How can I change a pulse into a state?

Question: Is there a logic gate that changes a pulse into a steady state?
Answer: You can do it through the follow design, where - or | is redstone and R is a repeater and I is your input.
I-------
  |    |
  ----R-

If you hate my amazing ACSII art :( (click for legend)

To turn it off, break the redstone and put it back or turn it into an RS-NOR gate
Design K is the easiest in my opinion
Answer: You have a range of options here, depending on how complex the functionality needs to be.
The simplest possible device is called an Input-Stabilizing Cell. It takes an input of any length, and once the input turns on, the output turns on and never turns off. Due to the simplicity of this device, it is a good solution if you only need bare basic functionality, but it has to be reset manually. Because of this, it's not very helpful in more complex circuits.
The next "level" of redstone data storage is called the RS NOR Latch. This is extremely helpful because it can be both turned off and on with redstone. This way, you can set the latch to the "on" state, and it will stay on, but you can then add some functionality that can later reset the latch to "off". Hence the name: Reset-Set NOR Latch. It's an RS Latch implemented with a NOR gate.
Moving on, we have the T Flip-Flop. The T Flip-Flop functions like a toggle button. The first pulse turns it on, the second turns it off. This is nice if you want toggleable functionality, since the RS NOR Latch requires two separate redstone lines for turning it on and off. Additionally, though pure-redstone implementations of a T Flip-Flop can be unwieldily, piston-based versions are extremely compact, and they don't even require sticky pistons.
While those three devices probably fill nearly all of your needs, sometimes, a couple of other, more complex gates can be helpful.
The D Flip-Flop and Gated D Latch circuits provide a way to use just a single input line. Coupled with this is a secondary input, known as the "clock" input. The way a Gated D Latch works is simple: when the clock is off, the output doesn't change. When the clock is on, the output will update to reflect the input. In this way, the Gated D Latch acts much like a "lock" on an input, since it will keep it from changing until you request an update from the input line.
The D Flip-Flop works in basically the same way, but it only updates the output when the clock line changes from off to on. This functions much the same as a Gated D Latch with a pulse limiter on the clock input.
Finally, we have the monstrous circuit that is the JK Flip-Flop. While highly impractical in most situations, the sheer amount of functionality it contains is entertaining. It is essentially a D Flip-Flop and T Flip-Flop combined into one. It has three inputs: J, K, and C (aka clock). Depending on the arrangements of J and K when C triggers, the output will either: turn on, turn off, do nothing, or toggle. Although this is not usually useful, if you ever have the need to have both toggling and manual setting in a single machine, this might be the way to go. If not, try and avoid the JK Flip-Flop.
Answer: You can make this able to be turned on or off simply by replacing the redstone that runs beside the repeater with a block, and have that block attached to a sticky piston that retracts to break the signal
Image referenced from above:
https://i.stack.imgur.com/cbf29.jpg
Comment: This is just repeating what Joe the Person's answer already said. I believe you're describing latch design K from his answer.

How can I change a pulse into a state?


How do 2 phones share ITunes and Calendar entries?

Question: My husband and I just got iPhones 4 running iOS 6, what we're looking for is to share ITunes and Calendar entries.  We have one computer and right now have Outlook for a calendar.  I want to be able to make one entry on the calendar and have it show up on both phones.  Can I use Outlook to accomplish this?  Or is there an app out there that will sync the computer calendar with both our phones?
Answer: My fiancé and I do this quite regularly simply using the iCal app that is part of the phone. We logged in under the same Apple ID and shared only the calendar & contacts (on her phone b/c she doesn't need/want my email). Doing this, we're able to sync calendar and contacts, which is really useful if one of our batteries die.
But you want to use Outlook, it is possible with Outlook, but I think it requires an Exchange service to sync. I'm not 100% sure on that, but I am pretty sure that's what it takes (without manually syncing).
I believe this post will be very useful for you :)


How do I turn on the battery full notification sound?

Question: Similar to this previous question, but I want to enable the battery full notification sound, not the opposite :)
My Galaxy S used to make a nice beep and switch the screen on briefly when the battery was fully charged, since the Gingerbread update (official Samsung 2.3.3 JV9 release) it still briefly switches the screen on, but no longer makes the noise.
This used to be great because the plug socket on my desk is right at one end, so (if I don't want a long cable getting tangled in everything) the phone's out of my line of sight and I rely on the audio notifications to tell me there's a reason to glance over at it. Now, since Gingerbread, I don't get any audible notification for the battery full, and never know when it's been fully charged.
Unlike a lot of people here I don't plug my phone in to charge overnight (why would I, when I have to pay for the electricity at home :D), so was never bothered by the noise waking me up.
I've hunted through all of the Sound and Notification settings that I can find, and already have System and Notification volumes turned up, so it's not those. I'm slightly worried that they've listened to the moaners and removed a useful feature, but hope that buried somewhere there's a nice setting that can keep everyone happy.
Answer: Try Batteryminder.
In the features section it claims to allow for "Low/Full Battery Alerts"
Hope this helps
Comment: Thanks, will give it a try, but hoping that there's some way to turn this previous functionality back on with needing a 3rd party app.


How can I produce this drop-shadow effect in Photoshop?

Question: More specifically, how do I produce this square background shadow:

Source
Answer: There are two effects you're asking here. One is the checkered background, and the other one is the drop shadow.
Let's start with the checkered background first.

Create a new image, make it 6px by 6px. Draw two black squares that are 3px by 3px. make sure you turn off the default background layer so the non black square spots are transparent. Hit cmd+A(select all) then go to Edit->Define Pattern. Name the pattern whatever you like.
Now we have the pattern defined, let's start the actual image. For my example, I'm using a 600px x 600px dimension.

From the screenshot above:

Fill the background layer with a light gray.
Start a new layer, use the elipse selection tool to draw a circle. I drew a circle that's about 300px in diameter.
While the selection is active, go to Select->Modify->Feather. In the Feather menu, enter 30px for feather radius.
Choose the Paintbucket tool, instead of filling with foreground color, choose "pattern." Then in the pattern dropdown, you will see the checker pattern that was just created. Now click on the circle selection to fill.

Then play with the opacity of the checkered circle layer until it's subtle enough for your liking.
Note: there are other ways to achieve this effect too. For example you can fill a solid circle(without feathering) with the pattern, then put a layer mask on it and use radial gradient. But I feel this way is simpler. The layer mask method gives you a lot more control though.
Now for the drop shadow:

Judging from your screenshot, I couldn't tell what type of cast shadow it is since it's cropped and dithered somewhat. But it looks like a fairly simple drop shadow in the layer effects. I drew a white block in a new layer, and set a few drop shadow options(click on the screenshot to see full res version). In your example it looks like the blur value is 0, but I can't tell if that's due to poor image quality. Feel free to play around with the values though.
Comment: +1 that's completely right answer, i thought he's asking about drop shadow only didn't noticed background.. ;|
Answer: The sharp-edged part is simple: set the size to 0 px. Use the Move tool to position the shadow where you want it and reduce the opacity as needed.
To get the cut-off at the bottom you need to take a couple of extra steps.
Right-click on the word "Effects" or "Drop Shadow" and choose "Create Layers" from the context menu. This will turn the drop shadow into its own layer, then you mask or delete the bottom edge of the drop shadow layer.
Answer: Create new document and go to layer style (double click on the layer), apply drop shadow as given in SS,Play with sliders if your not satisfied with this output,edit contour level to make it up to level....
hope it helps


Will I trap moisture in the concrete of basement walls and floors if I seal them with epoxy?

Question: I'm thinking about painting my basement's walls with some concrete sealer (before finishing the basement) and floor with epoxy paint. Many people are doing that, but I'm concerned about moisture in the concrete. The outside of the walls are covered with some black stuff which is a moisture barrier. If I paint the inside of the walls, I will lock moisture inside the concrete for ever - will it affect concrete in any way? What about mold in the concrete? I heard something about concrete breathing - does it apply here?
Answer: Would moisture be trapped? To some extent.
Is it bad? No, once concrete has been mixed and started to cure, keeping it moist helps it get stronger and this effect can go on for a long time. In no case will it ever make the concrete weaker. See the graph at the end of this URL:
http://wgbis.ces.iisc.ernet.in/energy/water/paper/drinkingwater/wellsconstruction/appendix4.html
There are some who say the concrete in the Hoover Dam is still curing.
There is no need for concrete to breath from the point of view of the strength of the material. There will be moisture retained from the original mix and it is possible some are confused about the chemistry of concrete and think it needs to dry, but the truth is the complete opposite of that.
Answer: Yes, you will trap moisture with the epoxy. Will that hurt? I'm not sure.
That said, it sounds like your foundation walls are already waterproofed on the outside...which is great, as that is where you want to do the waterproofing. As such, I wouldn't bother doing it internally.
And if you are interested in how to finish the basement, I put a detailed answer here: Should I use steel or wood studs for basement exterior walls?


Thick Chocolate Sauce

Question: Is it possible to do something with chocolate so that it does not get hard anymore so that its consistency gets something like the caramel in a mars/snickers bar? What precisley would I have to do to achieve that?
Comment: Are you asking for something different from chocolate syrup/fudge sauce? There are a lot of recipes for that (and in general it'd be easy to vary their thickness). I've certainly had homemade chocolate syrup that has that kind of consistency at room temperature, and flows more when hot.
Answer: Perhaps a thick ganache might meet your demands. It wouldn't be as sticky as caramel though. But take a look at this question. It mentions adding some honey and butter to the standard ingredients, cream and chocolate.
If you have questions on how to make ganache, watch this video or one of the many others available on the internet. Simply said, heat up heavy cream and pour it over chopped bittersweet chocolate. Stir till all the chocolate is melted and you have a nice, smooth consistency. For your purpose, use at maximum the same amount of cream as chocolate. Add some more chocolate if it's too runny. But know that the ganache will become harder/thicker when cooled.
Answer: I can think of a few ways of achieving this:
The easiest (although not strictly using pure chocolate) way is by combining some sort of syrup like honey, agave, golden or corn with cocoa powder. This would give the sticky texture your looking for, not go hard once cool and taste of chocolate. Here a recipe link, it's from a vegan blog, http://karensvegankitchen.blogspot.com/2010/09/thick-gooey-chocolate-sauce.html.
Another option is glacage chocolate which is a kind of fancy French chocolate glaze. It's probably not quite what your looking for but I'll explain anyway. It's made in its most basic form by combining chocolate with gelatine. This gives it a gloupy texture, although it is used in patisserie for it's mirror like shine. Here a recipe link if you do want to have a look: http://patissierjourney.blogspot.com/2010/08/glacage-chocolate.html.
Another final idea is making a typical ganache (chocolate and cream in a ratio of 1:1) and adding some corn syrup or other sugar based syrup. Which could give a sticky caramel texture and choclate flavour.

Thick Chocolate Sauce


Revision help URL contains wrong header

Question: There are two revision urls listed on the help page but they both seem to contain the same detailed information. The header for the non-guid version contains the url for the one that takes a guid. Should these be consolidated into one or the two help pages be changed to reflect their differences?
http://api.stackoverflow.com/0.8/help/method?method=revisions/{id}
vs
http://api.stackoverflow.com/0.8/help/method?method=revisions/{id}/{revisionguid}
Answer: Headers have been fixed.

Revision help URL contains wrong header


Ordering Post 2 Post admin meta box by meta from CPT

Question: I've just started using Posts 2 Posts (brilliant plugin by the way) in combination with a custom post type called "Artist" and the WP Event Manager Plugin. My aim is to use P2P to link Artists to Events so an Event Page can display who is performing.
I have created a recurring number of events in Event Manager which understandably have incrementing post ID's, e.g. 5,6,7. Let's say for argument these events are all on subsequent Mondays. I then created a recurring event, say on a Friday which has post ID's 8,9,10,11.....
Now I can order the order of the meta box in admin by post ID using something like:
function order_pages_by_ID( $args, $ctype, $post_id ) {
if ( 'posts_to_pages' == $ctype->name ) {
    $args['orderby'] = 'ID';
    $args['order'] = 'asc';
}

return $args;
}

add_filter( 'p2p_connectable_args', 'order_pages_by_ID', 10, 3 );

But this will now display all the Monday events and then all the Friday events when I'm trying to connect an Artist to an Event from the Artist's page. What I'd like is to have Monday, Friday, Monday, Friday etc.
The event CPT has a custcom field called _event_start_date which I can append to the label in the admin meta box by using:
function append_date_to_candidate_title( $title, $post, $ctype ) {
    if ( 'my_connection_type' == $ctype->name && 'event' == $post->post_type ) {
            $meta = get_post_meta( $post->ID, "_event_start_date" );
            $title .= " (" . reset($meta) . ")";
    }

    return $title;
}

Is there any possible way to order the query for the admin meta box to take this custom field into account and order by this (or even order it in PHP afterwards)? I've seen a few examples of ordering the posts when using 'get_connected()` from the frontend by nothing specifically related to the ordering in the admin meta box.
Answer: I finally found the answer. Turns out I just needed a better understanding of WPQuery.
It ended up being as simple as:
function order_pages_by_ID( $args, $ctype, $post_id ) {
    if ( 'posts_to_pages' == $ctype->name ) {
        $args['meta_key'] = '_event_start_date';
        $args['orderby']  = 'meta_value';
        $args['order']    = 'asc';
    }
}


Accelerating expansion of the universe and cyclic model

Question: So I've heard that recent observations indicated that the rate of expansion of the universe is accelerating. Does this indicate that the universe can NOT be cyclical, with an infinite series of big bangs and crunches extending forward and backward through time? Does it "prove" that the universe will not crunch?
Answer:
Does this indicate that the universe can NOT be cyclical

No.  See Conformal Cyclic Cosmology for an example cosmology that is cyclical with an infinite series of "Big Bangs" but no "Big Crunches".
Answer: The conformal cyclic cosmology Alfred mentioned is an idea from Roger Penrose. An unrelated idea is from Andrei Linde. See for example this paper or Google for more info. To be honest the details are beyond me, but Linde suggests that dark energy can change sign and cause a collapse. This means the universe could be cyclic even though dark energy is currently causing an accelerated expansion.
However, both Penrose's and Linde's ideas are highly speculative, and I suspect most of us believe that dark energy does indeed mean that the universe cannot be cyclic.
Answer: I was thinking that the expansion/contraction model of the old school is more likely than the big bang.  I have this idea that the electric theory of the universe;not sure where the power comes from but it seems to be time-varying,is more correct than other options.  Magnetic field in plasma create pressures not possible on Earth that make a black hole;aka LT starbody.
Over time the LT whose center has become a galactic nucleus loses pressure after it attains weight ratios (galactic nucleus/galaxy) stability, jets infrequently horizontally through the galactic center and not axially, off gassing from these events surrounding the entire galaxy with a gas cloud that cools and becomes a starless dark.  Cold gas clouds then become the majority of the universe dark energy;expansion allowing further cooling which then reaches $0K$ and phase changes.
here:https://www.llnl.gov/str/Schneider.html
Comment: Anser is not related to the question.


Some Mathematical Poker Questions (Discrete Math)

Question: Mathematically speaking, I have two questions:

How many ways can a 5-card draw-poker hand be dealt?
How many 5-card draw-poker hands exist?

Thanks!
Answer: I think your first question is asking, "How many ways can one be dealt 5 consecutive cards from a standard deck of 52 cards with the order of cards being distinct hands?" That is to say, a hand with the first card is Ace-of-spades then 2-of-spades then 3 other cards is distinct from being dealt first the 2-of-spades, next the Ace-of-spades, then the 3 other cards. In this case, you are looking for a permutation of the number of ways to order 5 cards from a set of 52 objects. Alternatively, this is asking for the number of ways to leave behind 47 (52-5) cards in a particular order from the deck box.
So the formula for a permutation of k items out of n items [notation for a Permutation is n_P_k]is n!/(n-k)!  (Look up the Factorial Function) to understand what n! means. To answer your question, we have n=52 cards in the deck, and k=5 cards per hand. So the answer is 52_P_5 = 52!/47! = 52*51*50*49*48 = 311875200.
Your second question, "How many 5-card poker hands exist?", is equivalent to asking how many ways to combine 5 cards together from a deck of 52. Note how this is different than the above permutations because in a Combination, order does not matter. So the formula (some like to say "n-choose-k") can be expressed with the Binomial Coefficient which is hard to demonstrate in this answer box, but it uses parentheses and almost looks like a fraction. Sometimes it is written n_C_k. The formula is n! / [k!(n-k)!]. You might also want to think of it as a permutation divided by k! to avoid repetition when it comes to the order of the items, so you could even say n_C_k = (n_P_k)/k! In your example, n=52 and k=5 like before so we calculate in two ways:
52_C_5 = (52!)/[5!*47!] = 2598960
52_C_5 = (52_P_5)/5! = 311875200/5! = 311875200/120 = 2598960
Answer: To calculate all the 5 Card draw poker hands that exist you have to calculate
51! / 47!
or explicite way  52 x 51 x 50 x 49 x 48 = 311,875,200
Now you need to know that for poker it doesn't matter wheter you have spades or clubs or whatever color. So in terms of value there are less hands. But that's the number of different hands you can have.
What do you mean by "how many ways can a 5-card draw-poker hand be dealt ?
like 5 cards to one person or first 1 than 2 than 3 cards etc?
as soon as I know, i calculate that.
Answer:
How many ways can a 5-card draw-poker hand be dealt?

In order to calculate the number of ways that a hand can be dealt, meaning that the order of the cards is significant, you need to multiply the possible number of cards at each step together.
For a particular hand (meaning 5 specific cards), when the first card is dealt, you have 5 possible cards that would "qualify".  When the second card is dealt, we assume that the first one has already been dealt, so there are now 4 possible cards.  When the third card is dealt, there are 3 cards left in the deck.  Continue this until there is only one possibility as the fifth card is dealt.  The math would look like this:  5*4*3*2*1 = 120 different ways when you consider all of the possible different orders.
In a similar manner, for a particular "group" of poker hands (meaning flush, straight, pair, etc) you would need to adjust the number of cards accordingly.  For instance, the number of ways to deal a flush would be:  52*12*11*10*9 = 617,760 (note that this would include straight flushes as well, so to have a "true" answer on how many ways are there to deal a flush, we would need to calculate the same for straight/royal flushes and subtract them from this answer).

How many 5-card draw-poker hands exist?

Now, a poker hand doesn't care about the order of the cards or the suit (unless they are all the same suit in which case the actual suit itself doesn't matter, just the fact that they are the same).  The total possible number of 5-card poker hands is 7,462 with 2,598,960 different variations (for instance, there is one possible hand in the royal flush group (T-J-Q-K-A, all in the same suit), but four different variations (suits) that it can be dealt in.)
Wikipedia also has a good explanation of the poker probabilities.


why does a mirror show what is in front of it?

Question: the only answer I can think of is that light is reflected from the objects in front of the mirror (visible color) and then reflects again off of the mirror to our eye, but im not quite satified with this answer
Comment: Did I provide a satisfactory answer?
Answer: Your answer is correct. A mirror is simply a surface that does not trap light, and is smooth enough that the light 'bounces' off of it, much like a ball bounces off of concrete.
Light from some outside source hits your body (or any object) and bounces off of it. Some of that will hit the mirror, and some of the reflected light will hit your eyes.
Answer: Think this way:
In order to light (an electromagnetic wave) to propagate through a material, the atoms of the structure should be "oscillating" in such a way that themselves are the resposibles for the creation of that electromagnetic wave (that's why it was "strange" to imagine light propagating in vaccuum)
Reflection is enhanced in the mirror by suppression of wave propagation.
Also, a mirror has a smooth surface that decreases the probability of light being absorbed by the material.
Furthermore, by the laws of reflection (wikipedia):

The incident ray, the reflected ray and the normal to the reflection surface at the point of the incidence lie in the same plane.
The angle which the incident ray makes with the normal is equal to the angle which the reflected ray makes to the same normal.
The reflected ray and the incident ray are on the opposite sides of the normal.

All this together, means exactly what you said: light (including the visible range) "bounces" in the surface of the mirror because the probability to propagate through the material and/or being absorved is minimized by the smooth surface and the material itslef.
If your question is actually related to: "Why do the laws of reflection are like that?", the answer is "Science".
Science means to do experiments, try to understand the what happens, and define patterns of behaviour (what's so called theories) that can be used to predict or analyse similar systems. That means "we are just telling how things work, not actually WHY".


Can You Use a Metamagic Rod and a Scroll/Staff/Wand Together?

Question: Are you allowed to use a Metamagic Rod (say Extend) with a spell cast from a scroll or other magical device?
Answer: No, unless the GM rules otherwise of course.  "The modifications made by these [metamagic] feats only apply to spells cast directly by the feat user. A spellcaster can't use a metamagic feat to alter a spell being cast from a wand, scroll, or other device."
And metamagic rods just let you act as if you have the feat a couple times a day.
Answer: My understanding has always been no.  A scroll is a spell which is already 99% "cast" and merely awaits the final word/gesture/etc to finish the casting.  A metamagic rod needs to be a part of the entire casting of the spell.
Answer: Artificers can use them while creating the scrolls. It's best if you have a wizard or cleric cast the spells as artificers often can't do the higher level metamagic such as persistent.


Post IDs missing on delete_postmeta action hook

Question: I'm using metadata in pages to store variables and do server side caching for dynamic generated PayPal buttons and other variables about the pages (page specific).
I'm trying to hook the page meta data's update, add and delete functions. The update and add works fine (both with same function) but I can't figure delete out.
In order to use delete_post_data I need the post_id but the do_action doesn't return that (at least what I can figure) from WordPress's source code it seems to return this:
do_action( 'delete_postmeta', $postmetaids ); //sometimes $post_meta_ids

where
$postmetaids = $wpdb->get_col( $wpdb->prepare( "SELECT meta_id FROM $wpdb->postmeta WHERE meta_key = %s", $post_meta_key ) );

or
$post_meta_ids = $wpdb->get_col( $wpdb->prepare( "SELECT meta_id FROM $wpdb->postmeta WHERE post_id = %d ", $postid ));

The code I'm using to call the hooks:
add_action ('update_postmeta', 'wp_paypal_meta_change',1,4);
add_action ('add_postmeta', 'wp_paypal_meta_change', 1, 4);
add_action ('delete_postmeta','wp_paypal_meta_delete', 1, 2);

The delete function:
function wp_paypal_meta_delete($post_id, $meta_keys){
/*
 handles cases where the user intentionally deletes the metadata, and deletes
 the appropriate hidden metadata variable in order to make sure the cached
 PayPal form matches the product data.
 */

    if ($meta_keys == 'image'){
    delete_post_meta($post_id,'_thumb');
    } else if ($meta_keys == 'shipping' || $meta_keys == 'cost' || $meta_keys== 'bookname'){
    delete_post_meta($post_id,'_paypal');
    }

}

I know the $post_id doesn't get into the function, but it's there because I hoped WordPress somehow references it - didn't work.
Answer: From the OPs edit moved to an answer.

I found an answer, also not a straight path, I'm sure there's a better way to do this.
  basically I use a little helper method that queries the SQL with the meta_id and return the post_id:

<?php
function get_key_by_id( $meta_ids )
{
    global $wpdb;

    $meta_ids = absint( $meta_ids );

    // Debugging:
    // $wpdb->show_errors();

    $meta_keys = $wpdb->get_var( $wpdb->prepare(
         "
            SELECT meta_key
            FROM {$wpdb->postmeta}
            WHERE meta_id = %d
         "
        ,$meta_ids
    ) );

    // Debugging:
    // $wpdb->print_error();

    if ( $meta_keys )
        return $meta_keys;

    // else:
    return false;
}


Groupoid structure on G/H?

Question: Let $G$ be a group and let $H$ be a subgroup. If $H$ is normal in $G$, then $G/H$ has a group structure. But in general, can there be a groupoid structure on $G/H$(left cosets or right cosets) that generalizes the normal case?
Answer: One answer to your question is that there is always the notion of an "action groupoid", although this does not reproduce the group structure on $G/H$ when $H$ is normal.
Let $G$ be a group acting on a set $X$.  (There are generalizations when both $X,G$ are groupoids.)  Then the action groupoid $X//G$ is the groupoid with objects $X$, and morphisms $X\times G$.  More precisely, if $x,y \in X$, then $\hom(x,y) = \{ g\in G \text{ s.t. } gx = y\}$.  The groupoid axioms are essentially obvious.
For example, let $X = G/H$ be the set of left $H$-cosets.  Then the action groupoid is very simple: it is connected (any object is isomorphic to any other), with $\text{aut}(e) = H$, where $e = eH$ is the trivial left $H$-coset.  And $\hom(e,g) = gH$, where $g = gH$ is a coset.
  So as a discrete groupoid, this action groupoid is equivalent to $\{\text{pt}\}//H$, also sometimes called the "classifying groupoid" $\mathcal B H$, because the geometric realization of the nerve of this groupoid is the usual classifying space of $H$.
In short-hand, we have the following equation:
$$ (G/H) //G \cong 1//H$$
where $\cong$ denotes equivalence of groupoids.  (Actually, since I'm talking about left actions, I should probably write $G \backslash \backslash X$ for the action groupoid, and so the equation really should be $G \backslash \backslash (G/H) = H \backslash \backslash 1$, but typing "/" is much faster than typing "\backslash", so I won't use the better notation.)
But you are probably asking a different question.  Recall that when $H$ is normal, then $G/H$ has a group structure, which is to say there is a groupoid with one object and whose morphisms are elements of $G/H$.  Of course, as you know, if $H$ is not normal, then $G/H$ does not have a natural group structure, because in general $g_1Hg_2H$ is not a left $H$-coset.
You can try to do the following.  Any set is naturally a groupoid with only trivial morphisms, and then the set $G/H$ is equivalent to the groupoid $G//H$, where $H$ acts on the set $G$ by translation = right multiplication.  (This is because $H$ acts on $G$ freely.)  But $G$ is actually more than a set: it is a group.  So let's think about it as a "groupal groupoid" or "2-group", i.e. a 2-groupoid with only one object; in this case, it will also only have identity 2-morphisms.
Then I guess you should try to form the "action 2-group" or something, by adding 2-morphisms for the translations by $H$.  But I think that if you do, you no longer have a groupal groupoid: I think that if $H$ is not normal, then the group multiplication is not a functor from the action groupoid $G//H$.
The other only thing I can think of is to define $K = \text{Norm}_GH$, the normalizer, and then $K/H$ is a group that embeds in $G/H$, so let the objects of your groupoid be cosets of $K$ and the morphisms given by $K/H$?
So, long story short: in the way that I think you are hoping, no, $G/H$ is not naturally a groupoid.
Comment: The second part of your comments is what I am looking for. I was thinking of forming a groupoid with objects given by conjugacy class of the subgroup $H$. But it is not clear how to define the morphisms.

I think your idea of forming a 2-group is quite interesting. But I don't quite understand it.

So i understand that $G$ can be seen as a 2-groupoid. But why are we expecting that the "action 2-group" you were mentioning has a groupal groupoid structure?
Comment: A 2-group is the same as a group object in groupoids.
Answer: When $G$ is finite, for arbitrary $H$ you can consider the quotient $G/H$ as an association scheme in the sense of [Zieschang, Paul-Hermann. An algebraic approach to association schemes. Lecture Notes in Mathematics, 1628. Springer-Verlag, Berlin, 1996. xii+189 pp. ISBN: 3-540-61400-1 MR1439253].
The notion of association scheme generalizes (at least...) that of group, that of distance regular graph, and that of (some types of?) building. When $H$ is normal, then the association scheme $G/H$ is the same thing as the association scheme associated to the group $G/H$, so in a sense, for $H$ non-normal considering the association scheme $G/H$ is a quite natural.
(In the infinite case, you can probably do more or less the same... I don't recall having seen infinite association schemes, though)
(Of course, this does not answer your question... but as Theo pointed in an earlier answer, I do not think $G/H$ is a groupoid in any sensible way in general, so the bit of structure I am mentioning might be a useful consolation prize!)
Answer: Any set on which a group acts can be given a natural groupoid structure, called the action groupoid.  Since $G/H$ admits a left $G$-action, the action groupoid construction applies.  You can find the definition in the wikipedia page de rigueur, under Group actions.
Answer: Another way of putting this is that there is an equivalences of categories between actions of a group (or groupoid) $G$ and covering morphisms of the group (or groupoid) $G$. See for example
Higgins, P.J. Notes on categories and groupoids, Mathematical Studies, Volume~32.
Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of
  Categories, No. 7 (2005) pp 1-195.
This links the notion of covering morphisms of groupoids  nicely with the theory of covering spaces.  There one finds that for "nice" spaces $X$ the fundamental groupoid $\pi_1$  gives an equivalence of categories  from covering maps of $X$ to covering morphisms of $\pi_1 X$. Thus a map of spaces is modelled by a morphism of groupoids, and this is convenient. For an exposition in these terms, see my book "Topology and groupoids", or earlier editions (1968, 1988).

Groupoid structure on G/H?


Do I need an LLC to self-publish a political book in Florida?

Question: Do I need to form an LLC to self-publish a political book out of Florida? Although, I have tried very hard not to break any laws, I am worried about frivolous or spiteful lawsuits. I am trying to do a cost benefit analysis regarding time and money.
The set up fee via a website that forms LLCs is $100 to $200.
Articles of Incorporation filing fee is $125.
Apparently there will be a yearly $138.75 annual report filing fee.
As a one person LLC, I would have to file yearly taxes on Schedule C of my personal income taxes.
Is that it? Would that be all the time and money involved?
I expect to make less than $500 per month with the business. I have very little free time to deal with paperwork. And filing every quarter would be a burden.
In Florida can they take your house? Other assets?
I am not wealthy and am not sure it is worth all the time and money to set up an LLC. But the legal protection would be nice.
Comment: Related: [What are the advantages of incorporation for a writer?](http://writers.stackexchange.com/q/1110/26)
Answer: IANAL.
From what I can see from this reference, an LLC does not protect you from lawsuits against you as the writer, such as with libel, defamation, or invasion of privacy torts.  That page does mention you can get media liability insurance, but it is not cheap (it states $3-4,000 and a $5k deductible).  You should check this against other information to see if this is the case.
Also, even if we were to assume an LLC could protect you, if you do not conduct your LLC completely properly (separate bank accounts, documenting your "meetings", filing all forms, sufficient capitalization, etc...), courts can find you are operating as an "alter ego" and pierce the corporate veil, affording you no personal protection anyway.  Given that you state that you have little time for paperwork/filing, you might want to think twice about whether you are up for what it takes to run an LLC properly.
Answer: First off, I am not a lawyer-
You should check with your lawyer but I believe that a LLC will not offer you any protection in a libel suit. If you knowingly print something that is untrue and causes injury to another party it is on you.
As far as your house goes, I believe Florida is one of the few states where your primary house (residence) is protected in bankruptcy. That may or may not help you in a libel suit.


Search has stopped working suddenly

Question: I am getting following error when I search for anything:
We did not find any results for KEYWORD.
Suggestions:
Ensure words are spelled correctly.
Try using synonyms or related searches.
Try broadening your search by searching from a different site.
Try removing search refinements or using more general keywords.

Additional resources:
Get additional search tips by visiting Search Help
If you cannot find a page that you know exists, contact your administrator.

I think crawler has stopped working and hence there is no indexing for search.
Additional info provided by @shobs :
Hello I am the administrator for this site Saumilm is referring to.
The crawl was working fine all along and when the mysite was added to the content source the crawl stopped working and it started crawling ohter test application in the farm.
Here is the error message:
The start address 'http://xxx.xxx.org' cannot be crawled.

Context: Application 'SharePoint_Server_Search', Catalog 'Portal_Content'

Details:

Item not crawled due to one of the following reasons: Preventive crawl rule;
Specified content source hops/depth exceeded; URL has query string parameter;
Required protocol handler not found; Preventive robots directive. (0x80040d07)

Answer: check whether search service is running.
Answer: You should also verify that the password has not expired on the Search and Crawl service accounts.
Answer: Unless you have removed your index it should still be there, check your Search Service Administration page and see if you have any crawl history.  If there are errors you can check them through the crawl log and that should tell you what is going on.

Search has stopped working suddenly


Unrelated kadish in a shiv'a house

Question: What should someone who would otherwise be saying kadish (whether as an obligation or not) do in a shiv'a (mourning) house? Should he say kadish along with the mourners (I'm asking according to the common custom that more than one person says kadish yasom at a time) as he would in a synagogue, or should he leave the kadish to the mourners only? I can't really think of a reason for the latter, but it seems somehow appropriate. Does anyone have a good argument either way, or a source for an answer?
Answer: The Shulchan Aruch rules (YD 393:1) that someone who is sitting Shiv'ah (who is generally not allowed to leave his house even for mitzva purposes) is allowed to attend the burial of someone else (even unrelated) in the community from the third day of Shiv'ah and onward, and if the deceased doesn't have enough people to act as pallbearers and gravediggers then he can attend even on the first day and in a different community.
The ruling states that when he attends he is placed with the mourners and not with the comforters, such that the comforters who attend end up comforting all the mourners.
It would seem from here that a given mourner does not have a monopoly on attention even in his own domain.
Comment: I don't intend 'monopoly on attention' to be a negative thing. It certainly is a reasonable _hava amina_ to think he gets that in his own Shiva house.
Answer: If you will permit me to draw on my own practice and experience. I try to avoid this but on those occasions where it was unpreventable, I said kaddish quietly along with the mourners. As against that, when my mother-in-law passed away, I, and another brother-in-law were in the 12 months for parents and the three of us (2 brothers-in-law and I) stood up together to say kaddish together - as it were within the extended family. I suspect that there are many customs for this, and if you don't have a defined custom, go for "darchey shalom" - one could even say that that is more of an "ilui neshama" than the kaddish itself (as per the halacha that one who yields to another in leading services of saying kaddish (in a one kaddish sayer place) is as though he himself lead the prayer/kaddish)
Comment: What's the point of saying a davar shebikdusha quietly? A Minyan doesn't hear you.
Comment: By quietly I mean not to dominate. Voiced, but letting the "major" mourner lead.

Unrelated kadish in a shiv'a house


sl(2,C) and the harmonic oscillator

Question: I've been studying the finite-dimensional representations of the lie algebra sl(2,C). I've read that these representations are related to the harmonic oscillator and the associated raising and lowering operators, but I'm not really sure how they are related. I've read you can generate a new energy state from an old one, but I'm not really sure how it works, and the energy states do not differ by 2 like the eigenspaces do for sl(2,C). I'm having trouble finding references about this, especially ones that are accessible. If anyone knows where I can find more info, or has some knowledge of the material, I would greatly appreciate their help.
Comment: Err...it looks like the commutation relations in physics involve $[a_+, a_-] = 1$, so it's not quite the same since $H$ doesn't act as the identity in the rep space.
Comment: My understanding is that the eigenvalues of the representation of $\text{sl}(2,\mathbb{C})$ are integer valued but the standard physics convention has an additional factor of $\hbar/2$ which may fix your issue.
Comment: I'm guessing you're using the basis $X_+,X_-,H$ with $[H,X_+] = 2X_+, [H,X_-] = -2X_-, [X_+,X_-] = H$?  Then I think in physics they use $\frac{1}{2}H$.  So if $v$ is an eigenvector of $\frac{1}{2}H$ with value $r$ then $X_+ v$ is an eigenvector of $\frac{1}{2}H$ with value $r+1$.
Comment: [Ryder](http://books.google.com/books?id=nnuW_kVJ500C&printsec=frontcover&dq=quantum+field+theory+ryder&hl=en&ei=0_YGTdeYO8L-8AaE85HnAg&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CCoQ6wEwAA#v=onepage&q&f=false) section 2.3.  [Hall](http://books.google.com/books?id=nnuW_kVJ500C&printsec=frontcover&dq=quantum+field+theory+ryder&hl=en&ei=0_YGTdeYO8L-8AaE85HnAg&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CCoQ6wEwAA#v=onepage&q&f=false) section 1.7.
Answer:
The Quantum Mechanical Harmonic Oscillator

If you look at the Wikipedia page on the quantum mechanical harmonic oscillator, you will see that it is not the Lie algebra $\mathfrak{sl}_2$ that enters the picture, but rather an algebra generated by
two operators $a$ and $a^{\dagger}$ satisfying $[a,a^{\dagger}] = 1$ (see also Eric's comment above).
This is a familiar algebra in disguise, the so called Weyl algebra of differential operators with polynomial coefficients $\mathbb C[x,\partial_x]$.  (Note that $[\partial_x,x] = 1$, by the Leibniz rule.)  The product $x\partial x$ is the so-called Euler operator (it acts on $x^n$ as multiplication by $n$); you can see that in the harmonic oscillator interpretation, it acts as the Hamiltonian (up to a shift of $1/2$, to add non-zero groundstate energy, and a rescaling by $\hbar \omega$).
This algebra has a natural representation, namely on the space of polynomials $\mathbb C[x]$, and we see that this is isomorphic to the representation on the span of the eigenstates that occurs in the harmonic oscillator picture.  (The polynomial $x^n$ corresponds to the eigenstate with energy $\hbar \omega(n + 1/2).$)

The Lie algebra $\mathfrak sl_2$ in quantum mechanics

If my memory serves, the first time $\mathfrak sl_2$ entered the picture (in my undergrad QM class, at least) is in the analysis of the hydrogen atom.  The point is that
the Schrodinger equation in this case has a rotational symmetry, and differentiating this action, one gets an action of the Lie algebra of $SO(3)$ on the space of states.  Complexifying, this gives an action of $\mathfrak sl_2$.
If you separate variables in the equation (using spherical coordinates, which are natural in view of the rotational symmetry), the radial part and the spherical part separate, the radial part is easily dealt with, and one is left
studying a certain differential equation on the space of functions of the
sphere $S^2$.
Mathematically, one is looking at $L^2(S^2)$, with its natural action of $\mathfrak sl_2$ (in the guise of the complexified Lie algebra of $SO(3)$).
The Shrodinger equation involves the Laplacian, which is also the Casimir in the
enveloping algebra of $\mathfrak sl_2$, and so if you fix the energy (eigenvalue of the Laplacian), one is trying to understand the corresponding eigenspace of the Casimir on $L^2(S^2)$.  This is a special case of the Peter--Weyl theorem,
which (in this case) also goes under the name of the theory of spherical harmonics.
The idea, roughly, is:  $S^2 = SO(3)/SO(2)$.  Now by Peter--Weyl,
$L^2(SO(3))$ is the Hilbert space direct sum of $V\otimes V^*$, where
$V$ runs over the irreps. of $SO(3)$.
Thus $L^2(S^2)$ is the Hilbert space direct sum of $V\otimes (V^*)^{SO(2)}$.
The space of $SO(2)$ invariants in each irrep. is one-dimensional, so in
fact $L^2(S^2)$ is the Hilbert space direct sum of $V$, as $V$ runs over
the irreps. of $SO(3)$.
Since $SO(3) = SU(2)/\{\pm 1\}$, these correspond on the Lie algebra level
to the odd-dimensional irreps. of $\mathfrak sl_2$.
Each $V$ is an eigenspace for the Casimir (if $V$ has dimension $n$ then
the Casimiar eigenvalue is some quadratic expression in $n$ that you can
easily figure out, or look up), and distinct $V$ give distinct eigenvalues.
So, going back to the hydrogen atom, if you fix the energy of the electron,
then the collection of states has dimension equal to the dimension of the
corresponding $V$. (The fact that this is typically greater than one-dimensional is referred to in physics as degeneracy; the energy does not uniquely determine the state.) This has a natural basis corresponding to the $SO(2)$
eigenvectors.  Physically, the derivative of this $SO(2)$ (the traditional
element $H$ in the Lie algebra $\mathfrak sl_2$, perhaps up to some scaling,
and often denoted $L_z$ in physics literature) is the operator corresponding to angular momentum around the axis in $S^2$ which is fixed by $SO(2)$ (often taken to be the $z$-axis, by convention, in the physics literature), and so the different eigenvectors for $SO(2)$ in the given $V$ are states with the same energy, but with different angular momenta around the $z$-axis.
Physically, one can apply a magnetic field along this axis which will then affect these different $L_z$-eigenstates differently (because its effect on
the electron depends on the angular momentum of the electron around the field
lines), and so one can split the degeneracy.  This is the Zeeman effect.  One sees the different energy lines in the spectrum of hydrogen split into a number of lines ($n$ lines when the energy is proportional to $n^2$).
I haven't looked, but this must be discussed in many, many places online, and is also in any standard intro. to quantum text.  (But probably with less appearance of the symbols $L^2(S^2)$ and $\mathfrak sl_2$ then a mathematician would use!)
Comment: Weyl's *The theory of groups and quantum mechanics* is a charming place where to read this :)
Comment: A nice reference for some of this are Peter Woit's notes: http://www.math.columbia.edu/~woit/LieGroups/
Answer: First, I think there is potentially more than one question here. The finite-dimensional version, addressed by Matt E, is about the hydrogen atom. The "harmonic oscillator" sometimes means a different thing, not about finite-dimensional repns of the Lie algebra $\mathfrak{sl}_2(\mathbb C)$, but about infinite-dimensional ones. There there is interaction with repns of SO(n) for all n. (This is also an example of "Segal-Shale-Weil" repns, ... also existing over p-adic fields, and abstractly, after Weil.)
I'm sure there are other worked-out versions on-line, but at least I can offer my own worked-exercise, with (a mathematician's notion of) a bibliography: http://www.math.umn.edu/~garrett/m/v/oscillator_repn.pdf

sl(2,C) and the harmonic oscillator


What's the difference between the different "generations" of iPod touches?

Question: I've been looking at buying a used ipod touch from craiglist but I'm confused as to what's what. The adds all say things like "2nd generation 8gb" or "3rd generation 16gb" and I'm not sure what the difference is.
Answer: The short answer is processing speed, screen resolution and memory. The newer ones are obviously faster, and the lastest generation (4g) includes both front and back cameras for shooting video and video calling.
For a full list of the differences check out this wikipedia page.
Comment: Even better see Apple;s technical specifications http://support.apple.com/specs/#ipodtouch
Answer: The long answer:
1st and 2nd have 128MB RAM while 3rd and 4th have 256.
3rd and 4th also come in 64GB versions as well as 32GB. The 4th added back an 8GB model.
2nd added Bluetooth (Only in iPhone OS 3.0)
3rd added voice control.
4th also has a Retina Display. Though not the same quality as the iPhone 4. Still it's nice and high-resolution. (It's an LCD, not an IPS for price reasons.)
And the 4th has two cameras as well. Still not as nice as the iPhone ones.
Processor is as follows: 412MHz, 533MHz, 600MHz ARM processors in first three versions. (Underclocked speed)
4th generation replaced the processor with an Apple-made 1GHz A4 processor underclocked to 800MHz.
4th also added a microphone built-in and a 3-axis Gyroscope.
The 5th generation will be released in September. This is 99.9% guaranteed. Most likely with some features from the to-be-announced iPhone 5. (iPods are always September every year. It's one of the only predictable things about Apple. iPhones are in June or July when WWDC takes place, but can change without notice.)
I could keep going but then I'm just into the useless techy stuff like GPU and weight.


What are the common techniques to close the gap in jeet kune do?

Question: I just started a jeet kune do classes a month ago, during the sparring parts I find it very hard to close the gap against people who have been doing this for much longer but I cannot understand (yet) how I can close the gap for going through their guard.
Answer: Practice at home footwork and angling. Speed is what will save you. When you move around a person they move faster because they just rotate on the pivot point. Timing your angled footwork lets you face their center point outside of there strikes if timed right. Control fighting distance. JKD secret is to make the other guy move first into range where you intercept his actions.  This is where you need to learn how to read a person. Also get rid of your own tells by shadowing boxing in a mirror and record your sparring sessions.
Comment: "JKD **secret** is to make the other guy move first into range where you intercept his actions."  Well now everybody knows!
Answer: A fake jumping kick works wonders. I am by no means a JKD expert, but I've found that faking a flying kick works well. Both in JKD and in Taekwondo. The idea is to sucker your opponent into counter-attacking, and then intercepting the counter.
Answer: I don't study JKD, but in other arts I've found that going from the outside (kicking or long punching range) to the clinch (wrestling or dirty boxing range) can be done with a front-leg kick or a jab that the opponent is not expecting to be followed up by a crash in.
Answer: There were three techniques which helped me a lot (and still do) when I started sparring in JKD:

progressive indirect attack
jamming their attack (e.g. striking their hand is it goes towards me, the essence of jeet kune do is to intercept, destroy or prevent an attack from an opponent).
attack by combination (for example a simple 1-2-3 or jab, cross, lead long hook) a couple of times followed by a straight blast (remember the JKD straight blast is the most effective entry or finish weapon once the opponent gives us an opening, its also the fastest way to punch)

Relating to footwork the fastest way to advance longer distances (e.g. when pursuing an opponent or recovering form a retreat is to use a slide and step or forward shuffle) to gain shorter distances is the half step/jump with the lead leg e.g. lunging jab.
In essence for all fighting speed and mobility is the key. Cover your self, move constantly both your body around the opponent and your tools (hand and head movement) and attack. The effectiveness of an compound attack is that it will show an opening which can be exploited by the attacker.
Stopping on a failed attack or stopping to think of what next is allowing time for the opponent to react, it also opens the possibility in your mind that the opponent is going to win.
One failed attack should be immediately be succeed by another attack on another angle, then another. Like water flowing to a wall flow around it till you find a hole you can pass trough.
And remember the entire human body is a target. There are no safe zone. We are no scoring points here. A hit on the arm can be effective and more achievable then a hit on the head since the arm is more exposed, especially if you can hit it with an elbow or leg. Hips, shoulders, joints, and lower body are all excellent areas.
But generally there is one principle to have in mind, to miss-quote Sifu Lee :

Be like water, flexible and adaptable yet strong and violent.
Answer:
during the sparring parts I find it very hard to close the gap against people who have been doing this for much longer

Great!  That suggests your school is teaching good technique such that more training will make you hard for people outside to hit.  Don't be discouraged by that, but by all means take it as a challenge to make up the ground as quickly as you can, as well as an indication that careful observation and contemplation of what those more experienced people are doing really - combined with similar attention to the instructor and diligent practice - will help you get there too.
As for techniques - closing the gap is largely about recognising how your opponent might attack you, and sliding forwards in a way that cuts off all the opponent's direct lines of attack, leaving them to make slower round-about attacking movements that you'll have time to see and react to.  A good way to start perceiving and relating to this "problem space" is to consciously study the lines from people's elbows through their fist to the place they might want to punch.  If their fist isn't directly pointing at you with their elbow behind, they'll have to swing their forearm towards you in an arc instead of being able to drive it in a straight-line jab/cross, if they try to move from such a position into a straight-ish punch and you put your arms or hands anywhere inside that arc as you move forwards they won't be able to complete their motion and actually get their fist moving towards their intended target (e.g. your face).  Once you're comfortably able to perceive and intercept the lines they must attack along, you can apply "intercepting fist" concepts to deflect them while punching/chopping/elbowing etc. in counterattack, as well as trap their limbs effectively.  You can accelerate your learning of these concepts by reading some JKD books and other sources, but be careful not to cement bad habits before your instructor can set you on the right path.  Some patience is required - just like nobody can age whiskey 20 years in a month.
A similar thing goes for kicks... see if you can be just "outside" the line they can comfortably front kick forwards to, such that they'll need to rock back a little while rotating towards you before delivering a kick - that way you're safer closing the gap, have some time to see their attack, and can either attack first or counter.
And given they're more experienced, watch how you get stuck... try to do what they do to you to them and see how they stop you, then try to stop them the same way.  Experiment and don't be afraid of getting an extra knock or two along the way... it's the only way to really learn what works, build your reflexes and expand your comfort zone and competencies.
Finally, study your timing, and look to close just after they've decided to do something.  Feint a little and if they tense up do it again and slide in behind it.  Don't be afraid to throw a few techniques to keep them busy even if you don't seriously expect them to be terribly effective... experience will distill what works eventually, and having them easily block is better than them being 100% focused on hitting you.
Answer: It is hard to do, especially against a younger or more experienced adversary.  I am 60 plus and not as fast as I used to be but in my dojo I tend to wait for my opponent's move and then counter attack to break in.  It's hard, no doubt about it but timing is the key. When he goes, you block and go like hell. Sometimes it works, sometimes it doesn't.


I have an Ubuntu Gutsy installation running on a server, how can I upgrade it to Karmic?

Question: How can I upgrade a really old Gutsy server to Karmic from the command line?
Answer: You can't do a direct upgrade from Gutsy to Karmic. You will have to Upgrade one version at a time to Karmic. Ubuntu actually reccomends a fresh install for a system that old, and honestly I do too, you'll run into too many odd issues and missing repositories (since Gutsy is no longer supported) to make in place upgrades really worth the time spent.
Ubuntu's Upgrade documentation has good documentation on going from one version to another.
From the above page here is the EOL upgrade instructions
Answer: It will probably be easier to simply make a full backup, reinstall clean, and then reconfigure the system and restore your data.
There is no supported direct upgrade path.  You will need to upgrade from gutsy -> hardy -> intrepid -> jaunty -> karmic.
The general steps you'll want to follow are

Do a sudo apt-get update
Do a sudo apt-get dist-upgrade (This will update the update-manager-core package which includes do-release-upgrade.
Run do-release-upgrade
Repeat, above until you are up to karmic

If you don't have a full backup you should make one before you start.  A lot of changes have been made between gutsy and karmic.   Things are likely to break, and you'll have to adjust configurations to fix things.
If you have used any non-official repositories you should disable those, and note any packages that came from the non-official locations.  If you have manually compiled anything you should make sure you have notes for that.
Answer: The official UpgradeNotes from Canonical recommend doing either:

Backup, followed by fresh install with new version
Upgrade through each successive version

They explicitly recommend against skipping versions during an upgrade.  Assuming that advice is sound, you have a long road ahead.  I would start with looking through the general UpgradeNotes, and then drill down into the individual steps you may have to take.
Answer: As long as you are going to be doing a full backup anyway, the better route may be to just do a fresh install of Karmic.
Answer: This doesn't actually answer your question, but if you upgrade to hardy and wait six months you'll be able to upgrade straight to Lucid. (LTS to LTS direct upgrades are supported)
Answer: I tend to do clean reinstalls (backup / format / install / restore) whenever possible. It's a great opportunity to get rid of all the crap that inevitably accumulates on any system.
Having a clean machine is great.


Dispensing beers at two different pressures?

Question: I'm working on getting a dual-tap kegerator.  Is it worth it to pay the extra to dispense two beers at two different pressures?  I can see dispensing a low-carbonation ESB and a high-carbonation weizen side-by-side, but I don't want to go overboard with the cost.
Comment: I'm now dispensing two beers at the same carbonation. Never been an issue, really, as I tend to brew in series (American, English, Lagers, Wheats) and similar beers tend to like similar carbonation levels.
Answer: In my opinion, if you're going to go the distance and convert a fridge, you should do it right and get the dual regulator setup so you can dispense at different pressures.  That is, if cost isn't an issue.
Well, cost would certainly be an issue for me, and I'm still debating whether or not I should do this, or even bother with converting a fridge at all--the picnic tap works fine for me and quick disconnects make it easy to switch back and forth between my two kegs in the fridge.
I guess the question is, how often are you going to be pouring?  Do you have a lot of people who come over to drink?  If so, then it might be worth it.
I generally only pour only a few glasses a night, so switching between kegs is easy enough for me.  But I think that if I were going to convert the fridge into a kegerator, it'd be for the "coolness factor" and I'd want to go all out with all the bells and whistles--including separate regulators for each keg to ensure appropriate carbonation and dispensing pressures.
Just my $.02
Comment: Assuming cost is not an option, split it into a Nitro side and a CO2 side. Its always interesting to taste the difference in your ESB when you push it with Nitrogen instead of CO2. But then, I almost always have a stout on one tap and an ESB or IPA on the other...
Answer: Not only can you dispense at different pressures, you can carbonate beers independently of your dispense pressure.  With a T off of one regulator you can gave your normal serving pressure, then you can use the other side for high pressure quick carbing if necessary on occasion.
If I had the funds I'd go with a dual stage set up just for the flexibility.
One other thing to think about though would be pouring stouts.  If you make stouts (or any style) that you want on nitrogen, maybe investing the money on a nitrogen hook up is a better investment.  Of course that would be more expensive because you'd need the stout faucet too.


Does every orientable surface embed in $\mathbb{R}^{3}$

Question: A topological surface can be pretty strange (consider, for instance, covers of $S^{2}-K$, where $K$ is a Cantor set.)  Can every orientable topological surface be topologically embedded in $\mathbb{R}^{3}$?
Comment: David -- if you are not requiring the embedding to be proper then this should follow from the classification theorem, see e.g. http://www.jstor.org/discover/10.2307/1993768?uid=3738936&uid=2129&uid=2&uid=70&uid=4&sid=21101300949803 My guess is that starting from this one can construct a proper embedding too.
Comment: Similar question was recently discussed here, but I don't now how to find it. Are you asking about topological surface (and topological embedding)? Or a Riemann surface (conformal embedding) or a surface
with some metric, or what? I believe that a topological surface,
and even a Riemann surface can be  so embedded.
Answer: Expanding slightly on my comment above: here is how one can get the embedding theorem from the classification theorem.
The classification theorem for non-compact surfaces (theorem 3 in http://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf; by the way, the comment links to a version that tries to charge one $30 unless one's reading this from a university account; sorry about that; this has been fixed now) states that
Every surface is homeomorphic to a surface formed from a sphere $\Sigma$ by first removing a closed totally disconnected set $X$ from $\Sigma$, then removing the interiors of a finite or infinite sequence $D_1,D_2,\ldots$ of nonoverlapping closed discs in $\Sigma - X$, and finally suitably identifying the boundaries of these discs in pairs. (It may be necessary to identify the boundary of one disc with itself to produce an odd "cross  cap.'')[...]
To get a proper embedding in $\mathbb{R}^3$ first note that one may assume that $X$ is non-empty; otherwise the surface will be compact (since we are throwing away something open and then identifying something), in which case everything is clear. Let $f:\Sigma\to [0,\infty)$ be a smooth function such that $f^{-1}(0)=X$.
Take an $x_0\in X$ and identify $\Sigma\setminus \{x_0\}=S^2\setminus \{x_0\}$ with $\mathbb{R}^2\subset\mathbb{R}^3$. First we properly embed $\Sigma-X=\mathbb{R}^2\setminus X$ in $\mathbb{R}^3$ as the graph of the function $\frac{1}{f}$.
Let $D_i,D_j$ be two disks whose boundaries are to be identified. There is an $\varepsilon(D_i,D_j)>0$ such that one can join $D_i$ and $D_j$ with a curve $\gamma(D_i,D_j)\subset \Sigma$ that misses $f^{-1}([0,\varepsilon(D_i,D_j)))$. Now delete the portion of the graph over the interiors of $D_i$ and $D_j$, attach vertical tubes ("chimneys") that reach at least as high as $1/\varepsilon(D_i,D_j)$ to the resulting boundaries, and then connect the tops of the chimneys with a horizontal tube along $\gamma(D_i,D_j)$.
If $D_k,D_l$ are two other disks that we must throw away and then glue the boundaries we proceed in a similar way but this time we may have to make the chimneys higher so that they miss the first tube. And so on. It is not hard to see that each ball centered at the origin intersects only finitely many tubes.
Answer: Here is a reference:
MR0304649 (46 #3781)
Rüedy, Reto A.
Embeddings of open Riemann surfaces.
Comment. Math. Helv. 46 (1971), 214–225.
He talks about Riemann surfaces, but every orientable topological surface (with a countable basis
of topology) has a structure of a Riemann surface.
I suppose there is no question about embedding of closed orientable surfaces.


On the image of a G_\delta set under a continuous bijection

Question: Let $X, Y$ be two metric spaces and $f$ be a continuous bijection (i.e. one-to-one map) from $X$ to $Y$.  Let $E$ be a $G_{\delta}$ subset of $X$. I want to know weather the image $f(E)$ is also a $G_{\delta}$ subset of $Y$.  Thanks!
Answer: Yes, if $X$ is $\sigma$-compact: $X\smallsetminus E$ is $F_\sigma$, hence a countable union of compact sets. Since a continuous image of a compact set is compact (and therefore closed, in a Hausdorff space), $f(X\smallsetminus E)$ is also $F_\sigma$, hence $f(E)$ is $G_\delta$.
EDIT: While ljjpfx seems to be happy with this answer as his or her spaces are compact, Nate Eldredge asked in a comment whether there is a counterexample with both spaces Polish, which I think is an interesting question. I believe the following example should work. Consider the function
$$f\colon\omega^\omega\to[0,1),\quad f(a_0,a_1,a_2,\dots)=0{.}\underbrace{1\cdots1}_{a_0}0\underbrace{1\cdots1}_{a_1}0\underbrace{1\cdots1}_{a_2}0\cdots,$$
where the right-hand side is meant to be the binary representation. (I will write $0{.}1^{a_0}01^{a_1}0\cdots$ to simplify the notation.) It is easy to see that $f$ is a continuous bijection, while $f^{-1}$ is not continuous. Let $U$ be the open set
$$\{a=(a_n)_{n\in\omega}\in\omega^\omega:\exists n\\,a_n\text{ is even}\},$$
I claim that $f(U)$ is not $G_\delta$. Assume for contradiction that $f(U)=\bigcap_nG_n$ for some open $G_n\subseteq[0,1)$. Note that $f(U)$ is the union of intervals of the form $[a2^{-k},(a+1)2^{-k})$, where $a,k\in\omega$, and $a$ has the binary representation $1^{a_0}01^{a_1}0\cdots01^{a_r}0$, where $a_i$ are odd for $i< r$, $a_r$ is even, and $k=\sum_{i\le r}(a_i+1)$. Since $G_n\supseteq f(U)$ is open, for every $n,k$ there exists $l_{n,k}$ such that
$$G_n\supseteq(a2^{-k}-2^{-(k+l_{n,k})},(a+1)2^{-k})$$
for every $a=1^{a_0}01^{a_1}0\cdots01^{a_r}0$ as above. Note that if $a_r>0$, the end-point of this interval has the form $0{.}1^{a_0}01^{a_1}0\cdots01^{a_{r-1}}01^{a_r-1}01^{l_{n,k}+1}$. Choose $a=(a_n)\in\omega^\omega$ so that every $a_n$ is odd, and $a_{n+1}>l_{n,k_n}$, where $k_n=1+\sum_{i\le n}(a_i+1)$. Then $a\notin U$, but the description above implies that $f(a)\in\bigcap_nG_n$, a contradiction.
Comment: A higher-level argument why $f(U)$ is not $G_\delta$ can be made as follows. Let $D$ be the set of the lower end-points $a2^{-k}$ of the intervals comprising $f(U)$, as described above. Then $P=\overline D$ is a Polish space. Since $D=P\cap f(U)$, $D$ is $G_\delta$ as a subset of $P$, and therefore comeager. However, being a countable set with no isolated points, $D$ is also meager, contradicting the Baire category theorem for $P$.
Answer: No, you have stated nowhere near enough conditions to achieve this.
For example, let $X$ and $Y$ both be your favourite metric space ($R$, (for example) but give $X$ the discrete metric (with $d(x,y)=1$ if $x\neq y$) and let $f$ be the identity (on points).
Then all of the subsets of $X$ are open and so also $G_\delta$. However, they need not have this property in $Y$.
Edit: I have removed, reconsidered and consolidated various other comments that I made about this problem.
My initial reaction was that ljjpfx had omitted to say that $f$ should be a homeomorphism (have a continuous inverse) and not just be continuous and bijective. There are well known conditions under which a continuous bijection $f:X\to Y$ is a homeomorphism, for example if $X$ is compact and $Y$ is Hausdorff.
I thought that all of this would render the original question rather trivial and likely to be closed, whereas I wanted to have the opportunity to develop it into something more interesting.
What I hadn't appreciated is that most subspaces that are defined in any reasonable fashion are $G_\delta$.  It seems that Nate didn't notice this either and nor am  I convinced by Emil's counterexample.
What is an example of a non-$G_\delta$ subspace of $\Bbb R$?
The direction in which I had wanted to develop this question was by introducing overt subspaces.  I asked ljjpfx what situation had given rise to the question because I was wondering whether s/he was actually working with overt subspaces rather than $G_\delta$ ones.
Formally, overtness is the lattice dual of compactness, in particular the direct image of an overt subspace is overt just as happens in the compact case.
The reason why you haven't heard of this notion before is that every (sub)space is overt in classical topology.  (The points mess things up, rather than excluded middle.)
The idea comes from various approaches to unifying topology with computation. Overt in English means explicit. In constructive topology it means having evidence.
The Bishop school of constructive analysis has used notions of locatedness and total boundedness to fill in constructive gaps in classical theorems.
Locale Theory and Formal Topology both develop the subject in terms of open subspaces instead of points, the former in a categorical setting (in a topos) and the latter in Martin-L&oumlautf type theory.  Peter Johnstone and André Joyal called a locale open if its terminal projection is an open map. Giovanni Sambin introduced positivity for formal topology.
My own approach to construcive/computable topology is called Abstract Stone Duality.  It started from categorical ideas but developed into a language that only expresses computable continuous functions. My paper A Lambda Calculus for Real Analysis is the best place to start with this; it emphasises the dual behaviour of overt and compact subspaces and applies them to the (constructive) intermediate value theorem.
Recently I have been working on a characterisation of overt subspaces in locally compact metric spaces and open maps between them. I will add a link from this page when I am ready to circulate a draft of this.
In particular, overt subspaces are $G_\delta$, which is why I got involved in this question. However, I now realise that all of the subspaces that I construct in ASD are $G_\delta$ too.
Comment: In case it's not obvious, $\mathbb Q$ is an example of a non-$G_\delta$ subspace of $\mathbb R$.
Comment: Is there a counterexample with $X,Y$ both Polish?
Comment: Many thanks.  I forgot to say both X and Y are compact metric spaces.
Answer: No; consider any bijection from $\mathbb{N}$ to $\mathbb{Q} \subset \mathbb{R}$.
Edit: I missed where the question said "bijection"; I saw the next sentence said "one-to-one map" and just thought "injection".  Any bijection from $\mathbb{N} \to \mathbb{Q}$ is a continuous injection $\mathbb{N} \to \mathbb{R}$ whose image is $\mathbb{Q}$, which is not $G_\delta$.  But of course it is not onto.  So this doesn't really answer the question, but I will leave it anyway.
Comment: I like this example on the grounds that $\Bbb Q$ is the one set on which it is natural to put two different topologies (the discrete and Euclidean ones).   Unfortunately, I think that every subset of $\Bbb Q$ is $G_\delta$ in the Euclidean topology (at least classically).
Comment: I don´t quite understand how this answers the question. Wouldn´t you need a continuous bijection from $\mathbb{R}$ onto itself sending $\mathbb{N}$ onto $\mathbb{Q}$? But there is no such a thing, unless you change the topology of your $\mathbb{R}$´s as in Paul´s answer.


Should secondary fermentation be the same temperature that primary was?

Question: Or is it okay for the secondary fermentation to be colder than primary?
Answer: If all you're doing with the secondary is conditioning the beer, i.e. letting the flavors develop and whatnot, it can be cooler. If you're trying to let the yeast continue to ferment out the last few gravity points, it's best to leave it around the same temperature. If the yeast are fermenting and you drop the temperature, they could fall out of suspension and you'll have a stalled fermentation.
Comment: +1. The distinction between "secondary fermentation" and "conditioning" is often misunderstood. Old homebrew books and many starter kits don't help the situation.
Answer: The suggested fermenting temperatures really are to create optimal conditions for the yeast to thrive in your fermenter. I usually let the gravity ferment out first before racking to secondary where it basically conditions. When racking, I leave much of the yeast sediment in the primary because it has done its job.


Misconception about the expectation of a quantum system

Question: For a two-level quantum system with energy eigenstates $|\phi_1\rangle$ and $|\phi_2\rangle$ at finite temperature, we can write a general state as $$|\Phi\rangle=c_1|\phi_1\rangle+c_2|\phi_2\rangle.$$
In quantum mechanics, we know that the expectation of an operator $\hat{O}$ is given by  $$\langle\Phi|\hat{O}|\Phi\rangle = |c_1|^2\langle\phi_1|\hat{O}|\phi_1\rangle +|c_2|^2\langle\phi_2|\hat{O}|\phi_2\rangle +c_1^*c_2\langle\phi_1|\hat{O}|\phi_2\rangle +c_2^*c_1\langle\phi_2|\hat{O}|\phi_1\rangle.$$
However, in statistical mechanics, the expectation value is instead
$$\text{probability}(\phi_1)\langle\phi_1|\hat{O}|\phi_1\rangle + \text{probability}(\phi_2)\langle\phi_2|\hat{O}|\phi_2\rangle,$$
which simplifies to only the first two terms of the expression obtained from quantum mechanics. What is wrong here?
Answer: Your formula for the expectation value in statistical mechanics is wrong.
A pure quantum mechanical state in this system is a sum $$|\Phi\rangle = c_1 |\lambda_1\rangle + c_2|\lambda_2\rangle$$ of two basis vectors, with complex coefficients.   Note that I'm using your notation $\Phi$ for a general state, and switching to $|\lambda_1\rangle, |\lambda_2\rangle$ for the basis vectors.  If we have a list of states $|\Phi_i\rangle$, we get a list of coefficients $c_{i,1}$ and $c_{i,2}$.
A thermal state in this system is an expectation value coming from an arbitrary sum (or integral) of pure quantum mechanical states:
$$
E(\hat{O}) = \sum_i d_i\langle \Phi_i | \hat{O} | \Phi_i \rangle.
$$
Here the coefficients $d_i$ are real.
You're getting confused because you're trying to equate a special two-state case of the latter formula with the expectation value coming from a pure state.  If you want to write the stat mech expectation value in terms of $|\lambda_1\rangle$ and $|\lambda_2\rangle$, the right formula is
$$
E(\hat{O}) = \sum_i d_i\big(|c_{i,1}|^2\langle\lambda_1|\hat{O}|\lambda_1\rangle +|c_{i,2}|^2\langle\lambda_2|\hat{O}|\lambda_2\rangle + c_{i,1}^*c_{i,2}\langle\lambda_1|\hat{O}|\lambda_2\rangle) +c_2^*c_1\langle\lambda_2|\hat{O}|\lambda_1\rangle\big)
$$
You get one kind of uncertainty from having a pure quantum state, and another from having a probability distribution on the space of quantum states.
Answer: Nothing is actually wrong.
The quantum answer includes $c_2^*c_1$ term which depends on the phase difference between $c_1$ and $c_2$.  But the phase difference is something that would vary randomly with even the tiniest coupling between states 1 and 2.  And thermodynamics requires such a coupling to exist to be applied.  So the stat mech answer "averages over" things that very randomly in thermal systems.  Since that phase varies, $$\langle c_2^*c_1\rangle = 0$$ averages out to zero, and the only terms that do not average out to zero in thermal equilibrium are the terms that both stat mech and quantum preserve.
The quantum answer applies when the states have no interaction, in which case it is not correct to apply stat mech because stat mech gives the answer "after we have reached thermal equilibrium" and the smaller the interaction between 1 and 2 is, the longer it takes to reach equilibrium, and when the interaction is vanishingly small, the time to thermal equilibrium approaches infinity.
Answer: Both of your formulae are correct. One critical point is that, in quantum mechanics, we are adding states with a particular phase. So, the combinations $|c_1\rangle + |c_2\rangle $ vs $|c_1\rangle + i|c_2\rangle $ vs $|c_1\rangle - |c_2\rangle $ are not terribly different, they're just different ways to combine states. In contrast, in statistical mechanics, you are adding real numbers (expectations values), so you had better use a $+$ and not a $-$.
It is also possible to do both at once. Take a look at the density matrix, which is a great way to handle that. There is a good introduction in Sakurai's Modern Quantum Mechanics, and this site looks decent.
Foregoing the density matrix formalism, this is how you would go about combining the two ideas. Take some states $|\Phi_i\rangle$ for $i=1,2$ (we'll stick to two levels) with a $[c_{ij}]$ matrix
$$ |\Phi_i\rangle = c_{i1} |\phi_1\rangle + c_{i2}|\phi_2\rangle  $$
This has an expectation value
$$ (A_i)_{quantum} = \langle \Phi_i | A | \Phi_i \rangle = (\sum_j c_{ij} \langle \phi_j |) A (\sum_j c_{ij} | \phi_j \rangle)$$
Now, we want to calculate some statistical quantity, and we say that our system has a weight $w_i$ of each quantum state, which has nothing to do with $c_{ij}$. For thermal samples, $w_i$ is proportional to the Boltzmann weight of the energy eigenstate $i$.
$$\langle A \rangle_{statistical} = \sum_{i} w_i \langle \Phi_i | A | \Phi_i \rangle = \sum_i w_i (A_i)_{quantum} $$
You might wonder how it would look in a different basis, that this, what weighting factors to use for $|\phi_i\rangle$. That is, which $v_i$ would I use to get the same answer for
$$\langle A \rangle_{statistical} = \sum_{i} v_i \langle \phi_i | A | \phi_i \rangle$$?
I'll leave it as an excercise for the reader. =)   Suffice to say, this transformation is one thing density matrices are useful for.


Does summoner heal reduce healing from other heals?

Question: When you use summoner heal, I know you get a status effect that reduces subsequent summoner heal effects, but by how much? and does it reduce healing effects from champion abilities/lifesteal/etc?
Answer: Heal applies a 35 second debuff that reduces all subsequent Summoner Heal spells, aside from your own, by 50%. This debuff does not affect Lifesteal, Spell Vamp, Health Regen or Champion Heal spells but only other allies' Summoner Heals.
Answer: No. The debuff from Heal only applies to the healing from the Summoner Spell Heal.
Back in Beta, it wasn't uncommon to see all 10 summoners running Heal, which horribly affected team fights. The debuff was added to discourage Heal stacking.


Need Help to "setup" a RESTful webservice

Question: I open this topic because I'm going to need a little help for rails and Webservices and also because i am a beginner.
I have now need to "build" a webservice that will enable a third party application (mobile application) to connect to a mySQL database server hosted on my ubuntu 10.04 server
So I have already committed some research and asked some questions and it seems that Rails 3.0 can help me materialize my project by creating a RESTful webservice type (that I don't know yet exactly what it is). Anyway, what I have understood is that I could make requests of kind 'http' in my DB via webservice that will return an XML I can parse.
So I have installed on my server ruby 1.8.7 and gem 1.8.21
I have then launched the famous gem install rails and everything seems ok ... I have Rails in its 3.2.3 version
Only now I 've no idea how to "create" my webservice (and where I must put it ? In /var/www/ to be reach from the web ?) Or how to parameter the 'linker' with a mySQL DB.
I have understood by reading different things on internet setting up a RESTful webservice with rails is not to complicate but i really don't find how to setup mine ... that's why i need help ! A lot of tutorials i've seen are too old. And I really would like advices to set it up !
So you have some tips, some ideas, or tips for m 'refer and help me to put all this in place.
Thanks for all !
Answer: A fantastic, up-to-date tutorial that is the very first thing you should look at (in my opinion) is the Rails Tutorial E-Book which you can go over for free here.  You're way in over your head right now.


What relations exist among (quasi-) modular forms of different levels?

Question: Given a (quasi-) modular form $f(\tau)$ for some congruence subgroup (say) $\Gamma(k)$, we know that $f(N\tau)$ is a (quasi-) modular form for $\Gamma(N k)$. Is there anything known about when we can do a partial reverse, that is, when we can take linear combinations of (quasi-) modular forms for some higher level subgroup to obtain one of stricktly lower level?
An example is the following:
Let $E(q) = \sum_{k=0}^\infty \sigma_1(2k+1)q^{2k+1}$ and $A(q) = \sum_{k=1}^\infty \sigma_1(k)q^k$. Then $E(q)$ is modular with respect to a non-trivial character, and both $A(q^2)$ and $A(q^4)$ are quasi-modular of level 2 and 4, respectively (though not of pure weight).
However: it turns out that
$$
E(q) + 3A(q^2) - 2A(q^4) = A(q)
$$
which shows that a linear combination of higher level terms (and one which is modular with respect to a non-trivial character) yields one of lower level.
Is this simply random chance? Are there known relations of this type?
Answer: There are (at least) two answers: there is a Galois theory of modular functions (as in G. Shimura's 1971 book, or in Lang's book on elliptic functions), and a theory of newforms" (Atkin-Lehner, and also Casselman in a representation-theoretic context).
Thinking in terms of general Galois theory, it ought not be so surprising that various sums of non-invariant things become invariant, I suppose, but the particulars are often non-trivial.