generators having potential applications for power generation in outer space. Have Stirling engines ever been sent into space?
Answer: Stirling Engine powered refrigeration units were first tested on STS-60 (on this trip the astronauts had ice-cream)
Have a look at this document from Sunpower for more information.
Now, I see the link you have is more around the use of Stirling Engines to provide high levels of output power from the heating from radioisotopes, but the principle is the same.


GIS Server for web service

Question: I am building an app that has some GIS requirements that go a little beyond (I think) what I can retrieve from existing online web services resources so I am attempting to build a server of my own to handle these requests.
The app is going to make requests to the GIS server specifying lat, lon and I want the server to return a JSON (or XML) document back containing information such as:
Habitat : Urban|SubUrban|Rural|Forest|Old Growth Forest|Desert|Alpine
<code>Desert|Lake|River|Sea|...
City/Town:
State/Province : Initially just for the US.
Country:
Alias/Also Known As: i.e. "The Big Apple" : "New York City"
</code>
I also want the server to respond to such queries as:
<code>"Can the Atlantic Ocean be seen from present location?"
"Can Mount Adams, Washington, USA be seen from present location?"
</code>
these will be formatted and POSTed in a well defined JSON document
I would also like the solution's stack to be cloud friendly, so I can utilize Heroku or Joyent or Amazon's cloud services and get any mapping datasets from publicly available sources. I do have the need to add specific location data points to these maps as well and have been able to prototype this out using a mongo db running a simple node.js server script (more out of curiosity than utility, python may be my server code in the end) this serving location information to an iOS client.
So far, I've just been able to insert & index documents with longitude & latitude and query with a given rectangle. I'm interested in performing the more complex tasks listed above. I am not beholden to mongodb or any other technology is something else makes more sense or is more compatible.
Comment: It sound like your requirements are that:
Answer: You have lots of options.  One way to go could be GeoDjango, if I have understood your requirements correctly.

GIS Server for web service


Where's my share of the loot?

Question: How do you handle loot disputes as a DM? Say we are doing a random loot drop from a felled enemy, and there happens to be one thing that two players really want. Do you make an initiative roll to see which character finds it first? or is it better to try tailor the loot drops from the campaign to the group? Any other ideas of how to avoid these situations?
Answer: As noted, this is a player decision rather than a GM one. However, here is the system that one of my groups used to good effect in the past:
Loot is pooled until the end of the night or adventure (although particularly powerful upgrades may be lent out on a temporary basis).
Calculate the total sell value of the pool of items (that is, how much the players would receive if they simply sold everything). Add any coins found during the adventure.
Divide the total value by the number of players. This is each player's "share" of the loot.
Players can "buy" loot out of the pool for its sell value.
If two players both want the same item, they bid on it as in an auction. The winner receives the item, and divides the extra gold they paid above the item's sell value evenly among the rest of the party members.
After everything is said and done, unclaimed items are sold to cover each player's remaining share.
The GM's main concern in this scheme is to avoid getting in the way of the players. Making selling items difficult or inconsistent will undercut this method.
Comment: This is pretty close to what we do as well.  We also let players "buy on credit" so to speak for expensive items and then they owe the party the extra amount which is taken out of their share before anything else during the next loot divy.  Either way, this is a player issue and the DM should keep strictly hands off in this.
Comment: This is by far the most fair system, in the case of a dispute. It ensures everyone in the party gets equal loot. A couple of games I've been in have done it this way. This is a great way to resolve the issue if one character is getting a super powered item, and the rest of the party is left out. It's also a little bit of math and bookkeeping, so if the players are more relaxed about it, they might settle for a less complicated solution
Comment: @Macona In my experience, this isn't particularly true. Keep in mind that buying from the party pool isn't the same as buying from a book: There are relatively few items, and each item has a history (as opposed to winnowing through hundreds of generic lines of a table). And items in the pool are at a steep discount (usually 50% or better) triggering the "it's on sale!" mentality. Sure, some oddball items get sold anyway... But in other loot systems I've seen people take items "just to sell" as well.
Comment: I see a problem with this method. There are a lot of quirky magic items that players will only use if they find it as loot. Players generally buy items that give straight & obvious bonuses (weapons, AC & saving throw increasing items, etc).
They party will just sell all their quirky stuff if they have to invest their own gold into it.
Answer: This is clearly something the master has no decision whatsoever. The characters need to find an agreement on that. Let them fight. Don't get involved until you see Coke bottles flying around.
Comment: If they fight for more than a couple minutes, then make it cursed, just to teach them.
Comment: Do intervene if it halts the game completely and other players are looking bored.
Comment: @Oddysey : then the issue is different. You don't have a loot fight. you have an attitute issue of your players. This requires a out-of-game approach.
Comment: This can also turn into a really serious problem in a system like 3e D&D where equipment is an important part of character power. If someone is consistently able to win these kinds of arguments they can completely derail the campaign. The DM does have a roll in adjudicating those kinds of social issues.
Comment: I agree that this should pretty much be settled "in character."  The DM does have a responsibility if it turns into problems, and that does happen, but his first instinct should be to look at the players and say, "What do you do next?"
Comment: Years ago I used to play on a group with this kind of childish arguments (although I think it never went really bad). Now I have moved on, play with completely different people, and it's awesome. Usually there are no arguments but in a 7th sea game one guy played a greedy pirate captain who always tried to get all the money and valuable items, but in a sane way. He just made the game more funny.
Comment: @DigitalJoel:  You are my kind of devious!  Cursed just to show them.
Comment: @Pulsehead: I think it comes from having 4 kids.  Sometimes, you just have to teach them it's not worth it :)
Answer: I notice that you didn't tag this question with a D&D tag, so I'm going to add an answer that doesn't assume D&D party dynamics.
If both players want their characters to get a particular piece of loot, awesome! They have their characters in a conflict, which is what makes stories interesting. Run with it. In this case, I would turn it over to the players, managing the situation as the GM like I would any other in-game situation.
First I would set it up:

"You both grab for the Orb. Make an opposed roll with [whatever system-relevant ability] to see who gets it first."

Then depending on the results, either:

"Tie? Awesome! Both your hands grasp the Orb. You tug a bit, neither managing to overcome the other. Now what are you going to do?"
"Angela wins the roll, snatching the Orb before Brand gets there. Angela, what's your attitude toward Brand now that you've got the prize? Brand, what are you going to do about that?"

Either case might lead to a follow-on contest, deepening the conflict and the immediate story. With the uncertainty of what happens next, everyone at the table will be suddenly paying attention and engaged with the game.
The follow-up depends on what the characters are willing to do for the loot and how they already relate to each other. They might engage in a reasoned/heated argument (using a social conflict system in games that support that). They might just break out into a fight. One of the PCs might make an impassioned speech about why their character needs that item for the greater good. Bystanding PCs might intervene in any argument or fight. The conflicting PCs might ask for a impartial third PC to act as arbiter. It might be an opportunity for wider roleplay where the arguing PCs are dressed down for infighting by the other PCs or a superior.
The end result is that you, with the player's help, have turned a minor logistical point into an engaging story event, given the players the chance to roleplay their characters' motivations and methods, and made the distribution of loot an interesting matter rather than just an entry in a spreadsheet.
Comment: Awesomeness in its purest form.
Comment: My thoughts exactly!
Answer: Hopefully, you've got one PC acting as group leader.  Assuming he's trying to be fair, all loot should be evenly divided at the end of a story arc.
Of course, if you've got this, you probably wouldn't need to ask the question.  As GM, you can give gold to each player while looting an area.  This may lead to PCs trying to loot areas first and keep everything they find.  A few spot/notice checks for the other PCs can help balance this.
Finally, it's almost impossible to divide the loot perfectly evenly.  As long as it isn't too unbalanced, I just try to move on with the game.
Answer: The players should be able to work this out on their own, with minimal DM involvement.  However, as the DM is generally held to be the moderator of the game, it may occasionally fall to them to resolve disputes.  Of course, there are probably as many ways to solve this as there are DMs in the world.
Probably the most commonly-agreeable method is a d20 roll-off.  Forget Initiative or any other bonuses, and let a fair and bare roll of the die decide.  Re-roll as necessary if both die come up the same.
Another thing one might wish to take into consideration, is whether the item in question is particularly better-suited for one character over the other, or whether one of the characters may already possess an item of similar quality and purpose.
If the argument stretches to a stubborn extreme, perhaps a time may come when a PvP battle could be considered.  I've seen players leave the table over less, so it wouldn't surprise me that this might happen from time to time.
Answer: Seems to me it's an issue between CHARACTERS, not players, and thus the DM shouldn't get involved.
Comment: But characters aren't real, so it's an issue between players isn't it?
Comment: @crom, neither is the loot.
Answer: If it's a one-off incident, then yeah, let them work it out on their own. If it turns into an extended conflict that's annoying the other players, tell them to drop it for now and work it out between sessions.
If you're having consistent issues of this kind, you need to talk to your players -- and get them to talk to each other. Designing treasure around the party can help, but it's also a lot of work, and if your players really are fighting and not able to work this stuff out on their own then there's probably some other issue going on. Either they have a personal problem with each other, or they just don't have good enough conflict resolution skills to handle it on their own. Ask questions (Why do your players think the problem keeps coming up?), suggest possible solutions (like rolling off, or trading off who gets to pick first), and maybe be willing to set boundaries. (If one or more players are disrupting the game consistently, you might need to take a break, or remove those players from the game entirely. This sucks, but some people can't play well with others.)
You might also consider switching to a system or a style of game where loot isn't as big of a deal. Or even stop handing out items entirely, and just give them gold and let them shop for the stuff they want. This won't work for every game, but in some situations it might be the easiest way to deal with the problem.
Answer: There could always be an in-game solution, where the PCs actually start fighting over it.
I admit this is unlikely in a normal party, but there's always the chance that someone is playing a chaotic, or evil character, who in-game would actually fight other PCs in the party for the item in question.
I'm not sure this is a good solution, as generally you want the characters in the party to get on, but if you've got some frictional relationships in-game in the party it could add to the internal conflict.
It's a different style of playing, and perhaps requires a different style of game or role play, but (in-game) internal party conflicts can make it that bit more interesting.
Answer: Related to the social contract question, in a game where loot is important this should be part of the social contract. Agreement upfront is always a good thing.
Answer: I prefer to tailor loot drops, because I know some of the groups I play with can't sanely handle the conflict. :)
Answer: This is a little weird for me.  Not the individual question, but coupled with the answers, I almost feel like I should look to make sure I am answering on the right board.  This is about an RPG, not WoW.
One person mentioned that it is between characters, but then said the GM should not get involved....What?
When you are running a roleplaying game that invovles treasure, if there is any immersion or verimilitude involved, the Characters should have this discussion long before 'grabs of opportunity'.  And it does not need to be part of the outside, social contract, though that is a reasonable option, especially in groups with a bad history, or in a short term group.
IN most cases, however, any formal grouping of exploerers, robbers, adventurers, tomb-raiders, orc-baby killers, expeditioners, etc, would realistically come up with a distribution contract.  I think every group I have run for decades has done this as a matter of course if they leave for an adventure.
Have an NPC ask what his share of the loot is, if you need to push it a long, or a shopkeeper asking what system the group is using out of 'professional curiousity'.
Answer: The old school response it to roll wandering monster checks while the players and their characters argue or fight.
If the game does not consist of hardcore dungeoncrawling, then using whatever mechanics for social conflict resolution the system includes would be a nice response.
If the game does not have useful social conflict rules and if the issue bothers players, then talking to them is required. I'd start with suggesting some sort of equal shares with auction, as in AceCalhoon's answer.
Answer: thief and barbarian dividing the loot....

thief: oh look, Angeline Jolie's over there and her top's fallen down
barb: no... she's not.
thief: oh well, I must have been mistaken.  Right, lets divide the loot up.
barb: hmm, weren't there 4 magic swords in there earlier?
thief: look, we've told you before.. counting does not go "1, 2, erm, 4, umm 8, ahh many"
barb: ok, but what's that bulge in your backpack?
thief: umm, I'm err, pleased to see you?

In other words, divide the loot up anyway your characters can argue amongst themselves. Keep it fun and don't even try to make it 'fair'. And absolutely don't turn it into an bean-counting accounts session (seriously, I have to do my own accounts in real life, dull dull dull)
Comment: From experience this approach doesn't work with some groups, and it can lead to serious resentment.  As commented by previous answers, you have to tailor how to do this to the people in question, and I do not believe there is a one size fits all approach.
Answer: The best I can offer is that it's a player issue unless you dabble in "destined" items or items granted by a class ability/advantage (Puzzle Sword in 7th Sea or Anointed Knight's weapon in D&D 3e).  In those cases they are only special to the individual and are specifically earmarked for them.  When it comes to loot, the only fair way to divvy magic items I can recall if players are particularly greedy is being in a particularly dangerous setting.  I was in a rather difficult AD&D campaign, and it involved a lot of magic items for survival and the party needed to put the items on the best person possible just to live.
Otherwise, only DM Magic can make them share.  If they hoard more than they can use, and they steal/strongarm from the rest of the party, only something like a party curse or geas or powerful NPC in party can solve the issue effectively.
Answer: From a gamist point of view, I believe loot should be divided by how if effects the overall party. If a melee Barbarian and a switch-hitter Ranger are both arguing over a +3 sword of awesomeness, it should go to the barbarian because he spends more time in melee and the party will do more damage overall if he has it.
As for your problem though, a DM can't really control how players divide loot, but you can control what loot they find. If the Barbarian & Ranger mentioned above also find a +3 Bow of Coolness lying next to the sword, everyone will be happy.

Where's my share of the loot?


Adding column to list view in particular order

Question: I have couple of sites and each contains a list. I'd like to add Title field to default view in each of those lists but I want them to be 3rd column in the view. Since I couldn't find anything similar to the web-based editor of sharepoint list views, where one can choose in what order should columns be displayed, I'm asking for some guidance and/or help with that.
Answer: If you are looking for code solution my approach would be to remove all fields from view and then add them in desired order.
Here is one simple example:
Change The Fields Order In The Sharepoint View Programmatically
Answer: Check out the article on my blog on Difference between SharePoint List Views and List Column Ordering at http://tad.co.in/?p=588


Conditional Probability

Question: An incoming freshman Mark believes that he has a 25% chance of earning a GPA in the 3.5 to 4 range, a 35% chance of graduating with a 3.0 to 3.5 GPA and 40% chance of finishing with a GPA less than 3.0. Also, Mark has an 8/10 chance of getting into med school if his GPA is above a 3.5, a 5/10 chance if his GPA is between 3 and 3.5 and a 1/10 chance if its below 3. Based on these estimates what is the probability that Mark gets into med school?
Answer: The law of total probability says that $Pr(A)=\sum_{n} Pr(B_n)Pr(A|B_n)$ as long as the $B_n$ are disjoint and the union of all the $B_n$ is the entire sample space. Applying that to this problem gives:
$$Pr(Med\, School)=Pr(GPA>3.5)*Pr(Med\, School|GPA>3.5)+Pr(3.0<GPA<3.5)Pr(Med\, School|3.0<GPA<3.5)+Pr(GPA<3.0)Pr(Med\, School|GPA<3.0)$$

Conditional Probability


Is a compactly generated Hausdorff space functionally Hausdorff?

Question: Question is the title. I suspect the answer is no, without some further conditions (clearly, normal is sufficient). Pointers to counterexamples would be appreciated, but not necessary.
Answer: There is an example at PlanetMath of a Hausdorff space which is not completely Hausdorff / functionally Hausdorff.  On the other hand it is second-countable, hence first-countable and hence compactly generated.
Comment: Thanks! I'd seen this example, but hadn't clicked to the fact it was compactly generated.
Answer: Since you implicitly ask for sufficient conditions for a compactly generated space to be completely Hausdorff, it might interest you that one such is that the family of compacta be countably generated.  As an appendix to the negative example given by Qiaochu Yuan, any Hausdorff space which is not completely Hausdorff provides a counterexample, since a space is completely Hausdorff
if and only if the associated compactly generated space is completely Hausdorff.  Both of these facts are in the 1969 paper "Topologies et compactologies" by Henri Buchwalter.


Finding the Universal Ideal of a (Covariant) Differential Calculus

Question: Let $(\Omega,d)$ be a differential calculus over an algebra $A$. It is easy to show that $\Omega$ is always equal to a quotient of $\Omega_u(A)$, the universal calculus over $A$, by some ideal $N$ of $\Omega_u(A)$. Now in practical terms, given a presentation of $\Omega$ in terms of generators and relations, how does one find the ideal $N$? I have some confused memory of there being a concrete recipe (due to Woronowicz?) for doing this when $A$ is a Hopf algebra and the calculus is left-covariant (or bi-covariant).
Comment: Ideal of what algebra?
Comment: Edited to make clearer. Does it make sense now?
Answer: I don't know if you still care, but I think I found the answer to your question.
Look at Proposition 1 in Chapter 14 of Quantum Groups and Their Representations by Klimyk and Schmudgen.  It shows that there is a bijection between left-covariant first-order differential calculi over a Hopf algebra $H$ and right ideals of the kernel of the counit of $H$, and it shows how the relations in a first-order calculus are obtained from the ideal.  I have never worked with these things, so I don't know how tractable the computations are, but as far as I can tell from quickly scanning it seems that all the maps are at least explicitly defined.
As for higher order calculi, I am not sure whether/how these results extend.  But at least maybe this is a good start?
Answer: I don't much about differential calculi, but back in my head I also remember somewhat like this....so as a HINT (?): Could N be the kernel of something like a "quantum shuffle map" or "quantum symmetrizer"? Then chances are you are talking about a Nichols algebra....and then one had a good description of N but NOT easily explicit relations....


Crystalline Characters

Question: Let $K$, $L$ be finite extensions of the $p$-adic numbers. Suppose $\chi:G_K\rightarrow L^{\times}$ is crystalline. It is my understanding that if either $K$ or $L=\mathbb{Q}_p$, then $\chi$ must be a Tate twist of an unramified character. Is there a classification of crystalline characters in general?
Answer: Hi Kevin,

Go to http://math.stanford.edu/~conrad/
Download "Grunwald--Wang for global character groups"
Read appendix A, especially prop A.3

The answer (note that $K$ and $L$ are switched in Brian's paper) is that once you've identified your character as a character of $K^\times$ via local class field theory, it should be "algebraic" on $O_K^\times$.
Comment: Update: The relevant statements are now in Appendix B of "Lifting global representations with local properties."
Comment: Aha, so the situation is completely analogous to the Hodge-Tate case. Thank you for the reference!
Answer: [EDIT: I misunderstood the question -- I thought the poster wanted to know why all crystalline characters are unramified twists of powers of cyclotomic when $L$ or $K$ is $\mathbb{Q}_p$, and wrote out a detailed proof. I realise now that the poster already knew this but wanted to understand the classification in the general case. I thought I'd leave this post here anyway in case anyone finds it useful.]
Step 1: Consider the space $\mathbb{D}_{\mathrm{dR}}(\chi)$. This is a free rank 1 module over $K \otimes_{\mathbb{Q}_p} L$, and the steps in the Hodge filtration are $K \otimes_{\mathbb{Q}_p} L$-submodules. By assumption, $K \otimes_{\mathbb{Q}_p} L$ is a field, there is only one Hodge-Tate weight. By twisting by a power of the cyclotomic character, we can assume without loss of generality that the Hodge-Tate weight is 0.
Step 2: Now consider the space $\mathbb{D}_{\mathrm{cris}}(\chi)$. This is a free rank 1 module over $L K_0$, with a Frobenius that is L-linear and $K_0$-semilinear. (Here $K_0$ is the maximal unramified subfield of $K$). If $[K_0 : \mathbb{Q}_p] = d$, then $\varphi^d$ is linear, so acts as multiplication by a scalar $\mu \in L$.
Step 3: Consider the unramified character $G_{K_0} \to L$ mapping geometric Frobenius to $\mu^{-1}$. Tensoring with this character, we may assume that $\mu = 1$.
Step 4: Now choose a basis of $\mathbb{D}_{\mathrm{cris}}(\chi)$ over $LK_0$, and suppose that (with respect to this basis) $\varphi$ acts as multiplication by $\alpha$. The condition that $\varphi^d = 1$ translates to $\alpha \varphi(\alpha) \dots \varphi^{d-1}(\alpha) = 1$. If $K = \mathbb{Q}_p$, then this is just $\alpha = 1$. If $K \ne \mathbb{Q}_p$, then $L = \mathbb{Q}_p$, and $\alpha \in K_0^\times$ satisfies $N_{K_0 / \mathbb{Q}_p} \alpha = 1$. Hence $\alpha = \varphi(x) / x$ for some $x \in K_0^\times$, by Hilbert 90; and after changing basis by $x$ we again have $\alpha = 1$.
So we've shown that $\mathbb{D}_{\mathrm{cris}}(\chi)$ is isomorphic (as a filtered $\varphi$-module) to $\mathbb{D}_{\mathrm{cris}}$ of the trivial character, i.e. $\chi$ is trivial (and hence crystalline!). QED.
Answer: By definition, any homomorphism from $G_K$ to an abelian group factorizes through $Gal(K^{ab}/K)$; by the local class field theory the latter group is the completion of $K^\times$. Hence it suffices to calculate $Hom (\overline{K^\times},L^\times)$. Lastly, one can note that up to torsion $K^\times\cong \mathbb{Z}_p^{[K:\mathbb{Q}_p]+1}$ and $L^\times\cong \mathbb{Z}_p^{[L:\mathbb{Q}_p]}\bigoplus \mathbb{Z}$ (via the corresponding logarithms).
Comment: It's not quite obvious (to me) from what you've written how this answers the question: which elements of this hom space correspond to crystalline characters?
Comment: (Also, I think you mean $\overline{K^\times} = \mathbb{Z}_p^{[K : \mathbb{Q}_p]} \oplus \overline{\mathbb{Z}}$, where the bar denotes profinite completion.)
Comment: Yes, you are right.

Crystalline Characters


How do 3D game engines render 3D environments to a 2D screen?

Question: For 2D sprite games, I can understand that sprites move on the X and Y axis. However, for 3D games, how do game engines calculate how to render a 3D environment on to a flat 2D computer monitor?
Comment: Math, lots of it.
Comment: Magic Imps design the land and then shoot the image onto the screen using their magic wands.
Comment: `Math.PI`, lots of it.
Answer: It's similar to how you can capture 2D pictures of our world on camera. The 3D environment is described in terms of X, Y, and Z coordinates, but ultimately you need to map that to a 2D screen with only X and Y axis. This is done by projection, where each 3D point is projected onto a 2D view plane.
To put it simply, you can imagine projection lines going from the virtual camera in 3D space to each 3D point by passing through the view plane. The point where a projection line intersects the 2D plane is the projection of the 3D point. In orthographic projection these projection lines are all parallel; the distance between the object and the camera doesn't make any difference for the projected shape. This type of projection isn't realistic but is useful in certain instances such as drawing 2D graphics using a 3D engine (the HUD for example). The other type of projection is perspective projection where these lines converge making more distant objects appear smaller and giving more realistic results. This page has some figures that could help you visualize the process.
Projection is one of several transformation objects undergo in the graphics pipeline, and is usually done using a projection matrix. Note that although projection implies discarding the Z coordinate, in reality a transformed Z value is still used a measure of screen depth so that we can tell which objects are visible and which ones are occluded by other objects. This is known as z-buffering.
Answer: 3D sprites get a new Z axis that stores depth, scaling size and movements in the X/Y plane. A sprite that is farther away has a bigger Z value. Now, divide sprite's size and X/Y axis movements by the Z axis. As others have noted, this is a 3D perspective projection. Here is the best illustration I could find:

This makes sense conceptually because things that are farther away look smaller. Their apparent movements also look smaller.
In addition, near objects need to be drawn on top of far objects. Again, this makes sense conceptually because near objects occlude the view of objects behind them.
These techniques can be implemented at the sprite level. However, for the most realistic projection each sprite has a 3D model and is rendered at the pixel level.
Comment: The diagram was what got to me. A picture is worth a thousand words.
Answer: It's all about projection, a geometric operation that lets you define a space from another space of superior dimension (basically, in your case, you're describing a 2D image from a 3D environment).
Simply put, this is how 3D rendering engines generate an image (given a 3D scene & a projection) :
For each pixel of the resulting image

Compute the looking direction (according to the projection), it's basically a 3D vector
Find the nearest object in your scene that is "seen" in this direction (simply guess what objects are intersected with the vector, and keep the nearest one)
Compute the color of this object, you have the color of the pixel !

Of course this is a simple model, I'm not talking about reflective/refractive/translucent objects of special effects such as smoke...
Answer: The Matrix. Or rather, lots of matrix math.
It's scary stuff for the uninitiated. There's typically three 4x4 matrices involved in turning a bunch of 3D coordinates in space into 2D coordinates on the screen (including the depth into the screen as Z).
3D matrices are a set of 16 floating point values arranged in a 4x4 grid. Algorithms are used to generate the required values and then using matrix-vector multiplication, these numbers transform a 3-dimensional vector (X, Y, Z).
<code>[ 1, 0, 0, 0 ]
[ 0, 1, 0, 0 ]
[ 0, 0, 1, 0 ]
[ 0, 0, 0, 1 ]
</code>
This is the identity matrix. It essentially does nothing to a vector's coordinates.

Model
The first is the world or model matrix. This matrix takes the vertices in an individual model (such as a crate) modelled around the origin (0, 0, 0) and transforms them into world vertex coordinates. This might include scaling the object, rotating it around it's origin and finally translating it across to where it's located in the scene.

View
The second matrix is the view matrix. This takes the world coordinates and transforms them so that they are within the context of the view. The concept of a camera is typically used to generate this matrix.
The camera usually contains a position vector, a direction or target vector, and an up vector. This up vector describes the 'spin' of the camera.

These vectors are used to generate what's called a look at matrix.
The affect this has is similar to the world matrix in the fact that it is essentially rotating and translating the world around the camera to "position" the camera.
A camera that is 10 units along the X axis will translate world coordinates 10 units in the other direction.

Projection
The final piece of the puzzle is the projection matrix. In a camera-like perspective projection, the matrix transforms the almost screen coordinates to give the illusion of a perspective with a field of view of x degrees.
If you think in terms of corners of the screen (for a 640x480 game), this is how projection "aligns" the coordinates along the Z axis within the 2D coordinates of the screen. For orthographic projection, no scaling takes place along the Z axis into the screen. For perspective, the further away the object, the smaller it becomes in relation to the screen's extents.


Probability density of detection of collinearly emitted photons in two detectors

Question: Update:
As proposed by @dmckee, I added equation numbers and improved the display of some equations.
The answer by @Trimok inspired me to look at coordinate systems which are not specific to the detector locations, but more general, and this indeed made things clearer since $\delta(\cos \theta + 1)$ is a $\delta(\cos \theta + \cos 0)$ in disguise.
We have a relation between a Dirac delta of a ray in spherical coordinates, $\delta^2$, and a Dirac delta of a point, $\delta^3 = \delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0)/(r^2 \sin \theta)$, such that
$\delta^2 = \int_0^\infty \delta^3 \operatorname{d}r_0 = \delta(\theta-\theta_0) \delta(\phi-\phi_0) / (r^2 \sin\theta) \tag{E1}$
(Using this definition, one can easily show that the sifting property of $\delta^2$ holds and the result is a line integral of some $f$.)
The major problem (of understanding) which I am having is now reduced to this: We have two expressions for $\delta^3$, which are
$\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0)/(r^2 \sin \theta) \tag{E2}$
and
$\delta(r-r_0)\delta(\theta-\theta_0)\delta(\phi-\phi_0)/(r_0^2 \sin \theta_0) \tag{E3}$
These are totally symmetric in $x$ and $x_0$ (note that $\delta$ is an even function).
However, for $\delta^2$ we have
$\delta(\theta-\theta_0)\delta(\phi-\phi_0)/(r^2 \sin \theta) \tag{E4}$
and
$\delta(\theta-\theta_0)\delta(\phi-\phi_0)/(r^2 \sin \theta_0) \tag{E5}$
which are not symmetric any longer. So what makes the $r$ stand out compared to $r_0$?
One explanation I worked out with the help of a colleague is directed towards the fact that there is no $r_0$ after integration, so having any $r_0$ would "overdetermine" the set of rays through the origin. Does anyone have a better (for example, more intuitive) explanation why the symmetry is lost here?
Thank you!
Original question:
This is a long one. I have been working on this for quite a while now, so I summarized most of what I found out. I am stuck at a certain point, however, and it would be great if someone could help me out.
Assume the following situation: Two perfectly collinear photons $p_1$ and $p_2$ are emitted in the origin of some Cartesian coordinate system. We consider two infinitesimal detectors $d_1$ and $d_2$, with surfaces $\operatorname{d}\!A_1$ and $\operatorname{d}\!A_2$ directed towards the origin so that we do not have to deal with angles. The solid angles of these detectors be $\operatorname{d}\!\Omega_1$ and $\operatorname{d}\!\Omega_2$.
I now express some basic detection probability densities, where $p_id_i$ means that "$p_i$ is emitted towards $d_i$".
$\frac{\operatorname{d}\!P(p_1d_1)}{\operatorname{d}\!\Omega_1}=\frac{1}{4\pi} \tag{1}$
$\frac{\operatorname{d}\!P(p_1d_1)}{\operatorname{d}\!A_1}=\frac{1}{4\pi (r_1)^2} \tag{2}$
$\frac{\operatorname{d}\!P(p_2d_2)}{\operatorname{d}\!\Omega_2}=\frac{1}{4\pi} \tag{3}$
$\frac{\operatorname{d}\!P(p_2d_2)}{\operatorname{d}\!A_2}=\frac{1}{4\pi (r_2)^2} \tag{4}$
I think there is no doubt that these probabilities are well-defined and the integrated detection probabilities in a spherical $4\pi$-detector equals 1, no matter what its radius is. So a photon is surely emitted towards a detector surrounding the origin.
Now, for some reason we want to calculate $P(p_2d_2 | p_1d_1)$, the conditional probability of $p_2$ emitted towards $d_2$, given that $p_1$ has been emitted towards $d_1$. Due to collinearity of the photons, the individual events are not independent: $P(p_2d_2 | p_1d_1) \neq P(p_2d_2)$.
I now propose that
$\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!\Omega_2} = \frac{\delta(\cos \theta_2(d_1)+1)}{2\pi}, \tag{5}$
where $\theta_2(d_1)$ is the polar angle of the position of $d_2$ expressed in spherical coordinates with polar axis from the origin through $d_1$. $\delta$ is the Dirac function, so obviously, this expression equals 0 if $\cos \theta_2(d_1) \neq -1$, that is, when $\theta_2(d_1) \neq \pi$.
We can verify that the integral of this probability over a spherical $4\pi$-detector $d_2$ equals 1: with $\operatorname{d}\!\Omega_2 = \operatorname{d}\!\cos \theta_2(d_1) \operatorname{d}\!\phi_2(d_1)$, we find
$\int \frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!\Omega_2} \operatorname{d}\!\Omega_2 = \int \frac{\delta(\cos \theta_2(d_1)+1)}{2\pi} \operatorname{d}\!\Omega_2  \tag{6}$
$= \int_0^{2\pi} \frac{1}{2\pi} \operatorname{d}\!\phi_2(d_1) \cdot \int_0^\pi \delta(\cos \theta_2(d_1)+1) \operatorname{d}\!\cos \theta_2(d_1) = 1. \tag{6}$
So $p_2$ is still surely emitted towards a detector surrounding the origin, no matter where $p_1$ is emitted towards.
Further, we would like to verify that the law of total probability is fulfilled by integrating $P(p_1d_1 \wedge p_2d_2) = P(p_2d_2 | p_1d_1) \cdot P(p_1d_1)$ over a spherical $4\pi$-detector $d_1$. This works regardless of which formulation of $\operatorname{d}\!P(p_1d_1)$ (with respect to $\operatorname{d}\!\Omega_1$ or $\operatorname{d}\!A_1$) we use -- I chose $\operatorname{d}\!A_1 = (r_1)^2 \operatorname{d}\!\cos \theta_1(d_2) \operatorname{d}\!\phi_1(d_2)$ here.
Note, however, that integration over $d_1$ changes the polar axis used in the formulation of $\operatorname{d}\!P(p_2d_2 | p_1d_1)$ above, so we need a different, but equivalent formulation. This is more than a change of coordinate system (from one polar axis to another), because $\theta_2(d_2)$ would not help us a lot (it's 0). Instead, one can use
$\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!\Omega_2}=\frac{\delta(\cos \theta_1(d_2)+1)}{2\pi}, \tag{7}$
which uses both a different variable in the argument of the $\delta$ as well as a different coordinate system: $\theta_1(d_2)$ is the polar angle of detector $d_1$'s location in spherical coordinates with polar axis from the origin through $d_2$. Note that both expressions of $\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!\Omega_2}$ are symmetric in the indices 1 and 2.
Finally, we arrive at
$\int \frac{\operatorname{d}\!P(p_1d_1 \wedge p_2d_2)}{\operatorname{d}\!A_1 \operatorname{d}\!\Omega_2} \operatorname{d}\!A_1 = \int_0^{2\pi} \int_0^\pi \frac{1}{4\pi (r_1)^2} \cdot \frac{\delta(\cos \theta_1(d_2)+1)}{2\pi} (r_1)^2 \operatorname{d}\!\cos \theta_1(d_2) \operatorname{d}\!\phi_1(d_2) \tag{8}$
$= \frac{1}{4\pi} \int_0^{2\pi} \frac{1}{2\pi} \operatorname{d}\!\phi_1(d_2) \cdot \int_0^\pi \delta(\cos \theta_1(d_2)-1) \operatorname{d}\!\cos \theta_1(d_2) = \frac{1}{4\pi} = \frac{\operatorname{d}\!P(p_2d_2)}{\operatorname{d}\!\Omega_2} \tag{8}$
So, the law of total probability is fulfilled: the probability of $p_2$ being emitted towards $d_2$ can be partitioned into all conditional probabilities of $p_2$ being emitted towards $d_2$ given that $p_1$ has been emitted towards $d_1$, weighted by the probability that $p_1$ is actually emitted towards $d_1$.
So far, so good. Congratulations for reading up to this point :)
As expressed above (and verifiable relatively easily), it does not matter which formulation of $\operatorname{d}\!P(p_1d_1)$ I use -- both $\operatorname{d}\!\Omega_1$ and $\operatorname{d}\!A_1$ work fine. However, when I try to base my calculations on $\frac{\operatorname{d}\!P(p_2d_2)}{\operatorname{d}\!A_2}$ instead of $\frac{\operatorname{d}\!P(p_2d_2)}{\operatorname{d}\!\Omega_2}$, hence, I try to calculate $\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!A_2}$ instead of $\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!\Omega_2}$, I arrive at the following problem.
So I started with
$\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!A_2}=\frac{\delta(\cos \theta_2(d_1)+1)}{2\pi (r_2)^2}, \tag{9}$
and (6) is still easily verified (integrating over $\operatorname{d}\!A_2$) since the two additional $(r_2)^2$ simply cancel out.
However, working out (8) is now more complex when it comes to changing from one variable/coordinate system to another. I figured that
$\frac{\operatorname{d}\!P(p_2d_2 | p_1d_1)}{\operatorname{d}\!A_2}=\frac{\delta(\cos \theta_1(d_2)+1)}{2\pi (r_2)^2}  \tag{10}$
is the expression which fulfills total probability (I can show this, if necessary), but now I wonder why this is not symmetric with the first expression in the indices 1 and 2 (as it was above). Specifically, I wonder where the connection to a Dirac of a ray along Oz in spherical coordinates is, which is $\frac{\delta(\cos \theta - 1)}{2\pi r^2}$ (compare http://www.physicsforums.com/showthread.php?t=215263)
More generally, I am working on expressing the probability density in 3D, so I expect to obtain something like
$\operatorname{d}\!\Omega_2 \cdot \delta_{L(x_1, x_2)}(x_0)$ or $\operatorname{d}\!A_2 \cdot \delta_{L(x_1, x_2)}(x_0)$,
where $L(x_1, x_2)$ is the line segment from $x_1$ and $x_2$, and $x_0$ is the point of emission.
So my specific questions are:

Why are the expressions involving $\operatorname{d}\!A_2$ not symmetric, while those involving $\operatorname{d}\!\Omega_2$ are?
Specifically, where is the connection to the Dirac of a ray in spherical coordinates, which involves an $r^2$ and which I would expect to be changed when converting from one coordinate system to another.
How (else) can I obtain a 3D probability density involving the three locations, which can be used to integrate over all three volumes in order to describe the number of coincidences detected from emissions from a volume in two finite detectors?
Answer: This is not an answer, but a suggestion. I am maybe completely mistaken, but it seems to me that you use a huge number of variables which should be simplified.
My feeling is that, putting $r_1=r_2=1$, your $dA_1$ and  $dA_2$ are simply disguised $ d\Omega_1$ and $d\Omega_2$, and that $p_1, d_1$ is the same thing that $\Omega_1$ and that $p_2, d_2$ is the same thing that $\Omega_2$
I consider here that the only interesting variables are $\Omega_1$ and $\Omega_2$ (so I worked at fixed energy for the photons, with $\vec p_1 = - \vec p_2$), where $\theta_2$ is defined relatively to  a $z$_axis opposed to the one  used for the definition of $\theta_1$ (to simplify the $\delta$ formulae).
If this is correct, one could write :
$$1 = \int ~d^2P(\Omega_1, \Omega_2)=\int ~dP(\Omega_1/\Omega_2)~dP(\Omega_2)\\=\int d\Omega_1 d\Omega_2~\frac{dP(\Omega_1/\Omega_2)}{d\Omega_1}~\frac{dP(\Omega_2)}{d\Omega_2}  \tag{1}$$
By interverting $1$ and $2$, we get also:
$$1 = \int d\Omega_1 d\Omega_2~\frac{dP(\Omega_2/\Omega_1)}{d\Omega_2}~\frac{dP(\Omega_1)}{d\Omega_1}  \tag{2}$$
Co-linearity, normalisation, and $\frac{dP(\Omega_1)}{d\Omega_1} = \frac{dP(\Omega_2)}{d\Omega_2} =\frac{1}{4 \pi}$ imply :
$$\frac{dP(\Omega_1/\Omega_2)}{d\Omega_1} =\frac{dP(\Omega_2/\Omega_1)}{d\Omega_2}= \delta(\Omega_1 - \Omega_2)\tag{3}$$
with $\delta(\Omega_1 - \Omega_2) = \delta(\phi_1 - \phi_2) \large \frac{\delta(\theta_1 - \theta_2)}{sin \theta_1}$
Comment: Thank for you for this suggestion. It has a lot of valid points, although my main point (the asymmetry with respect to the question which $r^2$ should be in the denominator) is kind of circumvented (instead of answered) by setting all $r=1$.
What is a very nice point is the idea of using a fixed coordinate system instead of my two systems rotated towards $d_1$ and $d_2$, respectively. I will try to use this idea tomorrow to simplify my original question and focus on the $d \Omega$ / $d A$ issue.


How to make the lock screen look like the login screen?

Question: Is there any way to make the login screen from a locked session (either by using Ctrl+Alt+L or from a screen timeout) look like the login screen when you're starting a new session? I hate that everything on the lock screen aside from the password input is black.
Comment: It's worth noting that in 11.10/Oneiric/GNOME3/whatever that the lock screen now uses your desktop background instead of a black screen.
Comment: In Gnome 3 things are a bit more complicated. Lock screen do use desktop background... But how to change it except replacing image files or editing xml's?
Answer: GDM
You can't make the lock screen look like the login screen.
However, you can use a theme for the lock screen. There are quite a few available at gnome-look.org. Here's an example.
Answer: GDM
The answer may be a little bit late, but aside from setting the theme of the actual unlock dialog you can also set the background (i.e. the black space around the dialog) to whatever image you like. Just run the following command in the terminal:
sudo gconftool-2 --direct --config-source xml:readwrite:/etc/gconf/gconf.xml.defaults --set /desktop/gnome/background/picture_filename --type string /usr/share/backgrounds/Holes_by_FireCobold.jpg

Replace the last path in the command with the path to the image you desire to use. Do not forget to restart your desktop session (or try to restart gconfd-2 by killing it, at least).
EDIT: unless you configured the background of the login screen (e.g. via Ubuntu Tweak), the above command will also change it.
Comment: I've searched for a solution applicable to Gnome 3. Unfortunately this one isn't.
Answer: LightDM
This was planned 12.04 -here is the bug report.  However, it did not make the final release.

How do I subscribe to a bug?
Answer: If you are somewhat proficient at C and a little bit of GTK, it's possible to have the lock screen look like (but not be!) the login screen -- since your main concern seems to be the ugliness of the lock box?
You'd do this by modifying the sources of the gnome-screensaver package, which provides the lock screen.
For tips, please see the How did you figure this out? section at the end of this answer. The original question was about repositioning the lock-box to the bottom-left instead of the center. For what you want, you'd move it to the left, make the grey background transparent, and change the sizes of the username and "Password:" text appropriately.
From my experience, the code you need to change is most probably only in two files: gs-lock-plug.c and gs-windows-x11.c.
If you try this and have any questions, please ask in a comment and I'll try my best to point you in the right direction.
Answer: Light DM
This tweak works by replacing call to OnScreen Virtual Keyboard with that of LightDM. A better tweak would be to replace the call to Gnome ScreenSaver Lockscreen(Ubuntu's default LockScreen till Ubuntu 13.10) with one to Unity Greeter(Ubuntu's default Login Screen), but I couldn't find any way to do that.

Method:
Open Dconf Editor,select the schema org.gnome.desktop.screensaver,edit the variable embedded-keyboard-command and replace it's contents with
dbus-launch gdmflexiserver -xnest

, and then check embedded-keyboard-enabled for the same schema.
Shortcut Commands:
For the Terminal lovers and lazy ones out there,you can do the above settings with just 2 commands:
gsettings set org.gnome.desktop.screensaver embedded-keyboard-command "dbus-launch gdmflexiserver -xnest"
gsettings set org.gnome.desktop.screensaver embedded-keyboard-enabled true

From now onwards, Ubuntu Login Screen (Unity Greeter) appears every time the system wake from either Screensaver (Lock Screen) or Suspend.

This method was provided by alina on the Ubuntu Discourse thread Wake from screensaver directly to the login screen. I'm using Ubuntu 12.04 LTS and this method works fine. May work on 12.10,13.04 and 13.10 too.
A limitation with this method is that you'll see the Gnome Screensaver LockScreen for a second before seeing Unity Greeter. Also, as Unity Greeter isn't designed for locking screen, it's not as fast and fluid as the default Lock Screen. Ubuntu 14.04 LTS will find a solution to these problems by using Light Locker instead of Gnome Screensaver LockScreen.


How big should a gap be between a built-in oven and a shelf for mounting a microwave?

Question: I want to put a microwave on a shelf above a built-in oven.
What gap should I leave between the top of the oven and the shelf?
Comment: What are the model numbers for both the oven and the micro?
Comment: Do you mean an [Oven](http://momsgoinggreenblog.com/wp-content/uploads/2010/09/4080steam_oven_open.jpg), or an [Oven](http://www.jerrysappliancerepairnova.com/wp-content/uploads/2012/10/oven-repair-virginia.jpg)?
Comment: is the microwave a counter top model or meant to be wall mounted?
Answer: All appliances should list minimum clearances in the technical specifications, make sure you follow these.  There is no general rule of thumb as each appliance is different.  Some don't require any clearance and others requires a lot more then you'd expect.
Answer: I assume your oven includes the cooktop (burners). Checking several appliance makers websites the general rule is 30 inches from the cooktop to the bottom of the cabinet. There are differences between various makes and models so check with the maker for exact dimensions. Since your shelf will act like a cabinet bottom I would use that as a guideline. You will notice that built-in microwaves generally require only 14 inches of clearance from the bottom of the microwave to the cooktop. The built-in microwave has a metal and heat resistant bottom panel. Your shelf assuming it is made out of wood is combustable and would require the 30 inch clearance.
Answer: Most counter top microwaves have an air gap below them and an internal fan to remove heat.  Unless the oven is very poorly insulated, the temperature above the oven shouldn't get too hot to put your hand comfortably. If it does, its too hot, period.
To answer your question directly, no gap other that what the microwave has, is necessary.


When *should* I use passive voice?

Question: Is there ever a good time to use passive voice in nonfiction?
Comment: related: http://writers.stackexchange.com/questions/523/should-i-avoid-using-a-passive-voice and meta discussion at http://meta.writers.stackexchange.com/questions/260/should-these-two-questions-be-merged
Answer: You can effectively use passive voice when the action is more important than the actor.
For example, if you were writing about the effect of pollutants on a river, you might introduce the facts in the passive voice, because in terms of the effects, it doesn't MATTER who did the action:
"The pollutants were released into the lake in 1979.  Significant amounts were detected in the last test of the area, in 2010."
If this was an expose on the SOURCE of the pollutants, maybe you would want to avoid the passive voice, and say:
"Company X, Company Z, and Farmer B released the pollutants into the lake in 1979.  While the govnerment said that it was all cleaned up immediately, independent labs detected significant amounts in the area in 2010."
Or there's the old "John Smith found the body on the courthouse steps," vs. "The body was found on the courthouse steps."  If John Smith is just a passerby, he's not really important to the story, and it's distracting to have him in there.
I think there's also a tradition of using passive voice in lab reports, to make things seem less personal.  "I turned the burner on" vs "The burner was turned on."  Again, I guess it's trying to de-emphasize the importance of the actor.  It doesn't matter WHO turned the burner on; the important thing is the act.
Answer: I agree with Kate, it is best used when you are trying to illustrate something where the actors are less important than the action being performed.
The best analogy I can come up with is "cold and impersonal". So it's probably best used when writing reports.
But, like Kate said, if you need to place emphasis on the actor(s) involved, then it is better to use the active voice. Then it becomes more a case of "They did it" instead of "Something happened".
Answer: I've struggled with this in my technical writing. When I read a technical article entirely in the passive voice, it feels like the writers are either: a) just doing what everyone else does, or b) trying to sound more authoritative than they are. In either case, it certainly comes across as dry.
On the other hand, if I'm writing a paper on something I alone did, it sounds egotistical to say, "I examined the data and found...". I much prefer "We examined the data and found..." if it's a team effort, or "When we look at the data..." using "we" to refer to myself and my readers.
Still wrestling with that one...

When *should* I use passive voice?


Resuming workouts after 2 months. Any beginner workout suggestions?

Question: I had to take a break from working out for 2 months.
When resuming after a long break, are there any good alternative workout plans that you follow to gain back some of the lost strength and stamina before they get back to your regular workout schedules? Or do you just do the same routines with lighter weights? Thanks.
Comment: Just to update, I'm starting out really slow, doing my routine workout with light weights. Will gradually increase after 2-3 weeks. Thanks, all.
Answer: Go very light with your weight.
Don't jump in there and try to push yourself to doing the weights you used to do, even with a spotter. If you do, you're going to be sore as hell for a very long time.
Do each body part in the same routine like you would before (chest on Monday, back on Wednesday, legs on Friday - or whatever), but go very light.
The second time you work that body part (next week, for example), go moderate with that weight.
Don't try to pop blood vessels in your forehead until you are sure your body isn't going to backfire on you.
The older you are, the longer it takes to get back into the swing of things.
Answer: I would recommend keeping with the same routine and reducing intensity through lighter weights or time (aerobic), put a little more time in warming up and cooling down and don't hesitate to push yourself a bit.  Typically you don't lose much strength/stamina within 2-3 months - but a lot has to do with why you took a break and how active (or not) you were during that time.
Comment: Agreed! Depending on your previous work capacity in the weight room and training in general he might bounce back to his previous shape within 2-3 weeks.  Concept called "old man's strength", where the body remembers a specific weight, body composition.  The human body is essentially a habit machine and that's what makes it hard to do technical improvements in sports mainly because you want to go back to old habits...hence athletes train 4-5 hours a day trying to overcome their techniqual habits.
Answer: My own rule of thumb for strength training is to start 2 months earlier - that is, in your example, the break period - in the workout routines and to proceed from there with a 1-2 week reboot period. After that and only if it feels ok, I try to proceed at the levels when the break period started. So I guess I am leaning towards the second part of your question, to stay with the routines but have lighter weights.
My own experience is that break periods should be part of the training cycle, it is natural to get bored and have other priorities from time to time. I see the reboot routines as opportunities to work on techniques for my favorite exercises (deadlift, squats and military press).
Comment: Doing just that. Thanks. Interesting point you make on including break periods in the training cycle.
Answer: I've taken lots of breaks over the last 45 years, some of them measured in years. You didn't say why you took the break so the following might not apply now but it will.
Every work out doesn't have to be part of a scientifically programmed routine. A random trip to the gym every week or so just 'messing around (with good form)' with some favorite exercises depending on time and energy might have eliminated this question completely. No gym? Try weird single leg exercises and one arm pushups - use furniture for support or leverage. This is not a criticism, just information. The fact that you didn't quit because you fell off the wagon tells me you'll do fine.
Answer: My basic suggestion is listen to your body, don't overdo it at first.  Workout for a week or two to get your body back into the routine before you start hitting it hard.


$f:(0,\infty)\rightarrow (0,\infty)$ is continuous, $\lim_{n\rightarrow\infty}f(nx)=0$ then $\lim_{x\rightarrow\infty}f(x)=0$

Question: Could any one tell me how to solve this one?
$f:(0,\infty)\rightarrow (0,\infty)$ is continuos, $\lim_{n\rightarrow\infty}f(nx)=0$ then $\lim_{x\rightarrow\infty}f(x)=0$
I have no clue how to solve this one. Thank you.
Comment: Is that limit for any (presumably positive) $x$?
Comment: This is a dupe (or probably I saw the dupe on MO).
Comment: for any $x$ I hope,
Answer: Lemma: If $U\subset(0,\infty)$ is an unbounded open set, then there exists an $x>0$ such that the sequence $(nx)_{n\geqslant0}$ falls into $U$ infinitely many times.
Using that lemma, you should be able to prove the contrapositive of what
you want: if $f(x)$ does not tend to $0$ when $x\to\infty$, then there exists an
$x>0$ such that $f(nx)$ does not tend to $0$ when $n\to\infty$.
Now it remains to prove the lemma. Consider the set
$$ X = \{x>0 : \{ nx \}_{n\geqslant0}\cap U~\text{is infinite} \}. $$
Assuming that $U$ is an unbounded open set, we want to prove that $X$ is not empty. Let's rewrite $X$ as follows:
\begin{align}
X &= \{x>0 : \forall N>0,~\exists n>N,~nx\in U \} \\
  &= \bigcap_{N>0}\bigcup_{n>N}\frac 1n U,
\end{align}
where $\frac 1n U$ is the set $\{u/n:u\in U\}$.
Now for any $N>0$, the open set $U_N=\bigcup_{n>N}\frac1nU$ is dense. To prove that, take some $y\in(0,\infty)$ and an $\varepsilon>0$, and let's find a $u\in U$ and an $n>N$ such that $|u/n-y|<\varepsilon$, or equivalently $|u-ny|<n\varepsilon$. The idea is to look at the intervals $I_n=(ny-n\varepsilon,ny+n\varepsilon)$, and to show that at some point one of them has to meet $U$. That's the case because if $n$ is large enough, two consecutive intervals $I_n$ and $I_{n+1}$ will start to overlap, so that $\bigcup I_n$ will cover a whole interval $[a,\infty)$.
More precisely, choose some $n_0>N$ such that $n_0>2y/\varepsilon$. Then $V=\bigcup_{n>n_0}(ny-2y,ny+2y)$ contains the interval $[n_0y,\infty)$, and since $U$ is unbounded, there is a point $u\in U\cap[n_0y,\infty)$. Now $u$ is in $V$, so there exists an $n>n_0$ such that $|ny-u|<2y$, which implies
$$ \left|\frac{u}{n}-y\right| < \frac{2y}{n} < \frac{2y}{n_0} < \varepsilon. $$
Finally, Baire's theorem allows us to conclude that $X$ is dense, so in particular it's not empty.
Comment: Good solution. Baire's theorem is essential.
Answer: That's a very hard problem. I got the proof from somebody and now, am just recalling. I'll write up the most significant part. Note that the condition of $f(x)\in(0,+\infty)$ is redundant.
$\newcommand\abs[1]{\left|#1\right|}$
It suffices to prove that, $\forall\epsilon>0,\exists b>a>0,\exists N,\forall x\in[a,b],\forall n\ge N: \abs{f(nx)}\le\epsilon$.
Suppose $\exists\epsilon_0>0,\forall b>a>0,\forall N,\exists x\in[a,b],\exists n\ge N: \abs{f(nx)}>\epsilon_0$. Fix $\epsilon_0>0$, and since $\lim_{n\to\infty}f(nx)=0$, we can suppose $g(x)=\max\left\{\;n\in\mathbb Z^+\;\big|\;\abs{f(nx)}>\epsilon_0\;\right\}$ (if there's no such $n$, $g(x)=0$, that's negligible). The condition is equivalent to: $g$ is unbounded on every non-empty interval.
Let $a_0=1$ and $b_0=2$, we have $g(x)\ge0$ on interval $[a_0,b_0]$. We'll construct intervals $[a_n,b_n]$ inductively (logically, the axiom of dependent choice is needed, I think, because we'll do infinite many choices), such that

$[a_0,b_0]\supset[a_1,b_1]\supset\cdots$
$g(x)\ge n$ on $[a_n,b_n]$, for $n=0,1,\cdots$

It's done for $n=0$. Suppose the intervals for $n=0,\ldots,m$ are constructed, let's construct one for $n=m+1$.
Since $g$ is unbounded on $(a_m,b_m)$, $\exists x_0\in(a_m,b_m): g(x_0)\ge m+1$, hence $\exists n\ge m+1: \abs{f(nx_0)}>\epsilon_0$. Since $f$ is continuous, $\exists\delta>0,\forall x\in[x_0-\delta,x_0+\delta]\subset[a_m,b_m]: \abs{f(nx)}>\epsilon_0$, therefore $g(x)\ge n\ge m+1$ on $[x_0-\delta,x_0+\delta]$. Let $a_{m+1}=x_0-\delta$ and $b_{m+1}=x_0+\delta$, constructed.
By Cantor's intersection theorem, $\bigcap_{n=0}^\infty [a_n,b_n]\neq\emptyset$, let $\xi\in\bigcap_{n=0}^\infty [a_n,b_n]$, we have $g(\xi)\ge n$ for all $n\ge0$, contradiction!